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Linkoping Studies in Science and Technology. DissertationsNo. 1044
Asymptotic analysis of solutions to
elliptic and parabolic problems
Peter Rand
Matematiska institutionenLinkopings universitet, SE-581 83 Linkoping, Sweden
Linkoping 2006
September 6, 2006 (12:43)
ii
Asymptotic analysis of solutions to elliptic and parabolicproblems
c© 2006 Peter Rand
Matematiska institutionenLinkopings universitetSE-581 83 Linkoping, [email protected]
ISBN 91-85523-04-6ISSN 0345-7524
Printed by UniTryck, Linkoping 2006
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iii
Abstract
In the thesis we consider two types of problems. In Paper 1, we studysmall solutions to a time-independent nonlinear elliptic partial differentialequation of Emden-Fowler type in a semi-infinite cylinder. The asymptoticbehaviour of these solutions at infinity is determined. First, the equationunder the Neumann boundary condition is studied. We show that anysolution small enough either vanishes at infinity or tends to a nonzero peri-odic solution to a nonlinear ordinary differential equation. Thereafter, thesame equation under the Dirichlet boundary condition is studied, the non-linear term and right-hand side now being slightly more general than in theNeumann problem. Here, an estimate of the solution in terms of the right-hand side of the equation is given. If the equation is homogeneous, thenevery solution small enough tends to zero. Moreover, if the cross-sectionis star-shaped and the nonlinear term in the equation is subject to someadditional constraints, then every bounded solution to the homogeneousDirichlet problem vanishes at infinity.
In Paper 2, we study asymptotics as t → ∞ of solutions to a lin-ear, parabolic system of equations with time-dependent coefficients inΩ×(0,∞), where Ω is a bounded domain. On ∂Ω×(0,∞) we prescribe thehomogeneous Dirichlet boundary condition. For large values of t, the coef-ficients in the elliptic part are close to time-independent coefficients in anintegral sense which is described by a certain function κ(t). This includesin particular situations when the coefficients may take different values ondifferent parts of Ω and the boundaries between them can move with t butstabilize as t → ∞. The main result is an asymptotic representation ofsolutions for large t. As a corollary, it is proved that if κ ∈ L1(0,∞), thenthe solution behaves asymptotically as the solution to a parabolic systemwith time-independent coefficients.
Acknowledgements
I would like to thank my supervisors Vladimir Kozlov and Mikael Langerfor all their support and invaluable hints during this work and Anders Bjornand Jonna Gill for helping me with LATEX. Thanks also to everyone elsewho has helped me in some way.
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Contents
Introduction 1
References 2
Paper 1: Asymptotic analysis of a nonlinear partial differentialequation in a semicylinder 7
1 Introduction 7
2 The Neumann problem 102.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Problem formulation and assumptions . . . . . . . . . . 112.3 The main asymptotic result . . . . . . . . . . . . . . . . 112.4 Corollaries of Theorem 2.1 . . . . . . . . . . . . . . . . . 122.5 The corresponding problem in C . . . . . . . . . . . . . 142.6 An auxiliary ordinary differential equation . . . . . . . . 162.7 The equation for v . . . . . . . . . . . . . . . . . . . . . 242.8 Asymptotics of small solutions of problem (2.10) . . . . 282.9 End of the proof of Theorem 2.1 . . . . . . . . . . . . . 31
3 The Dirichlet problem 323.1 Problem formulation and assumptions . . . . . . . . . . 323.2 The main asymptotic result . . . . . . . . . . . . . . . . 333.3 The corresponding problem in C . . . . . . . . . . . . . 333.4 End of the proof of Theorem 3.1 . . . . . . . . . . . . . 343.5 The case of a star-shaped cross-section . . . . . . . . . . 353.6 An estimate for solutions of a nonlinear ordinary differ-
ential equation . . . . . . . . . . . . . . . . . . . . . . . 38
A Some results from functional analysis 40A.1 Eigenvalues and eigenvectors of −∆ . . . . . . . . . . . 40A.2 Existence and uniqueness of bounded solutions of Pois-
son’s equation in C . . . . . . . . . . . . . . . . . . . . . 43A.3 A local estimate for solutions of Poisson’s equation . . . 50
Paper 2: Asymptotic analysis of solutions to parabolic sys-tems 57
1 Introduction 57
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2 Problem formulation and elementary properties 622.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
2.1.1 Spaces not involving time . . . . . . . . . . . . . 632.1.2 Spaces involving time . . . . . . . . . . . . . . . 64
2.2 Problem formulation and assumptions . . . . . . . . . . 652.3 An estimate for u . . . . . . . . . . . . . . . . . . . . . . 68
3 Spectral splitting of the solution u 69
4 Estimating the function v 714.1 A general estimate . . . . . . . . . . . . . . . . . . . . . 714.2 Estimate for v . . . . . . . . . . . . . . . . . . . . . . . . 82
5 Norm estimates for Rkl and gk(w) 84
6 Functions hJ+1, . . . , hM 866.1 Definition of functions v0, v1 and v2 . . . . . . . . . . . 866.2 Integro-differential system for hJ+1, . . . , hM . . . . . . . 876.3 A general estimate . . . . . . . . . . . . . . . . . . . . . 876.4 A particular case of equation (6.8) . . . . . . . . . . . . 966.5 Estimate for h . . . . . . . . . . . . . . . . . . . . . . . 996.6 A representation for h . . . . . . . . . . . . . . . . . . . 99
7 Functions h1, . . . , hJ 1007.1 Equation for h; existence and uniqueness results . . . . 1007.2 The homogeneous equation . . . . . . . . . . . . . . . . 1047.3 A particular solution of (7.15) . . . . . . . . . . . . . . . 116
8 Proof of Theorem 1.1 118
9 Corollaries of Theorem 1.1 120
A Eigenfunctions of a time-independent operator 124
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Introduction
Most differential equations and systems are not possible to solve exactly.Hence, it is important to develop other methods of analyzing the propertiesof solutions. One of these methods is based on asymptotic analysis. Al-though the solutions are unknown, it may be possible to find informationabout their behaviour as some variables tend to some finite value or toinfinity.
Asymptotic analysis is used to study time-dependent evolution prob-lems as well as time-independent stationary problems. Frequently, one isinterested in behaviour of solutions as time tends to infinity, for examplein questions concerning stability, periodicity, rate of growth etc. A surveyof evolution problems and a general theory of analyzing them, includingasymptotic analysis, can be found in Dautray, Lions [1], [2] or Lions, Ma-genes [8], [9], [10]. An important class of evolution problems are reaction-diffusion problems. Such occur frequently in biology and chemistry, see forexample Fife [3] or Murray [11]. Important contributions to asymptoticmethods for evolution problems can be found in Friedman [4], Pazy [12]and Vishik [13].
In this thesis, we use an approach developed in Kozlov, Maz’ya, [6],[7]. Starting with a linear or nonlinear equation or system of equations,the problem is reduced to first order ordinary differential equations withoperator coefficients. Then, by use of a spectral splitting, a finite dimen-sional system of first order ordinary differential equations perturbed by asmall integro-differential term is obtained for the leading term. The maindifficulty is to perform the above reduction and the study of the system ofordinary differential equations for the leading term. The main result is thatthe asymptotic behaviour of solutions of the initial system of equations isdescribed by solutions of the above finite dimensional system.
We use and extend this approach. In paper 1 we consider a non-linear Emden-Fowler type time-independent partial differential equationin a semi-infinite cylinder and study the asymptotics of solutions when the
September 6, 2006 (12:43)
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unbounded coordinate tends to infinity. The equation is complemented bythe Neumann or Dirichlet boundary condition. We analyze the asymptoticbehaviour of a given, small solution of the problem. In the Neumann case,we obtain a nonlinear ordinary differential equation for the leading termin the asymptotics of solution. We also find an estimate for the remainderterm. From this asymptotic formula it follows that the solution behavesasymptotically like a periodic solution. In the Dirichlet case we show thatsmall solutions decrease exponentially. We also consider the case of a star-shaped cross section and show that if the nonlinear term in the equation issubject to some additional constraints, then every bounded solution of thehomogeneous Dirichlet problem vanishes at infinity. The use of Pohozaev’sidentity is essential in the proof.
Paper 2 is devoted to the study of a linear parabolic system of equationsin a bounded domain under Dirichlet boundary conditions and with pre-scribed initial values. We consider the asymptotic behaviour of solutions astime tends to infinity. The elliptic part of the system is here considered asa perturbation of time-independent coefficients. We consider a larger classof perturbations than Kozlov, Maz’ya [6]. Smallness of the perturbationsis assumed only in integral sense. In particular, we include such situationswhen the leading coefficients may take different values on different partsof Ω and the boundaries between them can move with t but stabilize ast → ∞. Here we use another reduction than Kozlov, Maz’ya [6] to ob-tain the first order system of ordinary differential equations perturbed byan integro-differential term for the leading terms. Then an approach fromKozlov [5] is used to study the asymptotic behaviour of solutions to thissystem.
References
[1] R. Dautray, J-L Lions, Mathematical Analysis and NumericalMethods for Science and Technology. Volume 5. Springer-Verlag, 1992.
[2] R. Dautray, J-L Lions, Mathematical Analysis and NumericalMethods for Science and Technology. Volume 6. Springer-Verlag, 1993.
[3] P. C. Fife, Mathematical Aspects of Reacting and Diffusing Systems.Lecture Notes in Biomathematics, 28. Springer-Verlag, 1979.
[4] A. Friedman, Partial differential equations of parabolic type.Prentice-Hall, Inc., Englewood Cliffs, N.J., 1964.
[5] V. Kozlov, Asymptotic representation of solutions to the Dirichletproblem for elliptic systems with discontinuous coefficients near theboundary. Electron. J. Differential Equations 10 (2006), 46 pp.
[6] V. Kozlov, V. Maz’ya, Differential Equations with Operator Coef-ficients. Springer-Verlag, 1999.
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[7] V. Kozlov, V. Maz’ya, An asymptotic theory of higher-order oper-ator differential equations with nonsmooth nonlinearities. Journal ofFunctional Analysis 217 (2004), 448–488.
[8] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Prob-lems and Applications. Volume I. Springer-Verlag, 1972.
[9] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Prob-lems and Applications. Volume II. Springer-Verlag, 1972.
[10] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Prob-lems and Applications. Volume III. Springer-Verlag, 1973.
[11] J. D. Murray, Mathematical Biology. Springer-Verlag, 1993.
[12] A. Pazy, Semigroups of Linear Operators and Applications to PartialDifferential Equations. Springer-Verlag, 1983.
[13] M. I. Vishik, Asymptotic behaviour of solutions of evolutionary equa-tions. Cambridge University Press, 1992.
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Paper 1
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September 6, 2006 (12:43)
Asymptotic analysis of a nonlinear
partial differential equation in a
semicylinder
Peter Rand
Abstract
We study small solutions of a nonlinear partial differential equa-tion in a semi-infinite cylinder. The asymptotic behaviour of thesesolutions at infinity is determined. First, the equation under theNeumann boundary condition is studied. We show that any solutionsmall enough either vanishes at infinity or tends to a nonzero periodicsolution of a nonlinear ordinary differential equation. Thereafter, thesame equation under the Dirichlet boundary condition is studied, butnow the nonlinear term and right-hand side are slightly more generalthan in the Neumann problem. Here, an estimate of the solution interms of the right-hand side of the equation is given. If the equa-tion is homogeneous, then every solution small enough tends to zero.Moreover, if the cross-section is star-shaped and the nonlinear termin the equation is subject to some additional constraints, then everybounded solution of the homogeneous Dirichlet problem vanishes atinfinity. An estimate for the solution is given.
1 Introduction
Let Ω be a bounded domain in Rn−1 with C2-boundary. We define thesemi-infinite cylinder C+ = x = (x′, xn) : x′ ∈ Ω, xn > 0. In Section 2 westudy bounded solutions of the equation
∆U + q(U)U = H in C+ (1.1)
under the boundary condition
∂U
∂ν= 0 on ∂Ω× (0,∞). (1.2)
Our aim is to describe the asymptotic behaviour as xn → ∞ of solutionsU of problem (1.1), (1.2) subject to
|U(x)| ≤ Λ for x ∈ C+, (1.3)
where Λ is a positive constant.
7
8
We assume that q(u) > 0 if u 6= 0. Moreover, q is continuous and
|s|, |t| ≤ Λ ⇒ |q(s)s− q(t)t| ≤ CΛ|s− t| (1.4)
with CΛ < λ1. Here, λ1 is the first positive eigenvalue of the Neumannproblem for the operator
−∆′ = −n−1∑
k=1
∂2
∂x2k
in Ω.We set Ct = Ω × (t, t + 1) and define Lr
loc(C+), 1 ≤ r ≤ ∞, as thespace of functions which belong to Lr(Ct) for every t ≥ 0. We also supposeH ∈ Lp
loc(C+) and∫ ∞
0
(1 + s)‖H‖Lp(Cs) ds < ∞, (1.5)
where p > n/2 if n ≥ 4p = 2 if n = 2, 3.
(1.6)
The main result of Section 2 is Theorem 2.1, which states that one of twoalternatives is valid:
1. U admits the asymptotic representation
U(x) = uh(xn) + w(x) as xn → +∞,
where uh is a nonzero periodic solution of
u′′h + q(uh)uh = 0
and w → 0 as xn → ∞. An estimate for the remainder term w isgiven.
2. U → 0 as xn →∞. An estimate for U is given in the theorem.
If, for example, H = 0 and CΛ → 0 as Λ → 0, then Corollary 2.3 givesthe following estimate for U in the second case:
|U(x′, xn)| ≤ Cεe−√λ1−ε xn ,
where ε is an arbitrary small positive number and Cε is a constant depend-ing on ε.
In Sections 3.1-3.4 we study solutions U of (1.1) subject to (1.3) underthe Dirichlet boundary condition
U = 0 on ∂Ω× (0,∞). (1.7)
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Now we suppose that q is continuous and that
|q(v)| ≤ CΛ if |v| ≤ Λ,
where CΛ < λD. Here, λD is the first eigenvalue of the Dirichlet problemfor −∆′ in Ω. We assume also that ‖H‖Lp(Ct), with p as in (1.6), is abounded function of t, t ≥ 0. The main result is Theorem 3.1 which givesan explicit bound for ‖U‖L∞(Ct) in terms of the function ‖H‖Lp(Ct). Thisimplies in particular that ‖U‖L∞(Ct) → 0 as t → ∞ if the same is validfor ‖H‖Lp(Ct). If H = 0 and q(0)=0, then the estimate from Theorem 3.1implies that
|U(x′, xn)| ≤ Cεe−√λD−ε xn , (1.8)
where ε > 0 is arbitrary.In Section 3.5 we study all bounded solutions of (1.1), (1.7). Here we
suppose additionally that n ≥ 4, that the domain Ω is star-shaped withrespect to the origin, that the function q is continuous with q(0) = 0 andthat q(u) is positive for u 6= 0. We also assume that
n− 32
q(u)u2 − (n− 1)∫ u
0
q(v)v dv ≥ ε q(u)u2 (1.9)
for some ε > 0. Then Theorem 3.4 states that every bounded solutionof (1.1), (1.7) with H = 0 satisfies (1.8). Some examples of functionssatisfying (1.9) are
• q(u) = |u|p, p > 4n−3 ,
• q(u) = |u|pe|u|, p > 4n−3 ,
• q(u) = |u|p(e|u| − 1), p > 7−nn−3 ,
• linear combinations with positive coefficients of the functions above.
A natural question: under which conditions on q is it possible to removeε in the relation (1.8)? The aim of Section 3.6 is to study a similar questionfor the ordinary differential equation
u′′ − λu + q(u)u = 0, (1.10)
where λ > 0, q is continuous with q(0) = 0 but q(u) > 0 for u 6= 0 and
∫ 1
−1
q(u)|u| du < ∞.
Theorem 3.5 states that every solution u of (1.10) subject to u(t) → 0 ast →∞ satisfies
|u(t)|+ |u′(t)| = O(e−√
λt)
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for large positive t.
The problem (1.1) under the boundary conditions (1.2) or (1.7) withq(U) = |U |p−1, p > 1 has been studied in Kozlov [14]. There it is shownthat the restriction (1.3) is essential for Theorems 2.1 and 3.1. One of thegoals of this thesis is to extend some results from [14] to the equation (1.1).
The equation∆u− a|u|q−1u = 0 in C+, (1.11)
where q > 1, a > 0 and with the boundary condition (1.2) is considered inKondratiev [11]. Furthermore, the problem
Lu = 0 in C+
∂u
∂ν+ a|u|q−1u = 0 on ∂Ω× (0,∞),
where L is an elliptic partial differential operator, a > 0 and q > 1 are con-stants is studied in Kondratiev [12]. In both these cases it is proved that thesolutions of these problems have asymptotics of the form u(x′, xn) = Cx−σ
n
with σ > 0. This shows that the minus sign in (1.11) essentially changesthe asymptotic behaviour of solutions at infinity.
There is a lot of research on positive solutions of nonlinear problems inan infinite cylinder and other unbounded domains. We direct the reader toBandle and Essen [3], Berestycki [4], Berestycki, Caffarelli and Nirenberg[5], Berestycki, Larrouturou and Roquejoffre [6], Berestycki and Nirenberg[7] and Kondratiev [13] where also further references can be found.
Small global solutions of the equation
∆u + λu + f(u, ux, uy) = 0
in a two-dimensional strip with homogeneous Dirichlet boundary conditionsare studied in Amick, Toland [2] and Kirchgassner, Scheurle [10].
2 The Neumann problem
2.1 Notation
The Laplace operator and the gradient in Rn are denoted by ∆ and ∇,respectively. For the corresponding operators in Rn−1 we introduce
∆′ =n−1∑
k=1
∂2
∂x2k
and
∇′ =(
∂
∂x1, . . . ,
∂
∂xn−1
).
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By Ω we denote a bounded domain in Rn−1 with C2-boundary and (n−1)-dimensional Lebesgue measure |Ω|. We introduce the cylinder
C = (x′, xn) : x′ ∈ Ω and xn ∈ R
and the semicylinder
C+ = (x′, xn) : x′ ∈ Ω and xn > 0.
We let ν and ν′ denote the outward unit normals to ∂C and ∂Ω, respec-tively. Thus ν ∈ Rn while ν′ ∈ Rn−1 and ν = (ν′, 0). After introducingCt = Ω× (t, t + 1), we say that a function u : Rn → R belongs to Lr
loc(C)or W k,r
loc (C), 1 ≤ r ≤ ∞ and k = 0, 1, . . ., if it belongs to Lr(Ct) or W k,r(Ct)for every t ∈ R.
2.2 Problem formulation and assumptions
Assume that p is subject to (1.6). We study the asymptotic behaviour asxn → ∞ of solutions U ∈ W 2,p
loc (C+) of the problem
∆U + q(U)U = H in C+
∂U
∂ν= 0 on ∂Ω× (0,∞)
(2.1)
satisfying (1.3).We assume that q is continuous and positive for u 6= 0 and satisfies
(1.4). We suppose further that H ∈ Lploc(C+) is subject to (1.5)
In order to motivate (1.6), let us consider a bounded solutionU ∈ W 1,2
loc (C+) of (2.1). By Lemma A.16 in Section A.3 we get thatU ∈ W 2,p
loc (C+). Furthermore, it follows from well-known Sobolev inequali-ties, see for example Theorem 5.6 in Evans [8], that, since p > n/2, thereexists a positive γ such that either U ∈ C0,γ(Ct) or U ∈ C1,γ(Ct) for everyt > 0. Hence it is meaningful to assume that the studied solution belongsto W 2,p
loc (C+) and is bounded.
2.3 The main asymptotic result
The aim of Section 2 is to prove the following theorem concerning theasymptotic behaviour of solutions of (2.1) subject to (1.3):
Theorem 2.1 Suppose that U ∈ W 2,ploc (C+), where p satisfies (1.6), is a
solution of (2.1) subject to (1.3). Suppose also that q is continuous, q(u) > 0if u 6= 0 and that the Lipschitz condition (1.4) is fulfilled. Finally, assumethat H ∈ Lp
loc(C+) satisfies (1.5). Then one of the following alternatives isvalid:
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1. U(x) = uh(xn) + w(x), where uh is a nonzero periodic solution of
u′′h + q(uh)uh = 0
and
‖w‖L∞(Ct) ≤ C
(∫ ∞
t
s‖H‖Lp(Cs) ds
+t
∫ t
0
e−√
λ1−CΛ(t−s)‖H‖Lp(Cs) ds + te−√
λ1−CΛt
)
for t ≥ 1. The right-hand side tends to 0 as t →∞.
2. ‖U(·, xn)‖L∞(Ω) → 0 as xn → ∞. Furthermore, U(x) = u0(xn) +w(x), where
(u′0(t))2
2+
∫ u0(t)
0
q(v)v dv ≤ C
( ∫ ∞
t
‖H‖Lp(Cs) ds (2.2)
+∫ t
0
e−√
λ1−CΛ(t−s)‖H‖Lp(Cs) ds + e−√
λ1−CΛt
)
and
‖w‖L∞(Ct) ≤ C
(∫ ∞
0
e−√
λ1−CΛ|t−s|‖H‖Lp(Cs) ds + e−√
λ1−CΛt
)
(2.3)for t ≥ 1.
The proof of this theorem is contained in Sections 2.5 - 2.9.Theorem 2.1 is a generalization of Theorem 3 in Kozlov [14], where the
case q(U) = |U |p−1, p > 1 is studied. We use the same approach in thisthesis. Since most of the proofs in [14] are brief or absent, we present herecomplete proofs of all assertions. Our restriction (1.5) is different from thecorresponding restriction in Theorem 3 [14]. This rigorous analysis of theproofs indicates that possibly (1.5) is the right assumption also in [14].
In the next section we give some corollaries of Theorem 2.1.
2.4 Corollaries of Theorem 2.1
Corollary 2.2 Suppose, in addition to the conditions in Theorem 2.1, thatthe nonlinear term q has the property that the constant CΛ in (1.4) tendsto 0 as Λ tends to 0. Then the estimates (2.2) and (2.3) can be improved,namely, the constant
√λ1 − CΛ can be replaced by
√λ1 − ε where ε > 0
is arbitrary. In this case, the constant C appearing in (2.2) and (2.3) isdependent of ε.
Proof. Since ‖U(·, xn)‖L∞(Ω) → 0 as xn → ∞ we can apply the theoremfor the semicylinder Ω × (T,∞) where T is sufficiently large and Λ smallenough.
September 6, 2006 (12:43)
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Corollary 2.3 Suppose, in addition to the conditions in Corollary 2.2, thatH = 0. If case 2 in Theorem 2.1 occurs, then, for every ε ∈ (0, λ1), thereexists a constant Cε such that
|U(x′, xn)| ≤ Cεe−√λ1−ε xn for xn ≥ 1. (2.4)
Proof. We begin with proving that there exists a constant Aε such that
|U(x′, xn)| ≤ Aεe− 1
2
√λ1−ε xn for xn ≥ 1. (2.5)
From (2.2) it follows that
|u′0(t)| ≤ Ce−12
√λ1−CΛ t
andu0(t) → 0 as t →∞.
Sinceu0(t) = −
∫ ∞
t
u′0(s) ds
we get|u0(t)| ≤ Ce−
12
√λ1−CΛ t. (2.6)
Furthermore, (2.3) gives
‖w‖L∞(Ct) ≤ Ce−√
λ1−CΛt, (2.7)
where C does not depend on t. Since
U(x) = u0(xn) + w(x)
we get from (2.6) and (2.7) that
|U(x′, xn)| ≤ Ce−12
√λ1−CΛ xn . (2.8)
Using that CΛ → 0 as Λ → 0 and considering problem (2.1) in a semicylin-der Ω× (t0,∞), where t0 is sufficiently large, we can suppose that CΛ < ε.Then the estimate (2.5) follows from (2.8).
We setC′t = Ω× (t + 1/4, t + 3/4).
Lemma A.16 in Section A.3 implies that
‖U‖W 2,2(C′t) ≤ C(‖U‖L2(Ct) + ‖q(U)U‖L2(Ct))
so from (1.4) it follows that
‖U‖W 2,2(C′t) ≤ C‖U‖L2(Ct)
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14
and by using (2.5) we get the estimate
‖U‖W 2,2(C′t) ≤ Ce−12
√λ1−ε t.
Corollary 6.2.5 in Kozlov and Maz’ya [15], with the parameters k− = 0,k+ =
√λ1, m− = 2 and m+ = 1 together with the fact that R can be made
suitably small, implies that
‖U‖W 2,2(C′t) ≤ Ce−√
λ1−ε t. (2.9)
We can now use local estimates and an iteration procedure as in the proofof Lemma A.16 in Section A.3 to obtain
‖U‖L∞(C′′t ) ≤ C‖U‖W 2,2(C′t),
where C′′t = Ω × (t + 3/8, t + 5/8). Combination of the last estimate and(2.9) gives
‖U‖L∞(Ct) ≤ Ce−√
λ1−ε t
if t ≥ 3/8. This implies (2.4).
2.5 The corresponding problem in CWe now begin proving Theorem 2.1. Before turning to equation (2.1), westudy a solution u ∈ W 2,p
loc (C) of the problem
∆u + q(u)u = h in C∂u
∂ν= 0 on ∂C
(2.10)
satisfyingsupx∈C
|u(x)| ≤ Λ, (2.11)
where Λ is the same constant as in (1.3) and p is subject to (1.6). By theSobolev embedding theorem the solution is continuous. Thus we do notneed to use essential supremum.
As before, we assume that q is continuous and positive for u 6= 0 andsatisfies (1.4). We also suppose that h ∈ Lp
loc(C) and that
∫ ∞
−∞(1 + |s|)‖h‖Lp(Cs) ds < ∞. (2.12)
Theorem A.8 in Section A.1 states that there exists an ON-basis ofL2(Ω) consisting of eigenfunctions of the operator −∆′ for the Neumannproblem in Ω. Let φ0(·, xn) denote the eigenfunction with L2(Ω)-normequal to 1 corresponding to the eigenvalue λ0 = 0, i.e. φ0 = |Ω|−1/2. Set u
September 6, 2006 (12:43)
15
to the orthogonal projection of u onto the subspace of L2(Ω) spanned byφ0, that is
u(xn) =1|Ω|
∫
Ω
u(x′, xn) dx′,
and define v(x) by the equality
u(x) = u(xn) + v(x). (2.13)
Inserting (2.13) in (2.10) and integrating over Ω, we obtain
u′′(xn) +1|Ω|
∫
Ω
∆v(x′, xn) dx′ +1|Ω|
∫
Ω
f(u(x′, xn)) dx′ = h(xn), (2.14)
whereh(xn) =
1|Ω|
∫
Ω
h(x′, xn) dx′
andf(t) = q(t)t. (2.15)
Due to the homogeneous boundary condition in (2.10), Greens formulagives ∫
Ω
∆′u dx′ = 0.
Therefore
1|Ω|
∫
Ω
∆v dx′ =1|Ω|
∫
Ω
∆′v dx′ +d2
dx2n
(1|Ω|
∫
Ω
v dx′)
=1|Ω|
∫
Ω
∆′u dx′ = 0.
Equation (2.14) can now be written as
u′′(xn) +1|Ω|
∫
Ω
f(u(x′, xn)) dx′ = h(xn) (2.16)
and by defining
K(u, v)(xn) = f(u(xn))− f(u + v)(xn) (2.17)
we getu′′(xn) + f(u(xn)) = h(xn) + K(u, v)(xn). (2.18)
We will often write K(xn) or K instead of K(u, v)(xn).Using (2.10), the equation ∆v = h − f(u + v) − u′′ is obtained which,
together with (2.16), implies that
∆v = −M(u, v) + h1 in C∂v
∂ν= 0 on ∂C,
(2.19)
September 6, 2006 (12:43)
16
whereM(u, v)(x) = f(u(xn) + v(x))− f(u + v)(xn) (2.20)
andh1 = h− h. (2.21)
The equations (2.18) and (2.19) will play a central role in the sequel.
2.6 An auxiliary ordinary differential equation
In this section we study the equation
ξ′′(t) + q(ξ(t))ξ(t) = g(t), t ≥ t0, (2.22)
where t0 ≥ 1 is given. We assume that q(v) is positive for every v ∈ R,possibly except for v = 0, and continuous. We also assume that q(v)v isLipschitz continuous on every finite interval and that
∫ ∞
t0
|sg(s)| ds < ∞. (2.23)
Since solutions of (2.22) with g = 0 will play an important role in theasymptotic representation of ξ as t → ∞, we will now describe them. Wehave thus the equation
ξ′′(t) + q(ξ(t))ξ(t) = 0, t ≥ t0. (2.24)
Multiplying (2.24) by ξ′ and integrating, we obtain
12(ξ′(t))2 +
∫ t
t0
q(ξ)ξξ′ ds =12(ξ′(t0))2.
Using thatdG(ξ(t))
dt= q(ξ)ξξ′, (2.25)
whereG(v) =
∫ v
0
q(w)w dw,
we get12(ξ′(t))2 + G(ξ(t)) = c0,
where c0 is constant. If
c0 < min∫ ∞
0
q(s)s ds,
∫ −∞
0
q(s)s ds
, (2.26)
then ξ and ξ′ are bounded and ξ is periodic. Conversely, if we know inadvance that ξ is a bounded solution, then c0 satisfies (2.26). This fact andan even more general result will be deduced in the proof of Theorem 2.4.
September 6, 2006 (12:43)
17
We also note that if the right-hand side of (2.26) is infinite, then everysolution of (2.24) is periodic.
In the following theorem we describe the asymptotic behaviour ofbounded solutions of (2.22).
Theorem 2.4 Let ξ be a bounded solution of (2.22). Then one of the twofollowing alternatives occurs:
1. ξ(t) = ξh(t)+w(t), where ξh(t) is a nonzero periodic solution of (2.24)and
|w(t)|+ |w′(t)| = O
(∫ ∞
t
|sg(s)| ds
)(2.27)
as t →∞.
2. Both ξ(t) and ξ′(t) tend to 0 as t →∞ and
(ξ′(t))2
2+
∫ ξ(t)
0
q(v)v dv = O
(∫ ∞
t
|g(s)| ds
). (2.28)
The remaining part of this section is devoted to the proof of this theo-rem. We start with the following lemma:
Lemma 2.5 Let ξ be a bounded solution of (2.22). Then
12(ξ′(t))2 + G(ξ(t)) = c0 + O
(∫ ∞
t
|g(s)| ds
)(2.29)
as t →∞, where c0 is a nonnegative constant depending on t0 and ξ.
Proof. We begin with proving that ξ′ is bounded. Multiplying (2.22) byξ′ and integrating, we obtain
12(ξ′(t))2 −
∫ t
t0
g(s)ξ′(s) ds =12(ξ′(t0))2 −
∫ t
t0
q(ξ(s))ξ(s)ξ′(s) ds. (2.30)
Now (2.25) implies that∫ t
t0
q(ξ(s))ξ(s)ξ′(s) ds = G(ξ(t))−G(ξ(t0)),
where G(ξ(t)) is uniformly bounded in t due to the boundedness of ξ. Thismeans that also the left hand side of (2.30) is bounded in t.
SetCT = sup
t0≤t≤T|ξ′(t)|.
We have for t0 ≤ t ≤ T
∫ t
t0
gξ′ ds ≤ CT
∫ ∞
t0
|g| ds,
September 6, 2006 (12:43)
18
which is finite because of (2.23). Therefore, from (2.30) it follows that
12(ξ′(t))2 − CT
∫ ∞
t0
|g| ds ≤ C
with a constant C independent of t and T . Taking supremum overt0 ≤ t ≤ T we get
12C2
T − CT
∫ ∞
t0
|g| ds ≤ C
which gives an upper bound for CT independent of T . Thus ξ′(t) is boundedfor all t ≥ t0.
The equation (2.30) is equivalent to
12(ξ′(t))2 + G(ξ(t)) =
12(ξ′(t0))2 + G(ξ(t0)) +
∫ t
t0
gξ′ ds. (2.31)
From (2.23) and the boundedness of ξ′, we get∫ t
t0
gξ′ ds = C1 −∫ ∞
t
gξ′ ds = C1 + O
(∫ ∞
t
|g| ds
)
for some constant C1. This equality applied to (2.31) finally implies (2.29).By letting t →∞ it follows that c0 ≥ 0.
Proof of Theorem 2.4: Since ξ(t) is bounded, there exists a number Lsuch that
|ξ(t)| ≤ L, t ≥ t0. (2.32)
We rewrite (2.22) as the system of first order equations
y′1(t) = y2(t)y′2(t) = g(t)− q(y1(t))y1(t),
where y1 = ξ and y2 = ξ′. In polar coordinates, y1 = r cosφ, y2 = r sin φ,the above system takes the form
r′ cos φ− rφ′ sin φ = r sinφ
r′ sin φ + rφ′ cosφ = g − q(r cos φ)r cosφ.
This implies that
r′ = g sin φ + (1− q(r cosφ))r sin φ cosφ
φ′ =g cos φ
r− q(r cosφ) cos2 φ− sin2 φ.
(2.33)
Define the function ρ as
ρ =r2 sin2 φ
2+
∫ r cos φ
0
q(v)v dv (2.34)
September 6, 2006 (12:43)
19
and observe that ρ ≥ 0. Since
ρ(t) =12(ξ′(t))2 +
∫ ξ(t)
0
q(v)v dv,
it follows from Lemma 2.5 that
ρ(t) = c0 + O
(∫ ∞
t
|g(s)| ds
)(2.35)
with c0 ≥ 0. If c0 = 0, then the second alternative in the theorem is valid.Suppose now that c0 > 0.
It is more convenient to use the variables (ρ, φ) instead of (r, φ). Thesecond equation of (2.33) becomes
φ′ =g cos φ
r(ρ, φ)− F (ρ, φ), (2.36)
whereF (ρ, φ) = q(r(ρ, φ) cos φ) cos2 φ + sin2 φ. (2.37)
For readability we will make an abuse of notation and sometimes considerr as a function of t and sometimes of ρ and φ.
We now show that c0 satisfies (2.26). If both integrals in (2.26) areinfinite, this is obvious. Suppose that one of them is finite. First weconclude that there exists a positive number r0 and a t1 ≥ t0 such that
r(t) ≥ r0 (2.38)
for t ≥ t1. Indeed, if r(tj) → 0 for some sequence tj∞j=2 then alsoρ(tj) → 0 which contradicts (2.35). Thus (2.38) follows.
Integrating (2.36) from t1 to t, where t ≥ t1, we obtain
φ(t) = φ(t1) +∫ t
t1
gr−1 cosφ ds−∫ t
t1
q cos2 φds−∫ t
t1
sin2 φds. (2.39)
From (2.23) and (2.38) it follows that the first integral in (2.39) has a finitelimit as t → ∞. Let us show that one of the last two integrals tends toinfinity as t →∞. Suppose that
∫ ∞
t1
sin2 φds < ∞.
Then l(t ≥ t1 : sin2 φ(t) > 1/2) < ∞, where l(D) denotes the Lebesguemeasure of D. Thus l(E) = ∞ for E = t ≥ t1 : cos2 φ(t) ≥ 1/2. From(2.38) and (2.32) it then follows that
r0/√
2 ≤ |r cos φ| ≤ L
September 6, 2006 (12:43)
20
on E. Therefore, there exists a positive constant q0 such that q(r cosφ) ≥ q0
on E, which implies that∫ ∞
t1
q cos2 φds ≥ q0
2l(E) = ∞.
This proves thatφ(t) → −∞ (2.40)
as t →∞.Because of (2.38) and (2.40), we can find sequences τj, sj such
that τj , sj → ∞ as j → ∞ and r(τj) sin φ(τj) = r(sj) sin φ(sj) = 0,r(τj) cos φ(τj) < 0 but r(sj) cos φ(sj) > 0. Equations (2.34) and (2.32)then imply that
ρ(sj) =∫ r(sj) cos φ(sj)
0
q(v)v dv ≤∫ L
0
q(v)v dv <
∫ ∞
0
q(v)v dv.
Analogously,
ρ(τj) ≤∫ −L
0
q(v)v dv <
∫ −∞
0
q(v)v dv.
Since ρ(t) → c0 as t →∞, we have
c0 ≤ min ∫ L
0
q(v)v dv,
∫ −L
0
q(v)v dv
< min∫ ∞
0
q(v)v dv,
∫ −∞
0
q(v)v dv
and (2.26) is proved.By (2.26) there exists an ε such that
0 < ε < min∫ ∞
0
q(v)v dv,
∫ −∞
0
q(v)v dv
− c0.
Next, we prove that there exists positive constants A1, A2, B1 and B2 suchthat
A1 ≤ r(ρ, φ) ≤ A2 (2.41)
andB1 ≤ F (ρ, φ) ≤ B2 (2.42)
if c0 − ε ≤ ρ ≤ c0 + ε. By the same argument as in the proof of theexistence of r0 in (2.38), we can show that there exists a positive A1 suchthat r ≥ A1. To prove the right inequality in (2.41) we introduce r1(φ) asthe unique solution of the equation
ρ(r, φ) = c0 + ε.
That this equation has one solution r = r1(φ) for every φ is a consequencefrom these three facts which all follow from (2.34):
September 6, 2006 (12:43)
21
1. ρ(·, φ) is strictly increasing.
2. ρ(r, φ) → 0 as r → 0 for every φ.
3. We have for φ 6= nπ
ρ →∞ as r →∞
and for φ = nπ
ρ →∫ ±∞
0
q(v)v dv,
where both integrals are larger than c0 + ε.
Since ρ is continuous, so is r1. We can thus define
A2 = maxφ∈[0,2π]
r1(φ)
and conclude that the constant A2 satisfies the right inequality in (2.41).The existence of B2 follows from (2.37) together with the continuity of q
and the right inequality in (2.41). To show the existence of B1 we proceedas follows. Suppose that there exists sequences ρj∞j=1 and φj∞j=1 with
ρj ∈ [c0 − ε, c0 + ε], j = 1, 2, . . .
such that
F (ρj , φj) =q(ρj , φj)r(ρj , φj)2 cos2 φj + r(ρj , φj)2 sin2 φj
r(ρj , φj)2→ 0
as j →∞. Here we use the abbreviation q(ρ, φ) for q(r(ρ, φ) cos φ). Due to(2.41), this implies that
q(ρj , φj) cos2 φj → 0
sin2 φj → 0.
This can happen only if cosφj → 0 and sin φj → 0 which is impossible sothe left inequality in (2.42) is proved.
Let us show that
F (ρ, φ) = F (c0, φ) + O(|ρ− c0|) (2.43)
for |ρ − c0| < ε. In order to make the computations somewhat visuallyclearer, we set
r = r(ρ, φ) r0 = r(c0, φ)q = q(r cos φ) q0 = q(r0 cosφ)
September 6, 2006 (12:43)
22
and obtain
|F (ρ, φ)− F (c0, φ)| =∣∣q cos2 φ− q0 cos2 φ
∣∣
≤∣∣∣∣qr0r cosφ− q0r0r cosφ
rr0
∣∣∣∣
≤ r0|qr cos φ− q0r0 cos φ|+ |r0 − r||q0r0 cos φ|rr0
.
(2.44)
From (2.34) we get that∂r
∂ρ= (rF )−1
and by (2.41) and (2.42), the right-hand side is bounded uniformly in φ.Therefore
|r − r0| ≤ C|ρ− c0|, (2.45)
where C is a constant independent of r, ρ and φ. Using the Lipschitz con-tinuity of the function q(v)v and (2.45), we derive from (2.44) the relation(2.43).
If ρ and φ are considered as functions of t, the representation (2.35)shows that
F (ρ, φ) = F (c0, φ)− h(t),
where
h(t) = O
(∫ ∞
t
|g| ds
). (2.46)
Now choose t2 ≥ t0 such that
|ρ(t)− c0| ≤ ε
if t ≥ t2 and setg1 = r−1g cos φ. (2.47)
The equation (2.36) can now be rewritten as
φ′ = −F (c0, φ) + g1(t) + h(t).
This implies that
∫ φ(t)
φ(t2)
dψ
F (c0, ψ)=
∫ t
t2
(−1 +
g1(s)F (c0, φ(s))
+h(s)
F (c0, φ(s))
)ds
= −t + t2 +∫ t
t2
g1(s)F (c0, φ(s))
ds +∫ t
t2
h(s)F (c0, φ(s))
ds.
(2.48)
September 6, 2006 (12:43)
23
Using (2.41), (2.42) and (2.47), we get
∫ t
t2
g1(s)F (c0, φ(s))
ds = C1 −∫ ∞
t
g cosφ
rF (c0, φ)ds = C1 −O
(∫ ∞
t
|g(s)| ds
).
(2.49)Furthermore, the relations (2.42) and (2.46) imply that
∫ t
t2
h(s)F (c0, φ(s))
ds = C2 + O
(∫ ∞
t
|h(s)| ds
)= C2 + O
(∫ ∞
t
|sg(s)| ds
).
(2.50)By using (2.49) and (2.50), we derive from (2.48) the relation
∫ φ(t)
0
dψ
F (c0, ψ)= c1 − t + O
(∫ ∞
t
|sg(s)| ds
), t ≥ t2. (2.51)
Let us rewrite the equation (2.24) in the variables ρ and φ. The abovecalculations can be used with g = 0. Equations (2.35) and (2.51) become
ρh(t) = a0
∫ φh(t)
0
dψ
F (c0, ψ)= a1 − t, t ≥ t0.
(2.52)
We show that the homogeneous equation (2.24) has a solution which sat-isfies (2.52) with a0 = c0 and a1 = c1. First choose ρh(t0) and φh(t0) suchthat
ρh(t0) = c0
t0 +∫ φh(t0)
0
dψ
F (c0, ψ)= c1.
The last equation is solvable because of (2.42). We can now reconstructξh(t0) and ξ′h(t0) and take these as the Cauchy data for (2.24).
The equality (2.35) and the first equality in (2.52) (with a0 = c0) implythat
ρ(t) = ρh(t) + O
(∫ ∞
t
|g(s)| ds
)(2.53)
and by using (2.42) we obtain
|φ(t)− φh(t)| ≤ B2
∣∣∣∣∫ φ(t)
φh(t)
dψ
F (c0, ψ)
∣∣∣∣.
Furthermore, (2.51) together with the second equation in (2.52) (witha1 = c1) gives
∫ φ(t)
φh(t)
dψ
F (c0, ψ)= O
(∫ ∞
t
|sg(s)| ds
).
September 6, 2006 (12:43)
24
Therefore we have
φ(t) = φh(t) + O
(∫ ∞
t
|sg(s)| ds
). (2.54)
The next step is to study the relation between r and rh = r(ρh, φh). Ex-panding r(ρ, φ) near (ρh, φh) and using (2.53), (2.54), we obtain
r(t) = rh(t) + O
(∫ ∞
t
|sg(s)| ds
).
This, together with the fact that ξ = r cosφ, ξ′ = r sin φ, finally gives
ξ(t) = ξh(t) + O
(∫ ∞
t
|sg(s)| ds
)
ξ′(t) = ξ′h(t) + O
(∫ ∞
t
|sg(s)| ds
)
and the theorem follows. 2
2.7 The equation for v
We now study solutions v of (2.19). In Section 2.5 we introduced M andh1 by (2.20) and (2.21). It is straightforward to check that h1 ∈ Lp
loc(C),∫
Ω
h1(x′, xn) dx′ = 0, (2.55)
∫
Ω
M dx′ = 0, (2.56)
∫
Ω
v(x′, xn) dx′ = 0
andM = w − w, (2.57)
wherew = f(u)− f(u) (2.58)
and f is given by (2.15). Furthermore, by noting that ‖h1‖Lp(Ct) ≤2‖h‖Lp(Ct), we get from (2.12) that
∫ ∞
−∞(1 + |s|)‖h1‖Lp(Cs) ds < ∞. (2.59)
Clearly, the function v belongs to L∞(C) and we have the following result:
September 6, 2006 (12:43)
25
Lemma 2.6 The function v in (2.19) satisfies the estimate
‖v‖L∞(Ct) ≤ C
∫ ∞
−∞e−√
λ1−CΛ|t−s|‖h‖Lp(Cs) ds, (2.60)
where C depends on p, n, Ω, Λ and CΛ.
Proof. We start by proving the inequality
‖v(·, xn)‖L2(Ω) ≤1
2√
λ1 − CΛ
∫ ∞
−∞e−√
λ1−CΛ|xn−s|‖h1(·, s)‖L2(Ω) ds.
(2.61)Defining w as in (2.58), inequality (1.4) implies that |w| ≤ CΛ|v| ≤ 2ΛCΛ.It follows from (2.57) that |M | ≤ 4ΛCΛ, which together with (2.59) givesthat ∫ ∞
−∞e−√
λ1|s|‖ −M + h1‖L2(Cs) ds < ∞.
Also, the equations (2.55) and (2.56) imply that∫
Ω
(−M(u, v)(x′, xn) + h1(x′, xn)) dx′ = 0.
Because of the orthogonality between w − w and w in L2(Ω), we getfrom (2.57) and the Pythagorean theorem that ‖M(u, v)(·, xn)‖L2(Ω) ≤‖w(·, xn)‖L2(Ω). The condition (1.4) then gives
‖M(u, v)(·, xn)‖L2(Ω) ≤ CΛ‖v(·, xn)‖L2(Ω).
This and Lemma A.12 in Section A.2 show that (2.19) has a solution vfulfilling
‖v(·, xn)‖L2(Ω)
≤ 12√
λ1
∫ ∞
−∞e−√
λ1|xn−s|(‖h1(·, s)‖L2(Ω) + CΛ‖v(·, s)‖L2(Ω)
)ds. (2.62)
Inserting the right-hand side of this expression in the last occurrence of‖v(·, s)‖L2(Ω) and iterating, we obtain ‖v(·, xn)‖L2(Ω) ≤
∑∞k=0 Tk, where
Tk =Ck
Λ
(2√
λ1)k+1
×∫
Rk+1e−√
λ1(|xn−t0|+∑k−1
j=0 |tj−tj+1|)‖h1(·, tk)‖L2(Ω) dt0 dt1 · · · dtk.
The variable transformation tj = s + sj+1 for j = 0, . . . , k − 1, tk = sapplied to (2.62), gives
‖v(·, xn)‖L2(Ω) ≤∫ ∞
−∞G(xn − s)‖h1(·, s)‖L2(Ω) ds, (2.63)
September 6, 2006 (12:43)
26
where
G(t) =1
2√
λ1
e−√
λ1|t| +∞∑
k=1
CkΛ
(2√
λ1)k+1
×∫
Rk
e−√
λ1(|t−s1|+|s1−s2|+...+|sk−1−sk|+|sk|) ds1 ds2 . . . dsk. (2.64)
In order to calculate the function G(t) we consider the two differentialoperators− d2
dt2 +λ1−CΛ and− d2
dt2 +λ1. Clearly, their fundamental solutionsare
g(t) =1
2√
λ1 − CΛ
e−√
λ1−CΛ|t|
andh(t) =
12√
λ1
e−√
λ1|t|.
Therefore
−dg2
dt2+ λ1g(t) = δ(t) + CΛg(t),
which implies that g = h ∗ (δ + CΛg), i.e.
g(t) = h(t) + CΛ
∫ ∞
−∞h(t− s)g(s) ds.
Inserting this expression for g(t) into the right-hand side and repeating thisprocedure, we obtain
g(t) =1
2√
λ1
e−√
λ1|t| +∞∑
k=1
CkΛ
(2√
λ1)k+1
×∫
Rk
e−√
λ1(|t−s1|+|s1−s2|+...+|sk−1−sk|+|sk|) ds1 ds2 . . . dsk,
where the right-hand side coincides with the right-hand side of (2.64).Hence,
G(t) =1
2√
λ1 − CΛ
e−√
λ1−CΛ|t|
and (2.61) now follows from (2.63).In the remaining part of this section, let C denote a generic constant
depending only on p, n, Ω, Λ and CΛ. We define C′t = Ω× (t+1/4, t+3/4)and C ′′t = Ω × (t + 1/8, t + 7/8). From Corollary A.18 in Section A.3 itfollows that
‖v‖L∞(C′t) ≤ C(‖v‖Lp(C′′t ) + ‖M‖Lp(C′′t ) + ‖h‖Lp(C′′t )).
Furthermore, since |w| ≤ CΛ|v| and ‖w‖Lp(C′′t ) ≤ |Ω|−1/p‖w‖Lp(C′′t ), it fol-lows from (2.57) that
‖M‖Lp(C′′t ) ≤ CΛ(1 + |Ω|−1/p)‖v‖Lp(C′′t ).
September 6, 2006 (12:43)
27
Thus,‖v‖L∞(C′t) ≤ C(‖v‖Lp(C′′t ) + ‖h‖Lp(C′′t ))
and, after iterating as in the proof of Lemma A.16 in Section A.3, we get
‖v‖L∞(C′t) ≤ C(‖v‖L2(Ct) + ‖h‖Lp(Ct)). (2.65)
We are now in position to perform the last step of the proof of (2.60). Beginby looking at ‖h‖Lp(Ct) and set Ct,1 = Ω× (t, t + 1/2). For τ ∈ (t− 1/2, t)we have ‖h‖Lp(Ct,1) ≤ ‖h‖Lp(Cτ ) and integrating from t− 1/2 to t we get
‖h‖Lp(Ct,1) ≤ 2e√
λ1−CΛ/2
∫ t
t−1/2
e−√
λ1−CΛ|t−τ |‖h‖Lp(Cτ ) dτ.
With a similar estimate of ‖h‖Lp(Ct,2), where Ct,2 = Ω× (t + 1/2, t + 1), weget
‖h‖Lp(Ct) ≤ C
∫ ∞
−∞e−√
λ1−CΛ|t−τ |‖h‖Lp(Cτ ) dτ. (2.66)
To find an estimate for ‖v‖L2(Ct), we use (2.61) together with Minkowski’sinequality yielding
‖v‖L2(Ct) =(∫ t+1
t
‖v(·, τ)‖2L2(Ω) dτ
)1/2
≤ C
∫ ∞
−∞
(∫ t+1
t
e−2√
λ1−CΛ|τ−s| dτ
)1/2
‖h(·, s)‖L2(Ω) ds.
Using the inequality(∫ t+1
t
e−2√
λ1−CΛ|τ−s| dτ
)1/2
≤ Ce−√
λ1−CΛ|t−s|
and making the substitution s 7→ s + τ , we arrive at
‖v‖L2(Ct) ≤ C
∫ ∞
−∞
∫ 1
0
e−√
λ1−CΛ|t−s−τ |‖h(·, s + τ)‖L2(Ω) dτds.
From the inequality∫ 1
0
‖h(·, s + τ)‖L2(Ω) dτ ≤ ‖h‖L2(Cs),
we now obtain
‖v‖L2(Ct) ≤ C
∫ ∞
−∞e−√
λ1−CΛ|t−s|‖h‖L2(Cs) ds.
This, together with (2.65) and (2.66), gives the estimate (2.60) with‖v‖L∞(Ct) replaced by ‖v‖L∞(C′t). Since
‖v‖L∞(Ct) ≤ ‖v‖L∞(C′t−1/2)
+ ‖v‖L∞(C′t) + ‖v‖L∞(C′t+1/2)
,
the inequality for ‖v‖L∞(C′t) implies (2.60) and the proof is complete.
September 6, 2006 (12:43)
28
2.8 Asymptotics of small solutions of problem (2.10)
In this section we continue to study solutions of problem (2.10). Our goalis to find asymptotics of u as xn →∞.
Lemma 2.7 Let u ∈ W 2,ploc (C) be a solution of (2.10) subject to (2.11). Then
either
1. u(x) = uh(xn) + w(x), where uh is a nonzero periodic solution of
u′′h + q(uh)uh = 0 (2.67)
and
‖w‖L∞(Ct) ≤ C
(∫ ∞
t
s‖h‖Lp(Cs) ds
+t
∫ t
−∞e−√
λ1−CΛ(t−s)‖h‖Lp(Cs) ds
)(2.68)
for t ≥ 1
or
2. ‖u(·, xn)‖L∞(Ω) → 0 as xn →∞. If u = u + v as before, then
(u′(t))2
2+
∫ u(t)
0
q(v)v dv
≤ C
(∫ ∞
t
‖h‖Lp(Cs) ds +∫ t
−∞e−√
λ1−CΛ(t−s)‖h‖Lp(Cs) ds
)(2.69)
and
‖v‖L∞(Ct) ≤ C
∫ ∞
−∞e−√
λ1−CΛ|t−s|‖h‖Lp(Cs) ds. (2.70)
The right-hand sides of (2.68), (2.69) and (2.70) tend to 0 as t →∞.
Proof. We set α =√
λ1 − CΛ and, as before, represent u as u = u + v.Let us first study the term u which satisfies the equation (2.18). We denotethe right-hand side of (2.18) by g, i.e.
g(xn) = h(xn) + K(u, v)(xn)
with K given by (2.17). Since
K =1|Ω|
∫
Ω
(q(u)u− q(u)u) dx′,
it follows from (1.4) that |K(xn)| ≤ C‖v(·, xn)‖L1(Ω). From (2.61) we get
|g(xn)| ≤ C
(|h(xn)|+
∫ ∞
−∞e−α|xn−s|‖h1(·, s)‖L2(Ω) ds
). (2.71)
September 6, 2006 (12:43)
29
Let us find a bound for∫∞
t|sg(s)| ds for t ≥ 1. Obviously,
∫ ∞
t
|sh(s)| ds ≤ 1|Ω|
∫ ∞
t
s‖h(·, s)‖L1(Ω) ds ≤ C
∫ ∞
t
s‖h(·, s)‖L2(Ω) ds,
(2.72)and∫ ∞
t
s
∫ ∞
−∞e−α|s−τ |‖h1(·, τ)‖L2(Ω) dτ ds ≤ C
∫ ∞
−∞I1(t, τ)‖h(·, τ)‖L2(Ω) dτ,
(2.73)where
I1(t, τ) =∫ ∞
t
se−α|s−τ | ds. (2.74)
For τ ≤ t, a direct computation yields
I1 = eατ
(te−αt
α+
e−αt
α2
)≤ Cte−α(t−τ). (2.75)
If τ ≥ t, we make the substitution s− τ 7→ s in (2.74) and obtain
I1(t, τ) ≤∫ ∞
−∞|s|e−α|s| ds + τ
∫ ∞
−∞e−α|s| ds ≤ 2(α−2 + α−1)τ. (2.76)
The combination of (2.71), (2.72), (2.73) and the use of the estimates (2.75)and (2.76) for I1 give
∫ ∞
t
|sg(s)| ds ≤ C
(∫ ∞
t
s‖h(·, s)‖L2(Ω) ds
+t
∫ t
−∞e−α(t−s)‖h(·, s)‖L2(Ω) ds
). (2.77)
Next we show that it is possible to replace ‖h(·, s)‖L2(Ω) by ‖h‖L2(Cs)
in (2.77). Denote the right hand side of (2.77) by G(t) and fix a τ ∈ [0, 1].Making the variable substitution s 7→ s + τ we get
G(t) ≤ C
(∫ ∞
t−τ
(s + τ)‖h(·, s + τ)‖L2(Ω) ds
+t
∫ t−τ
−∞e−α(t−s)‖h(·, s + τ)‖L2(Ω) ds
).
Since τ ∈ [0, 1], this implies that
G(t) ≤ C
(t
∫ t
−∞e−α(t−s)‖h(·, s + τ)‖L2(Ω) ds
+∫ ∞
t
s‖h(·, s + τ)‖L2(Ω) ds
).
September 6, 2006 (12:43)
30
Therefore∫ ∞
t
|sg(s)| ds ≤ C
∫ ∞
−∞k(t, s)‖h(·, s + τ)‖L2(Ω) ds, (2.78)
where
k(t, s) =
te−α(t−s) if s < t
s if s ≥ t.
Integrating (2.78) with respect to τ over [0, 1] and using the inequality∫ 1
0
‖h(·, s + τ)‖L2(Ω) dτ ≤ C‖h‖Lp(Cs),
we obtain∫ ∞
t
|sg(s)| ds ≤ C
(∫ ∞
t
s‖h‖Lp(Cs) ds + t
∫ t
−∞e−α(t−s)‖h‖Lp(Cs) ds
).
(2.79)The right-hand side is bounded because of (2.12). Therefore the assumption(2.23) is verified and we can apply Theorem 2.4 on the function u. Henceeither
1. u(xn) = uh(xn) + w1(xn), where uh is a nonzero periodic solution of(2.67) and w1 satisfies (2.27)
or
2. u(xn) → 0 as xn →∞ and (2.28) is valid.
Let us estimate the right-hand sides in (2.27) and (2.28). From (2.79) weget that
|w1(t)| ≤ C
(∫ ∞
t
s‖h‖Lp(Cs) ds + t
∫ t
−∞e−α(t−s)‖h‖Lp(Cs) ds
)
and to obtain the estimate (2.69), we can use that∫ ∞
t
|g(s)| ds ≤ C
(∫ ∞
t
‖h‖Lp(Cs) ds +∫ t
−∞e−α(t−s)‖h‖Lp(Cs) ds
).
This follows from (2.71) and calculations similar to those done to obtainthe estimate (2.79).
Setting w(x) = w1(xn) + v(x) in the first case and using Lemma 2.6 toestimate ‖v‖L∞(Ct), we arrive at (2.68). The estimate (2.70) follows directlyfrom Lemma 2.6.
Let us show that the right-hand sides of (2.68), (2.69) and (2.70) tendto 0 as t →∞. In fact, it is enough to show this for (2.68). Due to (2.12),the first term in the right-hand side of (2.68) tends to 0 as t → ∞. Thesecond term is estimated by
September 6, 2006 (12:43)
31
t
∫ t
−∞e−α(t−s)‖h‖Lp(Cs) ds ≤ C
( ∫ M
−∞e−
α2 (t−s)(1 + |s|)‖h‖Lp(Cs) ds
+∫ t
M
e−α2 (t−s)(1 + |s|)‖h‖Lp(Cs) ds
)(2.80)
≤ C
( ∫ M
−∞e−
α2 (t−s)(1 + |s|)‖h‖Lp(Cs) ds +
∫ ∞
M
(1 + |s|)‖h‖Lp(Cs) ds
)
for any M ≤ t. Given ε > 0, we can choose M so large that the last integralis less than ε/2. The integral
∫ M
−∞e−
α2 (t−s)(1 + |s|)‖h‖Lp(Cs) ds
is then majorized by C exp(−αt/2) (where C depends on M and h) andthe expression in the right-hand side of (2.80) is less than ε if t is largeenough. Hence the right-hand side of (2.68) tends to 0 as t → ∞ and theproof is complete.
2.9 End of the proof of Theorem 2.1
We are now in position to complete the proof of Theorem 2.1.
Proof of Theorem 2.1: Let η(t) be a smooth function with 0 ≤ η ≤ 1,η(t) = 0 if t ≤ 1 and η(t) = 1 if t ≥ 2. We set u(x) = η(xn)U(x). Then
∆u + q(u)u = h in C
∂νu = 0 on ∂C,
whereh = ηH + q(u)u− ηq(U)U + 2η′Uxn + η′′U.
Obviously, ηH ∈ Lploc(C). The functions η′′U and χ = q(u)u− ηq(U)U are
bounded and equal to 0 for xn ≤ 1 and xn ≥ 2. Furthermore, we have thatthe function η′Uxn belongs to Lp(C) and is also equal to 0 for xn ≤ 1 andxn ≥ 2. Thus, the inequality (2.12) follows from (1.5). Now we can applyLemma 2.7 on u. In the first case we have the representation u = uh + w1
where uh and w1 are subject to (2.67) and (2.68), respectively. We set
w =
w1 if xn ≥ 2U − uh if 1 ≤ xn < 20 otherwise.
September 6, 2006 (12:43)
32
Then w ∈ L∞(C+) and for t > 2,
‖w‖L∞(Ct) ≤ C
(∫ ∞
t
s‖H‖Lp(Cs) ds + t
∫ t
2
e−α(t−s)‖H‖Lp(Cs) ds
+t
∫ 2
0
e−α(t−s)‖h‖Lp(Cs) ds
)
≤ C
(∫ ∞
t
s‖H‖Lp(Cs) ds + t
∫ t
0
e−α(t−s)‖H‖Lp(Cs) ds + te−αt
).
It is easy to see that this estimate is valid also for 1 ≤ t ≤ 2 so the firstcase in Theorem 2.1 is proved.
In the second case of Lemma 2.7, we have u = u+ v, where u and v aresubject to (2.69) and (2.70), respectively. We set u0 = u and
w =
v if xn ≥ 2U − u if 1 ≤ xn < 20 otherwise.
The inequalities (2.2) and (2.3) now follow from (2.69) and (2.70) and theproof is complete.
2
3 The Dirichlet problem
3.1 Problem formulation and assumptions
In Section 3 we study bounded solutions of the Dirichlet problem
∆U + q(U)U = H in C+
U = 0 on ∂Ω× (0,∞).(3.1)
We assume that U fulfills
supx∈C+
|U(x)| ≤ Λ (3.2)
and that q is continuous. Let λD be the first eigenvalue of the Dirichletproblem for −∆′ in Ω. We suppose that there is a constant CΛ < λD suchthat
|v| ≤ Λ ⇒ |q(v)| ≤ CΛ. (3.3)
Finally, we assume that H ∈ Lploc(C+), where p satisfies (1.6), and that
‖H‖Lp(Ct) is a bounded function of t, t ≥ 0.
September 6, 2006 (12:43)
33
3.2 The main asymptotic result
The main result of Section 3 is the following theorem concerning the asymp-totic behaviour of solutions U of (3.1).
Theorem 3.1 Assume that U ∈ W 2,ploc (C+), where p satisfies (1.6), is a
solution of (3.1) subject to (3.2). Assume further that q is continuous andthat (3.3) is satisfied. Also, assume that H ∈ Lp
loc(C+) and that ‖H‖Lp(Ct)
is a bounded function of t for t ≥ 0. Then
‖U‖L∞(Ct) ≤ C
(∫ ∞
0
e−√
λD−CΛ|t−s|‖H‖Lp(Cs) ds + e−√
λD−CΛt
), (3.4)
where C is independent of t. In particular, if H = 0 and q(0) = 0, then
‖U‖L∞(Ct) = O(e−√
λD−ε t)
, (3.5)
where ε > 0 is arbitrary.
Remark 3.2 If ‖H‖Lp(Ct) → 0 as t → ∞, then, by the same reasoning asin the end of the proof of Lemma 2.7, we see that the right-hand side of(3.4) tends to 0 as t →∞.
The proof of Theorem 3.1 is similar to the proof of Theorem 2.1 butshorter. It is contained in Section 3.3 and 3.4.
Theorem 3.1 is a generalization of Theorem 2(i) in Kozlov [14]. As inthe Neumann problem, the case q(U) = |U |p−1 is studied in [14].
3.3 The corresponding problem in CAs in the Neumann problem in Section 2, we first consider the problem inthe whole cylinder. Suppose that u ∈ W 2,p
loc (C) is a solution of the problem
∆u + q(u)u = h in Cu = 0 on ∂C (3.6)
satisfyingsupx∈C
|u(x)| ≤ Λ (3.7)
with the same Λ as in (3.2). We have the following:
Lemma 3.3 Let u be a solution of (3.6) fulfilling (3.7) and suppose thath ∈ Lp
loc(C) with ‖h‖Lp(Ct) bounded on R. Then we have the estimate
‖u‖L∞(Ct) ≤ C
∫ ∞
−∞e−√
λD−CΛ|t−s|‖h‖Lp(Cs) ds. (3.8)
September 6, 2006 (12:43)
34
Proof. We consider the boundary value problem−∆v = g in C
v = 0 on ∂C,
where ∫ ∞
−∞e−√
λ1|s|‖g‖L2(Cs) ds < ∞.
From Section A.2 it follows that it has a unique bounded solutionv ∈ W 1,2
loc (C) with
‖v(·, xn)‖L2(Ω) ≤1
2√
λD
∫ ∞
−∞e−√
λD|xn−s|‖g(·, s)‖L2(Ω) ds.
Applying this estimate to the solution u of the problem (3.6), we obtain
‖u(·, xn)‖L2(Ω)
≤ 12√
λD
∫ ∞
−∞e−√
λD|xn−s|(‖h(·, s)‖L2(Ω) + CΛ‖u(·, s)‖L2(Ω)
)ds.
Iterating this estimate in the same way as in the proof of Lemma 2.6 inSection 2.7, we get
‖u(·, xn)‖L2(Ω) ≤1
2√
λD − CΛ
∫ ∞
−∞e−√
λD−CΛ|xn−s|‖h(·, s)‖L2(Ω) ds.
(3.9)Now use the local estimate
‖u‖L∞(C′t) ≤ C(‖u‖L2(Ct) + ‖h‖Lp(Ct))
(compare with (2.65)) and estimate the term ‖u‖L2(Ct) by (3.9). Doing thesame calculations as in Section 2.7 we arrive at (3.8).
3.4 End of the proof of Theorem 3.1
We now complete the proof of Theorem 3.1.
Proof of Theorem 3.1: We use the same smooth function η as in theproof of Theorem 2.1 (page 31) and get, for u = ηU , the problem
∆u + q(u)u = h in C
u = 0 on ∂C
whereh = ηH + q(u)u− ηq(U)U + 2η′Uxn + η′′U.
Clearly, ‖h‖Lp(Ct) is bounded and the estimate (3.4) follows from (3.8).
September 6, 2006 (12:43)
35
Suppose now that H = 0 and q(0) = 0. From (3.4) it follows that
‖U‖L∞(Ct) = O(e−√
λD−CΛt)
.
By choosing T large enough and considering (3.1) in Ω× (T,∞) instead ofC+, we can make Λ and, since q(u) → 0 as u → 0, also CΛ arbitrary small.From this (3.5) follows. 2
3.5 The case of a star-shaped cross-section
In this section we show that under some special assumptions on Ω and q,every bounded solution of (3.1) with H = 0 will satisfy (3.5). This is ageneralization of Theorem 2(iii) in [14] where the case q(U) = |U |p−1 isstudied.
Theorem 3.4 Suppose that n ≥ 4 and Ω is star-shaped with respect tothe origin and has C2-boundary. Also assume that q is continuous withq(0) = 0, and q(u) > 0 otherwise, and that
n− 32
q(u)u2 − (n− 1)∫ u
0
q(v)v dv ≥ ε q(u)u2 (3.10)
for some ε > 0. Then every bounded solution of (3.1) with H = 0 is subjectto (3.5).
Before proving this theorem we give some examples of functions q sat-isfying (3.10). Let us check that all functions of the form
q(u) = f(|u|)|u|a+δ (3.11)
with a = 4/(n − 3), δ > 0 and f being a nondecreasing function satisfy(3.10). Obviously, the function q in (3.11) is even. Therefore also bothsides of the inequality (3.10) are even and we can assume that u ≥ 0. Wehave ∫ u
0
q(v)v dv ≤ f(u)∫ u
0
va+1+δ dv =f(u)ua+2+δ
a + 2 + δ
and by using this inequality we obtain
n− 32
q(u)u2 − (n− 1)∫ u
0
q(v)v dv ≥
n− 32
f(u)ua+2+δ − (n− 1)f(u)ua+2+δ
a + 2 + δ=
bf(u)ua+2+δ,
where
b =δ(n− 3)
2(a + 2 + δ).
September 6, 2006 (12:43)
36
Obviously, b > 0 for every δ > 0 since n ≥ 4. Choosing ε = b, we see that(3.10) is fulfilled.
Here are some examples of functions satisfying (3.11):
(i) q(u) = |u|p, p > 4n−3 .
(ii) q(u) = |u|pe|u|, p > 4n−3 .
(iii) q(u) = |u|p(e|u| − 1), p > 7−nn−3 .
(iv) Linear combinations with positive coefficients of functions from (i) -(iii).
Proof of Theorem 3.4: The function u is a solution of the problem
∆u + q(u)u = 0 in C+
u = 0 on ∂Ω× (0,∞).(3.12)
As before, we set G(u) =∫ u
0q(v)v dv. Using Pohozaev’s identity with
respect to the x′-variables, compare with Section 3.2, Kozlov [14], we get
0 = (∆′u + f(u))(x′ · ∇′u) + uxnxn(x′ · ∇′u)
= div′(∇′u(x′ · ∇′u)− x′
|∇′u|22
+ x′G(u))
+n− 3
2|∇′u|2
− (n− 1)G(u) + uxnxn(x′ · ∇′u).
This implies that
0 = div (∇u(x′ · ∇′u)) + div′(−x′|∇u|2
2+ x′G(u)
)+ u2
xn
+n− 3
2|∇u|2 − (n− 1)G(u).
(3.13)
Set CN = Ω× (1, N) and ΓN = ∂Ω× (1, N) and observe that
∇u =∂u
∂νν on ΓN .
This follows by representing the vector ∇u at a certain point on ΓN as thesum of a normal and a tangent vector. The tangent vector is then zerobecause of the homogeneous Dirichlet boundary condition.
Using that G(u) = 0 on ΓN and integrating (3.13) over CN , we obtain
0 =∫
CN
(n− 3
2|∇u|2 + u2
xn− (n− 1)G(u)
)dx
+12
∫
ΓN
(x · ν)(
∂u
∂ν
)2
dS + F1(N)− F1(1),(3.14)
September 6, 2006 (12:43)
37
where
F1(t) =∫
Ω
uxn(x′ · ∇′u) dx′∣∣∣∣xn=t
.
Multiplying (3.12) by u and using Green’s formula, we have
0 =∫
CN
(−|∇u|2 + q(u)u2)dx + F2(N)− F2(1), (3.15)
where
F2(t) =∫
Ω
uxnu dx′
∣∣∣∣xn=t
.
A linear combination of (3.14) and (3.15) gives
∫
CN
((∂u
∂xn
)2
+n− 3
2q(u)u2 − (n− 1)G(u)
)dx
+12
∫
ΓN
(∂u
∂ν
)2
(x · ν) dS = F (1)− F (N),
(3.16)
whereF (t) = F1(t) +
n− 32
F2(t).
Let us show that F is bounded on [1,∞). For given t > 0, choose r > n.Using the Sobolev embedding theorem, we obtain
‖∇u‖L∞(C′t) ≤ C‖∇u‖W 1,r(C′t) ≤ C‖u‖W 2,r(C′t).
Now, Lemma A.16 in Section A.3 gives
‖u‖W 2,r(C′t) ≤ C(‖q(u)u‖Lr(Ct) + ‖u‖L2(Ct)) ≤ C1.
Therefore, ‖∇u‖L∞(C′t) is a bounded function. This implies that F1 and F2
are bounded functions on [1,∞).Since Ω is star-shaped, it follows that x′ · ν′ ≥ 0. This and (3.10) imply
that every term in (3.16) is non-negative and, from the previous analysis,bounded in N . Thus ∫
C+q(u)u2 dx < ∞ (3.17)
and from (3.15) it therefore follows that∫
Ct
|∇u|2 dx → 0 as t →∞. (3.18)
Corollary A.18 in Section A.3 gives
‖u‖L∞(C′t) ≤ C(‖q(u)u‖Lp(Ct) + ‖u‖L2(Ct)
). (3.19)
September 6, 2006 (12:43)
38
From Poincare’s inequality it follows that
‖u‖L2(Ct) ≤ C‖∇u‖L2(Ct)
and using Holder’s inequality, we get
‖q(u)u‖pLp(Ct)
≤ ‖q(u)1/2u‖L2(Ct)‖q(u)p−1/2|u|p−1‖L2(Ct).
Since u is bounded and q continuous, the right-hand side is also bounded.Applying the last two inequalities for estimating the right-hand side in(3.19), we obtain
‖u‖L∞(C′t) ≤ C(‖q(u)u2‖1/(2p)
L1(Ct)+ ‖∇u‖L2(Ct)
).
Using (3.17) and (3.18) we see that ‖u‖L∞(Ct) → 0 as t →∞.Applying Theorem 3.1 on v(x′, xn) = u(x′, xn + N) for N large enough
completes the proof.2
3.6 An estimate for solutions of a nonlinear ordinarydifferential equation
The estimate (3.5) contains an arbitrary small parameter ε. Is it possibleto remove ε from this estimate? In order to see what kind of requirementswe need on the function q, we consider here the model equation
u′′ − λu + q(u)u = 0. (3.20)
Theorem 3.5 Let u be a solution of (3.20) subject to the condition
u(t) → 0 as t →∞. (3.21)
Suppose that λ > 0, that the function f(u) = q(u)u is Lipschitz continuouswith f(0) = 0 and that q(u) > 0 if u 6= 0. Suppose also that
∫ 1
−1
q(u)|u| du < ∞. (3.22)
Then|u(t)|+ |u′(t)| = O
(e−√
λt)
.
Proof. We may assume that u is not identically zero. Using (3.21), we seefrom (3.20) that u′′(t) → 0 as t →∞. Since
u(t + 1) = u(t) + u′(t)−∫ t+1
t
(s− t− 1)u′′(s) ds,
September 6, 2006 (12:43)
39
we have|u′(t)| ≤ |u(t)|+ |u(t + 1)|+ 1
2sup
[t,t+1]
|u′′(s)|.
Therefore also u′(t) → 0 as t →∞.As before, we use the notation
G(u) =∫ u
0
q(v)v dv.
We multiply (3.20) by u′ and obtain((u′)2 − λu2 + 2G(u)
)′ = 0.
Therefore (u′)2 − λu2 + 2G(u) is a constant. Since the functions u and u′
vanish at ∞ we conclude that this constant is 0. Thus
u′ = ±√
λu2 − 2G(u). (3.23)
Let us show that the function λu2(t)− 2G(u(t)) has no zeros for t largeenough. First, suppose that u(t0) = 0 for some t0 ∈ R. Then (3.23) impliesthat u′(t0) = 0 and we get u(t) ≡ 0 by the uniqueness of solutions to theCauchy problem of equation (3.20). Furthermore, define
Q(u) =1
λu2
∫ u
0
q(v)v dv.
Since
0 ≤ Q(u) ≤ 1λ
∫ u
0
q(v)v
dv,
we have, because of (3.21) and (3.22), that
Q(u(t)) → 0 as t →∞.
This implies that λu2− 2G(u) is positive for large t. From (3.23) it followsthat u′(t) has no zeros for large t.
Since u(t) → 0 as t →∞, we meet here one of two possibilities: Eitheru > 0 and u′ < 0 or else u < 0 and u′ > 0. We consider the first possibility.The second one is considered analogously. We have
u′ = −√
λu√
1− 2Q(u). (3.24)
Power series expansion yields that there exists an ε > 0 such that∣∣∣∣
1√1− 2x
− 1∣∣∣∣ ≤ 2|x| (3.25)
if |x| ≤ ε. Furthermore, since Q(v) → 0 as v → 0, there exists a δ > 0 suchthat |Q(v)| ≤ ε if |v| ≤ δ. Choose t so large that |u(t)| ≤ δ. Integrating(3.24), we obtain
1√λ
∫ δ
u
dv
v√
1− 2Q(v)= t− C1. (3.26)
September 6, 2006 (12:43)
40
The left-hand side can be written as
1√λ
∫ δ
u
dv
v+
1√λ
∫ δ
u
1v
(1√
1− 2Q(v)− 1
)dv. (3.27)
Furthermore, we get from (3.25) that∣∣∣∣∣∫ δ
u
1v
(1√
1− 2Q(v)− 1
)dv
∣∣∣∣∣ ≤ 2∫ 1
−1
Q(v)|v| dv < ∞, (3.28)
where the last inequality follows from (3.22). From (3.26), (3.27) and (3.28)it follows that
ln u = B(t)−√
λ t,
where B is a bounded function. Therefore u(t) = O(e−√
λt). Using (3.24),
we also obtain that u′(t) = O(e−√
λt).
A Some results from functional analysis
A.1 Eigenvalues and eigenvectors of −∆
Here we show that if ∂Ω is smooth enough there exists an ON-basis of L2(Ω)consisting of eigenvectors of the operator −∆. For the Dirichlet problemthis result is often proved in textbooks in partial differential equations sowe focus on the Neumann problem and prove it since in this case the proofis not easily found. On our way we need some lemmas which will be statedwithout proof.
Throughout this appendix, let H denote a Hilbert space and 〈·, ·〉H theinner product in H. If A is a bounded linear operator on H, we let σ(A)denote the spectrum of A and σp(A) the set of eigenvalues of A, i.e. thepoint spectrum.
The following lemmas are well-known results from functional analysis.
Lemma A.1 Let Ω be an open set in R. Then C∞0 (Ω) is dense in L2(Ω).
Definition A.2 If X is a normed space we say that a subset M ⊂ X istotal in X if the span of M is dense in X.
Lemma A.3 If M is total in X and for some x ∈ X we have that x⊥M ,then x = 0. Conversely, if X is complete and x⊥M implies x = 0, then Mis total in X.
Lemma A.4 Suppose K is a compact linear operator on H. Then N(I−K)is finite dimensional.
If dim H = ∞ then also
September 6, 2006 (12:43)
41
(i) 0 ∈ σ(K).
(ii) σ(K)− 0 = σp(K)− 0.
(iii) σ(K)−0 is either finite or countable with the unique limit point 0.
Lemma A.5 Suppose A is a linear, bounded and symmetric operator onH and define
m = inf〈Au, u〉H : u ∈ H, ‖u‖ = 1M =sup〈Au, u〉H : u ∈ H, ‖u‖ = 1.
Then m, M ∈ σ(A) ⊂ [m,M ].
Remark A.6 Since A is symmetric, 〈Au, u〉H is always real.
Lemma A.7 Suppose H is separable and A : H → H is linear, bounded,symmetric and compact. Then there exists a countable orthonormal basisof H consisting of eigenvectors of A.
Theorem A.8 Let Ω be an open, bounded region in Rn with C1-boundary.Then there exists an ON-basis φk∞k=0 of L2(Ω) consisting of eigenfunc-tions of the operator −∆ for the Neumann problem, i.e.
∫
Ω
∇φk · ∇v dx = λk
∫
Ω
φkv dx, ∀v ∈ H1(Ω), (A.1)
and the eigenvalues λk are subject to
0 = λ0 < λ1 ≤ λ2 ≤ λ3 ≤ . . . , λk →∞ as k →∞, (A.2)
where each eigenvalue is repeated according to its multiplicity.Moreover, φk∞k=0 is also an orthogonal basis of H1(Ω).
Proof. Given u ∈ H1(Ω), the functional v 7→ ∫Ω
vu dx is linear andbounded on H1(Ω). According to Riesz representation theorem, there ex-ists a unique Tu ∈ H1(Ω) such that
∫
Ω
vu dx = 〈v, Tu〉H1(Ω), ∀v ∈ H1(Ω). (A.3)
We therefore define the operator T : H1(Ω) → H1(Ω) by the relation (A.3).It is easy to see that T is linear and from the closed graph theorem it followsthat T is bounded. Namely, let Γ = (x, Tx) : x ∈ H1(Ω) be the graphof T and suppose that (xn, yn) ∈ Γ, (xn, yn) → (x, y) in H1(Ω) ×H1(Ω).From (A.3) it follows that 〈yn − Tx, yn − Tx〉H1(Ω) =
∫Ω(yn − Tx)(xn − x)
which, together with Holder’s inequality, shows that yn → Tx and thusy = Tx. Hence the graph is closed.
September 6, 2006 (12:43)
42
T is also symmetric, because the equality 〈Tu, v〉H1(Ω) = 〈u, Tv〉H1(Ω)
follows by combining the fact that 〈Tu, v〉H1(Ω) = 〈v, Tu〉H1(Ω) with therelation (A.3). Finally, T is compact: let i : H1(Ω) → L2(Ω) be theinclusion of H1(Ω) in L2(Ω) and T1 : L2(Ω) → H1(Ω) be the extensionof T to L2(Ω) as defined by the left-hand side of (A.3). Since H1(Ω) iscompactly embedded in L2(Ω) and T1, as T , is bounded, it follows thatT = T1 i is compact.
Lemma A.7 now gives that there exists a countable orthonormal ba-sis ψj∞j=0 of H1(Ω) consisting of eigenfunctions of T and we will nowprove that, after normalization, the same set is an ON-basis for L2(Ω).Let µj be the eigenvalue corresponding to ψj . From (A.3) we get that∫Ω
ψjψk dx = µk〈ψj , ψk〉H1(Ω) so the eigenfunctions are orthogonal to eachother also in L2(Ω). By setting
φj =ψj
‖ψj‖L2(Ω),
we thus get an orthonormal sequence φj∞j=0 in L2(Ω), each φj also aneigenfunction of T with the same eigenvalue as ψj . It remains to see thateach function in L2(Ω) can be written as a (finite or infinite) linear combi-nation of them.
Set M = φj∞j=0. The fact that M is a basis of H1(Ω) implies that M
is total in H1(Ω). Since H1(Ω) is dense in L2(Ω) according to Lemma A.1,M is also total in L2(Ω). For f ∈ L2(Ω) we construct
g =∞∑
k=0
〈f, φk〉L2(Ω)φk,
which is convergent in L2(Ω). We get immediately 〈f − g, φj〉L2(Ω) = 0 forevery j so Lemma A.3 gives that f = g. Thus φj∞j=0 is an ON-basis forL2(Ω) and the coordinates of f are its usual Fourier coefficients.
The final step is to investigate the eigenvalues of T . From (A.3) itfollows that
〈Tu, u〉H1(Ω) = ‖u‖2L2(Ω) (A.4)
so Lemma A.5 gives that σ(T ) ⊂ [0, 1] with 0, 1 ∈ σ(T ). (A.4) also showsthat 0 is not an eigenvalue. Furthermore, if λ is an eigenvalue, LemmaA.4 shows that N(λI − T ) = N(I − λ−1T ) is finite dimensional. But sincedim H1(Ω) = ∞ and the eigenfunctions form a basis, we conclude that Thas infinitely many eigenvalues. Lemma A.4 also gives that 1 really is aneigenvalue (with constants as eigenfunctions) and that we can label theeigenvalues in decreasing order as
1 = µ0 ≥ µ1 ≥ . . . > 0, µk → 0 as k →∞. (A.5)
We see that
〈u, v〉H1(Ω) =∫
Ω
fv dx, ∀v ∈ H1(Ω)
September 6, 2006 (12:43)
43
if and only ifu = Tf.
This shows that if φk is an eigenfunction of T with eigenvalue µk, then itis also a solution of (A.1) with
λk =1µk
− 1
and thus an eigenfunction of the operator −∆ for the Neumann problem.We also see from (A.1) that the only eigenfunctions to λ0 = 0 are theconstant functions. Hence λ0 has single multiplicity and (A.2) follows from(A.5).
Theorem A.9 Let Ω be a bounded domain in Rn. Then there exists anON-basis φk∞k=1 of L2(Ω) where φk ∈ H1
0 (Ω) are eigenfunctions of theoperator −∆ for the Dirichlet problem, i.e.
∫
Ω
∇φk · ∇v dx = λk
∫
Ω
φkv dx, ∀v ∈ H10 (Ω),
and the eigenvalues λk are subject to
0 < λ1 ≤ λ2 ≤ λ3 ≤ . . . , λk →∞ as k →∞,
where each eigenvalue is repeated according to its multiplicity.Also, φk∞k=0 is an orthogonal basis of H1
0 (Ω).
Proof. We prove only the last part of the theorem because the remainingparts of the complete proof can be found in Evans [8]. For u ∈ H1
0 (Ω) wehave the equality
〈u, φj〉H10 (Ω) = (1 + λj)〈u, φj〉L2(Ω)
so if u ∈ H10 (Ω) is orthogonal to φj∞1 in H1
0 (Ω), it is also orthogonal toφj∞1 in L2(Ω). From this and Lemma A.3 it now follows that φj∞1 istotal in H1
0 (Ω) and that it in fact is an orthogonal set.
A.2 Existence and uniqueness of bounded solutions ofPoisson’s equation in C
In this section we will see that under some conditions the problem−∆u = f in C
∂u
∂ν= 0 on ∂C
has a unique bounded solution with mean value 0 over Ω and also derive abound for its L2(Ω)-norm. We then turn to the problem
−∆u = f in C
u = 0 on ∂C (A.6)
September 6, 2006 (12:43)
44
and state that practically the same results hold in this case. We use thetheory of eigenvalues of the operator −∆′ in Ω presented in Section A.1and construct the solution in each case as an infinite linear combinationof the eigenfunctions. We denote the first positive eigenvalue by λ1 in theNeumann case and by λD in the Dirichlet case.
Lemma A.10 Suppose f ∈ L2loc(C) fulfills
∫ ∞
−∞e−√
λ1|s|‖f‖L2(Cs) ds < ∞.
Then also ∫ ∞
−∞e−√
λ1|s|‖f(·, s)‖L2(Ω) ds < ∞.
Proof. First note that if a ≥ 0 and |x− y| ≤ c for some c, then
e−a|y| ≤ eace−a|x|. (A.7)
This follows from the fact that |x| − |y| ≤ |x− y|.We write g(t) = ‖f(·, t)‖L2(Ω) and observe that for a ≥ 0, Holder’s
inequality gives∫ a
0
‖f(·, t + s)‖L2(Ω) ds = ‖g‖L1(t,t+a) ≤√
a‖f‖L2(Ω×(t,t+a)). (A.8)
It is easy to see that∫ ∞
−∞e−√
λ1|s|‖f(·, s)‖L2(Ω)ds =∫ 1
0
∫ ∞
−∞e−√
λ1|s+t|‖f(·, s + t)‖L2(Ω) ds dt
by noting that the value of the inner integral on the right-hand side isindependent of t. By applying (A.7) and (A.8) on the right-hand side wearrive at
∫ ∞
−∞e−√
λ1|s|‖f(·, s)‖L2(Ω)ds ≤ e√
λ1
∫ ∞
−∞e−√
λ1|s|‖f‖L2(Cs) ds
from which the lemma now directly follows.
Remark A.11 It follows from (A.7) with a =√
λ1, c = |xn|, x = s andy = s− xn that
e−√
λ1|xn−s| ≤ e√
λ1|xn|e−√
λ1|s|.
Under the same conditions as in Lemma A.10 we therefore also have that∫ ∞
−∞e−√
λ1|xn−s|‖f(·, s)‖L2(Ω) ds < ∞
for every xn ∈ R.
September 6, 2006 (12:43)
45
For functions v ∈ H1(Ω) subject to∫
Ω
v dx = 0 (A.9)
the following inequality holds:
‖∇′v‖2L2(Ω) ≥ λ1‖v‖2L2(Ω). (A.10)
In order to see this, let us consider the ON-basis φj∞j=0 of L2(Ω) consistingof eigenfunctions of the Neumann problem for −∆′ in Ω corresponding tothe eigenvalues 0 = λ0 < λ1 ≤ λ2 ≤ . . . (see Theorem A.8 in Section A.1).We have that
v =∞∑
j=0
vj ,
where vj = 〈v, φj〉L2(Ω)φj . Since φ0 = const, it follows from (A.9) thatv0 = 0. The set φj also forms an orthogonal basis of H1(Ω) so
∇′v =∞∑
j=1
∇′vj
and (A.10) now follows from Parseval’s identity together with the identity
〈∇′vi,∇′vj〉L2(Ω) = λi〈vi, vj〉L2(Ω)
obtained from Greens formula.
Lemma A.12 Suppose f ∈ L2loc(C) satisfies
∫
Ω
f(x′, xn) dx′ = 0 for a.e. xn ∈ R (A.11)
and ∫ ∞
−∞e−√
λ1|s|‖f‖L2(Cs) ds < ∞.
Then the problem −∆u = f in C
∂u
∂ν= 0 on ∂C
(A.12)
has a bounded solution u ∈ W 1,2loc (C) with
∫
Ω
u(x′, xn) dx′ = 0 for a.e. xn ∈ R (A.13)
and
‖u(·, xn)‖L2(Ω) ≤1
2√
λ1
∫ ∞
−∞e−√
λ1|xn−s|‖f(·, s)‖L2(Ω) ds, (A.14)
the last expression being finite. This solution is unique up to a constant.
September 6, 2006 (12:43)
46
Remark A.13 Since ∂Ω ∈ C2, it follows from Section A.3 that iff ∈ Lr
loc(C) for some r ∈ [2,∞), then the fact that u ∈ W 1,2loc (C) implies
that u ∈ W 2,rloc (C), i.e. there is no difference between a weak and a strong
solution of (A.12).
Proof. Lemma A.10 gives that the right hand side of (A.14) is finite (seethe remark). We begin by finding a solution of (A.12). According toTheorem A.8 in Section A.1 there exists an orthonormal basis φj∞j=0 ofL2(Ω) consisting of eigenfunctions of the Neumann problem in Ω, i.e.
−∆′φj = λjφj in Ω
∂φj
∂ν′= 0 on ∂Ω
(A.15)
with λ0 = 0 and all other λj > 0. Defining
fj =∫
Ω
fφj dx′, j = 1, 2, . . . ,
we get, since f0 = 0 by (A.11),
f(x′, xn) =∞∑
j=1
fj(xn)φj(x′).
We also define
uj(t) =1
2√
λj
∫ ∞
−∞e−√
λj |t−s|fj(s) ds, j = 1, 2, . . .
and set for N ≥ 1
u(N)(x′, xn) =N∑
j=1
uj(xn)φj(x′),
f (N) =N∑
j=1
fj(xn)φj(x′).
It then follows that
−∆u(N) = f (N).
Then, using Parseval’s identity, Minkowski’s inequality (see for exampleSection 6.3, Folland [9] or Section 2.4, Lieb and Loss [16]) and Parseval
September 6, 2006 (12:43)
47
again yields
‖u(N)(·, xn)‖L2(Ω) =( N∑
j=1
uj(xn)2)1/2
≤
N∑
j=1
(1
2√
λj
)2 (∫ ∞
−∞e−√
λj |xn−s||fj(s)| ds
)2
1/2
≤ 12√
λ1
∫ ∞
−∞
e−2
√λ1|xn−s|
N∑
j=1
fj(s)2
1/2
ds
≤ 12√
λ1
∫ ∞
−∞e−√
λ1|xn−s|‖f(·, s)‖L2(Ω) ds (A.16)
which shows that u(N)N is a Cauchy sequence in L2(Ω). When workingwith u′j instead of uj and noting that
u′j(t) ≤12
∫ ∞
−∞e−√
λ1|t−s||fj(s)| ds,
it can be proved in the same way that also ∂u(N)/∂xnN is Cauchy inL2(Ω).
Write
‖∇′u(N)(·, xn)‖L2(Ω) = 〈−∆′u(N)(·, xn), u(N)(·, xn)〉1/2L2(Ω)
=( N∑
j=1
λjuj(xn)2)1/2
≤
N∑
j=1
14
(∫ ∞
−∞e−√
λ1|xn||fj(s)| ds
)2
1/2
and proceed as before to obtain
‖∇u(N)(·, xn)‖L2(Ω) ≤12
∫ ∞
−∞e−√
λ1|xn−s|‖f(·, s)‖L2(Ω) ds. (A.17)
This proves that also ∇′u(N)N is Cauchy in L2(Ω). The conclusion isthat u(N) converges in W 1,2(Ω). To show the convergence in W 1,2
loc (C), weestimate ‖u(N)‖W 1,2(Ct) for a fixed t in the same way as above. Using
September 6, 2006 (12:43)
48
(A.16), Minkowski’s inequality and (A.7), we get
‖u(N)‖L2(Ct) =(∫ t+1
t
‖u(N)(·, xn)‖2L2(Ω) dxn
)1/2
≤ 12√
λ1
∫ ∞
−∞
(∫ t+1
t
e−2√
λ1|xn−s|‖f(·, s)‖2L2(Ω) dxn
)1/2
ds
≤ C
∫ ∞
−∞‖f(·, s)‖L2(Ω)
(∫ t+1
t
e−2√
λ1|t−s| dxn
)1/2
ds
≤ Ce√
λ1|t|∫ ∞
−∞e−√
λ1|s|‖f(·, s)‖L2(Ω) ds.
The right-hand side is finite from Lemma A.10 so u(N) is Cauchy also inL2
loc(C).In the same way it follows from (A.17) that ∇u(N) is Cauchy in
L2loc(C) and that u(N) converges to some function u in W 1,2
loc (C). Regularitytheory gives that u solves (A.12).
Furthermore, φ0 is constant so due to the orthogonality of φj we have∫Ω
u(N) dx′ = 0. Thus, the property (A.13) follows from the convergencein L2(Ω).
To prove the assertion about uniqueness, we now show that the onlybounded solutions of
∆u = 0 in C∂u
∂ν= 0 on ∂C
are the constant functions. Using the decomposition
u(x) =∞∑
j=0
uj(xn)φj(x′)
and (A.15), we get
0 = −∫
Ω
φj∆u dx′ = −∫
Ω
u∆′φj dx′ −∫
Ω
φjuxnxn dx′
= λjuj(xn)− u′′j (xn), j = 0, 1, 2, . . . . (A.18)
If j 6= 0, then λj > 0 and in these cases the only bounded solution of(A.18) is uj = 0. Furthermore, λ0 = 0, so in the case j = 0 we get theonly bounded solutions as u0(xn) = A, where A is constant. Since also φ0
is constant, we get that u is a constant function.We now prove (A.14). We assume that u is non-vanishing (so that
‖u(·, xn)‖L2(Ω) is twice differentiable). In other case, we can replace‖u(·, xn)‖L2(Ω) by (‖u(·, xn)‖2L2(Ω) + ε)1/2 in the calculations below andthen let ε → 0 to obtain the same result.
September 6, 2006 (12:43)
49
For clarity, we skip the index L2(Ω) in the norm and inner productnotations. Since u ∈ W 1,2
loc (C) we get that u(·, xn) ∈ W 1,2(Ω) for almostevery xn. When multiplying (A.12) by u, integrating over Ω and usingGreen’s theorem, we arrive at
∫
Ω
|∇′u|2 dx′ −∫
Ω
u∂2u
∂x2n
dx =∫
Ω
fu dx.
It follows, by (A.10) together with Cauchy-Schwartz inequality, that⟨
∂2u
∂x2n
, u
⟩≥ λ1‖u‖2 − ‖f‖‖u‖. (A.19)
Differentiating ‖u(·, xn)‖ twice gives
‖u‖d2‖u‖dx2
n
=⟨
∂2u
∂x2n
, u
⟩− ‖u‖−2
(⟨∂u
∂xn, u
⟩2
−∥∥∥∥
∂u
∂xn
∥∥∥∥2
‖u‖2)
.
The Cauchy-Schwartz inequality gives that 〈∂xnu, u〉2 − ‖∂xnu‖2‖u‖2 ≤ 0so from (A.19)
d2‖u‖dx2
n
≥ λ1‖u‖ − ‖f‖. (A.20)
Consider the equation
− d2
dx2n
‖u(·, xn)‖+ λ1‖u(·, xn)‖ = g(xn). (A.21)
By using a Green function and the fact that u is bounded, we get
‖u(·, xn)‖ =1
2√
λ1
∫ ∞
−∞e−√
λ1|xn−s|g(s) ds.
But from (A.20) and (A.21) we have g(xn) ≤ ‖f(·, xn)‖ so (A.14) nowdirectly follows.
We now turn to the Dirichlet problem (A.6) and have the followinganalogue of Lemma A.12:
Lemma A.14 Suppose f ∈ L2loc(C) is subject to
∫ ∞
−∞e−√
λ1|s|‖f‖L2(Cs) ds < ∞.
Then the problem (A.6) has a unique, bounded solution u ∈ W 1,2loc (C) and
‖u(·, xn)‖L2(Ω) ≤1
2√
λD
∫ ∞
−∞e−√
λD|xn−s|‖f(·, s)‖L2(Ω) ds,
where the last expression is finite.
September 6, 2006 (12:43)
50
Remark A.15 Since all eigenvalues of the Dirichlet problem are positive,we do not have the conditions that f and u are orthogonal to constants inL2(Ω) as in the Neumann problem.
Proof. The proof is similar to the proof of Lemma A.12. A difference ishowever that the estimate (A.10), with λ1 replaced by λD, is now valid forall v ∈ W 1,2
0 (C).
A.3 A local estimate for solutions of Poisson’s equa-tion
We say that D is a cylindrical type domain (CTD) if D = Ω× (a, b), where−∞ < a < b < ∞. Furthermore, if D = Ω× (a, d) and D′ = Ω× (b, c) aretwo CTD:s, we say that D′ is compactly contained in D in the xn-directionif a < b < c < d.
Given a solution u ∈ W 1,2loc (C) to some of the problems
∆u = f in C∂u
∂ν= 0 on ∂C
(A.22)
or ∆u = f in C
u = 0 on ∂C, (A.23)
where f ∈ Lploc(C), we will see that in fact u ∈ W 2,p
loc (C) and find a boundfor ‖u‖W 2,p(C′t) expressed in terms of ‖u‖L2(Ct) and ‖f‖Lp(Ct).
Lemma A.16 Suppose that p fulfills (1.6) and that u ∈ W 1,2loc (C) is a solu-
tion of (A.22) or (A.23). Suppose further that E1 and E2 are CTD:s andthat E1 is compactly contained in E2 in the xn-direction. Suppose also thatE2 ⊂ Ct for some t ∈ R. Then u ∈ W 2,p
loc (C) and
‖u‖W 2,p(E1) ≤ C0(‖u‖L2(E2) + ‖f‖Lp(E2)), (A.24)
for some constant C0 depending on p, n, E1 and E2 but not on t.
Remark A.17 The assumption that E2 ⊂ Ct for some t ∈ R is not essentialfor the result but indicates that the constant C0 is independent of t.
Proof. Given t ∈ R, make first the transformations v(x′, xn) =u(x′, xn + t), g(x′, xn) = f(x′, xn + t) and note that the equation ∆u = fon Ct is equivalent to the equation ∆v = g on C0. From Theorem 15.1” inAgmon, Douglis and Nirenberg, [1], we get the interior estimate
‖v‖W 2,r(D′) ≤ C(‖v‖Lr(D) + ‖g‖Lr(D)) (A.25)
if r ∈ (1,∞) and D′ ⊂⊂ D ⊂ C0. The constant C depends only onD, D′ and r. Furthermore, suppose x is a point on ∂Ω × (0, 1). After
September 6, 2006 (12:43)
51
straightening the boundary in a neighborhood of x, Theorem 15.3 [1] andthe remark immediately after it gives the estimate (A.25) for D′ and Dequal to hemispheres located at x. Using a partition of unity we finally getthe estimate
‖v‖W 2,r(D1) ≤ C(‖v‖Lr(D2) + ‖g‖Lr(D2)) (A.26)
for any D1, D2 such that D1 is compactly contained in D2 ⊂ C0 in thexn-direction. The constant C depends on r, D1 and D2 but not on v andg. We can therefore revert to the functions u and f and get the inequality
‖u‖W 2,r(D1,t) ≤ C(‖u‖Lr(D2,t) + ‖f‖Lr(D2,t)), (A.27)
where D1,t and D2,t are equal to D1 and D2 translated t steps in the xn-direction and C is the same constant as in (A.26), i.e. independent oft.
If n = 2 or 3, the lemma follows from (A.27) with r = 2, sincep = 2 in these cases. Now suppose ω is any bounded open subset ofRn with Lipschitz boundary and introduce φ(x) = nx/(n − 2x) with in-verse ψ(x) = nx/(n + 2x). A well-known Sobolev inequality states that ifr < n/2 and s = φ(r), then
‖u‖Ls(ω) ≤ C‖u‖W 2,r(ω) (A.28)
and if r > n/2‖u‖L∞(ω) ≤ C‖u‖W 2,r(ω). (A.29)
Set E01 = E1 and choose a sequence Ek
1 ∞k=1 such that Ek1 is compactly
contained in Ek+11 as well as in E2 in the xn-direction for k ≥ 0. We set
p1 = p, p2 = ψ(p) and notice that p2 < n/2. Using (A.27), (A.28) and then(A.27) again we obtain
‖u‖W 2,p(E1) ≤ C1(‖u‖Lp(E11) + ‖f‖Lp(E2))
≤ C2(‖u‖W 2,p2 (E11) + ‖f‖Lp(E2))
≤ C3(‖u‖Lp2 (E21) + ‖f‖Lp(E2)). (A.30)
We now continue the iteration indicated in (A.30) in the following way: fork ≥ 2, set pk+1 = ψ(pk) and combine again (A.27) and (A.28) to obtain
‖u‖Lpk (Ek1 ) ≤ C(‖u‖Lpk+1 (Ek+1
1 ) + ‖f‖Lp(E2)).
Obviously, p1 > p2 > . . . and eventually pK ≤ 2 for some K. We canassume pK > 1, since if pK = ψ(pK−1) ≤ 1, then pK−1 can be increased sothat the condition pK > 1 is met. We get
‖u‖W 2,p(E1) ≤ C(‖u‖LpK (EK1 ) + ‖f‖Lp(E2)),
from which (A.24) follows.
September 6, 2006 (12:43)
52
Corollary A.18 Under the same conditions as in Lemma A.16
‖u‖L∞(E1) ≤ C(‖u‖L2(E2) + ‖f‖Lp(E2)).
Proof. Since p > n/2, the corollary follows by combining Lemma A.16with the Sobolev inequality (A.29).
Remark A.19 The lemma as well as the corollary also hold with C and∂C replaced by C+ and ∂Ω × (0,∞), respectively, but with the additionalcondition t ≥ 0.
References
[1] S. Agmon, A. Douglis, L. Nirenberg, Estimates Near the Bound-ary for Solutions of Elliptic Partial Differential Equations SatisfyingGeneral Boundary Conditions. I. Communications on pure and appliedmathematics 12 (1959), 623–727.
[2] C. J. Amick, J. F. Toland, Nonlinear elliptic eigenvalue problemson an infinite strip - global theory of bifurcation and asymptotic bi-furcation. Math. Ann. 262 (1983), 313–342.
[3] C. Bandle, M. Essen, On the positive solutions of Emden equationsin cone-like domains. Arch. for Rat. Mech. 12 (1990), 319–338.
[4] H. Berestycki, Some nonlinear PDE’s in the theory of flame propa-gation. ICIAM 99 Proceedings of the Fourth International Congress onIndustrial and Applied Mathematics, Oxford university press (2000),13–22.
[5] H. Berestycki, L. Caffarelli, L. Nirenberg, Further Qualita-tive Properties for Elliptic Equations in Unbounded Domains. Ann.Scuola Norm. Sup. Pisa Cl. Sci. (4) 25 (1997), 69–94.
[6] H. Berestycki, B. Larrouturou, J. M. Roquejoffre, Stabilityof Travelling Fronts in a Model for Flame Propagation, Part I: LinearAnalysis. Arch. Rational Mech. Anal. 117 (1992), 97–117.
[7] H. Berestycki, L. Nirenberg, Travelling fronts in cylinders. Ann.Inst. H. Poincare, Analyse non lineaire 9 (1992), 497–572.
[8] L. C. Evans, Partial Differential Equations. American MathematicalSociety, 1998.
[9] G. B. Folland, Real Analysis: Modern Techniques and Their Appli-cations. Wiley-Interscience, 1999.
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53
[10] K. Kirchgassner, J. Scheurle, On the Bounded Solutions of aSemilinear Elliptic Equation in a Strip. Journal of differential equa-tions 32 (1979), 119–148.
[11] V. A. Kondratiev, Asymptotic behaviour of solutions of some non-linear parabolic or elliptic equations. Asymptot. Anal. 14 (1997), 117–156.
[12] V. A. Kondratiev, On some nonlinear boundary value problemsin cylindrical domains. Journal of Mathematical Sciences 85 (1997),2385–2401.
[13] V. A. Kondratiev, On the existence of positive solutions of second-order semilinear elliptic equations in cylindrical domains. Russ. J.Math. Phys. 10 (2003), 99–108.
[14] V. Kozlov, On Bounded Solutions of the Emden-Fowler Equation ina Semi-cylinder. Journal of differential equations 179 (2002), 456–478.
[15] V. Kozlov, V. Maz’ya, Differential Equations with Operator Coef-ficients. Springer, 1999.
[16] E. H. Lieb, M. Loss, Analysis. American Mathematical Society,1997.
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Paper 2
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September 6, 2006 (12:43)
Asymptotic analysis of solutions to
parabolic systems
Vladimir Kozlov Mikael Langer Peter Rand
Abstract
We study asymptotics as t → ∞ of solutions to a linear, parabolicsystem of equations with time-dependent coefficients in Ω × (0,∞),where Ω is a bounded domain. On ∂Ω× (0,∞) we prescribe the ho-mogeneous Dirichlet boundary condition. For large values of t, thecoefficients in the elliptic part are close to time-independent coeffi-cients in an integral sense which is described by a certain functionκ(t). This includes in particular situations when the coefficients maytake different values on different parts of Ω and the boundaries be-tween them can move with t but stabilize as t →∞. The main resultis an asymptotic representation of solutions for large t. As a corollary,it is proved that if κ ∈ L1(0,∞), then the solution behaves asymp-totically as the solution to a parabolic system with time-independentcoefficients.
1 Introduction
Let Ω denote an open, bounded region in Rn with Lipschitz boundary andintroduce Q = Ω× (0,∞). By x = (x1, . . . , xn) we denote the variables inΩ and by t the unbounded variable. We consider the parabolic system
ut −n∑
i,j=1
(Aijuxj )xi + Au = 0 in Q, (1.1)
where u = (u1, . . . , uN ) is a function from Q to CN and Aij , i, j = 1, . . . , n,and A are quadratic matrices of size N ×N whose elements are functionsfrom Q to C. We will assume that u satisfies the Dirichlet boundarycondition
u(x, t) = 0 if x ∈ ∂Ω, t > 0 (1.2)
and thatu(x, 0) = ψ(x), (1.3)
where ψ is a function from(L2(Ω)
)N .For general theory of parabolic equations and systems, which include in
particular solvability and uniqueness results, we refer to Ladyzenskaja et al[12], Dautray, Lions [1], Lions, Magenes [13] and Eidel’man [2]. Evolution
57
58
problems of the above type appear for example in biology and chemistrywhen studying reaction diffusion problems, see for example Murray [14]or Fife [3]. Another application can be found in multigroup diffusion inneutron physics, see Example 3, Chapter XVIIB, §3, Dautray, Lions [1]. Weare concerned only with the asymptotic behaviour of solutions as t → ∞.Therefore, we suppose that the matrices Aij and A can be written as
Aij(x, t) = A(0)ij (x) + A
(1)ij (x, t) (1.4)
andA(x, t) = A(0)(x) + A(1)(x, t), (1.5)
where A(1)ij and A(1) are considered as perturbations. We assume that the
relation (A
(0)ij
)∗ = A(0)ji , (1.6)
where A∗ denotes the adjoint matrix of A, holds for every pair (i, j) andthat A(0) is hermitian, i.e.
(A(0)
)∗ = A(0).
The matrices A(0)ij fulfill the two-sided inequality
ν
n∑
i=1
|ξi|2 ≤n∑
i,j=1
(A(0)ij ξj , ξi) ≤ ν−1
n∑
i=1
|ξi|2 (1.7)
for all ξi, ξj ∈ CN and some positive constant ν. Here we use the notations
(u, v) =N∑
k=1
ukvk
and |u| = (u, u)1/2 for u, v ∈ CN . The matrix A(0) is supposed to belong
to(Lq(Ω)
)N2
, where
q ∈ [n,∞] if n ≥ 3q ∈ (2,∞] if n = 2q ∈ [2,∞] if n = 1.
(1.8)
WritingA(0) = A
(0)+ −A
(0)− , (1.9)
where both matrices A(0)+ and A
(0)− are positive, we require that A
(0)− is
bounded. This means that there exists a constant ν1 such that
‖A(0)− ‖L∞(Q) ≤ ν1. (1.10)
September 6, 2006 (12:43)
59
Furthermore, we also assume that there exists a constant ν2 such that∫
Ω
(A(0)u, u
)dx ≤ ν2‖∇u‖2L2(Ω) (1.11)
for every u ∈ (L2(Ω)
)N .Under the above conditions on the matrices A
(0)ij and A(0), there ex-
ists an ON-basis of(L2(Ω)
)N consisting of eigenfunctions of the time-independent operator
−n∑
i,j=1
(A(0)ij uxj
)xi+ A(0)u
with the Dirichlet boundary condition. Let λk, k = 1, 2, . . . , denote theeigenvalues in increasing order and J be the multiplicity of λ1. This meansthat
λ1 = . . . = λJ < λJ+1 ≤ λJ+2 ≤ . . . , λk →∞ as k →∞.
Let furthermore φ1, φ2, . . . , φJ be a basis in the eigenspace of λ1 which isorthogonal in L2-sense. The conditions on A
(0)ij , A(0) and Ω imply that
φk ∈(W 1,p(Ω)
)N for some p > 2 and k = 1, 2, . . . , J , see Theorem A.2 inAppendix A.
We also assume some similar conditions on Aij and A, see Section 2.2.The main characteristic of our perturbation is the function
κ(t) =n∑
i,j=1
‖A(1)ij ‖Ls1,2(Ct) + ‖A(1)‖Ls2,1(Ct), (1.12)
whereCt = Ω× (t, t + 1),
s1 =2p
p− 2, (1.13)
and
s2 =
2 if n < p
s′2 if n = p2np
np−2(n−p) if n > p,
(1.14)
where s′2 denotes an arbitrary number in (2, n]. In (1.12) we have ex-tended the matrices A
(1)ij , i, j = 1, . . . , n, and A(1) to Ω × R by setting
A(1)ij (x, t) = A(1)(x, t) = 0 for t < 0, so κ(t) is defined for every t ∈ R. We
setκ0 = sup
t≥0κ(t), (1.15)
September 6, 2006 (12:43)
60
and consider perturbations subject to
κ0 ≤ κ, (1.16)
where κ is a sufficiently small constant depending on n, N , Ω, A(0)ij , A(0), p,
s′2, ν, ν1 and ν2. An exact value of κ is difficult to give but the requirementis that κ is so small that some inequality type conditions appearing in theproof of Theorem 1.1 are satisfied. Note that κ does not depend on A
(1)ij
or A(1).We define
(W 1,0;2
0,loc (Q))N as the space consisting of functions u vanishing
on ∂Ω × (0,∞) such that u has a weak derivate with respect to every xk
and every such derivate, together with u itself, belong to(L2(Ct)
)N forevery t ≥ 0. See further in Section 2.1.2. We let ∇ denote the gradientwith respect to the x-variables and define
(V 2
0,loc(Q))N as the subspace of(
W 1,0;20,loc (Q)
)N consisting of functions u such that
|u|Ct = ess supt<s<t+1
‖u(·, s)‖L2(Ω) + ‖∇u‖L2(Ct)
is finite for every t ≥ 0. It can be proved that problem (1.1)–(1.3) hasa unique solution in
(V 2
0,loc(Q))N . The main result of the paper is the
following theorem.
Theorem 1.1 If the constant κ introduced in (1.16) is small enough, thenthe unique solution u in
(V 2
0,loc(Q))N of (1.1)–(1.3) can be represented as
u(x, t) = e−λ1t+∫ t0 (−f(s)+Λ(s)) ds
(w0
J∑
k=1
θk(t)φk(x) + V (x, t))
, (1.17)
where w0 is a constant, Θ = (θ1, . . . , θJ) is an absolutely continuous unitvector and
f = (RΘ, Θ). (1.18)
Here, R denotes the J × J matrix with entry (k, l) equal to
Rkl =∫
Ω
[ n∑
i,j=1
(A(1)ij φlxj
, φkxi) + (A(1)φl, φk)
]dx. (1.19)
Furthermore, the following estimates are valid:
|w0| ≤ C‖ψ‖L2(Ω) (1.20)
and
‖Λ‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−b0(t−s)κ(s) ds + κ(t))
, (1.21)
‖Θ′‖L1(t,t+1) ≤ Cκ(t), (1.22)
|V |Ct ≤ C‖ψ‖L2(Ω)
(e−b0t +
∫ t
−1
e−b0(t−s)κ(s) ds + κ(t))
(1.23)
September 6, 2006 (12:43)
61
for t ≥ 0. Here, b0 = λJ+1 − λ1 − C1κ0 and C and C1 denote constantsdepending on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2.
Remark 1.2 Since we will study the asymptotic behaviour of solutions forlarge t, it suffices to require that
supt≥T
κ(t) ≤ κ
for some T ≥ 0, because the condition (1.16) will then be fulfilled after atranslation of the time variable.
In Corollary 9.2, we prove that the asymptotic formula (1.17) impliesthe estimate
|u|Ct ≤ C1‖ψ‖L2(Ω)e−λ1t+
∫ t0 (−f(s)+C2κ(s)2) ds.
If, in addition, κ ∈ L1(0,∞), Corollary 9.1 states that
u(x, t) = e−λ1t
( J∑
k=1
bkφk(x) + ω(x, t))
,
where bk, k = 1, . . . , J , are constants and |ω|Ct → 0 as t → ∞. We havehere the same leading term as in the case when A
(1)ij = 0, i, j = 1, . . . , n,
A(1) = 0. If, instead, A(1) = 0,
n∑
i,j=1
∫ ∞
0
∫
Ω
|A(1)ij (x, t)| dx dt < ∞
and p = ∞, i.e. the gradients of the eigenfunctions belong to L∞(Ω), weget from Corollary 9.3 that
u(x, t) = e−λ1t
(b
J∑
k=1
θk(t)φk(x) + ω(x, t))
,
where |ω|Ct → 0 as t →∞ and b is a constant which may depend on A(1)ij ,
i, j = 1, . . . , n, and ψ.As can be seen from (1.18), the function f in (1.17) is not given exactly
since its definition contains the unknown vector-valued function Θ. If theeigenvalue λ1 is simple, i.e. J = 1, and A and Aij are real-valued, thenΘ = 1 and we arrive at the following asymptotic expansion for u:
u(x, t) = e−λ1t+∫ t0 (−R(s)+Λ(s)) ds
(w0φ1(x) + V (x, t)
).
See Theorem 9.4.Asymptotics for solutions of (1.1) with a non-zero right-hand side
stabilizing at infinity has been studied by Friedman, [4], [5], [6], and
September 6, 2006 (12:43)
62
Pazy, [15], [16]. Ordinary differential equations with unbounded opera-tor coefficients which include parabolic ones are studied in Kozlov, Maz’ya[9]. In particular, asymptotic results from Part III can give the asymptoticformula (1.17) under the restriction that λ1 is simple and the quantity
n∑
i,j=1
‖A(1)ij ‖L∞(Ct) + ‖ρ−2A(1)‖L∞(Ct),
where ρ(x) denotes the distance to ∂Ω, is small.The proof of Theorem 1.1 can very briefly be outlined in the following
way. Using spectral splitting, we write
u(x, t) =J∑
k=1
hk(t)φk(x) + w(x, t), (1.24)
where hk =∫Ω(u, φk) dx and w(x, t) is the remainder term. The most part
of the proof is devoted to derivation of a system of first order ordinary dif-ferential equations for h1, . . . , hJ perturbed by a small integro-differentialterm and to estimation of w. An important role here plays a preliminaryspectral splitting with J in (1.24) replaced by M , where M is sufficientlylarge, see Section 3. After this, the proof is completed by study of asymp-totic behaviour of solutions to the perturbed system of ordinary differentialequations.
2 Problem formulation and elementary prop-erties
2.1 Notation
Throughout this paper Ω denotes an open, bounded region in Rn withLipschitz boundary. We let x = (x1, . . . , xn) denote a point in Ω and t bea real variable (time).
For two vectors u = (u1, u2, . . . , uN ), v = (v1, v2, . . . , vN ) ∈ CN (de-pending on the context, vectors will also be regarded as column vectors),we define the scalar product
(u, v) =N∑
i=1
uivi.
If ξ is a vector in CN where the k:th component is denoted by ξk, we write
|ξ| =( N∑
k=1
|ξk|2)1/2
.
September 6, 2006 (12:43)
63
For a matrix M with k columns, let mij denote the element of M onposition (i, j) and define |M | as the matrix norm corresponding to thevector norm | · |, i.e.
|M | = supu∈Ck
|u|=1
|Mu|.
The adjoint matrix of M is denoted by M∗. We say that M is positive if(Mu, u) ≥ 0 for every complex vector u of the appropriate dimension.
2.1.1 Spaces not involving time
Given a function u = (u1, . . . , uN ) : Ω → CN and a multiindexα = (α1, . . . , αn) with |α| = α1 + . . . + αn, we set
∂αu(x) =∂|α|u(x)
∂xα11 · · · ∂xαn
n,
where the differentiation acts on u componentwise. If α is the k:th unitvector, we often write uxk
or ∂xku. By ∇u we mean the collection of vectors
(∇u1, . . . ,∇uN ) and we set
‖∇u‖L2(Ω) =( N∑
k=1
n∑
j=1
∫
Ω
∣∣(uk)xj
∣∣2 dx
)1/2
.
We say that u belongs to(Lp(Ω)
)N or(W k,p(Ω)
)N , k = 0, 1, 2, . . . and1 ≤ p ≤ ∞, if every component of u belongs to Lp(Ω) or W k,p(Ω), respec-tively. The norms in these spaces are defined as
‖u‖Lp(Ω) =
(∫Ω|u|p dx
)1/p if 1 ≤ p < ∞ess supΩ|u| if p = ∞
and
‖u‖W k,p(Ω) =
(∑|α|≤k ‖∂αu‖p
Lp(Ω)
)1/p
if 1 ≤ p < ∞∑|α|≤k ess supΩ|∂αu| if p = ∞.
The space(W k,2(Ω)
)N becomes a Hilbert space, denoted by(Hk(Ω)
)N ,by introducing the scalar product
〈u, v〉Hk(Ω) =N∑
j=1
∑
|α|≤k
∫
Ω
(∂αuj
)(∂αvj
)dx.
By(W k,p
0 (Ω))N we denote the closure of
(C∞c (Ω)
)N in(W k,p(Ω)
)N .For a matrix M with elements belonging to Lp(Ω), we define ‖M‖Lp(Ω)
as ‖|M |‖Lp(Ω).
September 6, 2006 (12:43)
64
2.1.2 Spaces involving time
We introduceQ = Ω× (0,∞)
and, for a given T > 0,QT = Ω× (0, T ).
For t ∈ R, we setCt = Ω× (t, t + 1).
Let u denote a measurable function from Ct to CN . By(Lq,r(Ct)
)N wemean the space of all such functions with the norm
‖u‖Lq,r(Ct) =
(∫ t+1
t‖u(·, s)‖r
Lq(Ω) ds)1/r
if 1 ≤ r < ∞ess supt<s<t+1 ‖u(·, s)‖Lq(Ω) if r = ∞
(2.1)
being finite. Instead of(Lq,q(Ct)
)N we write(Lq(Ct)
)N .
Furthermore, we say that u belongs to the Hilbert space(W 1,0;2(Ct)
)N
if u has a weak derivative with respect to every xk and every such deriva-tive, together with u itself, belongs to
(L2(Ct)
)N . The scalar product in(W 1,0;2(Ct)
)N is defined as
〈u, v〉W 1,0;2(Ct) =∫
Ct
[(u, v) +
n∑
k=1
(uxk, vxk
)]
dx ds.
Analogously, u belongs to(W 1,1;2(Ct)
)N if u belongs to(W 1,0;2(Ct)
)N and
in addition has a weak derivative with respect to t, belonging to(L2(Ct)
)N .
The space(W 1,1;2(Ct)
)N is also a Hilbert space and the scalar product isgiven by
〈u, v〉W 1,1;2(Ct) =∫
Ct
[(u, v) +
n∑
k=1
(uxk, vxk
) + (ut, vt)]
dx ds.
We define(V 2(Ct)
)N as the subspace of(W 1,0;2(Ct)
)N consisting offunctions u with the norm
|u|Ct = ess supt<s<t+1
‖u(·, s)‖L2(Ω) + ‖∇u‖L2(Ct) (2.2)
being finite. Here, we define ‖∇u‖L2(Ct) as
‖∇u‖L2(Ct) =( N∑
k=1
n∑
j=1
∫ t+1
t
∫
Ω
∣∣(uk)xj
∣∣2 dx ds
)1/2
.
September 6, 2006 (12:43)
65
The spaces(W 1,0;2
0 (Ct))N ,
(W 1,1;2
0 (Ct))N and
(V 2
0 (Ct))N are defined as
the sets of functions in respective space without a zero subindex whichvanish on ∂Ω × (t, t + 1). If u ∈ (
Lq,r(Ct))N for every t ≥ 0, we say that
u ∈ (Lq,r
loc(Q))N . Other function spaces with the subscript “loc” are defined
similarly.All function spaces defined over Ct can equally well be defined over QT
for some T > 0 by changing the domain of integration in the norms andscalar products accordingly.
As in Section 2.1.1, the definition (2.1) can easily be generalized to bevalid also for matrices.
If f is a function of one variable and −∞ ≤ a < b ≤ ∞, we let ‖f‖Lp(a,b)
denote the Lp-norm of f on the interval (a, b).
2.2 Problem formulation and assumptions
We will study asymptotics of weak solutions of the equation
ut −n∑
i,j=1
(Aijuxj )xi + Au = 0 in Q. (2.3)
Here, u is a function from Q to CN and Aij , i, j = 1, . . . , n, and A arequadratic matrices of size N ×N whose elements are functions from Q toC. We assume that the boundary condition
u(x, t) = 0 if x ∈ ∂Ω, t > 0 (2.4)
and the initial conditionu(x, 0) = ψ(x), (2.5)
where ψ ∈ (L2(Ω)
)N , are valid. We suppose that the matrices Aij and A
can be written as (1.4) and (1.5), where A(1)ij and A(1) are small perturba-
tions in a sense described later.Let us now give the assumptions on A(0) and A
(0)ij . We require that
the symmetry condition (1.6) for matrices A(0)ij holds and the two-sided in-
equality (1.7) is fulfilled. The upper inequality in (1.7) implies in particularthat A
(0)ij all are bounded. The matrix A(0) is supposed to be hermitian and
belong to(Lq(Ω)
)N2
, where q is given by (1.8). We suppose also that thematrix A(0) admits the representation (1.9), where both matrices A
(0)+ and
A(0)− are positive, and that the conditions (1.10) and (1.11) are satisfied.
Let us introduce the operator
L(0)u = −n∑
i,j=1
∂xi
(A
(0)ij uxj
)+ A(0)u
September 6, 2006 (12:43)
66
defined on(W 1,2
0 (Ω))N . This operator has an infinite number of eigenvalues
λ1 ≤ λ2 ≤ . . . , λk →∞ as k →∞,
and the corresponding eigenfunctions φk∞k=1 form an ON-basis in(L2(Ω)
)N , see Theorem A.1 in Appendix A. We denote by p ∈ (2,∞]some number for which
φk ∈(W 1,p
0 (Ω))N
, k = 1, 2, . . . . (2.6)
Relation (2.6) is always true for p sufficiently close to 2, see Theorem A.2in Appendix A.
We suppose that the matrices Aij and A satisfy the following conditions.The relations
A∗ij = Aji, i, j = 1, . . . , n,
hold and the matrix A is hermitian. With ν the same as in (1.7), we assumefurther that
ν
n∑
i=1
|ξi|2 ≤n∑
i,j=1
(Aijξj , ξi) ≤ ν−1n∑
i=1
|ξi|2 (2.7)
for every set of N -dimensional vectors ξ1, . . . , ξn. We also assume that
A ∈ (Lq,r
loc(Q))N2
,
where q is the same as in (1.8) and
r =2q
2q − n.
Furthermore, writingA = A+ −A−, (2.8)
where A+ and A− are positive, we assume that A− is bounded by ν1 from(1.10), i.e.
‖A−‖L∞(Q) ≤ ν1. (2.9)
Let us show that the function κ(t) introduced in (1.12) is finite. It canbe checked that
2np
np− 2(n− p)≤ n
for n > p ≥ 2 and since n ≤ q, it follows from (1.14) that 2 ≤ s2 ≤ q.
Since A(1) ∈ (Lq,r
loc(Q))N2
and A(1)ij is bounded, the function κ(t) is finite
for every t.We define a weak solution of (2.3) under conditions (2.4) and (2.5). We
begin with studying the problem
ut −n∑
i,j=1
(Aijuxj )xi + Au = 0 in QT , (2.10)
September 6, 2006 (12:43)
67
u(x, t) = 0 if x ∈ ∂Ω, t ∈ (0, T ), (2.11)
u(x, 0) = ψ(x) (2.12)
for some fixed, positive T . We introduce
L1(u, η) =∫
Ω
[ n∑
i,j=1
(Aijuxj, ηxi
) + (Au, η)]
dx (2.13)
and say that u ∈ (V 2
0 (QT ))N is a weak solution of problem (2.10)–(2.12)
if
−∫
QT
(u, ηt) dx dt +∫ T
0
L1(u, η) dt =∫
Ω
(ψ(x), η(x, 0)
)dx
for all η ∈ (W 1,1;2
0 (QT ))N such that η(x, T ) = 0. It is a well-known
result that problem (2.10)–(2.12) has a unique weak solution u = uT from(V 2
0 (QT ))N . Indeed, this result for a single equation and a system where
Aij are scalars can be found in Ladyzenskaja et al [12], sections III.1-III.4and VII.1. The generalization to systems of equations where Aij are notnecessarily scalars is straightforward. Since this can be done for any T > 0,and the solution is unique, we obtain a unique function u defined on Q.We define this function as the weak solution from
(V 2
0,loc(Q))N of equation
(2.3) under conditions (2.4) and (2.5).Equivalently, one can say that u is a weak solution of the problem
(2.3)–(2.5) if u ∈ (V 2
0,loc(Q))N is the unique function satisfying the relation
−∫
Q
(u, ηt) dx dt +∫ ∞
0
L1(u, η) dt =∫
Ω
(ψ(x), η(x, 0)
)dx (2.14)
for all η ∈ (W 1,1;2
0 (Q))N with bounded support.
The symbol C, possibly with a subscript, is frequently used to denotea constant depending only on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2. The
lower-case letter c is used for constants depending on the same quantitiesbut in situations where the symbol denotes a specific constant which maybe referred to in some other part of the paper.
The remaining part of the paper except Section 9 is devoted to the proofof Theorem 1.1. Without loss of generality, we can assume that λ1 = 0.Namely, if λ1 6= 0, we set
U(x, t) = eλ1tu(x, t).
Then U satisfies the equation
Ut −n∑
i,j=1
(AijUxj )xi + A′U = 0 in Q (2.15)
and the initial and boundary conditions (2.4), (2.5). In (2.15) we haveA′ = A(0)′ + A(1), where A(0)′ = A(0) − λ1I. Clearly, the matrix A(0)′
satisfies (1.10) and (1.11), possibly with other constants ν1 and ν2.
September 6, 2006 (12:43)
68
2.3 An estimate for u
In this section, an estimate for u, to be used later in the paper, is derived.
Lemma 2.1 There exists a constant a0 depending on ν, ν1 and diam Ωsuch that the inequality
‖u(·, t)‖L2(Ω) ≤ ‖ψ‖L2(Ω)ea0t (2.16)
is valid a.e.
Proof. From (2.7) and Poincare’s inequality it follows for an arbitraryfunction w ∈ (
W 1,20 (Ω)
)N that
∫
Ω
n∑
i,j=1
(Aijwxj, wxi
) dx ≥ ν‖∇w‖2L2(Ω) ≥ C1‖w‖2L2(Ω). (2.17)
Furthermore, using the decomposition (2.8) and inequality (2.9) togetherwith the fact that A+ is positive, we get
(Aw, w) = (A+w, w)− (A−w,w) ≥ −ν1|w|2. (2.18)
From (2.13), (2.17) and (2.18) the inequality
L1(w, w) ≥ −a0‖w‖2L2(Ω) (2.19)
follows, where a0 = ν1 − C1.In the same way as in Ladyzenskaja et al [12] Section III.2, we can
derive the equality
12
∫
Ω
|u(x, s)|2 dx∣∣∣t
s=0+
∫ t
0
L1(u, u) ds = 0 (2.20)
for a.e. t > 0. This is equivalent to
12‖u(·, t)‖2L2(Ω) =
12‖ψ‖2L2(Ω) −
∫ t
0
L1(u, u) ds. (2.21)
Since Aij is bounded and u ∈ (V 2
0,loc(Q))N , we see by Holder’s inequality
that the term (Aijuxj , uxi) occurring in the expression of L1(u, u) is anelement of L1(Ω) for almost every t. The same is valid for (Au, u); this canbe seen from the derivation of (A.3) by replacing A(0) by A. Hence we candifferentiate (2.21) so that
12
d
dt
(‖u(·, t)‖2L2(Ω)
)= −L1(u, u)
and it follows from (2.19) that
12
d
dt
(‖u(·, t)‖2L2(Ω)
)≤ a0‖u(·, t)‖2L2(Ω).
September 6, 2006 (12:43)
69
Henced
dt
(‖u(·, t)‖2L2(Ω)e
−2a0t)≤ 0
so‖u(·, t)‖2L2(Ω) ≤ ‖ψ‖2L2(Ω)e
2a0t.
After taking the square root, inequality (2.16) follows.
3 Spectral splitting of the solution u
Let u denote the unique weak solution of (2.3) under the conditions (2.4)and (2.5) in Q as defined in Section 2.2. Setting
hk(t) =∫
Ω
(u(·, t), φk
)dx, k = 1, . . . , M,
we obtain the representation
u(x, t) =M∑
k=1
hk(t)φk(x) + v(x, t), (3.1)
where v(·, t) is orthogonal to φk in(L2(Ω)
)N for k = 1, 2, . . . , M . Theinteger M will be chosen later.
Recall that it is assumed that λ1 = 0 and let J denote the multiplicityof the eigenvalue 0. This means that
0 = λ1 = . . . = λJ < λJ+1 ≤ . . . . (3.2)
To get an equation for hk, we choose the functions η in (2.14) as
η(x, t) = ξ(t)φk(x),
where 1 ≤ k ≤ M and the scalar function ξ belongs C1c (0,∞). This gives,
after using the orthogonality between the elements φ1, . . . , φM and v, that
−∫ ∞
0
ξ′hk dt +∫ ∞
0
ξ
( M∑
l=1
Rklhl + gk(v))
dx = 0,
where
Rkl =∫
Ω
[ n∑
i,j=1
(Aijφlxj, φkxi
) + (Aφl, φk)]
dx
are known functions of t and
gk(v) =∫
Ω
[ n∑
i,j=1
(Aijvxj , φkxi) + (Av, φk)
]dx.
September 6, 2006 (12:43)
70
This implies that hk has a (distributional) derivate h′k and we get thesystem of equations
h′k +M∑
l=1
Rklhl + gk(v) = 0, k = 1, 2, . . . , M. (3.3)
Introducing
Rkl =∫
Ω
[ n∑
i,j=1
(A(1)ij φlxj
, φkxi) + (A(1)φl, φk)
]dx (3.4)
and
gk(v) =∫
Ω
[ n∑
i,j=1
(A(1)ij vxj , φkxi
) + (A(1)v, φk)]
dx, (3.5)
we see, by using (1.4) and (1.5), that
Rkl = B[φl, φk] +Rkl
andgk(v) = B[v, φk] + gk(v),
with B as given in (A.2). Since
B[φl, φk] =
0 if k 6= l
λk if k = l
and
B[v, φk] = B[φk, v] = λk
∫
Ω
(φk, v) dx = 0
for k = 1, . . . , M , it follows that (3.3) can be rewritten as
h′k + λkhk +M∑
l=1
Rklhl + gk(v) = 0, k = 1, 2, . . . , M. (3.6)
We also have the initial values
hk(0) =∫
Ω
(ψ, φk) dx, k = 1, 2, . . . , M. (3.7)
The system of equations (3.6), (3.7) will be analyzed further in Sections 6and 7.
In order to find an equation for v we use representation (3.1) in (2.14).This time we suppose that η(·, t) is orthogonal to φk, k = 1, . . . , M , in
September 6, 2006 (12:43)
71
(L2(Ω)
)N for almost every t ∈ (0,∞). This implies that the same orthog-onality relation holds between ηt(·, t) and φk and it follows that
−∫
Q
(v, ηt) dx dt +∫ ∞
0
[L1(v, η) +
M∑
k=1
hkL1(φk, η)]
dt =
∫
Ω
(ψ(x), η(x, 0)
)dx. (3.8)
Owing to decompositions (1.4) and (1.5), we can write
L1(φk, η) = B[φk, η] + L(1)1 (φk, η),
where
L(1)1 (φk, η) =
∫
Ω
[ n∑
i,j=1
(A(1)ij φkxj
, ηxi) + (A(1)φk, η)]
dx.
Because of the orthogonality between φk and η, it follows from (A.4) thatB[φk, η] = 0 for a.e. t and hence we can rewrite (3.8) as
−∫
Q
(v, ηt) dx dt +∫ ∞
0
[L1(v, η) +
M∑
k=1
hkL(1)1 (φk, η)
]dt =
∫
Ω
(ψ(x), η(x, 0)
)dx. (3.9)
4 Estimating the function v
4.1 A general estimate
We will now study solutions w ∈ (V 2
0,loc(Q))N of the equation
−∫
Q
(w, ηt) dx dt +∫ ∞
0
[L1(w, η) + L2(f , η)] dt =∫
Ω
(ψ(x), η(x, 0)
)dx
(4.1)subject to the orthogonality condition
∫
Ω
(w(x, t), φk(x)
)dx = 0 (4.2)
for k = 1, . . . ,M and a.e. t. The operator L1 was defined in (2.13) and
L2(f , η) =∫
Ω
[ n∑
i=1
(fi, ηxi) + (f, η)]
dx, (4.3)
September 6, 2006 (12:43)
72
where f ∈ (L2,1
loc(Q))N and fi ∈
(L2
loc(Q))N , i = 1, 2, . . . , n. Equal-
ity (4.1) should be satisfied for every η from the same class of functions(W 1,1;2
0 (Q))N with bounded support as in (2.14) such that
∫
Ω
(η(x, t), φk(x)
)dx = 0
for k = 1, . . . , M and a.e. t.In the proof of the next theorem we use the estimate
∫ t
0
|f(s)| ds ≤∫ t
−1
‖f‖L1(s,s+1) ds (4.4)
for a function f ∈ L1loc(R). This is obtained by noting that, for τ ∈ [0, 1],
we have∫ t
0
|f(s)| ds =∫ t−τ
−τ
|f(s + τ)| ds =∫ 1
0
∫ t−τ
−τ
|f(s + τ)| ds dτ
≤∫ t
−1
∫ 1
0
|f(s + τ)| dτ ds =∫ t
−1
‖f‖L1(s,s+1) ds.
Proposition 4.1 Suppose that f ∈ (L2,1
loc(Q))N and fi ∈ (
L2loc(Q)
)N
for i = 1, 2, . . . , n. Then, for any M ≥ 0, there exists a unique functionw ∈ (
V 20,loc(Q)
)N satisfying (4.1) and fulfilling (4.2) for k = 1, . . . , M anda.e. positive t. For every b > 0, there exists an integer M and a constantC, both depending on b, n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2, such that
the solution w is subject to the inequality
|w|Ct ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds + χ(t))
, (4.5)
where the norm in the left-hand side is defined by (2.2) and
χ(t) =n∑
i=1
‖fi‖L2(Ct) + ‖f‖L2,1(Ct). (4.6)
Here we have extended f and fi by 0 for negative values of t. By extendingalso w by 0 for t < 0, estimate (4.5) becomes valid for all t ≥ −1.
The existence and uniqueness of the solution can be proved in the sameway as in the case of a single equation, treated in sections III.3 and III.4in Ladyzenskaja et al [12]. We confine ourself to prove inequality (4.5) un-der an appropriate choice of M . The proof will be divided into several steps.
Proof of estimate (4.5): Step 1. Deriving a differential inequality for‖w(·, t)‖2L2(Ω). Analogously to (2.20), we can derive the equality
12
∫
Ω
|w(x, s)|2 dx∣∣∣t
s=t0+
∫ t
t0
[L1(w, w) + L2(f , w)] ds = 0 (4.7)
September 6, 2006 (12:43)
73
for any t0 and t such that 0 ≤ t0 < t < ∞. This corresponds to (2.13),Chapter III in Ladyzenskaja et al. Since L1(w, w) and L2(f , w) belong toL1
loc(0,∞), it follows that ‖w(·, t)‖2L2(Ω) is differentiable for t > t0 and
12
d
dt
(‖w(·, t)‖2L2(Ω)
)=
−∫
Ω
[ n∑
i,j=1
(Aijwxj , wxi
)+ (Aw, w) +
n∑
i=1
(fi, wxi) + (f, w)]
dx. (4.8)
We now estimate the terms in the right-hand side of (4.8). The de-composition A = A+ − A−, where A+ is positive and A− is bounded, andinequality (2.9) imply that
−∫
Ω
(Aw,w) dx = −∫
Ω
(A+w,w) dx +∫
Ω
(A−w,w) dx
≤ ν1‖w(·, t)‖2L2(Ω).
(4.9)
Together with (2.7) this yields
−∫
Ω
[ n∑
i,j=1
(Aijwxj , wxi
)+ (Aw, w)
]dx
≤ −ν‖∇w(·, t)‖2L2(Ω) + ν1‖w(·, t)‖2L2(Ω). (4.10)
Since w is orthogonal to φ1, . . . , φM , we have∫
Ω
[ n∑
i,j=1
(A
(0)ij wxj , wxi
)+
(A(0)w, w
) ]dx ≥ λM+1‖w(·, t)‖2L2(Ω),
cf. (A.5) in Appendix A. From this fact, together with the upper inequalityin (1.7) and property (1.11), we obtain
‖∇w(·, t)‖2L2(Ω) ≥ ν
∫
Ω
[ n∑
i,j=1
(A
(0)ij wxj , wxi
)+
(A(0)w,w
)]dx
− ν
∫
Ω
(A(0)w, w
)dx
≥ νλM+1‖w(·, t)‖2L2(Ω) − νν2‖∇w(·, t)‖2L2(Ω)
and hence‖∇w(·, t)‖2L2(Ω) ≥
νλM+1
1 + νν2‖w(·, t)‖2L2(Ω).
This gives together with (4.10) the estimate
−∫
Ω
[ n∑
i,j=1
(Aijwxj , wxi
)+ (Aw, w)
]dx
≤ −ν
2‖∇w(·, t)‖2L2(Ω) − b‖w(·, t)‖2L2(Ω),
September 6, 2006 (12:43)
74
where
b =ν2λM+1
2(1 + νν2)− ν1.
Since λM+1 →∞ as M →∞, the constant b can be made arbitrarily large.We continue to estimate the terms in the right-hand side of (4.8) and
get
−∫
Ω
n∑
i=1
(fi, wxi) dx ≤( n∑
i=1
‖fi(·, t)‖L2(Ω)
)‖∇w(·, t)‖L2(Ω)
≤ ν
2‖∇w(·, t)‖2L2(Ω) +
12ν
( n∑
i=1
‖fi(·, t)‖L2(Ω)
)2
,
(4.11)
where we have used the elementary inequality
ab ≤ α
2a2 +
12α
b2, α > 0,
in the last line. Also,
−∫
Ω
(f, w) dx ≤ ‖f(·, t)‖L2(Ω)‖w(·, t)‖L2(Ω)
and when combining all estimates with (4.8), we get
12
d
dt(‖w(·, t)‖2L2(Ω)) ≤ −b‖w(·, t)‖2L2(Ω)
+ ‖f(·, t)‖L2(Ω)‖w(·, t)‖L2(Ω) +12ν
( n∑
i=1
‖fi(·, t)‖L2(Ω)
)2
. (4.12)
Choose M so large that b becomes positive and set
h(t) = ‖w(·, t)‖2L2(Ω)
and
G(t) =1ν
( n∑
i=1
‖fi(·, t)‖L2(Ω)
)2
. (4.13)
Then, from (4.12), we obtain the differential inequality
h′(t) + 2bh(t) ≤ 2‖f(·, t)‖L2(Ω)h(t)1/2 + G(t), t > 0. (4.14)
Step 2. Finding an estimate for ‖w(·, t)‖2L2(Ω). We set
h+(t) =2h(0)e−2bt +32
∫ t
0
e−2b(t−τ)G(τ) dτ
+ 12(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
. (4.15)
September 6, 2006 (12:43)
75
Let us show that h(t) ≤ h+(t). We begin by proving that
h+(t) ≥ h(0)e−2bt +∫ t
0
e−2b(t−τ)(2‖f(·, τ)‖L2(Ω)h+(τ)1/2 + G(τ)
)dτ.
(4.16)Let us denote the right-hand side by r(t). When inserting the expressionfor h+ in r(t) and using the inequality
√a + b ≤ √
a +√
b for a, b ≥ 0, weobtain
r(t) ≤h(0)e−2bt +∫ t
0
e−2b(t−τ)
[2√
2h(0)‖f(·, τ)‖L2(Ω)e−bτ
+√
6‖f(·, τ)‖L2(Ω)
(∫ τ
0
e−2b(τ−s)G(s) ds
)1/2
+ 4√
3‖f(·, τ)‖L2(Ω)
∫ τ
0
e−b(τ−s)‖f(·, s)‖L2(Ω) ds + G(τ)]
dτ.
(4.17)
We estimate the terms in (4.17) one by one, beginning with
2√
2h(0)∫ t
0
e−2b(t−τ)‖f(·, τ)‖L2(Ω)e−bτ dτ
= 2√
2h(0)e−bt
∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
≤ h(0)e−2bt + 2(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
. (4.18)
Since τ ≤ t, we have
√6
∫ t
0
e−2b(t−τ)‖f(·, τ)‖L2(Ω)
(∫ τ
0
e−2b(τ−s)G(s) ds
)1/2
dτ
≤√
6(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)(∫ t
0
e−2b(t−s)G(s) ds
)1/2
≤ 3(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
+12
∫ t
0
e−2b(t−s)G(s) ds
(4.19)
and
4√
3∫ t
0
e−2b(t−τ)‖f(·, τ)‖L2(Ω)
(∫ τ
0
e−b(τ−s)‖f(·, s)‖L2(Ω) ds
)dτ
≤ 4√
3∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω)
(∫ t
0
e−b(t−s)‖f(·, s)‖L2(Ω) ds
)dτ
= 4√
3(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
.
(4.20)
September 6, 2006 (12:43)
76
Using (4.18), (4.19) and (4.20) in (4.17), we arrive at
r(t) ≤ 2h(0)e−2bt +32
∫ t
0
e−2b(t−τ)G(τ) dτ
+(5 + 4
√3) (∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
≤ h+(t).
This implies (4.16).We now prove that h(t) ≤ h+(t) and conclude first that both of the
functions h and h+ are continuous. For h+, this fact follows from (4.15)since G as well as ‖f‖L2(Ω) belong to L1
loc(0,∞). Furthermore, the conti-nuity of h follows from (4.7).
Suppose first that h(0) > 0. Then h+(0) = 2h(0) > h(0) and letus prove that h+(t) > h(t) for all t ∈ [0,∞). Namely, suppose thath(τ) < h+(τ) if τ ∈ [0, t) for some t > 0 (obviously, such a t exists be-cause of the continuity of h and h+). From (4.14) it follows that
(he2bt
)′ ≤ 2e2bt‖f(·, t)‖L2(Ω)h(t)1/2 + e2btG(t)
and hence
h(t) ≤ h(0)e−2bt +∫ t
0
e−2b(t−τ)(2‖f(·, τ)‖L2(Ω)h(τ)1/2 + G(τ)
)dτ.
(4.21)There are two possibilities. Either ‖f(·, τ)‖L2(Ω) = 0 for almost everyτ ∈ [0, t] or else ‖f(·, τ)‖L2(Ω) > 0 on a subset of [0, t] with positive measure.In the first case it follows from (4.21) that
h(t) ≤ h(0)e−2bt +∫ t
0
e−2b(t−τ)G(τ) dτ
and when comparing this with (4.15), we see that h(t) < h+(t).Let us now consider the case when ‖f(·, τ)‖L2(Ω) is not identically 0 on
[0, t], still assuming that h(0) > 0 and h < h+ on [0, t). From (4.21) and(4.16) we get that
h+(t)− h(t) ≥ 2∫ t
0
e−2b(t−τ)‖f(·, τ)‖L2(Ω)
(h+(τ)1/2 − h(τ)1/2
)dτ > 0,
so also in this case h(t) < h+(t).The conclusion is that the set
t ≥ 0 : h(τ) < h+(τ) for τ ∈ [0, t]
is non-empty, open (because of the continuity of h and h+) and closed in[0,∞). The last statement follows from the preceding analysis. Hence,
September 6, 2006 (12:43)
77
it must be equal to [0,∞) and we have proved that if h(0) > 0, thenh(t) < h+(t) for every t ≥ 0.
We now consider the case h(0) = 0. For a given ε > 0, introduce
hε(t) = h(t) + εe−2bt
and
hε+(t) =2hε(0)e−2bt +
32
∫ t
0
e−2b(t−τ)G(τ) dτ
+ 12(∫ t
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
)2
. (4.22)
Since h1/2 ≤ (hε)1/2, it follows from (4.21) that
hε(t) ≤ hε(0)e−2bt +∫ t
0
e−2b(t−τ)(2‖f(·, τ)‖L2(Ω)h
ε(τ)1/2 + G(τ))
dτ.
(4.23)The relations (4.22) and (4.23) are the same as (4.15) and (4.21) with theonly difference that h and h+ are replaced by hε and hε
+, respectively andalso (4.16) holds with the same substitutions. Obviously, hε(0) = ε > 0, sothe previous case h(0) > 0 shows that hε(t) < hε
+(t) for all t ≥ 0. The lastinequality says that
h(t) + εe−2bt < h+(t) + 2εe−2bt
and by letting ε tend to 0 we see that h(t) ≤ h+(t) for every t ≥ 0.We have thus obtained the inequality
‖w(·, t)‖2L2(Ω) ≤2‖ψ‖2L2(Ω)e−2bt +
32
∫ t
0
e−2b(t−s)G(s) ds
+ 12(∫ t
0
e−b(t−s)‖f(·, s)‖L2(Ω) ds
)2
. (4.24)
Step 3. Finding an estimate for ‖w(·, t)‖L2(Ω). Our next aim is to findan estimate for ‖w(·, t)‖L2(Ω). This will be done departing from (4.24).Before doing this, we set γ(t) =
√G(t) and prove that the inequality
(∫ t
0
e−2b(t−s)γ(s)2 ds
)1/2
≤ eb
∫ t
−1
e−b(t−s)‖γ‖L2(s,s+1) ds, (4.25)
where γ has been extended by 0 for t < 0, holds. It is readily verified thatif τ ∈ [0, 1], then
(∫ t
0
e−2b(t−s)γ(s)2 ds
)1/2
≤( dte∑
k=0
∫ k+1−τ
k−τ
e−2b(t−s)γ(s)2 ds
)1/2
,
September 6, 2006 (12:43)
78
where dte denotes the least integer larger than or equal to t. Since theleft-hand side is independent of τ , we integrate the inequality from 0 to 1with respect to τ and obtain
(∫ t
0
e−2b(t−s)γ(s)2 ds
)1/2
≤∫ 1
0
( dte∑
k=0
∫ k+1−τ
k−τ
e−2b(t−s)γ(s)2 ds
)1/2
dτ.
By using the inequality√
a + b ≤ √a +
√b for a, b ≥ 0, we can move the
sum outside the parentheses. This gives the inequality
(∫ t
0
e−2b(t−s)γ(s)2 ds
)1/2
≤ eb
dte∑
k=0
∫ 1
0
e−b(t+τ−k)‖γ‖L2(k−τ,k−τ+1) dτ.
(4.26)In the last integral we make the substitution s = k − τ . This yields
dte∑
k=0
∫ 1
0
e−b(t+τ−k)‖γ‖L2(k−τ,k−τ+1) dτ
=dte∑
k=0
∫ k
k−1
e−b(t−s)‖γ‖L2(s,s+1) ds =∫ dte
−1
e−b(t−s)‖γ‖L2(s,s+1) ds. (4.27)
The left-hand side of (4.25) shows that we can set γ(s) = 0 for s > t. Thus,(4.25) follows from (4.26) and (4.27).
In the remaining part of the proof, we let C denote a constant whichmay depend on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2 and, additionally,
b. We use (4.24) and get the estimate
‖w(·, t)‖L2(Ω) ≤√
2‖ψ‖L2(Ω)e−bt +
√32
(∫ t
0
e−2b(t−s)G(s) ds
)1/2
+ 2√
3∫ t
0
e−b(t−s)‖f(·, s)‖L2(Ω) ds.
An application of (4.25) yields, after using (4.13), that
‖w(·, t)‖L2(Ω) ≤ C
(‖ψ‖L2(Ω)e
−bt +n∑
i=1
∫ t
−1
e−b(t−s)‖fi‖L2(Cs) ds
+∫ t
0
e−b(t−s)‖f(·, s)‖L2(Ω) ds
).
(4.28)
By using inequality (4.4), we obtain
∫ t
0
e−b(t−s)‖f(·, s)‖L2(Ω) ds ≤ eb
∫ t
−1
e−b(t−s)‖f‖L2,1(Cs) ds (4.29)
September 6, 2006 (12:43)
79
and (4.28) becomes
‖w(·, t)‖L2(Ω) ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds
), (4.30)
with χ as defined in (4.6). The convention that w(x, t) = 0 for t < 0 makes(4.30) valid for all t ≥ −1.
Step 4. Completing the proof of (4.5). We now go on with deriving theestimate (4.5). Fix a value of t > −1 and introduce the matrices
Aij(x, s) =
Aij(x, s) if s ≤ t + 10 if s > t + 1
and
A(x, s) =
A(x, s) if s ≤ t + 10 if s > t + 1.
The matrices A(0)ij , A
(1)ij , A(0), A(1) and the functions fi, f are defined
analogously. By replacing Aij , A, fi, f with Aij , A, fi, f in (4.1) and(4.3), we see that the solution w is unchanged on the interval [0, t + 1].
Introduce
χ(s) =n∑
i=1
‖fi‖L2(Cs) + ‖f‖L2,1(Cs).
For ε ∈ [0, 1], inequality (4.30) yields that
‖w(·, t + ε)‖L2(Ω) ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds
+∫ t+ε
t
e−b(t−s)χ(s) ds
).
Since obviously χ(s) ≤ χ(t) for t ≤ s ≤ t + 1, it follows that
ess supt<s<t+1
‖w(·, s)‖L2(Ω) ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds + χ(t))
.
(4.31)In estimating ‖∇w‖L2(Ct), we use that it follows from (4.7) that
‖w(·, t + 1)‖2L2(Ω) − ‖w(·, t)‖2L2(Ω) + 2∫ t+1
t
[L1(w, w) + L2(f , w)] ds = 0.
This, together with (2.7), gives that
2ν
∫ t+1
t
‖∇w(·, s)‖2L2(Ω) ds ≤ 2∫ t+1
t
∫
Ω
n∑
i,j=1
(Aijwxj , wxi) dx ds
≤ ‖w(·, t)‖2L2(Ω) − 2∫ t+1
t
∫
Ω
(Aw,w) dx ds
− 2∫ t+1
t
∫
Ω
n∑
i=1
(fi, wxi) dx ds− 2∫ t+1
t
∫
Ω
(f, w) dx ds.
(4.32)
September 6, 2006 (12:43)
80
As in (4.9) and (4.11), we obtain the inequalities
−2∫ t+1
t
∫
Ω
(Aw, w) dx ds ≤ 2ν1‖w‖2L2(Ct)(4.33)
and
−2∫ t+1
t
∫
Ω
n∑
i=1
(fi, wxi) dx ds ≤ ν‖∇w‖2L2(Ct)+
1ν
( n∑
i=1
‖fi‖L2(Ct)
)2
. (4.34)
Furthermore, from Holder’s inequality it follows that
−2∫ t+1
t
∫
Ω
(f, w) dx ds ≤ 2‖f‖L2,1(Ct) ess supt<s<t+1
‖w(·, s)‖L2(Ω)
≤ ‖f‖2L2,1(Ct)+
(ess supt<s<t+1
‖w(·, s)‖L2(Ω)
)2
. (4.35)
By inserting (4.33), (4.34) and (4.35) in (4.32) and using that
‖w(·, t)‖2L2(Ω) ≤ ess supt<s<t+1
‖w(·, s)‖2L2(Ω),
it follows that
2ν‖∇w‖2L2(Ct)≤2ν1‖w‖2L2(Ct)
+ 2(
ess supt<s<t+1
‖w(·, s)‖L2(Ω)
)2
+ ν‖∇w‖2L2(Ct)+
1ν
( n∑
i=1
‖fi‖L2(Ct)
)2
+ ‖f‖2L2,1(Ct),
or, equivalently,
ν‖∇w‖2L2(Ct)≤2ν1‖w‖2L2(Ct)
+ 2(
ess supt<s<t+1
‖w(·, s)‖L2(Ω)
)2
+1ν
( n∑
i=1
‖fi‖L2(Ct)
)2
+ ‖f‖2L2,1(Ct).
After taking the square root, this yields
‖∇w‖L2(Ct) ≤ C(‖w‖L2(Ct) + ess sup
t<s<t+1‖w(·, s)‖L2(Ω) + χ(t)
). (4.36)
We now estimate the terms in the right-hand side and begin by studying‖w‖L2(Ct). Since
‖w‖2L2(Ct)=
∫ t+1
t
‖w(·, s)‖2L2(Ω) ds,
September 6, 2006 (12:43)
81
we get from (4.24) and (4.13) the estimate
‖w‖2L2(Ct)≤ C
(‖ψ‖2L2(Ω)
e−2bt
2b
+n∑
i=1
∫ t+1
t
∫ s
0
e−2b(s−τ)‖fi(·, τ)‖2L2(Ω) dτ ds
+∫ t+1
t
(∫ s
0
e−b(s−τ)‖f(·, τ)‖L2(Ω) dτ
)2
ds
).
(4.37)
We get an estimate for ‖w‖L2(Ct) by taking the square root of each term.Indeed,
( ∫ t+1
t
∫ s
0
e−2b(s−τ)‖fi(·, τ)‖2L2(Ω) dτ ds
)1/2
≤( ∫ t+1
t
∫ t+1
0
e−2b(t−τ)‖fi(·, τ)‖2L2(Ω) dτ ds
)1/2
=(∫ t+1
0
e−2b(t−τ)‖fi(·, τ)‖2L2(Ω) dτ
)1/2
≤ eb
∫ t+1
−1
e−b(t−τ)‖fi‖L2(Cτ ) dτ
≤ eb
(∫ t
−1
e−b(t−τ)‖fi‖L2(Cτ ) dτ + eb‖fi‖L2(Ct)
),
(4.38)
where we have used (4.25) with γ = ‖fi‖L2(Ω) in the second inequality. Byextending the interval of integration as above, we have that
( ∫ t+1
t
( ∫ s
0
e−b(s−τ)‖f(·, τ)‖L2(Ω) dτ
)2
ds
)1/2
=∫ t+1
0
e−b(t−τ)‖f(·, τ)‖L2(Ω) dτ
≤ eb
(∫ t
−1
e−b(t−τ)‖f‖L2,1(Cτ ) dτ + ‖f‖L2,1(Ct)
),
(4.39)
where (4.29) was used in the last inequality. We use the inequalities (4.38)and (4.39) in (4.37) and get
‖w‖L2(Ct) ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds + χ(t))
.
This, together with (4.31), used in (4.36) gives that
‖∇w‖L2(Ct) ≤ C
(‖ψ‖L2(Ω)e
−bt +∫ t
−1
e−b(t−s)χ(s) ds + χ(t))
. (4.40)
By combining (4.31) and (4.40), we finally obtain (4.5). 2
September 6, 2006 (12:43)
82
4.2 Estimate for v
A consequence of Proposition 4.1 is that equation (3.9) has a unique so-lution in
(V 2
0,loc(Q))N . Let us fix b = 2λJ+1, with J defined in (3.2).
According to Proposition 4.1, it is possible to find an integer M such thatthe solution satisfies the estimate (4.5). Observe that after this choice ofM , the constant C depends only on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and
ν2. We introduce
H(t) =
ess supt<s<t+1
∑Mk=1 |hk(s)| if t ≥ 0
ess sup0<s<t+1
∑Mk=1 |hk(s)| if − 1 < t < 0
0 if t ≤ −1.
(4.41)
Then Proposition 4.1 yields the following corollary.
Corollary 4.2 For k = 1, . . . , M , let hk be arbitrary functions fromL∞loc(0,∞). Then equation (3.9) has a unique solution v ∈ (
V 20,loc(Q)
)N ,orthogonal to φ1, . . . , φm. After extending v by 0 for t < 0, the estimate
|v|Ct ≤ C
(‖ψ‖L2(Ω)e
−2λJ+1t +∫ t
−1
e−2λJ+1(t−s)κ(s)H(s) ds + κ(t)H(t))
(4.42)is valid for all t ≥ −1.
Proof. We apply Proposition 4.1 on (3.9). With b = 2λJ+1, we get, from(4.5), the estimate
|v|Ct ≤ C
(‖ψ‖L2(Ω)e
−2λJ+1t +∫ t
−1
e−2λJ+1(t−s)χ(s) ds + χ(t))
, (4.43)
where
χ(t) =n∑
i=1
‖fi‖L2(Ct) + ‖f‖L2,1(Ct), (4.44)
for
fi =n∑
j=1
M∑
k=1
A(1)ij hkφkxj
and
f =M∑
k=1
A(1)hkφk.
Using that φk ∈(W 1,p
0 (Ω))N for some p > 2, see (2.6), we derive that
‖fi‖L2(Ct) ≤ Cκ(t)H(t), (4.45)
September 6, 2006 (12:43)
83
where κ(t) was introduced in (1.12). Namely, let a denote an element ofA
(1)ij and ϕ an element of φkxj
Mk=1. With s1 as defined in (1.13), Holder’s
inequality implies that
‖aϕ‖L2(Ω) = ‖a2ϕ2‖1/2L1(Ω) ≤ ‖a2‖1/2
Ls1/2(Ω)‖ϕ2‖1/2
Lp/2(Ω)
= ‖a‖Ls1 (Ω)‖ϕ‖Lp(Ω)
and it is now easy to show (4.45).
Moreover, Sobolev’s embedding theorem implies that φk ∈(Lp1(Ω)
)N ,where
p1 =
∞ if n < p
p′1 if n = pnp
n−p if n > p,
where p′1 is an arbitrary number in [1,∞), and the estimate
‖φk‖Lp1 (Ω) ≤ C‖φk‖W 1,p(Ω)
is valid. The constant C depends on n, N , Ω, p and p′1. It follows that
‖f(·, t)‖L2(Ω) ≤ C‖A(1)(·, t)‖Lp2 (Ω)
M∑
k=1
|hk(t)|‖φk‖W 1,p(Ω),
where
p2 =
1 if n < pp′1
p′1−1 if n = pnp
np−n+p if n > p
is the conjugate exponent to p1. In the case n = p, we set
p′1 =s′2
s′2 − 1,
where s′2 was introduced in (1.14). This means in particular that p2 = s′2.From (1.14) it then follows that p2 ≤ s2 for every value of n, so
‖f‖L2,1(Ct) ≤ Cκ(t)H(t). (4.46)
Using the estimates (4.45) and (4.46) in (4.44) we get
χ(t) ≤ Cκ(t)H(t)
and (4.42) follows from (4.43).
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84
5 Norm estimates for Rkl and gk(w)
In this section we find norm estimates for Rkl and gk(w), introduced in(3.4) and (3.5).
Lemma 5.1 Suppose that w is a function in V 2loc(Q). There exist constants
c1 and c2 such that‖Rkl‖L1(t,t+1) ≤ c1κ(t) (5.1)
and‖gk(w)‖L1(t,t+1) ≤ c2κ(t)|w|Ct
. (5.2)
Here, c1 and c2 depend only on n, p, s′2 and Ω.
Proof. We begin with the case n > p. As in the proof of Corollary 4.2, itfollows that φk ∈
(Lp1(Ω)
)N with p1 = np/(n− p) and the estimate
‖φk‖Lp1 (Ω) ≤ C‖φk‖W 1,p(Ω) (5.3)
is valid. By applying Holder’s inequality on (3.4), the estimate
|Rkl(τ)| ≤n∑
i,j=1
‖A(1)ij (·, τ)‖Ls0 (Ω)‖φlxj
‖Lp(Ω)‖φkxi‖Lp(Ω)
+ ‖A(1)(·, τ)‖Lr0 (Ω)‖φl‖Lp1 (Ω)‖φk‖Lp1 (Ω) (5.4)
is obtained, where s−10 + 2p−1 = 1 and r−1
0 + 2p−11 = 1, i.e. s0 = p/(p− 2)
andr0 =
np
np− 2(n− p).
Integrating (5.4) from t to t + 1 and using (5.3), we obtain
‖Rkl‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)‖φl‖W 1,p(Ω)
×( n∑
i,j=1
‖A(1)ij ‖Ls0,1(Ct) + ‖A(1)‖Lr0,1(Ct)
). (5.5)
Analogously, we get from (3.5) that
∣∣(gk(w))(τ)
∣∣ ≤n∑
i,j=1
‖A(1)ij (·, τ)‖Ls1 (Ω)‖wxj (·, τ)‖L2(Ω)‖φkxi
‖Lp(Ω)
+ ‖A(1)(·, τ)‖Ls2 (Ω)‖w(·, τ)‖L2(Ω)‖φk‖Lp1 (Ω),
with the same s1 and s2 as in (1.13), (1.14). After integrating from t tot + 1, we obtain
‖gk(w)‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)|w|Ct
×( n∑
i,j=1
‖A(1)ij ‖Ls1,2(Ct) + ‖A(1)‖Ls2,1(Ct)
). (5.6)
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85
Since r0 = s2/2 and s0 = s1/2, Lemma 5.1 follows for n > p from (5.5) and(5.6).
In the case where n < p, it follows immediately from Sobolev’s embed-ding theorem that φk ∈
(L∞(Ω)
)N with the estimate
‖φk‖L∞(Ω) ≤ C‖φk‖W 1,p(Ω).
Using this, we get analogously to the case where n > p the estimates
‖Rkl‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)‖φl‖W 1,p(Ω)
×( n∑
i,j=1
‖A(1)ij ‖Ls0,1(Ct) + ‖A(1)‖L1(Ct)
)
and
‖gk(w)‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)|w|Ct
×( n∑
i,j=1
‖A(1)ij ‖Ls1,2(Ct) + ‖A(1)‖L2,1(Ct)
)
with the same s0 as before and the lemma follows in the case when n < p.For n = p, the fact that φk ∈
(W 1,p(Ω)
)N implies that
‖φk‖Lp2 (Ω) ≤ C‖φk‖W 1,p(Ω) (5.7)
for any p2 ∈ [1,∞), where C depends on n, p, p2 and Ω. We obtain in thesame way as before
‖Rkl‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)‖φl‖W 1,p(Ω)
×( n∑
i,j=1
‖A(1)ij ‖Ls0,1(Ct) + ‖A(1)‖L2,1(Ct)
),
which implies (5.1). Finally,
∣∣(gk(w))(τ)
∣∣ ≤n∑
i,j=1
‖A(1)ij (·, τ)‖Ls1 (Ω)‖wxj (·, τ)‖L2(Ω)‖φkxi
‖Lp(Ω)
+ ‖A(1)(·, τ)‖Ls′2 (Ω)
‖w(·, τ)‖L2(Ω)‖φk‖Ls3 (Ω), (5.8)
where s′2 was introduced in (1.14) and
s3 =2s′2
s′2 − 2.
September 6, 2006 (12:43)
86
By using (5.7) with p2 = s3 and integrating (5.8) from t to t+1, we obtain
‖gk(w)‖L1(t,t+1) ≤ C‖φk‖W 1,p(Ω)|w|Ct
×( n∑
i,j=1
‖A(1)ij ‖Ls1,2(Ct) + ‖A(1)‖
Ls′2,1(Ct)
)
and (5.2) follows. The proof is complete.
6 Functions hJ+1, . . . , hM
6.1 Definition of functions v0, v1 and v2
For k = 1, . . . ,M , let zk denote an element of L∞loc(0,∞) and introduce thevectors
z = (z1, . . . , zM ),
z = (z1, . . . , zJ) (6.1)
andz = (zJ+1, . . . , zM ). (6.2)
We define v0(z), v1(z) and v2 as solutions of the equations
−∫
Q
(v0, ηt) dx dt +∫ ∞
0
[L1(v0, η) +
M∑
k=J+1
zkL(1)1 (φk, η)
]dt = 0,
−∫
Q
(v1, ηt) dx dt +∫ ∞
0
[L1(v1, η) +
J∑
k=1
zkL(1)1 (φk, η)
]dt = 0
and
−∫
Q
(v2, ηt) dx dt +∫ ∞
0
L1(v2, η) dt =∫
Ω
(ψ(x), η(x, 0)
)dx,
respectively. Here we mean solutions in the sense of Section 4.1.For a given z, the existence and uniqueness of v0(z), v1(z) and v2, con-
sidered as elements of(V 2
0,loc(Q))N , are guaranteed by Proposition 4.1. Fur-
thermore, the elements v0 and v1 are linear operators on(L∞loc(0,∞)
)M−J
and(L∞loc(0,∞)
)J , respectively. Also gk, as defined in (3.5), is a linear
operator but on(V 2
loc(Q))N . It follows that
Γk(z) = gk
(v0(z)
)(6.3)
is a linear operator on(L∞loc(0,∞)
)M−J .
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87
6.2 Integro-differential system for hJ+1, . . . , hM
We introduce
h = (h1, . . . , hJ) and h = (hJ+1, . . . , hM ).
Because of linearity, we can split the function v in (3.1) as
v = v0(h) + v1(h) + v2. (6.4)
Since also gk in (3.5) is linear, the last M − J equations in (3.6) can berewritten as
h′k + λkhk +M∑
l=J+1
Rklhl + Γk(h) = Fk, k = J + 1, . . . , M, (6.5)
where
Fk = −J∑
l=1
Rklhl − γk(h)− gk(v2) (6.6)
andγk(h) = gk
(v1(h)
),
which obviously is linear with respect to h. Together with the boundaryconditions
hk(0) =∫
Ω
(ψ, φk) dx, k = J + 1, . . . ,M (6.7)
this yields a new linear system of equations for h with h considered as agiven vector.
6.3 A general estimate
For some given functions Fk ∈ L1loc(0,∞), not necessarily coinciding with
(6.6), consider the linear system of equations
z′k + λkzk +M∑
l=J+1
Rklzl + Γk(z) = Fk, k = J + 1, . . . , M (6.8)
on the positive real half axis together with the initial conditions
zk(0) = ak, k = J + 1, . . . ,M. (6.9)
We introduce
a =M∑
k=J+1
|ak| (6.10)
September 6, 2006 (12:43)
88
and
F (t) =M∑
k=J+1
‖Fk‖L1(t,t+1). (6.11)
Under a few assumptions on Fk, we will prove existence and uniqueness ofsolutions of the problem (6.8), (6.9) in an appropriate Banach space. Wewill also find an estimate of the solution. But let us first introduce thespace in which we are going to work.
Suppose that ξk, k = J + 1, . . . , M , are measurable functions on (0,∞)and introduce
(Ξ2ξ)(t) =
ess supt<s<t+1
∑Mk=J+1 |ξk(s)| if t ≥ 0
ess sup0<s<t+1
∑Mk=J+1 |ξk(s)| if − 1 < t < 0
0 if t ≤ −1.
For a given Υ ≥ 0, set
‖ξ‖AΥ = supt≥−1
e−Υt(Ξ2ξ)(t).
The reason for the subindex “2” will appear later. We define the Banachspace AΥ as the set of measurable functions ξ = (ξJ+1, . . . , ξM ) on (0,∞)such that ‖ξ‖AΥ < ∞.
Before proving next result, we observe that if f ∈ L1loc(R), it follows
from (4.4) that∫ t+1
0
|f(s)| ds =∫ t
0
|f(s)| ds + ‖f‖L1(t,t+1)
≤∫ t
−1
‖f‖L1(s,s+1) ds + ‖f‖L1(t,t+1)
(6.12)
for t ≥ −1.
Proposition 6.1 For k = J + 1, . . . ,M , let ak be arbitrary real numbersand assume that Fk ∈ L1
loc(R) where Fk(t) = 0 for t < 0. Suppose furtherthat, for some positive Υ, the relations
supt≥−1
e−Υt‖Fk‖L1(t,t+1) < ∞, k = J + 1, . . . , M, (6.13)
hold. If the constant κ in (1.16) is small enough, then there exists a uniquesolution z ∈ AΥ of (6.8), (6.9) and there exist constants C and c0 such that
(Ξ2z)(t) ≤ C
(ae−µt +
∫ t
−1
e−µ(t−s)F (s) ds + F (t))
, t ≥ −1, (6.14)
whereµ = λJ+1 − c0κ0 (6.15)
and a is given by (6.10). The constants C and c0 do only depend on n, N ,Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2 and not on Υ.
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89
Proof. We first prove the existence and uniqueness of solutions. Let us fixk ∈ J + 1, . . . , M and consider the equation corresponding to k in (6.8).It follows immediately that
(zkeλkt
)′= eλkt
(Fk −
M∑
l=J+1
Rklzl − Γk(z))
and after integrating from 0 to t we obtain
zk(t) = ake−λkt +∫ t
0
e−λk(t−s)
(Fk −
M∑
l=J+1
Rklzl − Γk(z))
ds. (6.16)
We construct a linear operator S acting on AΥ so that the k:th componentof S becomes
(Sz)k(t) =∫ t
0
e−λk(t−s)
( M∑
l=J+1
Rklzl + Γk(z))
ds (6.17)
for t ≥ 0. If t < 0 we set (Sz)(t) = 0. The fact that Sz ∈ AΥ if z ∈ AΥ
will appear from the following analysis.Equation (6.16) can be written as
zk(t) = −(Sz)k(t) + αk(t), (6.18)
where
αk(t) = ake−λkt +∫ t
0
e−λk(t−s)Fk(s) ds. (6.19)
From this, the existence and uniqueness of zk follows if we can prove that‖S‖ < 1 and that
α = (αJ+1, . . . , αM ) (6.20)
belongs to AΥ.Let us prove that ‖S‖ < 1 if κ is small enough. We analyze (6.17) term
by term and introduce S1l through the relation
(S1lz)k(t) =∫ t
0
e−λk(t−s)Rklzl ds, t ≥ 0. (6.21)
Let us prove that there exists a constant C such that
‖S1lz‖AΥ ≤ Cκ‖z‖AΥ . (6.22)
To make clear that C depends only on n, N , Ω, A(0)ij , A(0), p, s′2, ν, ν1
and ν2 and not on, for example, Υ, we will write out the exact values of all
September 6, 2006 (12:43)
90
constants occurring in the calculations below. Using (6.12), it follows from(6.21) that, for t ≥ −1,
ess supt<s<t+1
M∑
k=J+1
|(S1lz)k(s)| ≤∫ t+1
0
e−λJ+1(t−s)
( M∑
k=J+1
|Rkl|)|zl| ds
≤∫ t
−1
∥∥∥∥e−λJ+1(t−·)( M∑
k=J+1
|Rkl|)
zl
∥∥∥∥L1(s,s+1)
ds
+∥∥∥∥e−λJ+1(t−·)
( M∑
k=J+1
|Rkl|)
zl
∥∥∥∥L1(t,t+1)
,
(6.23)
where we have extended zl by 0 for t < 0. After a use of Holder’s inequalityand (5.1), this expression can be majorized by
c1κ0(M − J)eλJ+1
(∫ t
−1
e−λJ+1(t−s)‖zl‖L∞(s,s+1) ds + ‖zl‖L∞(t,t+1)
).
Using the fact that‖zl‖L∞(s,s+1) ≤ eΥt‖z‖AΥ
for −1 ≤ s ≤ t and the inequality∫ t
−1
e−a(t−s) dτ ≤ 1a, a > 0,
with a = λJ+1, we finally get that
ess supt<s<t+1
M∑
k=J+1
|(S1lz)k(s)| ≤ c1(M − J)eλJ+1
(1 +
1λJ+1
)κ0e
Υt‖z‖AΥ .
We remind that the constant M depends only on n, N , Ω, A(0)ij , A(0),
p, s′2, ν, ν1 and ν2, see Proposition 4.1 and the beginning of Section 4.2.Therefore, denoting by C the constant before κ0 in the right-hand side, wearrive at (6.22).
We next introduce S2 so that
(S2z)k(t) =∫ t
0
e−λk(t−s)Γk(z) ds
for t ≥ 0 and prove the existence of a constant C satisfying the relation
‖S2z‖AΥ ≤ Cκ2‖z‖AΥ . (6.24)
From Corollary 4.2 it follows that
|v0(z)|Ct ≤ C
(∫ t
−1
e−2λJ+1(t−s)κ(s)Z(s) ds + κ(t)Z(t))
, (6.25)
September 6, 2006 (12:43)
91
where Z = Ξ2z. Using (5.2) together with this fact on Γk(z), as defined in(6.3), we obtain
‖Γk(z)‖L1(t,t+1) ≤ c2κ(t)|v0(z)|Ct
≤ c3κ(t)(∫ t
−1
e−2λJ+1(t−s)κ(s)Z(s) ds + κ(t)Z(t))
(6.26)
for some constant c3.As in (6.23), we get the estimate
ess supt<s<t+1
M∑
k=J+1
|(S2z)k(s)|
≤ eλJ+1
M∑
k=J+1
( ∫ t
−1
e−λJ+1(t−s)‖Γk(z)‖L1(s,s+1) ds
+ ‖Γk(z)‖L1(t,t+1)
).
(6.27)
We use (6.26) to estimate the terms in (6.27). For the integral, it followsthat
∫ t
−1
e−λJ+1(t−s)‖Γk(z)‖L1(s,s+1) ds
≤ c3κ20
∫ t
−1
e−λJ+1(t−s)
(∫ s
−1
e−2λJ+1(s−τ)Z(τ) dτ + Z(s))
ds
≤ c3κ20‖z‖AΥ
λJ+1 + Υ
(1
2λJ+1 + Υ+ 1
)eΥt,
where we have used that Z(t) ≤ eΥt‖z‖AΥ . In the same way we get that
‖Γk(z)‖L1(t,t+1) ≤ c3κ20‖z‖AΥ
(1
2λJ+1 + Υ+ 1
)eΥt.
With these estimates, inequality (6.27) implies (6.24) with
C = c3eλJ+1(M − J)
(1 +
12λJ+1
)(1 +
1λJ+1
).
Since
S =M∑
l=J+1
S1l + S2,
it follows from (6.22) and (6.24) that S : AΥ → AΥ and that ‖S‖ < 1, i.e.S is a contraction on AΥ if κ is small enough.
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92
We now prove that α, as defined in (6.19) and (6.20), is an element ofAΥ. Obviously, the term ake−λkt belongs to AΥ since it is bounded fort ≥ −1. What remains is to check that β = (βJ+1, . . . , βM ) ∈ AΥ, where
βk(t) =∫ t
0
e−λk(t−s)Fk(s) ds.
We have
ess supt<s<t+1
M∑
k=J+1
|βk(s)|
≤ eλJ+1
M∑
k=J+1
( ∫ t
−1
e−λJ+1(t−s)‖Fk‖L1(s,s+1) ds + ‖Fk‖L1(t,t+1)
)
≤ beλJ+1(M − J)(
1λJ+1
+ 1)
eΥt,
where
b = max
supt≥−1
e−Υt‖Fk‖L1(t,t+1) : k ∈ J + 1, . . . , M
is finite because of (6.13). Hence β ∈ AΥ so the same is true for α.We sum up what has been done: Since the relation (6.18) is equivalent
to(I + S)z = α, (6.28)
where α ∈ AΥ and ‖S‖ < 1, it follows that (6.28) has exactly one solutionz ∈ AΥ. Hence, the existence and uniqueness of solutions in AΥ is proved.
We continue by proving the estimate (6.14). From (6.16) it follows that
|zk(t + ε)| ≤ ake−λkt +∫ t+1
0
e−λk(t−s)
(|Fk|+
M∑
l=J+1
|Rklzl|+ |Γk(z)|)
ds
for ε ∈ [0, 1] (if −1 < t < 0, we require that ε ∈ [|t|, 1]). Making the
same estimates as in the computations of ‖S1lz‖AΥ and ‖S2z‖AΥ and usingLemma 5.1, we obtain
Z(t) ≤ ae−λJ+1t + C
(∫ t
−1
e−λJ+1(t−s)B1(s) ds + B1(t))
, (6.29)
whereB1(t) = F (t) + κ(t)
(Z(t) + |v0(z)|Ct
).
The quantities a and F (t) were defined in (6.10) and (6.11). Using (6.25),we get
Z(t) ≤ ae−λJ+1t + C ′(∫ t
−1
e−λJ+1(t−s)B2(s) ds + B2(t))
, (6.30)
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93
with
B2(t) = F (t) + κ(t)(
Z(t) +∫ t
−1
e−2λJ+1(t−s)κ(s)Z(s) ds
). (6.31)
In the last parenthesis we have omitted a term κ(t)Z(t), since, for smallvalues of κ, its contribution is covered by making C ′ larger than C in (6.29).
We rewrite the estimate contained in (6.30), (6.31). By changing theorder of integration and using that κ(s) ≤ κ0, it follows that
∫ t
−1
e−λJ+1(t−s)κ(s)∫ s
−1
e−2λJ+1(s−τ)κ(τ)Z(τ) dτ ds
≤ κ0e−λJ+1t
∫ t
−1
e2λJ+1τκ(τ)Z(τ)∫ t
τ
e−λJ+1s ds dτ
≤ κ0
λJ+1
∫ t
−1
e−λJ+1(t−τ)κ(τ)Z(τ) dτ.
(6.32)
By inserting (6.31) into (6.30) and using (6.32), we get
Z(t) ≤ ae−λJ+1t +C2
2
(∫ t
−1
e−λJ+1(t−s)B3(s) ds + B3(t))
(6.33)
for some constant C2 and
B3(t) = F (t) + κ(t)Z(t).
Assume that κ fulfills the inequality
κ ≤ 1C2
. (6.34)
It then follows from (6.33) that
Z(t) ≤ 2ae−λJ+1t + C2
(κ0
∫ t
−1
e−λJ+1(t−s)Z(s) ds
+∫ t
−1
e−λJ+1(t−s)F (s) ds + F (t))
. (6.35)
We also assume thatκ <
λJ+1
2C2(6.36)
and setε = 2C2κ0 and µ = λJ+1 − ε. (6.37)
Observe that, because of (6.36), we have µ > 0. For t ≥ −1, we define
Z+(t) = b1ae−µt + b2
∫ t
−1
e−µ(t−s)F (s) ds + C2F (t), (6.38)
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94
where b1 = 4eε andb2 = 2C2(C2κ0 + 1). (6.39)
We are going to prove that Z(t) ≤ Z+(t) and first step is to show that
2ae−λJ+1t + C2
(κ0
∫ t
−1
e−λJ+1(t−s)Z+(s) ds
+∫ t
−1
e−λJ+1(t−s)F (s) ds + F (t))≤ Z+(t) (6.40)
if t ≥ −1. We expand the first integral in (6.40) and obtain∫ t
−1
e−λJ+1(t−s)Z+(s) ds = b1ae−λJ+1t
∫ t
−1
eεs ds
+ b2
∫ t
−1
e−λJ+1(t−s)
∫ s
−1
e−µ(s−τ)F (τ) dτ ds
+ C2
∫ t
−1
e−λJ+1(t−s)F (s) ds.
(6.41)
Since ∫ t
−1
eεs ds ≤ eεt
ε
and∫ t
−1
e−λJ+1(t−s)
∫ s
−1
e−µ(s−τ)F (τ) dτ ds
= e−λJ+1t
∫ t
−1
eµτF (τ)∫ t
τ
eεs ds dτ ≤ 1ε
∫ t
−1
e−µ(t−τ)F (τ) dτ,
we get from (6.41) the inequality∫ t
−1
e−λJ+1(t−s)Z+(s) ds ≤ b1a
εe−µt +
(C2 +
b2
ε
) ∫ t
−1
e−µ(t−s)F (s) ds.
Using this estimate, we obtain
2ae−λJ+1t + C2κ0
∫ t
−1
e−λJ+1(t−s)Z+(s) ds
+ C2
( ∫ t
−1
e−λJ+1(t−s)F (s) ds + F (t))
≤(
2eε +b1
2
)ae−µt +
(C2
2κ0 +b2
2+ C2
) ∫ t
−1
e−µ(t−s)F (s) ds
+ C2F (t)= Z+(t)
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95
and (6.40) is proved.We now prove that Z(t) ≤ Z+(t) for t ≥ −1 and use an approach
similar to the one used in Step 2 of the proof of Proposition 4.1 whenproving that h(t) ≤ h+(t). It follows immediately from (6.38) thatZ+(−1) ≥ 4a, so a = Z(−1) ≤ Z+(−1). We divide the problem into thecases when Z+(−1) > Z(−1) and Z+(−1) = Z(−1). Suppose first thatZ+(−1) > Z(−1). Since Z and Z+ are continuous, it follows that thereexists a t > −1 such that Z(τ) < Z+(τ) if τ ∈ [−1, t). Then (6.35) and(6.40) imply that
Z+(t)− Z(t) ≥ C2κ0
∫ t
−1
e−λJ+1(t−s)(Z+(s)− Z(s)
)ds > 0
and the consequence is that the set
t ≥ −1 : Z(τ) < Z+(τ) for τ ∈ [−1, t]
is non-empty, open and closed in [−1,∞), i.e. equal to [−1,∞). HenceZ(t) < Z+(t) for every t ≥ −1.
If Z+(−1) = Z(−1), we introduce, for a given δ > 0, the functions
Zδ(t) = Z(t) + δe−λJ+1t
andZδ
+(t) = Z+(t) + b1δe−µt
for t ≥ −1. From this, it is easy to see that Zδ satisfies the inequality(6.35) with a replaced by a + δ and every occurrence of Z replaced by Zδ.It is also seen that, for t ≥ −1, the function Zδ
+ satisfies (6.38) with areplaced by a + δ. As was the case for Z+, we can hence prove that (6.40)is fulfilled with Z+ replaced by Zδ
+. From the facts that Z(−1) = Z+(−1)and b1 = 4eε it follows that
Zδ(−1) = Z(−1) + δeλJ+1 = Z+(−1) + δeµ+ε
< Z+(−1) + b1δeµ = Zδ
+(−1).
The previously treated case gives that Zδ(t) < Zδ+(t) for t ≥ −1, i.e.
Z(t) + δe−λJ+1t < Z+(t) + b1δe−µt.
By letting δ → 0, we see that Z(t) ≤ Z+(t) for t ≥ −1 also in the casewhere Z(−1) = Z+(−1). Since b1 < 4eλJ+1 and b2 ≤ 4C2, where we haveused that κ0 ≤ κ and (6.34) in (6.39), we have thus proved the estimate
Z(t) ≤ C
(ae−µt +
∫ t
−1
e−µ(t−s)F (s) ds + F (t))
,
i.e. (6.14). From (6.37), we see that c0 in (6.15) can be chosen as 2C2.
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96
6.4 A particular case of equation (6.8)
For k = 1, . . . , J , let ξk be a measurable function on (0,∞). We introduce
(Ξ1ξ)(t) =
ess supt<s<t+1
∑Jk=1 |ξk(s)| if t ≥ 0
ess sup0<s<t+1
∑Jk=1 |ξk(s)| if − 1 < t < 0
0 if t ≤ −1(6.42)
and, for a given Υ ≥ 0, the Banach space BΥ consisting of functionsξ = (ξ1, . . . , ξJ) measurable on (0,∞) such that the norm
‖ξ‖BΥ = supt≥−1
e−Υt(Ξ1ξ)(t)
is finite. Let z and z be as defined in (6.1), (6.2) and set
Fk(z) = −J∑
l=1
Rklzl − γk(z)− gk(v2), k = J + 1, . . . , M. (6.43)
We consider the system of equations
z′k + λkzk +M∑
l=J+1
Rklzl + Γk(z) = Fk(z), k = J + 1, . . . , M, (6.44)
on (0,∞) together with the boundary conditions
zk(0) =∫
Ω
(ψ, φk), k = J + 1, . . . , M, (6.45)
and have the following result.
Corollary 6.2 Assume that z ∈ BΥ for some Υ > 0. Then, the system(6.44), (6.45) has a unique solution z from AΥ. The solution satisfies theestimate
(Ξ2z)(t) ≤ C
(‖ψ‖L2(Ω)e
−µt +∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds
+ κ(t)(Ξ1z)(t))
, t ≥ −1,
(6.46)
with the same µ as in Proposition 6.1.
Proof. We prove that the assertion (6.13) is true for given Υ. Inequality(5.1) implies that
J∑
l=1
‖Rklzl‖L1(t,t+1) ≤ Cκ(t)(Ξ1z)(t), (6.47)
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97
where zl in the left-hand side has been extended by 0 for t < 0. From theassumption that z ∈ BΥ, it follows that
(Ξ1z)(t) ≤ ‖z‖BΥeΥt (6.48)
and when using this in (6.47), we obtain
J∑
l=1
‖Rklzl‖L1(t,t+1) ≤ C‖z‖BΥκ(t)eΥt. (6.49)
From (5.2), we get
‖γk(z)‖L1(t,t+1) ≤ c2κ(t)|v1(z)|Ct ,
which together with Corollary 4.2 shows that
‖γk(z)‖L1(t,t+1)
≤ Cκ(t)(∫ t
−1
e−2λJ+1(t−s)κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
. (6.50)
We use (6.48) in (6.50) and obtain the inequality
‖γk(z)‖L1(t,t+1) ≤ Cκ‖z‖BΥκ(t)eΥt. (6.51)
When considering the term gk(v2), we see from Corollary 4.2 that
|v2|Ct ≤ C‖ψ‖L2(Ω)e−2λJ+1t,
so (5.2) implies that
‖gk(v2)‖L1(t,t+1) ≤ Cκ(t)‖ψ‖L2(Ω)e−2λJ+1t. (6.52)
When using (6.34) in (6.51) and combining (6.49), (6.51) and (6.52), itfollows that Fk(z), as defined in (6.43), is subject to
‖Fk(z)‖L1(t,t+1) ≤ C(‖z‖BΥ + ‖ψ‖L2(Ω)
)κ(t)eΥt.
Since κ(t) ≤ κ, we can again use (6.34) and conclude that Fk = Fk(z)fulfills (6.13).
From (6.47), (6.50) and (6.52) we find the estimate
F (t) ≤ Cκ(t)(‖ψ‖L2(Ω)e
−2λJ+1t
+∫ t
−1
e−2λJ+1(t−s)κ(s)(Ξ1z)(s) ds + (Ξ1z)(t))
, (6.53)
where
F (t) =M∑
k=J+1
‖Fk(z)‖L1(t,t+1).
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98
Furthermore, from (6.45) it follows that the quantity
a =M∑
k=J+1
|zk(0)|
occurring in (6.14) can be majorized by a constant times ‖ψ‖L2(Ω). Hencewe obtain from (6.14) the estimate
(Ξ2z)(t) ≤ C
(‖ψ‖L2(Ω)e
−µt +∫ t
−1
e−µ(t−s)F (s) ds + F (t))
(6.54)
and we will now find an estimate for the term∫ t
−1
e−µ(t−s)F (s) ds
by using (6.53). Since
∫ t
−1
e−(2λJ+1−µ)s ds ≤ e2λJ+1−µ
2λJ+1 − µ<
e2λJ+1
λJ+1,
we see that ∫ t
−1
e−µ(t−s)e−2λJ+1s ds ≤ Ce−µt.
As in (6.32) we also get the inequalities
∫ t
−1
e−µ(t−s)
∫ s
−1
e−2λJ+1(s−τ)κ(τ)(Ξ1z)(τ) dτ ds
≤ 12λJ+1 − µ
∫ t
−1
e−µ(t−τ)κ(τ)(Ξ1z)(τ) dτ
<1
λJ+1
∫ t
−1
e−µ(t−τ)κ(τ)(Ξ1z)(τ) dτ.
Combining these results with (6.53), it follows that
∫ t
−1
e−µ(t−s)F (s) ds
≤ C
(κ0‖ψ‖L2(Ω)e
−µt +∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds
).
From this inequality, it is now easy to see that (6.46) follows from (6.54).
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6.5 Estimate for h
We return to the study of h and introduce
H1 = Ξ1h and H2 = Ξ2h. (6.55)
Corollary 6.3 The vectors h and h are elements of Aa0 and Ba0 , respec-tively, where a0 was introduced in Lemma 2.1, and the estimate
H2(t) ≤ C
(‖ψ‖L2(Ω)e
−µt +∫ t
−1
e−µ(t−s)κ(s)H1(s) ds
+ κ(t)H1(t))
, t ≥ −1,
with the same µ as in Proposition 6.1, is valid.
Proof. We are going to use Corollary 6.2 applied on the system (6.5), (6.7)with Fk(h) being the functions Fk, k = J + 1, . . . , M , as defined in (6.6).Since
hk =∫
Ω
(u, φk) dx
and ‖φk‖L2(Ω) = 1, it follows from Holder’s inequality and (2.16) that
|hk(t)| ≤ ‖ψ‖L2(Ω)ea0t. (6.56)
This implies that h ∈ Aa0 and h ∈ Ba0 . Since h obviously is a solution of(6.5), (6.7), Corollary 6.3 is a consequence of Corollary 6.2.
6.6 A representation for h
The terms Rklhl and γk(h) occurring in Fk are linear functions of h whilegk(v2) is independent of h. This makes us introduce the vectors
h0 = (h0,J+1, . . . , h0,M ) and h1 = (h1,J+1, . . . , h0,M )
as the unique solutions in Aa0 of the systems of equations
h′0,k + λkh0,k +M∑
l=J+1
Rklh0,l + Γk(h0) = −J∑
l=1
Rklhl − γk(h),
h0,k(0) = 0 (6.57)
and
h′1,k + λkh1,k +M∑
l=J+1
Rklh1,l + Γk(h1) = −gk(v2), (6.58)
h1,k(0) =∫
Ω
(ψ, φk) dx (6.59)
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100
for k = J + 1, . . . ,M . We observe that the the solution h0 depends lin-early on h while h1 is independent of h. This motivates us to write h0(h)instead of h0. We also realize that, in account of linearity and uniquenessof solutions in Aa0 , the following representation for h holds:
h = h0(h) + h1. (6.60)
For a given z ∈ BΥ, introduce
G0(z) = Ξ2
(h0(z)
)
and consider the equations
h′0,k + λkh0,k +M∑
l=J+1
Rklh0,l + Γk(h0) = −J∑
l=1
Rklzl − γk(z) (6.61)
for k = J + 1, . . . , M . By comparing (6.61) with (6.44), we see that Corol-lary 6.2 can be used with ψ = 0 to prove that the system (6.61), (6.57) hasa unique solution h0(z) in AΥ (and hence h0(h) ∈ Aa0). We also obtainthe estimate
(G0(z)
)(t) ≤ C
(∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
(6.62)
for t ≥ −1.Let us now consider the system (6.58), (6.59) and introduce
G1 = Ξ2h1.
Corollary 6.2 can be used with z = 0 to prove uniqueness of the solution.We also obtain the estimate
G1(t) ≤ C‖ψ‖L2(Ω)e−µt. (6.63)
7 Functions h1, . . . , hJ
7.1 Equation for h; existence and uniqueness results
We now consider the system of equations (3.6), (3.7) for k = 1, . . . , J .Equation (3.6) obviously becomes
h′k +M∑
l=1
Rklhl + gk(v) = 0
and using the decompositions (6.4) and (6.60) gives the system
h′k +J∑
l=1
Rklhl +Mk(h) = βk, (7.1)
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hk(0) = ak (7.2)
for k = 1, . . . , J , where
Mk(h) =M∑
l=J+1
Rklh0,l(h) + Γk
(h0(h)
)+ γk(h), (7.3)
βk = −M∑
l=J+1
Rklh1,l − Γk(h1)− gk(v2) (7.4)
and
ak =∫
Ω
(ψ, φk) dx. (7.5)
We concluded in Section 6 that the functions h0 and γk depend linearlyon h while Γk depends linearly on h and the functions h1 and gk(v2) areindependent of h. It follows that Mk depends linearly on h while βk isindependent of h. Let us prove the following estimates for the operatorsMk and functions βk, k = 1, . . . , J .
Lemma 7.1 (i) For every Υ > 0, the operator Mk is defined on BΥ andsatisfies
‖Mk(z)‖L1(t,t+1) ≤ Cκ(t)( ∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds
+ κ(t)(Ξ1z)(t))
, t ≥ −1.
(7.6)
The space BΥ and the operator Ξ1 were introduced on page 96.
(ii) The function βk, given by (7.4), satisfies
‖βk‖L1(t,t+1) ≤ C‖ψ‖L2(Ω)κ(t)e−µt, t ≥ −1. (7.7)
The proofs of (i) and (ii) are similar, compare the right-hand sides in(7.3) and (7.4).
Proof. (i): Let z ∈ BΥ. From (5.1) and (6.62) it follows that
‖Rklh0,l(z)‖L1(t,t+1) ≤ c1κ(t)(G0(z)
)(t)
≤ Cκ(t)(∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
.(7.8)
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We get an estimate for Γk
(h0(z)
)from (6.26). After using (6.62) again and
changing the order of integration, we obtain
‖Γk
(h0(z)
)‖L1(t,t+1)
≤ Cκ(t)( ∫ t
−1
e−2λJ+1(t−s)κ(s)(G0(z)
)(s) ds + κ(t)
(G0(z)
)(t)
)
≤ Cκ(t)(∫ t
−1
e−µ(t−s)κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
.
(7.9)
As in (6.50) it finally follows that
‖γk(z)‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−2λJ+1(t−s)κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
(7.10)and by combining (7.8), (7.9) and (7.10), inequality (7.6) follows.
(ii): The proof is similar to that of (i) but instead of (6.50) and (6.62),we use (6.52) and (6.63).
We are going to study problem (7.1), (7.2) with a general right-hand sideβ = (β1, . . . , βJ) from BΥ for some appropriate Υ and arbitrary complexnumbers ak, k = 1, . . . , J . We will need a two-sided estimate for the numberµ introduced in (6.15). In the remaining part of the paper we assume thatκ fulfills the inequality
κ ≤ λJ+1
8c0. (7.11)
This, together with (6.15), gives
7λJ+1
8≤ µ ≤ λJ+1. (7.12)
In the following lemma we hence assume that βk, k = 1, . . . , J , arearbitrary functions, not necessarily the same as in (7.4).
Lemma 7.2 Assume that there exist positive constants C1 and Υ such that
‖βk‖L1(t,t+1) ≤ C1eΥt, k = 1, . . . , J (7.13)
and let a1, . . . , aJ be arbitrary complex numbers. Provided that κ is smallenough, there exists a unique solution of (7.1), (7.2) in BΥ.
Proof. The proof is similar to the corresponding part of Proposition 6.1.It follows that (7.1), (7.2) is equivalent to
hk = −(Sh)k + αk, k = 1, . . . , J,
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103
where
(Sh)k(t) =∫ t
0
( J∑
l=1
Rklhl +Mk(h))
ds (7.14)
for t ≥ 0 and
αk(t) = ak +∫ t
0
βk ds.
We set (Sh)(t) to 0 for t < 0. The aim is to prove that S is a bounded,linear operator from BΥ to BΥ with norm less than 1 if κ is small enough.So suppose that z ∈ BΥ. We estimate ‖Sz‖L∞(t,t+1) for t ≥ −1 by using(7.14). This yields that
ess supt<s<t+1
|(Sz)k(s)| ≤∫ t
−1
( J∑
l=1
‖Rklzl‖L1(s,s+1) + ‖Mk(z)‖L1(s,s+1)
)ds
+J∑
l=1
‖Rklzl‖L1(t,t+1) + ‖Mk(z)‖L1(t,t+1).
Here we use Lemma 7.1 to estimate the terms containing Mk and (5.1)to estimate the terms containing Rkl. Furthermore, it follows from (7.12)that
∫ t
−1
κ(s)∫ s
−1
e−µ(s−τ)κ(τ)(Ξ1z)(τ) dτ ds ≤ κ0
µ
∫ t
−1
κ(τ)(Ξ1z)(τ) dτ
≤ 8κ0
7λJ+1
∫ t
−1
κ(τ)(Ξ1z)(τ) dτ.
It is now easy to derive the inequality
ess supt<s<t+1
J∑
k=1
|(Sz)k(s)| ≤ C0
(∫ t
−1
κ(s)(Ξ1z)(s) ds + κ(t)(Ξ1z)(t))
,
from which it follows that
e−Υt ess supt<s<t+1
J∑
k=1
|(Sz)k(s)| ≤ C0‖z‖BΥ
(e−Υt
∫ t
−1
κ(s)eΥs ds + κ(t))
≤ C0
(1Υ
+ 1)κ‖z‖BΥ ,
i.e.
‖Sz‖BΥ ≤ C0
(1Υ
+ 1)κ‖z‖BΥ .
The condition ‖S‖ < 1 is thus met if
κ <1
C0(1 + 1/Υ).
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104
In order to complete the proof, it suffices to prove thatα = (α1, . . . , αJ) ∈ BΥ. Since ak is constant, it is sufficient to show thatf = (f1, . . . , fJ), where
fk(t) =∫ t
0
βk ds,
belongs to BΥ. It follows from (7.13) that
ess supt<s<t+1
J∑
k=1
|fk(s)| ≤ CeΥt
(1 +
1Υ
),
so f ∈ BΥ and thus α ∈ BΥ.
We write problem (7.1), (7.2) in matrix form
h′ +Rh +M(h) = β, (7.15)
h(0) = a, (7.16)
where R is the J × J-matrix with entry (k, l) equal to Rkl and M(h), β
and a are vectors of dimension J with k:th component equal to Mk(h), βk
and ak, respectively, as given in (7.3)–(7.5). Taking Υ = a0 in Lemma 7.2,where a0 was defined in Lemma 2.1, we arrive at the following corollary.
Corollary 7.3 With βk and ak given in (7.4), (7.5), there exists a uniquesolution h of (7.15), (7.16) in Ba0 .
7.2 The homogeneous equation
We now study the homogeneous version of (7.15), (7.16), i.e. the system
h′ +Rh +M(h) = 0, (7.17)
h(0) = a. (7.18)
We use here an approach for studying the above problem suggested inSection 16, 17, Kozlov, Maz’ya [10] and Section 4, Kozlov [7]. See alsoKozlov, Langer [8].
Suppose that a 6= 0. The question is whether it is possible to haveh(t0) = 0 for some t0 ≥ 0. The answer is no, as follows from the followinglemma.
Lemma 7.4 Assume that h is a solution of (7.17). If h(t0) = 0 for somet0 ≥ 0, then h is identically 0.
Proof. By translating the t-variable, it follows immediately from Corollary7.3 that h(t) = 0 for t ≥ t0. Set
f(t) = eµt2 H1(t). (7.19)
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Obviously, f(t) = 0 for t < −1, and, by continuity, there exists at1 ∈ [−1, t0] such that
A := f(t1) = maxt∈[−1,t0]
f(t). (7.20)
Since f is increasing on [−1, 0], it follows that t1 ≥ 0.From (7.17) we write
h(t) =∫ t0
t
(Rh +M(h)
)ds
and obtain the estimate
|h(t)| ≤ Cκ0
∫ t0
t−1
(H1(s) + κ0
∫ s
−1
e−µ(s−τ)H1(τ) dτ
)ds. (7.21)
Here we estimate∫ t0
t−1
H1(s) ds ≤ A
∫ t0
t−1
e−µs2 ds ≤ 2A
µe
µ2 e−
µt2
and∫ t0
t−1
∫ s
−1
e−µ(s−τ)H1(τ) dτ ds ≤ A
∫ t0
t−1
e−µs
∫ s
−1
eµτ2 dτ ds ≤ 4Ae
µ2
µ2e−
µt2 .
Using these estimates in (7.21) and the fact that µ ≥ 7λJ+1/8 from (7.12),we obtain
H1(t) ≤ C1κ0Ae−µt2
and henceA ≤ C1κA.
By requiring κ < C−11 , it follows that A = 0. But then (7.19) and (7.20)
imply that h(t) = 0 for 0 ≤ t ≤ t0 so h is identically 0.
We assume that h(t) is not identically 0. From Lemma 7.4 it thenfollows that |h(t)| > 0 for all t ≥ 0. This means that we can represent h as
h(t) = ρ(t)Θ(t), (7.22)
whereρ(t) = |h(t)|, |Θ(t)| = 1
and ρ and Θ are both absolutely continuous. We insert (7.22) in (7.17) andmultiply by Θ. This yields
((ρΘ)′, Θ
)+ ρ
(RΘ,Θ)
+(M(ρΘ),Θ
)= 0. (7.23)
Obviously, ((ρΘ)′, Θ
)= ρ′
(Θ,Θ
)+ ρ
(Θ′, Θ
)
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106
and (Θ, Θ) = |Θ|2 = 1. This implies that
0 =d
dt(Θ, Θ) = (Θ′, Θ) + (Θ, Θ′) = 2R(Θ′,Θ),
so (Θ′, Θ) is purely imaginary. This shows that
R((ρΘ)′,Θ
)= ρ′.
The ellipticity and symmetry properties of the matrices A(1)ij and A(1) imply
that R is hermitian, which in particular means that (RΘ, Θ) is real. Aftertaking the real part of (7.23), we obtain
ρ′ + fρ +M0(ρ) = 0 (7.24)
on (0,∞), wheref = (RΘ,Θ) (7.25)
andM0(ρ) = R
(M(ρΘ), Θ).
Here we temporarily consider Θ as a given function and the reason is thatwe want an equation for ρ. We also have the initial condition
ρ(0) = ρ0, (7.26)
whereρ0 = |a| > 0,
with a from (7.18). Since M is linear, it also follows that M0 is linear soequation (7.24) is linear.
Suppose that Υ is a given positive number. If ξ ∈ L∞loc(0,∞), we set
(Ξ3ξ)(t) =
‖ξ‖L∞(t,t+1) if t ≥ 0‖ξ‖L∞(0,t+1) if − 1 < t < 00 if t ≤ −1.
and say that ξ ∈ DΥ if
‖ξ‖DΥ = supt≥−1
e−Υt(Ξ3ξ)(t) < ∞.
We also introduce the operator Z on functions defined on [0,∞) as
(Zy)(t) = y(t) exp(∫ t
0
f(s) ds
)
and setz = Zρ.
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107
From (7.24), (7.26) we obtain the problem
z′ +N (z) = 0 on (0,∞), (7.27)
z(0) = z0, (7.28)
whereN (z) = ZM0(Z−1z) (7.29)
and z0 = ρ0. We will need the following estimate for the operator N .
Lemma 7.5 N is a linear operator on DΥ with the estimate
‖N (g)‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−µ1(t−s)κ(s)(Ξ3g)(s) ds + κ(t)(Ξ3g)(t))
,
(7.30)where
µ1 = µ− c3κ0 (7.31)
for some constant c3.
Proof. The linearity of N follows from (7.29) since every operator thereis linear. Furthermore, it follows that
‖N (g)‖L1(t,t+1) ≤ ess supt<s<t+1
exp
(∫ s
0
f(τ) dτ
)‖M0(Z−1g)‖L1(t,t+1).
(7.32)As before, we carry out this analysis for t ≥ −1. This means that if−1 ≤ t < 0, we take supremum over (0, t + 1) instead of (t, t + 1) in (7.32).
From Lemma 5.1 it follows that f ∈ L1loc(0,∞) with the estimate
‖f‖L1(t,t+1) ≤ c3κ(t) (7.33)
for some constant c3. Hence, the function
s 7→ exp(∫ s
0
f(τ) dτ
)
is continuous and the essential supremum in (7.32) is attained for somet0 ∈ [t, t + 1]. By writing h = Z−1g, it follows that
‖M0(Z−1g)‖L1(t,t+1) ≤ ‖M(hΘ)‖L1(t,t+1)
and since |Θ| = 1, we get an estimate for the last norm from (7.6). Usingall this in (7.32), we obtain
‖N (g)‖L1(t,t+1) ≤ Cκ(t) exp(∫ t0
0
f(τ) dτ
)
×(∫ t
−1
e−µ(t−s)κ(s)(Ξ3h)(s) ds + κ(t)(Ξ3h)(t))
. (7.34)
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108
We furthermore have that
(Ξ3h)(s) = ess sups<u<s+1
exp
(−
∫ u
0
f(τ) dτ
)|g(u)|
≤ (Ξ3g)(s) sups<u<s+1
exp
(−
∫ u
0
f(τ) dτ
)
= (Ξ3g)(s) exp(−
∫ u0(s)
0
f(τ) dτ
),
(7.35)
for some u0(s) ∈ [s, s + 1].We consider the term
exp(∫ t0
0
f(τ) dτ
) ∫ t
−1
e−µ(t−s)κ(s)(Ξ3h)(s) ds (7.36)
appearing in (7.34). According to (7.35), the factor
exp(∫ t0
0
f(τ) dτ
)(Ξ3h)(s)
can be majorized by
(Ξ3g)(s) exp( ∫ t0
u0(s)
f(τ) dτ
).
In turn, ∫ t0
u0(s)
f(τ) dτ ≤∫ b
a
‖f‖L1(τ,τ+1) dτ, (7.37)
wherea = mint0, u0(s) − 1, b = maxt0, u0(s).
Since t0 ∈ [t, t + 1] as well as u0(s) ∈ [s, s + 1] and s ≤ t, we see thatb− a ≤ t− s + 2. Hence, it follows from (7.33) that
∫ b
a
‖f‖L1(τ,τ+1) dτ ≤ (t− s + 2)c3κ0. (7.38)
Summing this up, we get the estimate
exp(∫ t0
0
f(τ) dτ
)(Ξ3h)(s) ≤ e2c3κ0(Ξ3g)(s)ec3κ0(t−s). (7.39)
For the term in (7.36), it then yields the inequality
exp(∫ t0
0
f(τ) dτ
) ∫ t
−1
e−µ(t−s)κ(s)(Ξ3h)(s) ds
≤ C
∫ t
−1
e−µ1(t−s)κ(s)(Ξ3g)(s) ds, (7.40)
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109
with µ1 = µ− c3κ0. By setting s = t in (7.39), we get the inequality
exp(∫ t0
0
f(τ) dτ
)(Ξ3h)(t) ≤ C(Ξ3g)(t). (7.41)
By combining (7.40) and (7.41), estimate (7.30) follows from (7.34).
As was the case for µ, we will also need a two-sided estimate for µ1.Assuming that κ ≤ λJ+1/(8c3) and using (7.31) and (7.12), we obtain
3λJ+1
4≤ µ1 ≤ λJ+1. (7.42)
We now prove uniqueness of solutions of at most exponential increaseof problem (7.27), (7.28).
Lemma 7.6 Let Υ be an arbitrary, positive number. If z0 = 0 and κ issmall enough, then z = 0 is the only solution in DΥ of (7.27), (7.28).
Proof. We assume that z is a solution of problem (7.27), (7.28) with z0 = 0and prove that z = 0. The problem is equivalent to the integral equation
z(t) = −∫ t
0
N (z) ds. (7.43)
Equation (7.43) and estimate (7.30) imply that
(Ξ3z)(t) ≤∫ t
−1
‖N (z)‖L1(s,s+1) ds + ‖N (z)‖L1(t,t+1)
≤ C1κ0
( ∫ t
−1
∫ s
−1
e−µ1(s−τ)κ(τ)(Ξ3z)(τ) dτ ds
+∫ t
−1
κ(s)(Ξ3z)(s) ds + κ(t)(Ξ3z)(t))
.
After changing the order of integration and using (7.42), we see that theinequalities
∫ t
−1
∫ s
−1
e−µ1(s−τ)κ(τ)(Ξ3z)(τ) dτ ds ≤ 1µ1
∫ t
−1
κ(τ)(Ξ3z)(τ) dτ
≤ 43λJ+1
∫ t
−1
κ(τ)(Ξ3z)(τ) dτ
hold, so
(Ξ3z)(t) ≤ C1κ0
((1 +
43λJ+1
) ∫ t
−1
κ(s)(Ξ3z)(s) ds + κ(t)(Ξ3z)(t))
.
(7.44)
September 6, 2006 (12:43)
110
We now assume that κ is so small that the inequality 1 − C1κ2 ≥ 1/2 isfulfilled. It then follows from (7.44) that
(Ξ3z)(t)2
≤ Cκ0
∫ t
−1
κ(s)(Ξ3z)(s) ds
and Gronwall’s inequality gives that (Ξ3z)(t) = 0 for every t ≥ −1, i.e.z = 0.
Lemma 7.6 deals with the uniqueness of solutions of (7.27), (7.28). Theexistence of solutions follows from the following lemma.
Lemma 7.7 For z0 6= 0 and κ small enough, there exists a solution of(7.27), (7.28) of the form
z(t) = z0 exp(∫ t
0
Λ(s) ds
). (7.45)
Here, the real-valued function Λ satisfies the inequality
‖Λ‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−µ2(t−s)κ(s) ds + κ(t))
, t ≥ 1, (7.46)
whereµ2 = µ− c4κ0 (7.47)
for some positive constant c4.
Proof. It follows from (7.45) and the fact that z is real-valued that thesame is true for Λ. Because of the linearity, we can without loss of generalityassume that z0 = 1. Inserting
z(t) = exp(∫ t
0
Λ(s) ds
)
in (7.27), we obtain the equation
T Λ = Λ, (7.48)
where
(T Λ)(t) = −N[s 7→ exp
(−
∫ t
s
Λ(τ) dτ
)](t). (7.49)
The existence of a solution of the given form is proved if we can find a fixedpoint of the operator T . We introduce
‖Λ‖B = ess supt≥−1
‖Λ‖L1(t,t+1)
and define the Banach space
B =Λ ∈ L1
loc(R) : Λ(t) = 0 for t < 0 and ‖Λ‖B < ∞.
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The quantity ‖ · ‖B is a norm in this space. We will prove, provided κ issmall, that T maps the set
B = Λ ∈ B : ‖Λ‖B ≤ κ0 ,
where κ0 is the same number as in (1.15), into itself and is a contractionon B.
We estimate the expression given in (7.49) by using (7.30). We firsthave to estimate
G(s) = ess sups<τ<s+1
exp
(−
∫ t
τ
Λ(u) du
)
for s ≤ t. Note that in the function which N is acting on in (7.49), thesymbol t is considered a fixed quantity. Similarly to (7.37), we obtain
G(s) ≤ exp( ∫ b
a
‖Λ‖L1(u,u+1) du
),
for some a, b such that 0 ≤ b − a ≤ t − s + 2. Since Λ ∈ B, we can makethe estimate G(s) ≤ e2κ0eκ0(t−s). When using this in (7.30), we see from(7.49) that
‖T Λ‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−µ2(t−s)κ(s) ds + κ(t))
, (7.50)
where µ2 = µ1 − κ0 = µ − (c3 + 1)κ0 with c3 as in (7.31). Here we alsosee that the constant c4 in (7.47) is equal to c3 + 1 and by requiring thatκ ≤ λJ+1/4, it follows that µ1 − µ2 ≤ λJ+1/4. Together with (7.42), itgives the bounds
λJ+1
2≤ µ2 ≤ λJ+1. (7.51)
By estimating κ in (7.50) by κ0 and computing the resulting integral,we obtain
‖T Λ‖L1(t,t+1) ≤ Cκ20
(1µ2
+ 1)≤ C
(2
λJ+1+ 1
)κκ0 = C ′κκ0.
We see that‖T Λ‖L1(t,t+1) ≤ κ0
if κ ≤ 1/C ′, so T maps B into itself, as desired.We go on and prove that T is a contradiction on B, i.e. that there exists
a constant C < 1 such that
‖T Λ1 − T Λ2‖B ≤ C‖Λ1 − Λ2‖Bif Λ1, Λ2 ∈ B. Because of the linearity of N , it follows from (7.49) that
(T Λ1 − T Λ2)(t) = N (g2 − g1)(t), (7.52)
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112
where
gk(s) = exp(−
∫ t
s
Λk(u) du
), k = 1, 2.
We also set
ζk(s) = −∫ t
s
Λk(u) du, k = 1, 2
and introduceG(s) = ess sup
s<τ<s+1|g2(τ)− g1(τ)|.
Since |ea − eb| ≤ |a− b|emaxa,b, we obtain
G(s) ≤ ess sups<τ<s+1
(|ζ2(τ)− ζ1(τ)| emaxζ1(τ),ζ2(τ)
). (7.53)
As in (7.37) and (7.38) we get, with τ ∈ [s, s + 1] and s ≤ t, the estimates∫ t
τ
∣∣Λ1(u)− Λ2(u)∣∣ du ≤ ‖Λ1 − Λ2‖B(t− s + 2)
and
−∫ t
τ
Λk(u) du ≤ ‖Λk‖B(t− s + 2).
Since Λ1, Λ2 ∈ B, it follows from (7.53) that
G(s) ≤ ‖Λ1 − Λ2‖B(t− s + 2) eκ0(t−s+2).
We use this in (7.30) with g = g2−g1 and obtain from (7.52) the inequality
‖T Λ1−T Λ2‖L1(t,t+1) ≤ C1κ20‖Λ1−Λ2‖B
(∫ t
−1
(t− s + 2)e−µ2(t−s) ds + 1)
.
(7.54)Since ∫ t
−∞(t− s + 2)e−µ2(t−s) ds =
1µ2
2
+2µ2
,
it follows from (7.54) that
‖T Λ1 − T Λ2‖L1(t,t+1) ≤ C1κ20
(1µ2
2
+2µ2
+ 1)‖Λ1 − Λ2‖B
= C1κ20
(1µ2
+ 1)2
‖Λ1 − Λ2‖B≤ C2‖Λ1 − Λ2‖B,
where (7.51) is used to calculate
C2 = C1κ2
(2
λJ+1+ 1
)2
.
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113
We assume that κ so chosen so small that this value is less than 1. Hence,T is a contraction on B and, since B is closed in B, it follows from theBanach fixed point theorem that there exists a unique function Λ in Bsatisfying relation (7.48). As a consequence, we see that estimate (7.50) isvalid also for Λ, so (7.46) follows.
We now summarize what has been done and get the following result.
Lemma 7.8 Provided that κ is small enough, there exists a unique solutionh of (7.17), (7.18) in Ba0 . This solution can be represented in the form
h(t) = |a| exp( ∫ t
0
(− f(s) + Λ(s))ds
)Θ(t), (7.55)
wheref = (RΘ, Θ), (7.56)
|Θ(t)| = 1, t ≥ 0, (7.57)
‖Λ‖L1(t,t+1) ≤ Cκ(t)(∫ t
−1
e−µ2(t−s)κ(s) ds + κ(t))
, t ≥ −1, (7.58)
and‖Θ′‖L1(t,t+1) ≤ Cκ(t), t ≥ −1. (7.59)
The functions Λ and Θ′ are here extended by 0 for t < 0.
Proof. According to Lemma 7.2, there exists a unique solution of (7.17),(7.18) in Ba0 . If a = 0, the solution is obviously 0 and we are done. Supposethat a 6= 0. Then it follows from Lemma 7.4 that h is non-vanishing. Hencewe can write h = ρΘ as in (7.22) and obtain the problem (7.24), (7.26).Here, the function f is introduced in (7.25), which is identical to (7.56).
We make one more transformation, namely z = Zρ, and see that zsolves the problem (7.27), (7.28) with z0 = |a|. It is easy to realize thatthis is equivalent to (7.24), (7.26).
It follows that z ∈ DΥ for some Υ with DΥ defined on page 106. Namely,since
z(t) = exp(∫ t
0
f(s) ds
)|h(t)|
and estimate (7.33) is valid, we see that
(Ξ3z)(t) ≤ C1eC2κ0tH1(t),
where H1 was defined in (6.55). By using (6.56), this implies that
(Ξ3z)(t) ≤ C1‖ψ‖L2(Ω)e(C2κ0+a0)t,
so z ∈ DΥ for Υ = C2κ0 + a0. According to Lemma 7.6, this is theonly solution of problem (7.27), (7.28) in DΥ. By Lemma 7.7, it can be
September 6, 2006 (12:43)
114
written in the form (7.45) with Λ satisfying (7.46). The conclusion is thath = (Z−1z)Θ and (7.55) follows.
It remains to prove (7.59). We write h(t) = |a| exp( ∫ t
0g(s) ds
)Θ(t),
where g = −f + Λ. Differentiating this formula with respect to t, weobtain
|a|Θ′(t) = exp(−
∫ t
0
g(s) ds
)h′(t)− |a|g(t)Θ(t).
From this, the norm inequality
|a|‖Θ′‖L1(t,t+1) ≤‖h′‖L1(t,t+1) supt≤s≤t+1
exp
(−
∫ s
0
g(τ) dτ
)
+ |a|‖g‖L1(t,t+1)
(7.60)
follows.We proceed with estimating ‖g‖L1(t,t+1). Using estimate (7.33) for f
and the estimate ‖Λ‖L1(t,t+1) ≤ Cκ0κ(t) for Λ, which follows from (7.46),we obtain
‖g‖L1(t,t+1) ≤ ‖Λ‖L1(t,t+1) + ‖f‖L1(t,t+1) ≤ c6κ(t), (7.61)
where c6 is a constant.We next estimate ‖h′‖L1(t,t+1). From (7.17) we obtain the equality
h′ = −(Rh +M(h)).
Using (5.1) and (7.6), it follows that
‖h′‖L1(t,t+1) ≤ Cκ(t)(
H1(t) +∫ t
−1
e−µ(t−s)κ(s)H1(s) ds
). (7.62)
We search for a convenient expression for H1(t). Because of continuity, thefollowing is valid for some t0 ∈ [t, t + 1]:
supt≤s≤t+1
|hk(s)| ≤ supt≤s≤t+1
|a| exp( ∫ s
0
g(τ) dτ
)
= |a| exp( ∫ t0
0
g(τ) dτ
)
= |a| exp( ∫ t
0
g(τ) dτ
)exp
( ∫ t0
t
g(τ) dτ
).
(7.63)
From (7.61) it follows that
exp( ∫ t0
t
g(τ) dτ
)≤ exp
( ∫ t+1
t−1
‖g‖L1(τ,τ+1) dτ
)≤ e2c6κ0 ,
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115
i.e. a constant, so from (7.63) we get
H1(t) ≤ C|a| exp( ∫ t
0
g(s) ds
).
We use this in (7.62) to obtain
‖h′‖L1(t,t+1) ≤ C|a|κ(t)(
exp( ∫ t
0
g(τ) dτ
)
+∫ t
−1
e−µ(t−s)κ(s) exp( ∫ s
0
g(τ) dτ
)ds
).
Since
exp( ∫ s
0
g(τ) dτ
)≤ Cec6κ0(t−s) exp
( ∫ t
0
g(τ) dτ
),
where c6 is the same constant as in (7.61), we get, with µ4 = µ− c6κ0, theestimate
‖h′‖L1(t,t+1) ≤ C|a|κ(t) exp( ∫ t
0
g(τ) dτ
) (1 +
∫ t
−1
e−µ4(t−s)κ(s) ds
).
(7.64)When assuming that κ ≤ λJ+1/(4c6) and using (7.12), we conclude thatµ4 ≥ 5λJ+1/8. Therefore, the last integral in (7.64) is estimated by aconstant depending on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2. Hence
‖h′‖L1(t,t+1) ≤ C|a|κ(t) exp( ∫ t
0
g(τ) dτ
). (7.65)
A use of (7.61) and (7.65) in (7.60) yields
|a|‖Θ′‖L1(t,t+1) ≤ C|a|κ(t),
which implies (7.59).
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7.3 A particular solution of (7.15)
We start the subsection with the following lemma.
Lemma 7.9 Letµ3 = λJ+1 − 2c0κ0, (7.66)
where c0 is the same constant as in (6.15). Suppose that the right-handside β in (7.15) is a measurable vector function satisfying
‖β‖L1(t,t+1) ≤ bκ0e−µ3t, t ≥ 0 (7.67)
for some constant b. Then equation (7.15) has a solution h subject to theestimate
|h(t)| ≤ Cbκ0e−µ3t, t > 0, (7.68)
provided that κ is small enough.
Proof. We introduce the Banach space E consisting of functions
z = (z1, . . . , zJ) ∈ L∞loc(0,∞)
such that‖z‖E = sup
t≥−1eµ3t(Ξ1z)(t) < ∞,
where Ξ1, was introduced in (6.42). If h is a solution from E of (7.15), thenobviously
h = Kh, (7.69)
where(Kh)(t) =
∫ ∞
t
(Rh +M(h)− β)ds,
since the integrand belongs to L1(0,∞). We prove the lemma by showingthat K has a fixed point in E . Let us first prove that K maps the set
E = z ∈ E : ‖z‖E ≤ b
into itself so assume that h ∈ E. By using estimates (5.1), (7.6) and (7.67),we obtain, after having extended β by 0 for t < 0, the estimate
|(Kh)(t)| ≤ Cκ0
∫ ∞
t−1
(H1(s) + κ0
∫ s
−1
e−µ(s−τ)H1(τ) dτ + be−µ3s
)ds.
(7.70)We estimate the terms in the right-hand side. Since h ∈ E, it follows that
H1(t) ≤ be−µ3t. (7.71)
By using the inequality
3λJ+1
4≤ µ3 ≤ λJ+1, (7.72)
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117
obtained when combining (7.11) and (7.66), we get∫ ∞
t−1
be−µ3s ds ≤ beµ3
µ3e−µ3t ≤ 4beλJ+1
3λJ+1e−µ3t, (7.73)
Furthermore,∫ ∞
t−1
κ0
∫ s
−1
e−µ(s−τ)H1(τ) dτ ds ≤ beµ3κ0
µ3(µ− µ3)e−µ3t ≤ 4beλJ+1
3λJ+1c0e−µ3t,
(7.74)where we have used (7.71) in the first inequality and (6.15), (7.66) and(7.72) in the second. Combining (7.70), (7.71), (7.73) and (7.74), we finallyobtain
|(Kh)(t)| ≤ Cbκ0e−µ3t. (7.75)
From this it follows that ‖Kh‖E ≤ b if κ is small, so K maps E into itself,as desired.
In the same way it follows that
‖Kh1 −Kh2‖E ≤ α‖h1 − h2‖Ewith α < 1 if h1, h2 ∈ E and κ being small enough. Hence K is a contrac-tion on the closed set E and has a unique fixed point there. This functionis a solution of (7.15) and the estimate (7.68) is obtained from (7.69) and(7.75).
Corollary 7.10 With β = (β1, . . . , βJ), where βk is given by (7.4), thereexists a solution hp of (7.15) such that
|hp(t)| ≤ C‖ψ‖L2(Ω)e−µ3t. (7.76)
Proof. From (7.7) it follows that (7.67) is fulfilled with b = C‖ψ‖L2(Ω) so(7.76) follows from (7.68).
We combine what has been proved in Section 7 to obtain the followingsplitting of the solution of problem (7.15), (7.16).
Lemma 7.11 The unique solution h in Ba0 of (7.15), (7.16) from Corol-lary 7.3, can be written as
h = hh + hp, (7.77)
where hp is the same as in Corollary 7.10. The vector function hh is asolution of (7.15) with β = 0 and can be represented as
hh(t) = |hh(0)| exp( ∫ t
0
(− f(s) + Λ(s))ds
)Θ(t), (7.78)
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118
where the function f is given by (7.56), Θ is satisfying (7.57), the inequality
|hh(0)| ≤ C‖ψ‖L2(Ω) (7.79)
is valid and estimates for Λ and Θ′ are found in (7.58) and (7.59).
Proof. The function hh satisfies the problem
h′h +Rhh +M(hh) = 0,
hh(0) = a− hp(0),
where a is the same vector as in (7.16). The representation (7.78) followsfrom Lemma 7.8. Inequality (7.76) implies that |hp(0)| ≤ C‖ψ‖L2(Ω) andfrom (7.5) it follows that |ak| ≤ ‖ψ‖L2(Ω). This proves (7.79).
8 Proof of Theorem 1.1
As mentioned in Section 2.2, it is sufficient to prove the theorem under theassumption that λ1 = 0. Using the representation (7.77), we can rewrite(3.1) in the form
u(x, t) =J∑
k=1
hh,k(t)φk(x) +J∑
k=1
hp,k(t)φk(x) +M∑
k=J+1
hk(t)φk(x) + v(x, t),
(8.1)where hh,k and hp,k denote the k:th component of hh and hp, respectively.We set
V (x, t) = exp( ∫ t
0
(f(s)− Λ(s)
)ds
)
×( J∑
k=1
hp,k(t)φk(x) +M∑
k=J+1
hk(t)φk(x) + v(x, t))
.
By using the representation (7.78), it follows that (8.1) implies (1.17) withw0 = |hh(0)|. The estimates (1.20), (1.21) and (1.22) follow from (7.79),(7.58) and (7.59), respectively, with b0 ≤ µ2 in (1.21). We will return tothe question how b0 should be chosen.
It remains to prove (1.23). We start with estimating H2(t), where H2
is given by (6.55). Set g = −f + Λ and H1,h = Ξ1hh, where Ξ1 is definedin (6.42). From (7.78) and (7.79), it follows that
H1,h(t) ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
). (8.2)
An estimate for H1,p(t), defined analogously to H1,h(t), follows from (7.76)as
H1,p(t) ≤ C‖ψ‖L2(Ω)e−µ3t,
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119
where µ3 is given by (7.66). Since ‖g‖L1(t,t+1) ≤ c6κ0, where c6 is the sameas in (7.61), we can rewrite this estimate as
H1,p(t) ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
)e−µ5t, (8.3)
where µ5 = µ3 − c6κ0. When combining (8.2) and (8.3), we get
H1(t) ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
), (8.4)
where H1 is given by (6.55). Using (8.4) and Corollary 6.3, we arrive atthe inequality
H2(t) ≤ C‖ψ‖L2(Ω)
(e−µt +
∫ t
−1
e−µ(t−s)κ(s) exp( ∫ s
0
g(τ) dτ
)ds
+ κ(t) exp( ∫ t
0
g(s) ds
)).
This implies that
H2(t) ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
)
×(
e−µ4t +∫ t
−1
e−µ4(t−s)κ(s) ds + κ(t))
, (8.5)
where µ4 = µ− c6κ0.We now turn to estimating the function H given by (4.41). Since
H(t) ≤ H1(t) + H2(t), it follows from (8.4) and (8.5) that
H(t) ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
). (8.6)
In order to obtain (1.23), we also need an estimate for the function v.The inequalities (4.42) and (8.6) imply that
|v|Ct ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
)
×(
e−λJ+1t +∫ t
−1
e−λJ+1(t−s)κ(s) ds + κ(t))
. (8.7)
We are now in position to complete the derivation of estimate (1.23) butlet us first find an estimate for |W |Ct , where
W (x, t) =J∑
k=1
hp,k(t)φk(x) +M∑
k=J+1
hk(t)φk(x) + v(x, t).
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120
From (7.76) it follows that
|hp,kφk|Ct≤ C‖ψ‖L2(Ω) exp
( ∫ t
0
g(s) ds
)e−µ5t (8.8)
and (8.5) implies that
|hkφk|Ct ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
)
×(
e−µ4t +∫ t
−1
e−µ4(t−s)κ(s) ds + κ(t))
(8.9)
for k = J + 1, . . . , M . By combining (8.7), (8.8) and (8.9), we obtain
|W |Ct ≤ C‖ψ‖L2(Ω) exp( ∫ t
0
g(s) ds
)
×(
e−µ5t +∫ t
−1
e−µ4(t−s)κ(s) ds + κ(t))
and, since
V (x, t) = exp(−
∫ t
0
g(s) ds
)W (x, t),
it follows that
|V |Ct ≤ C‖ψ‖L2(Ω)
(e−µ5t +
∫ t
−1
e−µ4(t−s)κ(s) ds + κ(t))
.
The estimate (1.23) follows if we choose b0 = minµ2, µ4, µ5. Using thedefinitions of µ2 and µ given by (7.47) and (6.15), respectively, and thedefinitions of µ4 and µ5, we see that b0 = λJ+1 − C1κ0 for some constantC1 depending only on n, N , Ω, A
(0)ij , A(0), p, s′2, ν, ν1 and ν2. This
completes the proof of Theorem 1.1.
9 Corollaries of Theorem 1.1
In the next corollary we see that if κ ∈ L1(0,∞), then the leading term inthe asymptotics is the same as in the case when the perturbation is zero.
Corollary 9.1 If κ ∈ L1(0,∞), then
u(x, t) = e−λ1t
( J∑
k=1
bkφk(x) + ω(x, t))
, (9.1)
where bk, k = 1, . . . , J , are constants and |ω|Ct → 0 as t → ∞. Theconstants bk may depend on A
(1)ij , i, j = 1, . . . , n, A(1) and ψ.
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121
Proof. We begin by proving that
|V |Ct → 0 as t →∞ (9.2)
for the function V in (1.17). We first show that
κ(t) → 0 as t →∞. (9.3)
Similar to (5.43), Kozlov, Maz’ya [9], it can be proved that
κ(t) ≤ 2∫ t+1/2
t−1/2
κ(s) ds. (9.4)
Since κ ∈ L1(0,∞), this implies (9.3). Using (9.3), we check that
∫ t
−1
e−b(t−s)κ(s) ds → 0 as t →∞
for any positive number b. This, together with (1.23) and (9.3), proves(9.2).
It follows from (7.59) that∫ ∞
−1
‖Θ′‖L1(t,t+1) dt < ∞.
This implies that the vector function Θ(t) has a limit Θ0 as t →∞ and
|Θ(t)−Θ0| ≤∫ ∞
t−1
‖Θ′‖L1(s,s+1) ds. (9.5)
Analogously, ‖g‖L1(t,t+1), where g(t) = −f(t)+Λ(t), is estimated by Cκ(t).Therefore, ∫ t
0
g(s) ds = C1 −∫ ∞
t
g(s)ds
and ∫ ∞
t
g(s) ds ≤ C
∫ ∞
t−1
κ(s) ds.
This implies that
exp( ∫ t
0
g(s) ds
)− eC1 → 0 as t →∞. (9.6)
We apply (1.17), (9.2), (9.5) and (9.6) and arrive at (9.1).
We give an estimate for the function u which is of particular interest ifκ ∈ L2(0,∞).
September 6, 2006 (12:43)
122
Corollary 9.2 The estimate
|u|Ct ≤ C1‖ψ‖L2(Ω)e−λ1t+
∫ t0 (−f(s)+C2κ(s)2) ds (9.7)
is valid. If, in particular, κ ∈ L2(0,∞), then we have
|u|Ct≤ C‖ψ‖L2(Ω)e
−λ1t−∫ t0 f(s) ds.
Proof. We begin by proving the inequality∫ t
0
|Λ(s)| ds ≤ C
∫ t
−1
κ(s)2 ds. (9.8)
From (7.58), we obtain
∫ t
0
|Λ(s)| ds ≤ C
(∫ t
−1
κ(s)∫ s
−1
e−b0(s−τ)κ(τ) dτ ds +∫ t
−1
κ(s)2 ds
).
(9.9)Using Holder’s inequality to estimate the double integral in the right-handside and boundedness of convolution in L2-norm, it follows that
∫ t
−1
κ(s)∫ s
−1
e−b0(s−τ)κ(τ) dτ ds ≤ C
∫ t
−1
κ(s)2 ds.
This, together with (9.9), gives (9.8).From (1.23), it follows that
|V |Ct ≤ C‖ψ‖L2(Ω). (9.10)
Estimate (9.7) follows from (1.17), (1.20), (9.8) and (9.10).
Corollary 9.3 Assume, in addition to the conditions in Theorem 1.1, thatthe matrix A(1) = 0, A
(1)ij ∈ L1(Q) for i, j = 1, 2, . . . , n and that p, as given
by (2.6), is equal to ∞. Then it follows that
u(x, t) = e−λ1t
(b
J∑
k=1
θk(t)φk(x) + ω(x, t))
, (9.11)
where |ω|Ct → 0 as t →∞ and b is a constant which may depend on A(1)ij ,
i,j = 1, . . . , n, and ψ.
Proof. Since p = ∞, the number s1 given by (1.13) is equal to 2 and thefunction κ is in this case defined by (1.12) as
κ(t) =n∑
i,j=1
‖A(1)ij ‖L2(Ct).
September 6, 2006 (12:43)
123
By changing the order of integration, it follows for i, j = 1, . . . , n that∫ ∞
0
‖A(1)ij ‖2L2(Ct)
dt =∫ ∞
0
∫ t+1
t
∫
Ω
|A(1)ij (x, s)|2 dx ds dt
≤∫ ∞
0
∫
Ω
|A(1)ij (x, s)|2 dx ds.
Since |A(1)ij | is bounded and A
(1)ij ∈ L1(Q), we have
∫ ∞
0
κ(t)2 dt < ∞. (9.12)
Since
κ(t)2 ≤ 4∫ t+1/2
t−1/2
κ(s)2 ds,
compare with (9.4), it follows that κ(t) → 0 as t → ∞. Using that∇φk ∈
(L∞(Ω)
)N , k = 1, . . . , J , we derive from (1.18) and (1.19) that
∫ ∞
0
|f(t)| dt ≤ C
n∑
i,j=1
∫ ∞
0
∫
Ω
|A(1)ij (x, t)| dx dt < ∞. (9.13)
In the same way as in Corollary 9.1, we prove that
|V |Ct → 0 as t →∞ (9.14)
and that
exp( ∫ t
0
(− f(s) + Λ(s))ds
)→ C as t →∞. (9.15)
In the derivation of (9.15), we use the estimates (9.8), (9.12) and (9.13).The formulae (1.17), (9.14) and (9.15) give (9.11).
The asymptotic formula (1.17) contains two unknown functions, Θ andΛ. If the first eigenvalue is simple, i.e. J = 1, and the coefficients are real,then Θ can be chosen as 1. Thus, we arrive at the following theorem.
Theorem 9.4 If J = 1 and the matrices Aij, i, j = 1, . . . , n and A arereal-valued, then the solution u in
(V 2
0,loc(Q))N of problem (1.1)–(1.3) can
be written as
u(x, t) = e−λ1t+∫ t0 (−R(s)+Λ(s)) ds
(w0φ1(x) + V (x, t)
),
where
R =∫
Ω
[ n∑
i,j=1
(A(1)ij φ1xj
, φ1xi) + (A(1)φ1, φ1)
]dx,
w0 is a constant and the estimates (1.20), (1.21) and (1.23) are valid.
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124
A Eigenfunctions of a time-independent op-erator
In this section we consider the boundary value problem
L(0)u = f in Ωu = 0 on ∂Ω,
(A.1)
where
L(0)u(x) = −n∑
i,j=1
(A
(0)ij (x)uxj
(x))xi
+ A(0)(x)u(x)
and f ∈ (L2(Ω)
)N . The main result is that there exists an ON-basis of(L2(Ω)
)N consisting of eigenfunctions of L(0).
We define the sesquilinear form B :(H1
0 (Ω))N × (
H10 (Ω)
)N → C as
B[u, η] =∫
Ω
[ n∑
i,j=1
(A
(0)ij uxj , ηxi
)+
(A(0)u, η
)]dx (A.2)
and say that u ∈ (H1
0 (Ω))N is a weak solution of the problem (A.1) if
B[u, η] =∫
Ω
(f, η) dx, ∀η ∈ (H1
0 (Ω))N
.
Before studying the eigenfunctions of L(0), let us first prove the inequal-ity ∫
Ω
(A(0)u, η
)dx ≤ C‖A(0)‖Lq(Ω)‖u‖H1(Ω)‖η‖H1(Ω), (A.3)
with q given in (1.8). In the case when n ≥ 3, Holder’s inequality withs = 2n/(n− 2) implies that
∫
Ω
(A(0)u, η
)dx ≤ C‖A(0)‖Ln/2(Ω)‖u‖Ls(Ω)‖η‖Ls(Ω).
Since n/2 < q and(H1(Ω)
)N is compactly embedded in(Ls(Ω)
)N , in-equality (A.3) follows.
If n ≤ 2, we set s = 2q/(q−1) and observe that s ∈ [2, 4]. From Holder’sinequality it follows that
∫
Ω
(A(0)u, η
)dx ≤ C‖A(0)‖Lq(Ω)‖u‖Ls(Ω)‖η‖Ls(Ω).
In this case, the space(H1(Ω)
)N is compactly embedded in every(Lr1(Ω)
)N for 1 ≤ r1 < ∞ so (A.3) follows.For convenience, we will use the notation ‖ · ‖H instead of ‖ · ‖H1(Ω).
We have the following result.
September 6, 2006 (12:43)
125
Theorem A.1 Let Ω be an open, bounded region in Rn with Lipschitzboundary. Then there exists an ON-basis φk∞k=1 of
(L2(Ω)
)N consistingof eigenfunctions of the operator L(0), i.e.
B[φk, η] = λk
∫
Ω
(φk, η) dx, ∀η ∈ (H1
0 (Ω))N
, (A.4)
where the eigenvalues λk satisfy
λ1 ≤ λ2 ≤ . . . , λk →∞ as k →∞.
Furthermore, φk ∈(H1
0 (Ω))N and
B[u, u] ≥ λ1‖u‖2L2(Ω). (A.5)
Proof. We first prove the theorem under the extra assumption thatA
(0)− = 0, i.e. when the matrix A(0) is positive. We note that B[·, ·]
defines a scalar product on(H1
0 (Ω))N . In fact, condition (1.7) implies that
all elements in the matrices A(0)ij are bounded. From this and (A.3) we
conclude that there exists a constant C not depending on u or η such that∣∣B[u, η]
∣∣ ≤ C‖u‖H‖η‖H . (A.6)
Hence B[u, η] is finite. The linearity in the first argument of B is straight-forward. The relation B[η, u] = B[u, η] follows from (1.6) and that A(0) ishermitian. By using (1.7) and the extra assumption that A(0) is positivewe see that
B[u, u] ≥ ν‖∇u‖2L2(Ω). (A.7)
Thus B[u, u] ≥ 0 with equality only for u = 0, showing that B is a scalarproduct on
(H1
0 (Ω))N .
Our next step is to show that the norms ‖ · ‖H and ‖ · ‖B , defined as
‖u‖B =(B[u, u]
)1/2,
are equivalent. It follows immediately from (A.6) that there exists a con-stant C independent of u such that ‖u‖B ≤ C‖u‖H . Conversely, we canuse Poincare’s inequality and (A.7) to obtain
‖u‖2H = ‖u‖2L2(Ω) + ‖∇u‖2L2(Ω) ≤ C1‖∇u‖2L2(Ω) ≤C1
ν‖u‖2B ,
where C1 does not depend on u. Together this gives that ‖ · ‖H and ‖ · ‖B
are equivalent on(H1
0 (Ω))N .
For u fixed in(H1
0 (Ω))N , we introduce the linear and bounded func-
tional η 7→ ∫Ω(η, u) dx on
(H1
0 (Ω))N . It follows from Riesz representation
theorem that there exists a unique element Ku ∈ (H1
0 (Ω))N such that
∫
Ω
(η, u) dx = B[η,Ku], ∀η ∈ (H1
0 (Ω))N
.
September 6, 2006 (12:43)
126
In the same way as in the proof of Theorem A.8, Section A.1 in Part 1 ofthis thesis, we can show that K is linear, bounded, symmetric and compactbut this time all computations are made with respect to the scalar productB. Hence there exists an, with respect to B, orthogonal basis φk∞k=1 of(H1
0 (Ω))N consisting of eigenvectors of K with corresponding eigenvalues
µk∞k=1. We can assume that
‖φk‖L2(Ω) = 1, k = 1, 2, . . .
and we will soon see that this set is orthogonal even in(L2(Ω)
)N .We introduce
m = infB[Ku, u] : u ∈ (H1
0 (Ω))N
, ‖u‖B = 1
andM = supB[Ku, u] : u ∈ (
H10 (Ω)
)N, ‖u‖B = 1 (A.8)
and get from Lemma A.5 in Section A.1 in Part 1 of the thesis thatm,M ∈ σ(K) ⊂ [m,M ] where σ(K) is the spectrum of K. We observethat
B[Ku, u] = B[u,Ku] =∫
Ω
(u, u) dx = ‖u‖2L2(Ω)
and, from the equivalence of norms, that there exists a constant c0 suchthat
‖u‖2L2(Ω) ≤ ‖u‖2H ≤ c0‖u‖2B ,
so M ≤ c0. On the other hand, it follows from Lemma A.4(i), Section A.1,Part 1, that 0 ∈ σ(K). Thus m = 0.
From Lemma A.4(ii), Part 1, we see that the point spectrum σp(K) isequal to σ(K), except possibly for 0. But 0 can not be an eigenvalue of Ksince Ku = 0 would imply that
∫
Ω
(η, u) dx = B[η, 0] = 0, ∀η ∈ (H1
0 (Ω))N
,
which with η = u shows that u = 0. As in the proof of Theorem A.8,Part 1, we conclude that
M = µ1 ≥ µ2 ≥ . . . > 0, µk → 0 as k →∞.
The relation Kφk = µkφk implies that
B[φk, η] = λk
∫
Ω
(φk, η) dx, ∀η ∈ (H1
0 (Ω))N
,
where λk = µ−1k is an eigenvalue of L(0). Since λk 6= 0, it follows, by
setting η = φl, that the sequence φk∞k=1 is orthogonal also in(L2(Ω)
)N .Furthermore, it follows from (A.8) that B[Ku, u] ≤ M‖u‖2B and, since
September 6, 2006 (12:43)
127
M = µ1 = λ−11 and B[Ku, u] = ‖u‖2L2(Ω), the inequality (A.5) follows.
This concludes the proof in the case A(0)− = 0.
Suppose that A(0)− 6= 0. Since A
(0)− is bounded, there is a constant c such
that the matrix −A(0)− +cI is positive. Then the matrix G = A
(0)+ −A
(0)− +cI
is positive, so we can apply the proof above on the form
B′[u, η] =∫
Ω
[ n∑
i,j=1
(A
(0)ij uxj , ηxi
)+ (Gu, η)
]dx
to obtain an ON-basis φk∞k=1 of(H1
0 (Ω))N with corresponding scalars
λ′k∞k=1 satisfying
B′[φk, η] = λ′k
∫
Ω
(φk, η) dx, ∀η ∈ (H1
0 (Ω))N
and λ′k →∞ as k →∞. But
B[φk, η] = B′[φk, η]− c
∫
Ω
(φk, η) dx = (λ′k − c)∫
Ω
(φk, η) dx,
so φk is an eigenfunction of L(0) with eigenvalue λk = λ′k − c. Also (A.5)follows from the last line and the proof is complete.
Actually, more can be said about the eigenfunctions.
Theorem A.2 There exists a p ∈ (2,∞] such that φk ∈(W 1,p
0 (Ω))N for
k = 1, 2, . . ..
Proof. If a > −λ1, then the operator
L(0) + a :(W 1,2
0 (Ω))N → (
W−1,2(Ω))N
is isomorphic. Furthermore, it can be checked, using the assumptions onA
(0)ij and A(0), that the operator
L(0) + a :(W 1,p
0 (Ω))N → (
W−1,p(Ω))N (A.9)
is continuous for any p ∈ [1,∞]. By Theorem 3.5, Triebel [18],(W 1,p
0 (Ω))N
and(W−1,p(Ω)
)N are interpolation spaces. Therefore, using Shneiberg [17]or Krugljak, Milman [11], we derive the existence of ε > 0 such that theoperator in (A.9) is isomorphic for p ∈ [2−ε, 2+ε]. This implies in particularthat φk ∈
(W 1,2+ε(Ω)
)N for all k, completing the proof.
September 6, 2006 (12:43)
128
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September 6, 2006 (12:43)
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