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ISRAEL JOURNAL OF MATHEMATICS 194 (2013), 259–275
DOI: 10.1007/s11856-012-0098-9
POINTWISE ESTIMATES FOR SOLUTIONSTO THE POROUS MEDIUM EQUATIONWITH MEASURE AS A FORCING TERM
BY
Vitali Liskevich
Department of Mathematics, Swansea University, Swansea SA2 8PP, UK
e-mail: [email protected]
AND
Igor I. Skrypnik
Institute of Applied Mathematics and Mechanics, Donetsk 83114, Ukraine
e-mail: [email protected]
ABSTRACT
For porous media equations with a Radon measure on the right-hand side,
we derive pointwise estimates for solutions via the Riesz potentials.
1. Introduction and main results
Due to applications and interesting mathematical properties the porous media
equations attracted much attention during the last decades. The results of the
theory are well documented in monographs [1], [14], [15], [16] (see also [3, 4]).
One of the important issues of the research is basic qualitative properties of
solutions such as local boundedness, summability properties and alike.
In this note we study weak solutions to porous media equations with measure
as a forcing term. We provide pointwise estimates for solutions. The estimates
are expressed in terms of the Riesz potential of the measure. For the case of the
heat equation the corresponding result was recently given in [5]. The estimates
obtained in this paper allow us to give a sharp condition of local boundedness
Received March 10, 2011 and in revised form August 10, 2011
259
260 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
and for unbounded solutions they are useful for studying Lp properties, mem-
bership in other functional spaces etc. As an application we give a supremum
estimate for the solution of a model initial boundary value problem.
Let Ω be a domain in Rn, T > 0. Let μ be a Radon measure on Ω. We are
concerned with pointwise estimates for a class of non-homogeneous divergence
type quasi-linear parabolic equations of the type
(1.1) ut −Δ(|u|m−1u) = μ in ΩT = Ω× (0, T ), Ω ⊂ Rn, m > 1.
Before formulating the main results, let us remind the reader of the definition
of a weak solution to equation (1.1). We rewrite the equation in the divergence
form
ut −m div(|u|m−1∇u) = μ.
We say that u is a weak solution to (1.1) if
u ∈ C([0, T ];L2loc(Ω)) ∩ L2
loc(0, T ;W1,2loc (Ω)) |u|m ∈ L2
loc(0, T ;W1,2loc (Ω)),
and for any compact subset K of Ω and any interval [t1, t2] ⊂ (0, T ) the integral
identity
(1.2)∫Kuϕdx
∣∣∣t2t1+
∫ t2
t1
∫K
{−uϕτ +m|u|m−1∇u∇ϕ}dx dτ =
∫ t2
t1
∫Kϕμ(dx) dτ
holds for any ϕ ∈ C1(ΩT ).
The crucial role in the first main result is played by the truncated version of
the Riesz potential defined by
(1.3) Iμ2 (x,R) =
∫ R
0
μ(Br(x))
rn−2
dr
r.
In the sequel, γ stands for a constant depending only on n and m, which may
vary from line to line.
Let (x0, t0) ∈ ΩT . Consider the cylinder
(1.4) Q(r, τ) = Br(x0)× (t0 − τ, t0 + τ).
Further on, we assume that Q(r, τ) � ΩT .
The first main result of this paper is the following theorem.
Theorem 1.1: Let u be a weak solution to equation (1.1). For every λ ∈(0, 1/n) there exists γ > 0 depending on n,m and λ, such that for almost all
Vol. 194, 2013 POROUS MEDIUM EQUATION 261
(x0, t0) ∈ ΩT such that Q0 = Q(r0, θ0) � ΩT we have
(1.5)
u(x0, t0) ≤ γ
{(1
rn+20
∫∫Q0
um+λ+ dx dτ
) 11+λ
+ supt∈(t0−θ,t0+θ)
1
rn0
∫Br0
u+ dx
+
(r20θ0
) 1m−1
+
(1 +
θ0r20
)Iμ2 (x0, 2r0)
}.
Also, the following version of the above estimate holds:
u(x0, t0) ≤ γ
{(1
rn+20
∫∫Q0
um+λ+ dx dτ
) 11+λ
+
(1
rn+20
∫∫Q0
um−1+2λ+ dx dτ
) 12λ
+1
θ0rn0
∫∫Q0
u+dx dτ +
(r20θ0
) 1m−1
+
(1 +
θ0r20
)Iμ2 (x0, 2r0)
}.
The estimate above is not homogeneous in u which is usual for such type
of equations [3, 14]. The proof of Theorem 1.1 is based on a suitable modi-
fications of De Giorgi’s iteration technique [2] following the adaptation of the
Kilpelainen–Maly technique [7] to parabolic equations with ideas from [10, 13].
A pointwise estimate for the evolutional p-Laplace equation was recently ob-
tained in [11].
An immediate application of the above result is the following corollary.
Corollary 1.2: Assume that there exists r0 > 0 such that for every Ω � Ω
the measure μ satisfies the condition
(1.6) supx∈Ω
Iμ2 (x, r0) < ∞.
Then every weak solution to (1.1) is locally bounded, that is u ∈ L∞loc(ΩT ).
In particular, if dμ(x) = f(x)dx with f ∈ Ln/2, 1(Ω) (Lorenz space) then (1.6)
is satisfied.
(See, e.g., [12] for relevant properties of the Riesz potential.)
Condition of local boundedness (1.6) is sharp, as one can easily see looking at
stationary solutions. If (1.6) does not hold, using local properties of the Riesz
potential (see, e.g., [12]) one can deduce summability properties of solutions to
(1.1) such as membership in Lp-spaces, Orlicz spaces etc. We will not dwell
upon this.
262 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
As another application of Theorem 1.1 we consider the following model initial
boundary value problem:
(1.7)
ut −Δ(um) =μ, (x, t) ∈ Q = BR × (0, T ),
u(x, t) =0, (x, t) ∈ S = ∂BR × (0, T ),
u(x, 0) =0, x ∈ BR.
First, let us remark on the well posedness of problem (1.7) in the sense of
weak solutions as defined above. The uniqueness can be verified in the same way
as in [9, Chap. 1, Sec. 12]. However, the most general result on the existence
of weak solutions to (1.7), known to the authors, requires dμ(x) = f(x)dx with
f ∈ L(m+1)/m (cf. [9, Theorem 12.2]). So the next result is a global a priori
bound on the solution.
Corollary 1.3: Let u be the solution to problem (1.7). Then
(1.8) supQ
u ≤ γ
(R2
T
) 1m−1
+ γ
(1 +
T
R2
)supx∈BR
Iμ2 (x, 2R).
Remark 1.4: 1. Based on the cases of the Laplacian and p-Laplacian it
is plausible to conjecture that (1.7) is well posed under the assumption
that supx∈BRIμ2 (x, 2R) < ∞. To identify the optimal class of measures
μ guaranteeing well posedness of problem (1.7) seems an interesting
open question.
2. Note that if dμ(x) = f(x)dx with f ∈ Ln/2,1(BR) and n ≥ 4, then by
the above-mentioned result of J.-L. Lions, (1.7) is well posed and (1.8)
becomes a posteriori supremum bound.
The approach we undertake in the proof of Theorem 1.1 allows for an ex-
tension of the result to the case of measures depending on (x, t) ∈ Rn+1. To
formulate the result we first restrict the geometry of the cylinders, compara-
tively to (1.4), namely, we introduce
Qr = Br(x0)× (t0 − r2, t0 + r2).
The time-dependent analogue of the truncated Riesz potential is introduced by
Iμ2 (x, t, R) =
∫ R
0
μ(Qr)
rndr
r.
Theorem 1.5: Let u be a weak solution to equation (1.1). For every λ ∈(0, 1/n) there exists γ > 0 depending on n,m and λ, such that for almost all
Vol. 194, 2013 POROUS MEDIUM EQUATION 263
(x0, t0) ∈ ΩT such that Q0 = Qr0 � ΩT we have
(1.9) u(x0, t0) ≤
γ{( 1
rn+20
∫∫
Q0
um+λ+ dx dτ
) 11+λ
+ supt∈(t0−r20,t0+r20)
1
rn0
∫
Br0
u+ dx+1+Iμ2 (x0, t0, 2r0)
}.
The rest of the paper contains the proofs of the above statements.
2. Proofs
Proof of Theorem 1.1. Let Q(r, τ) be as above and ξ(x, t) be a cut-off function
such that
1Q(r/2,τ) ≤ ξ ≤ 1Q(r,τ), |∇ξ| ≤ γr−1, |ξt| ≤ γτ−1.
Further on, we assume that ut ∈ L2loc(ΩT ), since otherwise we can pass to
Steklov averages.
We start with some auxiliary integral estimates for the solutions of (1.1)
which are formulated in the next lemma.
Define
G(u) =
⎧⎨⎩u for u > 1,
u2−2λ for 0 < u ≤ 1.
Lemma 2.1: Let the conditions of Theorem 1.1 be fulfilled. Let u be a solution
to (1.1). Then there exists a constant γ > 0 depending only on n,m such that
for any l, δ > 0, any cylinder Q(r, τ) and any ξ as above,
(2.1)
δ2∫L(t)
G(u(x, t)− l
δ
)ξ(x, t)kdx
+
∫∫L
um−1(1 +
u− l
δ
)−1+λ(u− l
δ
)−2λ
|∇u|2ξ(x, t)kdx dt
≤γδ2
τ
∫∫L
(u− l
δ
)ξk−1dxdt+γ
δ2
r2
∫∫L
um−1(1+
u− l
δ
)1−λ(u− l
δ
)2λξk−2dxdt
+ γ τδμ(Br(x0)),
where L = Q(r, τ) ∩ {u > l}, L(t) = (Br(x0)× {t}) ∩ {u > l} and λ ∈ (0, 1/n),
k > 2.
264 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
Proof. First, note that
(2.2)
∫ u
l
(1 +
s− l
δ
)−1+λ(s− l
δ
)−2λ
ds ≤ γδ,
and
(2.3)
∫ u
l
dw
∫ w
l
(1 +
s− l
δ
)−1+λ(s− l
δ
)−2λ
ds
=
∫ u
l
(1 +
s− l
δ
)−1+λ(s− l
δ
)−2λ
(u − s)ds
≥1
2(u− l)
∫ u+l2
l
(1 +
s− l
δ
)−1+λ(s− l
δ
)−2λ
ds
=δ2
2
(u− l
δ
)∫ u−l2δ
0
(1 + z)−1+λz−2λdz
≥γδ2G(u− l
δ
).
Let η be the standard mollifier in Rn and as usual ησ(x) = η(x/σ). Set
uσ = ησ ∗ u. Let ε > 0. Test (1.2) by ϕ defined by
(2.4) ϕ(x, t) =
[ ∫ uσ(x,t)
l
(1 +
s− l
δ
)−1+λ(ε+
s− l
δ
)−2λ
ds
]+
ξ(x, t)k,
and t1 = t0 − τ , t2 = t.
Using first (2.2) on the right-hand side, then passing to the limit σ → 0 on
the left, after applying the Schwarz inequality we obtain for any t ∈ (0, t0)
∫L(t)
∫ u
l
dw
∫ w
l
(1 +
s− l
δ
)−1+λ(ε+
s− l
δ
)−2λ
dsξkdx
+
∫∫L
(1 +
u− l
δ
)−1+λ(ε+
u− l
δ
)−2λ
um−1|∇u|2ξkdxdt
≤γ
∫∫L
∫ u
l
dw
∫ w
l
(1 +
s− l
δ
)−1+λ(ε+
s− l
δ
)−2λ
ds|ξt|ξk−1dxdt
+ γr−2δ2∫∫
L
um−1(1+
u− l
δ
)1−λ(ε+
u− l
δ
)2λξk−2dx dt+γδτμ(Br(x0)).
Passing to the limit ε → 0 on the right-hand side of the above inequality and
Vol. 194, 2013 POROUS MEDIUM EQUATION 265
using the Fatou lemma on the left, by (2.3) we obtain the required (2.1).
Let us set
v =1
δ
[ ∫ u
l
(1 +
s− l
δ
)− 1−λ2(s− l
δ
)m−1−2λ2
ds
]+
,(2.5)
w =1
δ
[ ∫ u
l
(1 +
s− l
δ
)− 1−λ2(s− l
δ
)−λ
ds
]+
.(2.6)
With this notation the next lemma is a consequence of Lemma 2.1.
Lemma 2.2: Let the conditions of Lemma 2.1 be fulfilled. Then
(2.7)∫L(t)
G
(u− l
δ
)ξkdx + δm−1
∫∫L
|∇v|2ξkdxdt + lm−1
∫∫L
|∇w|2ξkdxdt
≤γτ−1
∫∫L
(u− l
δ
)ξk−1dxdt
+ γr−2
∫∫L
um−1
(1 +
u− l
δ
)1−λ(u− l
δ
)2λ
ξk−2dxdt
+ γ τδ−1μ(Br(x0)).
Let (x0, t0) ∈ ΩT , R0 ≤ 12 dist(x0, ∂Ω), θ0 ≤ 1
2 min{t0, T − t0}.For j = 0, 1, 2, . . . set rj = R02
−j, θj = θ02−j, Bj = Brj(x0).
We define the sequence (lj)j∈N inductively, and then the sequences
(δj)j∈N∪{0}, (Δj)j∈N∪{0} by setting δi = li+1 − li and Δi = max{li, δi}. Set
l0 = 0. Suppose that l1, . . . , lj have been chosen. Let us show how to choose
lj+1.
For l > lj we define
(2.8) δj(l) = l − lj, Δj(l) = max{δj(l), lj}, τj(l) = min
{θj ,
r2j
Δm−1j (l)
}.
Next we introduce the cylinder Qj and the set Lj by
Qj = Bj × Ij(l), Ij(l) = (t0 − τj(l), t0 + τj(l)),
Lj = Qj ∩ ΩT ∩ {u(x, t) > lj}, Lj(t) =((Bj ∩ Ω)× {t}) ∩ {u > l}.
266 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
Let ξj ∈ C∞0 (Qj) be such that
ξj(x, t) = 1 for (x, t) ∈ Bj+1 × (t0 − 1
2τj(l), t0 +
1
2τj(l)),
|∇ξj | ≤ γr−1j , |∂ξj/∂t| ≤ γτj(l)
−1.
For l ≥ lj set
(2.9)
Aj(l) =1
τ1+ n
2j (l)Δ
(1+n2 )(m−1)
j (l)
∫∫˜Lj
um−1
(u− ljδj(l)
)1+λ
ξk−2j dxdt
+ supt∈Ij(l)
1
τn2
j (l)Δn2 (m−1)j (l)
∫˜Lj(t)
G
(u− ljδj(l)
)ξkj dx.
Note that Aj(l) is continuous as a function of l and Aj(l) → 0 as l → ∞.
Fix a number κ ∈ (0, 1) depending on n,m, c1, c2, which will be specified
later. For j = 0, 1, 2, . . . , if
(2.10) Aj
(lj +
(r2jθj
) 1m−1
)≤ κ,
we set
lj+1 = lj +
(r2jθj
) 1m−1
.
If, on the other hand,
(2.11) Aj
(lj +
(r2jθj
) 1m−1
)> κ,
then there exists
l > lj +
(r2jθj
) 1m−1
such that Aj(l) = κ. In this case we set lj+1 = l and δj(lj+1) = lj+1 − lj in
both cases. Hence
Δj := Δj(lj+1) ≥(r2jθj
) 1m−1
so that
τj := τj(lj+1) =r2j
Δm−1j
.
Vol. 194, 2013 POROUS MEDIUM EQUATION 267
After choosing lj+1 we have from (2.9)
(2.12) Aj(lj+1) =
1
rn+2j
∫∫Lj
um−1
(u− ljδj
)1+λ
ξk−2j dxdt + sup
t∈Ij
1
rnj
∫Lj(t)
G
(u− ljδj
)ξkj dx,
where
Ij = (t0 − τj , t0 + τj), Qj = Bj × Ij , Lj = Qj ∩ΩT ∩ {u > lj}.
First, we note that Δj ≥ Δj−1 and therefore τj ≤ τj−1, which implies that
Qj ⊂ Qj−1.
Also note that our choices guarantee that
(2.13) Aj(lj+1) ≤ κ.
The following lemma is a key in the Kilpelainen–Maly technique [7].
Lemma 2.3: Let the conditions of Theorem 1.1 be fulfilled. There exists γ > 0
depending on the data, such that for all j ≥ 1 we have
(2.14) δj ≤ 2−1
m−1 δj−1 + γ
(r2jθj
) 1m−1
+ γ
(1 +
θ0r20
)1
rn−2j
μ(Bj).
Proof. Fix j ≥ 1. Without loss assume that
(2.15) δj > 2−1
m−1 δj−1, δj >
(r2jθj
) 1m−1
,
since otherwise (2.14) is evident. The second inequality in (2.15) guarantees
that Aj(lj+1) = κ and Qj = Qj.
Next we claim that
(2.16)1
rn+2j
∫∫Lj
um−1dx dt ≤ γκ.
Indeed, for (x, t) ∈ Lj one has
(2.17)u(x, t)− lj−1
δj−1= 1 +
u(x, t)− ljδj−1
≥ 1.
268 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
Hence we have, by (2.17),
(2.18)
1
rn+2j
∫∫Lj
um−1dx dt ≤ 1
rn+2j
∫∫Lj
um−1
(u− lj−1
δj−1
)1+λ
ξk−2j−1 dx dt
≤γ1
rn+2j−1
∫∫Lj−1
um−1
(u− lj−1
δj−1
)1+λ
ξk−2j−1 dx dt
≤γAj−1(lj) ≤ γκ.
Let us estimate the terms in the right-hand side of (2.12). For this we de-
compose Lj as Lj = L′j ∪ L′′
j ,
(2.19) L′j =
{(x, t) ∈ Lj :
u(x, t)− ljδj
< ε
}, L′′
j = Lj \ L′j,
where ε ∈ (0, 1), depending on n, p, c1, c2, is small enough to be determined
later.
By (2.16) we have
(2.20)
1
rn+2j
∫∫L′
j
um−1
(u− ljδj
)1+λ
ξk−2j dx dt ≤ γ
ε1+λ
rn+2j
∫∫L′
j
um−1dx dt ≤ γε1+λκ.
Let vj and wj be defined by (2.5) and (2.6), respectively, with δj in place of
δ and lj in place of l. Note that λ ≤ 1/n due to the assumption.
The following inequalities are easy to verify
(2.21)
vj ≤γ
(u(x, t)− lj
δj
)m−λ2
for (x, t) ∈ Lj, and
vj ≥γ
(u(x, t)− lj
δj
)m−λ2
for (x, t) ∈ L′′j .
Similarly,
(2.22)
wj ≤γ
(u(x, t)− lj
δj
) 1−λ2
for (x, t) ∈ Lj, and
wj ≥γ
(u(x, t)− lj
δj
) 1−λ2
for (x, t) ∈ L′′j .
Vol. 194, 2013 POROUS MEDIUM EQUATION 269
Hence we have
(2.23)
1
rn+2j
∫∫L′′
j
um−1
(u− ljδj
)1+λ
ξk−2j dxdt
≤ γδm−1j
rn+2j
∫∫L′′
j
(u− ljδj
)m+λ
ξk−2j dxdt+ γ
lm−1j
rn+2j
∫∫L′′
j
(u− ljδj
)1+λ
ξk−2j dxdt
≤ γ(ε)δm−1j
rn+2j
∫∫L′′
j
v2m+λ
m−λ
j ξk−2j dxdt+ γ(ε)
lm−1j
rn+2j
∫∫L′′
j
w2 1+λ
1−λ
j ξk−2j dxdt.
The integrals on the right-hand side of (2.23) are estimated by using the
Gagliardo–Nirenberg inequality in the form given in [8, Chapter II, Theo-
rem 2.2], as follows:
(2.24)
RHS of (2.23) ≤γ
(supt∈Ij
1
rnj
∫Lj(t)
v2
m−λ
j dx
)2n(δm−1j
rnj
∫∫Lj
∣∣∣∇(vjξ
k−22
j
)∣∣∣2dxdt)
+γ
(supt∈Ij
1
rnj
∫Lj(t)
w2
1−λ
j dx
)2n(lm−1j
rnj
∫∫Lj
∣∣∣∇(wjξ
k−22
j
)∣∣∣2dxdt).
Let us estimate separately the first factors in both terms of the right hand
side of (2.24):
(2.25)
supt
1
rnj
∫Lj(t)
v2
m−λ
j dx+ supt
1
rnj
∫Lj(t)
w2
1−λ
j dx
by (2.21), (2.22)
≤ c−1 supt
1
rnj
∫Lj(t)
u− ljδj
dx
by (2.15)
≤ 2c−1 supt
1
rnj
∫Lj(t)
u− lj−1
δj−1ξkj−1 dx
by (2.17), (2.12)
≤ γ supt
1
rnj−1
∫Lj−1(t)
G
(u− lj−1
δj−1
)ξkj−1 dx
by (2.13)
≤ γκ.
Combining (2.23), (2.24) and (2.25) we obtain
1
rn+2j
∫∫L′′
j
um−1
(u− ljδj
)1+λ
ξk−2j dxdt
≤ γ(ε)κ2n
(δm−1j
rnj
∫∫Lj
∣∣∣∇(vjξ
k−22
j
)∣∣∣2 dxdt+ lm−1j
rnj
∫∫Lj
∣∣∣∇(wjξ
k−22
j
)∣∣∣2 dxdt).
270 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
To estimate the right hand side of the above inequality we start with the terms
containing ∇ξj :
(2.26)δm−1j
rnj
∫∫Lj
v2j |∇ξj |2 ξk−4j dx dt+
lm−1j
rnj
∫∫Lj
w2j |∇ξj |2 ξk−4
j dx dt
≤γ1
rn+2j
(δm−1j
∫∫Lj
(u− ljδj
)m−λ
dx dt+ lm−1j
∫∫Lj
(u− ljδj
)1−λ
dx dt
)
( by (2.15))
≤γ1
rn+2j
∫∫Lj
um−1
(u− lj−1
δj−1
)1−λ
ξk−2j−1 dx dt
(by using the decomposition (2.19) and (2.16))
≤γε1−λκ + γ
1
rn+2j−1
∫∫Lj−1
um−1
(u− lj−1
δj−1
)1+λ
ξk−2j−1 dx dt ≤ γκ.
For the terms with ∇vj and ∇wj we use Lemma 2.2:
(2.27)
δm−1j
rnj
∫∫Lj
|∇vj |2ξk−2j dx dt+
lm−1j
rnj
∫∫Lj
|∇wj |2ξk−2j dx dt
≤γ1
τjrnj
∫∫Lj
u− ljδj
ξk−3j dxdt
+ γ1
rn+2j
∫∫Lj
um−1
(1 +
u− ljδj
)1−λ(u− ljδj
)2λ
ξk−4j dxdt
+ γτjrnj
δ−1j μ(Bj).
For the first term on the right-hand side of (2.27) we have
(2.28)
1
τjrnj
∫∫Lj
u− ljδj
ξk−3j dxdτ ≤γ
1
rnjsupt∈Ij
∫Lj(t)
u− ljδj−1
dx
≤γ1
rnjsupt∈Ij
∫Lj−1(t)
G
(u− lj−1
δj−1
)ξkj−1dx
≤γκ.
The second term of the right-hand side of (2.27) is estimated by γκ using
the decomposition (2.19).
Vol. 194, 2013 POROUS MEDIUM EQUATION 271
Thus we obtain the following estimate for the first term of Aj(lj+1):
(2.29)1
rn+2j
∫∫Lj
um−1
(u− ljδj
)1+λ
ξk−2j dxdτ
≤ γε1+λκ + γ(ε)κ
2n
(κ +
τjrnj
δ−1j μ(Bj)
).
Let us estimate the second term in the right-hand side of (2.9). By Lemma 2.2
with l = lj, δ = δj , using the decomposition (2.19), (2.16) and finally estimate
(2.29), we have
(2.30) supt∈Ij
1
rnj
∫Lj(t)
G
(u− ljδj
)ξkj dx
≤ γε1+λκ + γ(ε)κ
2n
(κ + δ−1
j
τjrnj
μ(Bj)
)+ γδ−1
j
τjrnj
μ(Bj).
Combining (2.12), (2.29) and (2.30), and choosing ε appropriately, we can
find γ1 and γ such that
(2.31) κ ≤ γ1κ2n
(κ + δ−1
j
τjrnj
μ(Bj)
)+ γδ−1
j
τjrnj
μ(Bj).
Now choosing κ < 1 such that κp/n = 1/2γ1 we have
(2.32) δj ≤ γτjrnj
μ(Bj) = γ1
rn−2j Δm−1
j
μ(Bj).
Taking into account that
Δm−1j ≥ Δm−1
0 = δm−10 ≥ r20
θ 0,
we obtain
δj ≤ γθ0r20
1
rn−2j
μ(Bj),
which completes the proof of the lemma.
In order to complete the proof of Theorem 1.1 we sum up (2.14) with respect
to j from 1 to J − 1
(2.33)
lJ ≤γδ0 + γ
∞∑j=1
(r2jθj
) 1m−1
+ γ
∞∑j=1
r2−nj μ(Bj)
≤γ
(δ0 + γ
(r20θ0
) 1m−1
+ γ
(1 +
θ0r20
)Iμ2 (x0, r0)
).
272 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
Let us estimate δ0. There are two cases to consider. Either
l1 = δ0 =
(r20θ0
) 1m−1
,
or l1 and δ0 are defined by A0(l1) = κ.
In this case
θ0Δm−10 > r20 .
Then by (2.12) we have
(2.34)
κ =1
rn+20
∫∫Q0
um−1
(u+
δ0
)1+λ
ξk−20 dx dτ + sup
tr−n0
∫B0
G
(u+
δ0
)ξk0 dx
≤ 1
rn+20
∫∫Q0
um−1
(u+
δ0
)1+λ
ξk−20 dx dτ + sup
tr−n0
∫B0
u+
δ0ξk0 dx.
Hence we obtain that
(2.35) δ0 ≤ γ
(1
rn+20
∫∫Q0
um+λ+ dxdt
) 11+λ
+ γ1
rn0supt
∫B0
u+dx.
This proves the first assertion.
By Lemma 2.2
(2.36)
r−n0
∫L0(t)
G
(u
δ0
)ξk0dx ≤γτ−1
0 r−n0
∫∫L0
u+
δ0ξk−10 dxdτ
+ γr−2−n0
∫∫L0
um−1(1 +
u
δ0
)1−λ(u+
δ0
)2λξk−20 dxdτ
+ γθ0r20
r2−n0 δ−1
0 μ(B0(x0))
≤γτ−10 r−n
0
∫∫L0
u+
δ0ξk−10 dxdτ
+ γr−2−n0
∫∫Q0
um−1(u+
δ0
)2λξk−20 dxdτ
+ γr−2−n0
∫∫Q0
um−1(u+
δ0
)1+λ
ξk−20 dxdτ
+ γθ0r20
r2−n0 δ−1
0 μ(B0(x0)).
Vol. 194, 2013 POROUS MEDIUM EQUATION 273
Hence we obtain from (2.12) that
(2.37)
δ0 ≤γ
(1
rn+20
∫∫Q0
um+λ+ dxdτ
) 11+λ
+ γ
(1
rn+20
∫∫Q0
um−1+2λ+ dxdτ
) 12λ
+r20θ0
1
rn+20
∫∫Q0
u+dxdτ + γθ0r20
r2−n0 μ(B0(x0)).
This completes the proof of the second assertion.
Proof of Corollary 1.3. Direct analysis of the proof of Theorem 1.1 shows that
for the solution to (1.7), estimate (1.5) holds for any (x0, t0) ∈ Q = BR× (0, T ).
Let M := supQ u. Hence from Theorem 1.1 we have
(2.38)
M ≤γ
(1
Rn+2
∫∫Q
um+λdx dt
) 11+λ
+ γ supt
∫BR
udx
+ γ
(R2
T
) 1m−1
+ γ
(1 +
T
R2
)supx∈BR
Iμ2 (x, 2R).
Using
ϕ =u
u+ ε, ε > 0,
as a test function and subsequently passing to the limit ε → 0, we obtain
(2.39) supt
1
Rn
∫BR×{t}
udx ≤ T
Rnμ(BR) ≤ T
R2Iμ2 (0, 2R).
Let us estimate the first integral on the right-hand side of (2.38). By the
Poincare inequality we have
1
Rn+2
∫∫Q
um+λdx dt ≤ 1
Rn
∫∫Q
um+λ−2|∇u|2dx dt.
Multiplying the equation by u(u + ε)λ−1 with ε > 0, integrating by parts and
passing to the limit as ε → 0, we obtain∫∫Q
um+λ−2|∇u|2dx dt ≤∫∫
Q
uλdμ dt ≤ MλTμ(BR).
Taking the above into account and applying the Young inequality we see that(
1
Rn+2
∫∫Q
um+λdx dt
) 11+λ
≤γMλ
λ+1
(T
R2Iμ2 (0, 2R)
) 1λ+1
≤1
2M + γ
T
R2Iμ2 (0, 2R).
This completes the proof of (1.8).
274 V. LISKEVICH AND I. I. SKRYPNIK Isr. J. Math.
Proof of Theorem 1.5. The proof follows the same lines as the proof of Theo-
rem 1.1. So we only explain the required modifications.
In Lemmas 2.1 and 2.2 the obvious modification is to replace τμ(Br(x0)) by
μ(Qr). The sequence (lj) will start from l0 = 1, and the formula for τj(l) in (2.8)
is replaced by τj(l) = r2j /Δm−1j (l). In the alternative (2.10), (2.11) governing
the choice of lj+1 the quantity(r2j /θj
) 1m−1 should be replaced by μ(Qrj ). In
the formulation of Lemma 2.3 inequality (2.14) is replaced by
δj ≤ 2−1
m−1 δj−1 + γ1
rnjμ(Qrj ).
The last terms in the right-hand sides of (2.27), (2.29), (2.30) require obvious
replacement of τjμ(Qrj ). Inequality (2.33) will be rewritten as
lJ ≤ γ(δ0 + Iμ2 (x0, t0, r0)).
One can also provide the analogue of the second assertion of Theorem 1.1
with obvious changes in (2.36) and (2.37). We leave this to the reader.
Acknowledgement. This paper was completed when the first-named author
was visiting the Department of Mathematics at the Technion. The support
of the British Technion Society is gratefully acknowledged. The first-named
author would like to thank the Department of Mathematics at the Technion for
its warm hospitality. Special thanks go to Moshe Marcus and Yehuda Pinchover
for useful discussions.
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