Oblique derivative problems for elliptic and parabolic ...lieb.public.iastate.edu/China 2011/6.pdf · 1 The second boundary value problem for Monge-Amp ere ... The equation Du() =

Obliquederivativeproblems

Gary M.Lieberman

Outline

The secondboundaryvalue problem

Obliquederivativeproblem

First estimate

Secondderivativeestimate

Improvedobliqueness

Preliminaries

Tangentialderivativeestimates

Oblique derivative problems for elliptic andparabolic equations, Lecture VI

Gary M. Lieberman

Iowa State University

August 1, 2011


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


1 The second boundary value problem for Monge-Ampereequations

2 Conversion to an oblique derivative problem

3 Estimate of the solution and its gradient

4 Second derivative estimateAn improved obliqueness conditionPreliminary second derivative estimatesTangential derivative estimates


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We now look at the second boundary value problem forMonge-Ampere equations. This means we try to solve

detD2u = f (x , u,Du) in Ω, Du(Ω) = Ω∗,

where Ω and Ω∗ are bounded domains in Rn and f > 0.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


The equation Du(Ω) = Ω∗ is not a boundary condition in theusual sense; however, it’s a natural kind of condition to placeon u because we want to consider only solutions at which theequation is elliptic. This requires the eigenvalues of D2u to beeverywhere positive, so Du is a diffeomorphism onto its image.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Many authors have studied this problem, including Caffarelli,Delanoe, Pogorelov, Urbas, and Wolfson. We’ll report here onUrbas’s results. We assume that f has Lipschitz continuousfirst derivatives with respect to the variables (x , z , p) with

limz→∞

inf(x ,p)∈Ω×Ω∗

f (x , z , p) =∞,

limz→−∞

sup(x ,p)∈Ω×Ω∗

f (x , z , p) = 0

and f (x , z , p) > 0 for all (x , z , p) ∈ Ω× R× Ω∗. We alsoassume that Ω and Ω∗ are uniformly convex with H3 boundary.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


To see how this problem relates to oblique derivative problems,we first observe that our assumptions imply that Du maps theboundary of Ω onto the boundary of Ω∗. Since Ω∗ is uniformlyconvex, there is a uniformly concave H3 function h such thath(p) > 0 if and only if p ∈ Ω∗. Moreover |Dh| is never zero onΩ∗. We may assume that |Dh| ≡ 1 on ∂Ω.It follows that h(Du) = 0 on ∂Ω. Here is our boundarycondition for u.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Since H, defined by H(x) = h(Du(x)) is positive in Ω and zeroon ∂Ω, it follows that

hkDkτu = DτH = 0 on ∂Ω

for any tangential vector field τ and

hkDkνu = DνH ≥ 0 on ∂Ω,

where ν is the unit inner normal. It follows that

hkDiku = DiH = DνHνi .


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We now set χ = hkνk and write [uij ] for the inverse matrix toD2u (which exists because u is strictly convex). Then

χ = hkDkiuuijνj = DνHνiu

ijνj .

Since DνH ≥ 0 and uij is positive definite, we have χ ≥ 0, sothe boundary condition is degenerate oblique.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


ButhkhiDkiu = DνHνih

i = DνHχ.

Since the left hand side is positive, so are the terms on theright. In particular χ > 0, and the boundary condition isoblique.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


If we integrate the differential equation over Ω and use thechange of variables formula, we have

|Ω∗| =

∫Ω

det(D2u) dx =

∫Ωf (x , u,Du) dx .

Then there are constants C1 and C2 such that

f (x , z , p) > |Ω∗|/|Ω|

for all (x , p) ∈ Ω× Ω∗ if z > C1 and

f (x , z , p) < |Ω∗|/|Ω|

for all (x , p) ∈ Ω× Ω∗ if z < C2.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Thereforesup u > C2, inf u < C1.

Since Du(Ω) = Ω∗ and Ω∗ is bounded, there is a constant M1

such that |Du| ≤ M1 in Ω. It follows that

sup u − inf u ≤ M1R,

where R is the diameter of Ω.It follows that

C2 −M1R ≤ u ≤ C1 + M1R.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Our next step is to show that there is a positive constant c0

(determined only by the data of the problem) such that

χ ≥ c0.

To estimate this number c0, we let x0 be a point on ∂Ω atwhich χ attains its minimum on ∂Ω, and realize that χ(x0)need not be the minimum value of χ over Ω.For our calculations, it will be useful to define the functionv = χ+ AH for a positive constant A to be chosen.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We now claim that Dνv(x0) ≥ −C for a constant C if wechoose A appropriately. The proof of this inequality is via abarrier argument. We define F (r) = ln det r and g = ln f .Differentiating the equation the equation in the direction of xkgives

uijDijku =∂g

∂xk+∂g

∂zDku +

∂g

∂pi,

Using superscripts to denote derivatives with respect to p thengives

uijDijv − g iDiv ≤ hijkDijuνk + AhijDiju + C1(1 + A) + C2T ,

where T is the trace of [uij ] and C1 and C2 are constantsdetermined by the data.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Since all the eigenvalues of [hij ] are less than −σ1 for someknown σ1 > 0 and D2u is positive definite, we can choose A sothat

hijkDijuνk + AhijDiju ≤ 0,

and the arithmetic-geometric mean inequality tells us that

T ≥ n(det(uij)

)1/n= nf −1/n ≥ σ2.

It follows thatuijDijv − g iDiv ≤ CT .


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Now let h be a uniformly convex H3 function such thath(x) > 0 if and only if x ∈ Ω. If f satisfies the condition

f (n−1)/n)|fp| ≤ δ0nR1,

where R1 is a positive number such that the eigenvalues of D2hare all less than −R1, then we set w = v(x0)− Bh. Then wecan choose B so large that

uijDijw − g iDiw ≥ CT in Ω.

Since w ≤ v on ∂Ω, it follows that w ≤ v in Ω, and therefore

Dνv(x0) ≥ Dνw(x0) ≥ −C .


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


More generally, it is possible to find a function w defined inΩε = B(x0, ε) ∩ Ω (for some ε > 0) which is positive inΩε \ x0 with w(x0) = 0 and which satisfies

uijDijw − g iDiw ≥ CT in Ωε.

Our argument that χ ≥ 0 on ∂Ω implies that χ ≥ 0 on all levellines h = k with k sufficiently close to 1, it follows thatv(x0)− Bw ≤ v in Ωε so we obtain

Dνv(x0) ≥ −C

in this case, too.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Now we rotate axes so that the inner normal to ∂Ω points alongthe positive xn axis and the xα axes are tangential to ∂Ω at x0.Then hkDkαu = 0 at x0 for α = 1, . . . , n − 1, and hkDknu ≥ 0at x0. In addition we have Dαv(x0) = 0 if α = 1, . . . , n− 1 andDnv(x0) ≥ −C . It follows that

hkDkv(x0) ≥ −Cχ(x0).

Here we note that χ(x0) = hn(x0).


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


If we write the derivatives of v out in terms of h and itsderivatives (using the formula χ = hkνk), we find that

hkhniDiku + hkhiDkνi + AhkhiDkiu ≥ −Cχ

at x0. Our formulas give

hkhniDiku = hkhnnDnku ≤ 0

because hnn ≤ 0 and hkDnku ≥ 0.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Therefore

ADijuhihj ≥ −Cχ− Dkνih

ihj ≥ −Cχ+ c1 at x0

for some positive number c1. Note that the minimumeigenvalue of [−Dkνi ] is bounded from below by a positiveconstant and hp(x0) is a unit vector.If χ(x0) ≥ c1/(2C ), we can take c0 = c1/(2C ). Otherwise, weconclude that

Dikuhihj ≥ c1

2Aat x0.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We now recall the Legendre transform u∗, which is defined onΩ∗ by

u∗(p) = x · Du(x)− u(x),

where x is the unique point in Ω such that Du(x) = p. If wewrite Du∗ for ∂u∗/∂p, we find that Du∗ = x andDiju

∗(p) = uij(x). Hence u∗ satisfies the conditions

detD2u∗ = f ∗(p, u∗,Du∗) in Ω∗, Du∗(Ω∗) = Ω,

where

f ∗(p, t, q) =1

f (q, p · x − t, p)

and (as usual) x is the unique point in Ω with Du(x) = p.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We can now argue as before with the function u∗ in place of uto find that

uνν(p0) ≥ c2

for some known positive number c2. (Here p0 = Du(x0).) Ittherefore follows (since χ2 = uννDijuh

ihj) that

χ(x0) ≥ c0

with

c0 = min c1

2C,

√c1c2

2A.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We are finally ready to estimate D2u. We start by setting

M = supΩ|D2u|.

(For later use, we use the norm

|D2u(x)| = sup|ξ|=1

Dijuξiξj

since we know that Dijuξiξj ≥ 0.) We also set

β = hp.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


The easy estimate is that

Dτβu = 0 on ∂Ω

for any tangential directional τ . Also, because H = 0 on ∂Ωand H > 0 in Ω, it follows that

Dνβu = DνH ≥ 0 on ∂Ω.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


By differentiating the differential equation for u in the formlog detD2u = g and recalling that [uij ] is the inverse matrix toD2u, we find that

uijDijH = hijDiju + hk [gk + gzDku + g iDiku]

≥ −C [1 + |D2u|].

Since detD2u is bounded from above by a known constant, itfollows that, for any ε > 0, there is a constant Cε such that

uijDijH ≥ −[Cε + εM]T .


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


The previously described barrier argument now shows that

Dνβu ≤ Cε + εM on ∂Ω.

Along with our equation Dτβu = 0, this estimate implies that

Dββu ≤ Cε + εM on ∂Ω.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Note we observe that we can decompose any unit vector ξ intoa tangential component τ(ξ) and a β component. Then

ξ = τ(ξ) +ν · ξβ · ν

β,

where

τ(ξ) = ξ − (ν · ξ)ν − ν · ξβ · ν

βT ,

βT = β − (β · ν)ν.

It follows that

|τ(ξ)|2 ≤ 1 + C (ν · ξ)2 − 2(ν · ξ)βT · ξβ · ν

.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Now, we suppose that the maximum over all second tangentialderivatives of the form Dξξu occurs (after rotation andtranslation) at 0 ∈ ∂Ω with ξ = (1, 0, . . . , 0). We may alsoassume that the vector (0, . . . , 0, 1) is the inner normal to ∂Ωat 0. For simplicity, we write τ for τ(ξ). Then

D11u = Dττu +2ν1

β · νD1βu +

ν21

(β · ν)2Dββu

on ∂Ω.Using our estimates for Dτβu and Dββu, and recalling that|τ |2D11u(0) ≥ Dττu, we find that

D11u ≤(

1 + Cν21 −

2ν1βT1

β · ν

)D11u(0) + [Cε + εM]ν2

1 .


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


To continue, we define

w =D11u

D11u(0)+

2ν1βT1

β · ν

and conclude that

w ≤ 1 +

(C +

Cε + εM

D11u(0)

)|ν1|2

on ∂Ω.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


Now, on ∂Ω,we have

Dξξu = Dτ(ξ)τ(ξ)u +

(ν · ξβ · ν

)2

Dββu

(because Dτβu = 0) and hence

max∂Ω|D2u| ≤ C [D11u(0) + Cε + εM].

The standard estimate for the Monge-Ampere equation impliesthat

M ≤ C [max∂Ω|D2u|+ 1],

and therefore, for all sufficiently small ε, we have

M ≤ Cε + CD11(0).


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


If D11u(0) ≤ 1, we have our second derivative bound.Otherwise, we conclude from our bound for M that

Cε + εM ≤ CεD11u(0)

which means that

w ≤ 1 + C (ε)|x ′|2 on ∂Ω.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


We now temporarily pursue a different estimate. By directcomputation, we have

uijDijw − g iDiw ≥ [Cε + εM]T .

Our previous barrier argument then shows thatDβw(0) ≤ C [Cε + εM], so

D11βu(0) ≤ [Cε + εM]D11u(0).


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


On the other hand, if we differentiate the boundary conditiontwice in the tangential direction (1, 0, . . . , 0) at 0, we obtain

D11βu + hijDi1uDj1u + κ1Dνβu = 0 at 0,

where κ1 is the normal curvature of ∂Ω in the direction(1, 0, . . . , 0) at 0. Using our upper bounds for D11βu(0) andDνβu gives

−hijDi1uDj1u ≤ [Cε + εM]D11u at 0.


Gary M.Lieberman

Outline



First estimate


Improvedobliqueness

Preliminaries


The uniform convexity of h implies that

−hijDi1uDj1u ≤ c3(D11u)2

for some positive constant c3 and hence

D11u(0) ≤ Cε + εM.

Since we have also shown that

M ≤ Cε + CD11u(0),

it follows thatM ≤ C

once we have chosen ε sufficiently small.

Documents

Oblique derivative problems for elliptic and parabolic ...lieb.public.iastate.edu/China 2011/6.pdf · 1 The second boundary value problem for Monge-Amp ere ... The equation Du() =