oblique derivative problems for elliptic and parabolic ...lieb.public.iastate.edu/china...
TRANSCRIPT
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
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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 .
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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 .
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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 .
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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).
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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 .
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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
.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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 .
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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).
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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).
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.
Obliquederivativeproblems
Gary M.Lieberman
Outline
The secondboundaryvalue problem
Obliquederivativeproblem
First estimate
Secondderivativeestimate
Improvedobliqueness
Preliminaries
Tangentialderivativeestimates
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.