Oblique derivative problems for elliptic and parabolic equations, Lecture II

Transcription

1 of the for elliptic and parabolic equations, Lecture II Iowa State University July 22, 2011

2 of the 1 2 of the

3 of the As a preliminary step in our further we now look at a special situation for elliptic. We begin by taking α 1, ε, and ω 0 to be positive constants with α 1 < 1 and ε < 1. We then set A = 2 1+2ε ε 4ε, and we define G : R n (0, ) R by ( y 2 ) (1+ε)/2 G(y, r) = r 2 + α1 2 + (y n ) 2 (Aω 0 r 2 ). We also introduce, for any x 1 R n, the set E(x 1, r) = {x : G(x x 1, r) < 1}.

4 of the Let R > 0 and suppose x n > ω 0 x and ω 0 β (1 ε)β n on Ω B(0, R). Let r (0, R) and suppose x 1 = (0,..., x1 n) satisfies the s ) x1 n (A ε6 ω 0 r 4 and E(x 1, r) B(0, R). Then η, defined by η(x) = 1 G(x x 1, r) satisfies the inequality βn β Dη ε 4 Aω 0 r on S, the subset of Ω on which G < 1.

5 of the To prove this estimate, we first observe that ε 6 /4 < 1 and A > 2, so A ε 6 /4 > 1. Then x n < x1 n on S because ) x1 n x n (A ε6 ω 0 r ω 0 x 4 and x < r on S. We now set N = x 1 n x n, P = x Aω 0 r r, α = ε6 4.

6 Then of the N 1 α A P A, (P2 + α 2 1) (1+ε)/2 + N 2 1. Combining these inequalities yields and hence ( P 1+ε + 1 α A P ) 2 1, A P 1+ε 2P A 2α A.

7 of the It follows that because otherwise ( 2 + 2ε 2 P A ) 1/ε P 1+ε 2P A = [APε 2] P A > [(2 + 2ε2 ) 2] P A = 2ε2 P A, and P (2/A) 1/ε = ε 4 /4, so 2ε 2 P A 2ε6 4A = 2α A.

8 of the To get a lower bound for N, first observe that where Now we compute where α + P ε6 4 + ε4 4 H(ε), H(s) = (1 + s 2 ) 1/s. H (s) = H(s) g(s2 ) s 2, g(σ) = 2σ ln(1 + σ). 1 + σ Since g (σ) = (1 σ)/(1 + σ) 2, it follows that g is increasing on [0, 1]. But g(0) = 0, so g is nonnegative on [0, 1] and hence H takes its maximum at s = 1.

9 of the Since H(1) = 2, it follows that P + α ε6 4 + ε4 2 ε4. The inequality A 2 yields so N 1 ε4 2 β Dη = 2β n (x n 1 x n ) (Aω 0 r) 2 (1 + ε)β x r 2 (P2 + α 2 1) (ε 1)/2 2 βn Aω 0 r (N 1 2 (1 ε2 )AP ε ) 2 βn Aω 0 r (1 1 2 ε4 1 + ε 4 ) = ε 4 βn Aω 0 r.

10 of the We are now ready to prove the corresponding estimate for the parabolic case. We now suppose that Ω satisfies an interior parabolic cone at X 0. Specifically, we assume that ω 0 β (1 ε)β n, x n < ω 0 x + ω 1 t 1/2 on SΩ Q(0, R). To simplify the arguments, we shall assume that b = 0, c = 0, and β 0 = 0. With G and η as before, we define A 0 = ω2 1 ε 12 ω0 2.

11 of the We also assume that the eigenvalues of [a ij ] are in the interval [λ, Λ] for positive constants λ Λ, and we suppose that Lu f in Ω, 0 on SΩ. (We make no assumptions about u on BΩ.) Also, we set F (r) = r n/(n+1) f n+1, with r so small that G(x x 1, r) 1 for X SΩ with x R and t ( r 2, 0) and x n 1 ) (A ε6 ω 0 r. 8

12 of the Then, for any numbers α 0, θ and θ 1 in (0, 1), there is a positive constant C 1 such that if u(x ) 0 for all X Ω with G(x x 1 ) < 1 and t ( r 2 /A 0, 0) u(x ) + F (r) h for all X Ω with G(x x 1 ) < θ and t = r 2 /A 0 (here h is a nonnegative constant), then u(x ) + F (r) C 4 h for all X Ω with G(x x 1 ) < θ 1 and t = (α 0 1)r 2 /A 0.

13 of the We now define three functions: ψ 0 = 1 θ2 α 0 (A 0 t + r 2 ) + θ 2 r 2, ψ 1 = max{ψ 0 r 2 G(x x 1, r), 0} and, for q 2 a constant to be chosen, ψ = ψ 2 1/ψ q 0. By observing that DG 2 Cψ 0 wherever ψ ψ 0, we find that there are positive numbers N 1, N 2, N 3 (determined by this number C and by Λ and λ) such that Lψ ψ 1 q 0 [qn 1 1 θ 2 α 0 ξ 2 N 2 ξ + N 3 ], where ξ = ψ 1 /ψ 0. Hence, we can choose q so that Lψ 0 in Q = {X Ω : G < ψ 0, r 2 /A 0 < t < (α 0 1)r 2 /A 0 }.

14 of the Then, on SQ Ω, we have ψ = 0, while ψ 0 = θ 2 r 2 on BQ Ω, so ψ = (θ 2 r 2 r 2 G) 2 (θ 2 r 2 ) q (θr) 2q+4 there. Also, because β G C 2 /r (for some positive constant C 1 ) and ψ 1 r 2 in Q, β Dψ C 2 r ψ on SQ SΩ. Finally, when t = (α 0 1)r 2 /A 0, we have ψ 0 = r 2, so ψ (1 θ 2 1) 2 r 2q+4 if t = (α 0 1)r 2 /A 0 and G θ 1.

15 of the Now we set w = (θr) 2q 4 hψ ū and note that where Lw f in Q, β Dw w on SQ, β = w 0 on BQ, { r β on SΩ, C2 0 elsewhere. Our maximum principle gives w C 3 F (r) in Q. Rewriting this inequality for G < θ 1 and t = (α 0 1)r 2 /A 0, we see that ū C 4 h C 3 F (r), so (1 + C 3 )ū C 4 h. Hence, we have our result with C 1 = C 4 /(1 + C 3 ).

16 of the We are now in a position to prove a weak Harnack inequality, but it has some additional geometric complications compared to the usual one. We start by fixing a point X 0 SΩ and a positive number ρ and we write Q 1 for the intersection of Ω with the cylinder Q(X 0, ρ). Then there are constants B 1 1 and θ 1 (0, 1) and x 1 with x 1 = x 0 such that G(x x 1, B 1 ρ) < θ 1 in B(x 0, ρ) and x n 1 x n 0 ) (A ε6 ω 0 B 1 ρ. 4 We now write E 1 for the set of all X Ω on which G(x x 1, B 1 ρ) < θ 1 and t 0 (4A 0 /9)ρ 2 < t < t 0.

17 of the We now set h = inf Q1 u + F (ρ) and apply the estimate for each time level, noting that we can replace r in our with any convenient multiple of r. In this way, we obtain inf u + F (ρ) C(inf u + F (ρ)) E 2 Q 1 where E 2 is the set of all X Ω with G(x x 1, B 1 ρ) < θ and t 0 A 0 ρ 2 < t < t 0 (5A 0 /9)ρ 2, and θ (0, 1) is chosen so that there is a positive constant K such that d Kρ on E 2.

18 of the The standard weak Harnack inequality for parabolic equations then yields positive constants B 2, C, and κ such that (ρ n 2 Q 2 u κ dx ) 1/κ C(inf Q 1 u + F (ρ)) for Q 2 = Q((x 1, t 0 2A 0 ρ 2 ), B 2 ρ). We must assume for these arguments that t < t 0 (2A 0 + 4B 2 2 )ρ2 ) on BΩ, but it s possible to modify the arguments to replace this assumption with one on the behavior of u on BΩ.

19 of the The usual argument then gives a Hölder estimate for u near SΩ if Lu = f in Ω and Mu = 0 on SΩ. It s straightforward to modify the arguments to allow Mu = g on SΩ with g bounded (and to allow b, c, and β 0 to be nonzero), but that will take too long.

20 of the To prove Hölder continuity of the gradient of the solution, we introduce a different kind of for SΩ. We call any set of the form {X : T < t < 0, x ( t) 1/2 x 0 < R( t) 1/2 } (with R and T positive constants and x 0 R n ) a tusk. We say that SΩ satisfies a uniform exterior tusk at X 0 if there is a tusk such that, for each point X 0 SΩ, if the tusk is translated (so that the origin is translated to X 0 ) and then rotated in space, then the tusk is exterior to Ω. Note that the exterior tusk is weaker than the exterior parabolic cone.

21 of the A key fact, which we will not prove, is that, for any R and T and any positive numbers λ Λ, there are constants α > 0 and C determined only by R, T, x 0, λ, and Λ along with a function w with continuous second spatial s and continuous time exterior to the tusk (at least for t ( T, 0) satisfying the : X α w(x ) C X α, DX C X α 1, w t + a ij D ij w X α 2.

22 The of the Hölder gradient proceeds in a number of steps: 1 Assume that a ij and β are constant and all other coefficients are zero. of the

23 of the The of the Hölder gradient proceeds in a number of steps: 1 Assume that a ij and β are constant and all other coefficients are zero. 2 Show that is Hölder continuous.

24 of the The of the Hölder gradient proceeds in a number of steps: 1 Assume that a ij and β are constant and all other coefficients are zero. 2 Show that is Hölder continuous. 3 Prove an oscillation estimate for u P for a suitable polynomial P.

25 of the The of the Hölder gradient proceeds in a number of steps: 1 Assume that a ij and β are constant and all other coefficients are zero. 2 Show that is Hölder continuous. 3 Prove an oscillation estimate for u P for a suitable polynomial P. 4 Show that this oscillation estimate gives a Hölder gradient estimate.

26 of the The of the Hölder gradient proceeds in a number of steps: 1 Assume that a ij and β are constant and all other coefficients are zero. 2 Show that is Hölder continuous. 3 Prove an oscillation estimate for u P for a suitable polynomial P. 4 Show that this oscillation estimate gives a Hölder gradient estimate. 5 Use a perturbation argument to get the full result.

27 of the To be more specific, we shall look only in a parabolic neighborhood of some boundary point X 0. The neighborhood is then Q(X 0, r) for some r > 0. When a ij and β are constant, and all other coefficients are zero, then v = satisfies the s v t + a ij D ij v = 0 in Ω Q(X 0, r), v = 0 on SΩ Q(X 0, r).

28 of the Since SΩ satisfies a uniform exterior tusk, for each Y SΩ Q(X 0, r), there is a function w Y satisfying the preceding s (with X replaced by X Y but C independent of Y ), so we have ( ) Lv L sup v 2α w Y r α in Ω[Y, r/2], v sup v 2α w Y r α where Ω[Y, r/2] = Ω Q(Y, r/2). on PΩ[Y, r/2],

29 of the The maximum principle implies that in Ω[Y, r/2], and hence v sup v 2α w Y r α d(x )α v(x ) C sup v r α for X Ω[X 0, r/2]. A similar argument with v in place of v then implies that v C sup v d α r α, which says that v is Hölder continuous up to SΩ.

30 of the Note that the preceding result holds without assuming that β is oblique, in particular, the interior parabolic cone is not needed here. The next step will use these assumptions.