The continuity method

Transcription

1 The continuity method The method of continuity is used in conjunction with a priori estimates to prove the existence of suitably regular solutions to elliptic partial differential equations. One crucial ingredient of the method relies on global a priori estimates of solutions (see the Schauder estimates and the L p regularity theory and this provides one important application of the regularity theory for such equations. 1. The general framework Theorem 1.1. Let B be a Banach space and let T : B B. Assume that there exists θ (0, 1 such that T u T v θ u v. Then there is a unique fixed point u B such that T u = u. Proof. For any u 0 B, define the sequence {u i } by We have that for j > i u j u i j k=i+1 j k=i+1 u i = T u i 1. u k u k 1 = θ k 1 u 1 u 0 θi 1 θ u 1 u 0. j k=i+1 T u k 1 T u k 2 Thus the sequence {u i } is a Cauchy sequence and hence it converges to some u B. By the continuity of T, we have that u = T u. To show the uniqueness, suppose that there exists two solutions u = T u and v = T v. Then, u v = T u T v θ u v < u v and it follows that u = v. Theorem 1.2. Let B be a Banach spaces and V be a normed vector space. Let L 0, L 1 : B V be two bounded linear operators. Set L t := (1 tl 0 + tl 1 for t [0, 1]. 1

2 Assume that there exists a constant C > 0 such that for all u B and t [0, 1], we have u B C L t u V. Then L 0 is surjective if and only if L 1 is surjective. Proof. Suppose that L 0 is surjective. Then L 0 is a bijection, hence the inverse L 1 0 : V B exists. For t [0, 1] and f V, the equation is equivalent to L t u = f L 0 u = f + L 0 u L t u = f + t(l 0 L 1 (u. By the invertibility of L s, it is also equivalent to u = L 1 0 (f + t(l 0 L 1 (u := T u. Hence solving L t u = f is equivalent to find a fixed point to the operator T : B B defined by T u := L 1 0 (f + t(l 0 L 1 (u. We claim that T is a contraction mapping. Indeed T u T v B = L 1 0 (f + t(l 0 L 1 (u L 1 0 (f + t(l 0 L 1 (v B = t L 1 0 (L 0 L1(u v B tc (L 0 L 1 (u v B tc( L 0 + L 1 u v B 1 and hence if tc( L 0 + L 1 < 1 that is if t < 2δ = C( L 0 + L 1, then T is a contraction. Hence for t δ = t 1, T admits a fixed point. Then starting from t 1, we get t 2 and so forth. So we can divide the interval [0, 1] into subintervals of length less than δ until we rich 1. Also L 1 u = f is solvable. This completes the proof of the theorem. 2. Existence of classical solutions Schauder estimates for general equations. Theorem 2.1. Let be a bounded C 2,α domain, f C α ( and g C 2,α ( for some α (0, 1. Then, the Dirichlet problem { u = f in (1 u = g on. posses a unique solution u of class C 2,α (. 2

3 Proof. The uniqueness follows from the maximum principle. The existence follows from approximating f by smooth functions, using the regularity for C rhs and the a priori estimates of Schauder which gives a compactness result. Let be a bounded domain, let a ij, b i and c be defined in with a ij symmetric. Consider the second-order elliptic operator (2 Lu = ij and assume L is uniformly elliptic a ij (xd i D j u + i λ ξ 2 ij a ij ξ i ξ j Λ ξ 2 b i D i u + cu a ij, b i, c C α ( and a ij C α ( + i ij b i C α ( + c C α ( M. Our aim is to prove a general existence result for solutions of Dirichlet boundary value problem with C 2,α boundary values involving the operator L with C α coefficients. First we need a priori estimates (we suppose that a solution exists. Theorem 2.2 (Weak maximum principle. Let u C 2 ( C( be a solution to (2 with c 0. Then u L ( C( f L ( + sup u where C = depends on n, λ, Λ, and the L norm of the coefficients. Proof. First we prove that sup u sup u + + C f L (. First we suppose that c(x c 0 > 0 and consider v := u u +. We have that v satisfies Lv = f csup u + f in and v 0 on the boundary. If v attains a positive maximum at an interior point x 0, then c 0 v(x 0 c(x 0 v(x 0 f(x 0 f L (. It follows that sup v f L (, and hence c 0 sup u sup u + + f L ( c 0. 3

4 In the general case c 0, we write v = z v = zw for an appropriate z z > 0 to be determined. We have that w satisfies a ij D ij w + ( b i 2 a ij D j z D i w z ij i j ( + c + 1 z ( b i D i z a ij D ij z w f z. i i,j Choosing z bounded and such that z > 0 and 1 z ( i b id i z a ij D ij z > i,j c 0, we see that w satisfies and equation with c > 0. Hence we can aplly the previous arguments to show that sup w C f/z L i nfty( and recalling that w = v/z we get the desired result for v and consequently sup u sup u + + C f L (. Applying this for u and u we get the result. Proposition 2.3. Let be a C 2,α domain and let u C 2,α (. Then given ε > 0 there exists a constant C(ε, such that for k = 0, 1 and β (0, 1 and k = 2 and β < α, we have u C k,β ( ε D2 u C 2,α ( + C u L (. We now try to extend our results on the Poisson equation to the case of non-constant coefficients, which was in fact our original goal. Theorem 2.4 (Schauder estimates for general elliptic equations. Let α (0, 1. Let be a bounded C 2,α domain and let L be a uniformly elliptic operator of the form (2 with C α coefficients. Assume also that f C α ( and g C 2,α (. If u C 2,α ( is a solution to (2, then we have the estimate u C 2,α ( ( f C C α ( + g C 2,α ( + u L (, where the constant C depends on n, λ, Λ,, α and the Hölder norms of a ij, b i and c. Proof. Thanks to the interpolation proposition we only need to bound the C 2,α semi-norm with the above rhs. Without loss of generality we can assume that g = 0 (convider v = u g which solves Lv = f Lg = f. We can also reduce the problem by dropping the lower order term. Indeed we can write (2 as ij a ij (xd ij u = f 4

5 where f := f i b id i u cu. If we can show that u C 2,α ( C f then using the interpolation inequality: for all ε > 0 and C α ( u C 2,α ( we have we get that u C 1,α ( ε u C 2,α ( + C(n,,, ε u L ( u C 2,α ( C ( f C α ( + u L ( and using the maximum principle (u = 0 on the boundary u L ( C f L (, we have u C 2,α ( ( f C C α (, Hence the problem is reduced to prove the estimate for the special case, (3 ij a ij (xd ij u = f We use the so called method of freezing coefficients. Th idea is to fix a point x 0 and to rewrite (3 as ij a ij (x 0 D ij u(x = f(x + (a ij (x a ij (x 0 D ij u(x =: h(x. Now a ij (x 0 is positive definite matrix with constant coefficients. After a change of variable, we can reduce the equation to a Poisson equation û = ĥ and use the result of the previous chapter concerning Schauder estimates for the Poisson equation. Note B R a ball centered at x 0 and suppose that u is supported in B R. First we have ĥ C α (B R C( f C α (B R + R α D 2 u C α (B R + u C 2 (B R. Using that a ij (x a ij (x 0 CR α (coefficients are Hölder continuous and the interpolation proposition, we have ĥ C α (B R C( f C α (B R + R α D 2 u C α (B R + u L (B R. ĥ L (B R C( f L (B R + R α D 2 u L (B R + u L (B R. From Theorem 5.13 in the lecture notes, we have ( D 2 u C α (B R C ĥ C α (B R + 1 ĥ + 1 R α L (B R R u 2+α L (B R 5

6 ( 1/α 1 So if we take R 0 =, then for 0 < R R 0, 2C ( D 2 u C α (B R C f C α (B R + u L (B R. The interpolation proposition gives the estimate for the full norm. This Theorem gives still just an interior Schauder type estimate by taking a cut-off function and covering any subdomain by finitely many balls of radius R 0 /2. Next we extend it to the boundary (proof omitted see the books of Wu, Yin, Wang. Existence of classical solutions. Idea: we can solve Dirichlet problems for general elliptic operators with Hölder continuous coefficients, provided that we can solve the equation for the Laplacian. Theorem 2.5. Let be a bounded C 2,α domain and let L be a uniformly elliptic operator of the form (2 with C α coefficients and c 0. Then for any f C α ( and g C 2,α (, there exists a unique solution u C 2,α ( of the Dirichlet problem { Lu = f in (4 u = g in We shall prove the solvability of the boundary value problem (4 if the same is true for the boundary value problem with L 0 = i.e., for Poisson s equation. Of course, the latter is a basic known result and so Theorem follows accordingly. Proof. Since the problems are linear, without loss of generality, we assume g = 0; otherwise, we consider Lv = f Lg with v = 0 on the boundary. Consider the family of equations: L t u := (1 t( u + (1 tlu. We note that L 0 = and L 1 = L. If we write L t u = ij we can easily verify that a t ijd i D j u + i b t id i u + c t u max(1, Λ ξ 2 ij a t ijξ i ξ j min(1, λ ξ 2 for all x and ξ R n and that a t ij C α (, b t i C α (, c t C α ( M 6

7 independently of t. Thus, L t u C α ( C u C 2,α (, where C is a positive constant depending only on n, α, λ, Λ and M. Then for each t [0, 1], L t : B V is a bounded linear operator, where B := { u C 2,α ( u = 0 on } is a Banach space and V := C α ( is a normed vector space. We know that L 0 is solvable (L 0 is surjective, thus if we show the a priori estimates u B = u C 2,α ( C L tu V = C L t u C α ( we are done since the existence of a classical solution (the surjectivity of L 1 is a direct consequence of Theorem 1.2. The uniqueness can be proved by the maximum principle. Since u = 0 on the boundary, the maximum principle implies that u L ( L tu L ( and the global Schauder estimates implies that (since u solves the equation with L t u as a right hand term u C 2,α ( C L tu C α (. 7