Regularity for Poisson Equation

Regularity for Poisson Equation OcMountain Daylight Time. 4, 20 Intuitively, the solution u to the Poisson equation u= f () should have better regularity than the right hand side f. In particular one expects u to be twice more differentiable than f. The validity of this conjecture depends on the function spaces we are looking at. Note. Schauder Theory in fact denotes the similar results for the general linear elliptic PDE u aij (x) + b i (x) u +c(x) u(x) =0. (2) x i x j x i Nevertheless we use it (instead of C 2,α estimates ) as the title of this lecture to make it easy to display on the web.. Counter-examples. The most natural conjecture one would make is f C() u C 2 (). Anyway, it is indeed true in D. However it cease to be true when the dimension is bigger than. Example. (f L but u C, ). We compute (in x, x 2 > 0) Thus x 2 = x [ u(x, x 2 )= x x 2 log ( x + x 2 ). (3) x 2 log (x + x 2 ) + x ] x 2 = 2 x 2 x x 2 x + x 2 x +x 2 (x +x 2 ) 2; (4) 2 x = 2 x x x 2 2 x + x 2 (x + x 2 ) 2. (5) u=2 2 x x 2 (x +x 2 ) 2 (6) and the RHS is a bounded function. However, we compute = log (x + x 2 ) + x x 2 x x 2 (x + x 2 ) 2 L. (7) Example 2. (f continuous but u C, ). [ u= f(x) x 2 2 2 x n +2 2 x 2 ( log x ) + /2 2( log x ) 3/2 ], x B R R n. (8) f(x) is continuous after setting f(0)=0. However, the solution has Therefore u C,. u(x)=(x 2 x 2 2 ) ( log x ) /2 (9) x 0. (0) 2 x. One can show that there is no classical solution to this problem. Assume otherwise a classical solution v exists, then the difference u v is a bounded harmonic function in B R \{0}. One can show that such functions can be extended as a harmonic function in the whole B R which means 2 u must be bounded, a contradiction.

2 Regularity for Poisson Equation 2. C α regularity. The right space to work on are the Hölder spaces. Definition 3. (Hölder continuity) Let f: R, x 0, 0 < α <. The function f is called Hölder continuous at x 0 with exponent α if f(x) f(x sup 0 ) x x x 0 α <. () f is called Hölder continuous in if it is Hölder continuous at each x 0 (with the same exponent α), denoted f C α (). When α =, f is called Lipschitz continuous at x 0, denoted f Lip() or f C 0, (). C k,α ( ) contains f C k ( ) whose kth derivatives are uniformly Hölder continuous with exponent α over, that is f(x) f(y) sup x,y x y α <. (2) C k,α () contains f C k () whose kth derivatives are uniformly Hölder continuous with exponent α in every compact subset of. Example 4. The functions f(x) = x α, 0 < α <, is Hölder continuous with exponent α at x = 0. It is Lipschitz continuous when α =. Remark 5. When k =0, we usually use C α for C 0,α since there is no ambiguity for 0 < α <. We can define the seminorm and the norms where f(x) f(y) f C α ( ) sup x,y x y α, (3) f C α ( ) = f C0 () + f C ( ), (4) α f C k,α ( ) = α f C0 () + α f C ( ). (5) α α k α =k The following property is important. In short, C α is an algebra. f C ( )= sup f. (6) 0 x Lemma 6. If f, f 2 C α (), then f f 2 C α (G), and ( ) ( ) f f 2 C α sup f f 2 C α + sup f 2 f C α. (7) Proof. Left as exercise. Theorem 7. Let R d be open and bounded, u(x) Φ(x y) f(y)dy, (8) where Φ is the fundamental solution. Then a) If f C α 0 ( ), 0 <α <, then u C 2,α ( ), and u C 2,α ( ) c f C ( ). (9) α b) If f L () (α =0 case), then u C,α ( ) for any 0<α<, and u C,α ( ) c f L ( ). (20)

OcMountain Daylight Time. 4, 20 3 c) If f Lip( ) (α = case) with support contained in, then u C 2,α ( ) for any 0 <α <, and Proof. a) Recall that Φ(x y)=c log x y for n =2 and Φ(x y) =C. We first show u C. Formally differentiating we obtain xi u= ( xi Γ(x, y)) f(y) dy =C u C 2,α ( ) c f Lip( ). (2) x y n 2. for n 3. x i y i x y n f(y)dy. (22) It is easy to check that the integrand is integrable. Therefore by the theorem regarding differentiating with respect to a parameter for Lebesgue integrals, we see that the formal relation x xi u=c i y i x y n f(y)dy (23) indeed holds. 2. Next we show u C 2,α. In the following we will omit the constant factor C. In this step we do some preparations. Again formally differentiating, we obtain ( xix j u = δij x y n n(x i y i )(x j y j ) x y n+2 ) f(y) dy. (24) But this time the integrand is not automatically integrable and therefore this equality is dubious. To overcome this difficulty, we first work in the weak sense. By extending f outside to be 0 (resulting in a distribution with compact support), we can write in the sense of distributions. Thus we have [ xi x j u= xi u= x i x n f (25) xj ( xi x n )] f (26) ( xi in the sense of distributions. We compute the distributional derivative xj now. Take any φ C 0 (R n ), we know [ ( )] xi x xj x n (φ)= i x n ( x j φ)(x)dx = lim εց0 x i x n ( x j φ)dx. (27) Now integrate by parts, we have x i R n \B ε x n ( x j φ)dx = = where S ij (x) = B ε φ(x) x i x n x =ε φ(x) x i x j ε n+ + R n \B ε x n ) ( x ) j + S ij (x) φ(x)dx x R n \B ε R n \B ε S ij (x) φ(x)dx. (28) δ ij x y n n(x i y i )(x j y j ) x y n+2 (29) is the formal derivative. For the boundary term, we write φ(x) x ix j x =ε ε x =ε n+ = φ(0) x i x j ε x =ε n+ + [φ(x) φ(0)] x i x j εn+. (30)

4 Regularity for Poisson Equation Note that since φ C 0, φ(x) φ(0) = O( x ) = O(ε) which makes the second term an O(ε) quantity. For the first term, a symmetry argument shows that the integral vanishes when i j. When i = j, we use symmetry and the fact that x =ε xi x i x n+ = x =ε ε n = ω n, (3) where ω n is the surface area of the n dimensional unit sphere, to conclude that the limit is c φ(0) for some constant c. Therefore we have shown that [ xj ( xi x n )] (φ) = lim εց0 R n \B ε S ij (x) φ(x)dx +cδ (32) As a consequence, we have xix j u(x)= lim εց0 S ij (x y) f(y)dy + c f(x). \B ε (33) We now show directly that the second derivative xi x j u is Hölder continuous with power α. Since f(x) C α, we only need to show that [ ] lim S ij (x y) f(y)dy S ij (x 2 y) f(y) dy x x 2 α <. (34) εց0 \B ε(x ) \B ε(x 2) uniformly for x, x 2. 3. xix j u C α. Inspection of S ij reveals that for any 0 <R <R 2 : R y R 2 S ij (x y) dy =0. (35) To make things simple, we extend f to be 0 outside. The resulting function is in C α 0 (R n ) 2. We have S ij (x y)f(y)dy = S ij (x y) [f(y) f(x)] dy. (36) R n \B ε R n \B ε Note that since f C α, the integrand is integrable now, which means u C 2 has been proved. To show u C 2,α we need more refined analysis of the integral. First note that in writing the quantity as lim S ij (x y) [f(y) f(x)] dy (37) εց0 R n \B ε the singularity has been removed and we can take the limit and write xi x j u(x)= S ij (x y)[f(y) f(x)] dy. (38) R n For any x, x 2, setting δ = 2 x x 2, we have xix j u(x ) xix j u(x 2 ) = S ij (x y) [f(y) f(x )] S ij (x 2 y)[f(y) f(x 2 )] R n = + A +B. (39) B δ(x ) R n \B δ(x ) For A, we bound f(y) f(x ) f C α y x α and f(y) f(x 2 ) f C α y x 2 α, and get A C f C α δ α = C f C α x x 2 α. (40) 2. Let f be the extended function. Then one notices that f (x) f (y) = f(x) f(y) when x, y, vanishes when x, y, and equals f(x) f(y ) when x and y, where y is the intersection of and the line connecting x, y.

OcMountain Daylight Time. 4, 20 5 For B, we have B = = R n \B δ (x ) R n \B δ (x ) + R n \B δ(x ) S ij (x y)[f(y) f(x )] S ij (x 2 y)[f(y) f(x 2 )] S ij (x y)[f(x 2 ) f(x )] dy [S ij (x y) S ij (x 2 y)] [f(y) f(x 2 )] dy B + B 2. (4) It is easy to see that B = 0. For B 2, we estimate 3 S ij (x y) S ij (x 2 y) S ij (x 3 y) x x 2 C x x 2 x 3 y n+ (42) for some x 3 lying on the line segment connecting x, x 2. We have x B 2 C f x 2 C α R n \B δ (x ) x 3 y n+ y x 2 α C f C α x x 2 x y α (n+) dy R n \B δ(x ) = C f C α x x 2 x x 2 α = C f C α x x 2 α. (43) where we have used the fact that x i y are all comparable (i =, 2, 3) for y B δ (x ). b) We prove the stronger statement xi u is Log-Lipschitz, that is It is easy to get xi u(x ) xi u(x 2 ) C sup f x x 2 log( x x 2 ). (44) xi u(x ) xi u(x 2 ) sup f (x y) i x y n (x 2 y) i x 2 y n dy. (45) We extend f by 0 and break the integral to + with δ = 2 x B δ (x ) R n \B δ (x ) x 2. For the first term we obtain a bound C sup f x x 2, for the second we use (x y) i x y n (x 2 y) i x 2 y n C x x 2 x 3 y n (46) with a uniform C. Now note that for R big enough, R n \B δ integration can be carried out explicitly and yields the bound = B R \B δ (x ) B R \B δ/2 (x 3 ). The Thus ends the proof (the details are left as exercise). C x x 2 (log R log x x 2 ). (47) c) This part is the same as b). Omitted. Remark 8. The techniques involved in the above proof is standard in the theory of singular integrals and are applied extensively in equations arising from fluid mechanics, mathematical biology, etc. Remark 9. One may notice that when f L, one cannot reach xi u Lip (that is xix j u L ). The reason is that the operator xi x j ( ) does not map L into L. Details can be found in any textbook in real harmonic analysis. 3. Note that the intermediate value theorem gives x 3 depending on y. But when we are working outside B δ (x ), ξ y are all comparable for any ξ between x and x 2.

6 Regularity for Poisson Equation When f does not have compact support, we cannot obtain uniform bounds for u over the whole, but we can obtain estimates on any smaller set 0. 4 Theorem 0. Let R n be open and bounded, and 0. Let u solve u= f in. a) If f C 0 (), then u C,α () for any α (0, ), and b) If f C α () for 0<α<, then u C 2,α (), and u C,α ( 0) c( f C0 () + u L2 ()). (48) Here u C 2,α ( 0) c( f C α () + u L 2 ()). (49) ( ) /2 u L 2 () = u 2. (50) Proof. We just give an outline of the proof here. Set η be a cut-off function and consider φ= η u. We have where the RHS has compact support. This gives and Next we show that for any ε > 0, there is N(ε)>0 such that and φ =F η f + 2 u η + u η (5) F L c(η) f L +C(η) u C ] (52) F C α c(η) f C α + C(η) u C,α]. (53) u C N(ε) u L 2+ε u C,α (54) u C,α N(ε) u L 2 +ε u C 2,α. (55) This is shown via reductio ad absurdum using the Arzela-Ascoli theorem. Thus we obtain u C,α ( 0 ) C(η)[ε u C,α () + u L2 ()] + c(η)n(ε) f C0 () (56) (and a similar estimate for u C 2,α ( 0)) with the problem that the C,α norm on the LHS is on 0 while that on the RHS is on a bigger set and therefore cannot be absorbed into the LHS. This difficulty is overcome by the following technical trick. Consider the case when 0 =B r, =B R2, we have u C,α (B r) C(η) [ ε u C,α (B R2 ) + u L2 (B R2 )] + c(η)n(ε) f C0 (B R2 ). (57) Now set 5 A sup for some R > R 2. Now choose R such that This gives (R r) 3 u C,α (B r). (58) 0 r R A 2 (R R ) 3 u C,α (B R ), (59) A 2 (R R ) 3 u C,α (B R ) 2 (R R ) 3[ ] ε C(η) u C,α (B R2 ) + C(η) u L 2 (B R2 ) +2 (R R ) 3 c(η)n(ε) f C 0 (B R2 ). (60) 4. Meaning: The closure 0 is a compact subset of. 5. Here it seems we need to assume the finiteness of this quantity.

OcMountain Daylight Time. 4, 20 7 Now observe that C(η) and c(η), we have, using the definition of A (R 2 R ) 2, A C (R R ) 3 (R R 2 ) 3 ε (R R 2 ) 2 A + C N(ε) (R R ) 3 (R 2 R ) 2 u L 2 (B R2 ) + C (R R ) 3 f C0 (B R2 ). (6) Now for fixed R,R, one can choose R 2 and ε appropriately so that the coefficient of A on the RHS is less than. Thus we obtain the desired estimate for u C,α (B r) (R r) 3 A. (62) Now we can cover 0 by balls B r, and set R = r + d where d = dist( 0, ), and finish the proof. Corollary. If u solves u = f with f C k,α () for k N and 0 < α <, then u C k+2,α ( 0 ) for any 0 and In particular, u C when f C. u C k+2,α ( 0) c ( f C k,α () + u L 2 ()). (63) 3. Regularity and existence: method of continuity. We briefly discuss why the regularity estimates matter. Consider two bounded linear operators L, L from Banach spaces X to Y. 6 Assume that we know that L is surjective and wish to establish that L is also surjective, in other words the solvability of for arbitrary y Y. Define a family of operators L x= y. (64) L t = ( t)l+tl. (65) Thus L 0 = L and L = L. Assumption. We have uniform (that is, independent of t) a priori (that is, assuming the existence of solutions) estimates Under this assumption, one has Theorem 2. If L 0 is surjective, so is L. u X c L t u Y. (66) Proof. The idea is to show that there is ε independent of t, such that if L τ is surjective, so is L t for all t (τ,τ +ε). To see this, note that the estimate u X c L t u Y implies that all L t s are injective. Thus the inverse L τ is well-defined and bounded. We write into This gives L t u= f (67) L τ u = f + (L τ L t )u = f + (t τ)(l 0 L )u. (68) u=l τ f +(t τ)l τ (L 0 L )u. (69) Therefore all we need to do is to show the existence of a fixed point of the mapping (from X to X): u Tu L τ f + (t τ)l τ (L 0 L )u. (70) It is clear that if we take t τ small enough, we can find 0<r <, such that Tu Tv X r u v X. (7) 6. For example, in the case L = i,j aij 2 with a ij C α, X = C 2,α, Y = C α. xi xj

8 Regularity for Poisson Equation Now set v 0 =0 and v n =Tv n, we see that {v n } is a Cauchy sequence in X and therefore has a limit v which is a fixed point. An application of this theorem is to show the existence of the solutions to L u = i,j a ij (x) x i x j + i b i (x) u x i + c(x)u(x)= f (72) for Hölder continuous a ij, b i, c starting from the existence of the Poisson equation which can be shown by explicitly construct the solutions.