Boundedness, Harnack inequality and Hölder continuity for weak solutions

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1 Boundedness, Harnack inequality and Hölder continuity for weak solutions Intoduction to PDE We describe results for weak solutions of elliptic equations with bounded coefficients in divergence-form. The ideas of proofs come from DeGiorgi, Nash and oser. References abound; we mostly use Gilbarg and Trudinger. We discuss equations Lu = g + i f i (1) where We assume uniform ellipticity Lu = i (a ij j u + b i u) + c j j u + du () a ij (x)ξ i ξ j λ ξ (3) with λ > 0 good for all x considered. We assume that a ij = a ji are measurable and bounded, aij(x) Λ. (4) We assume that the coefficients b, c, d are bounded λ ( b i (x) + c i (x) ) + λ 1 d(x) Γ (5) and that the right hand side g L q, f (L q ) n, for some q > n. Denoting by A i (x, z, p) = a ij (x)p j + b i (x)z + f i, B(x, z, p) = c j (x)p j + d(x)z g(x), we say that u is a W 1, () weak subsolution in if ( i v)a i (x, u, u) + vb(x, u, u)dx 0 (6) 1

2 holds for all v C0(), 1 v 0. inequality is reversed We say that u is a supersolution if the ( i v)a i (x, u, u) + vb(x, u, u)dx 0. (7) We will quote theorems in full generality but we will present the ideas of proofs only, and for that purpose we will take b = c = d = f = g = 0. 1 Local boundedness, Harnack inequality We denote k(r) = λ 1 ( R 1 n q f L q + R (1 n q ) g L q ) Theorem 1. If u is a W 1, () subsolution of (1) then there exists a constant C = C(n, Λ, q, p, ΓR) such that, for all balls B(y, R), and p > 1 we λ have ( ) sup B(y,R) u C R n p u + L p (B(y,R)) + k(r). (8) If u is a W 1, () supersolution then sup ( u) C B(y,R) ( R n p u L p (B(y,R)) + k(r) ). (9) The idea of proof is to use v = η u β as test function, where η is a cutoff function, β is arbitrary, positive, and deduce bounds of the type u u β 1 η dx C η u β+1 dx This, together with a Sobolev embedding produces bounds for higher L p norms on the left hand side, depending on lower L p norms on the right hand side on larger domains. An iteration, due to oser, finishes the proof. Unfortunately, because u β is not an admissible test function we have to trim it first. We consider k > 0 a small positive constant, a large positive constant. We take u = u + + k and set u = u if u <, u = + k if u. We take now v = η (u β 1 u kβ ).

3 where η is a nonnegative smooth cutoff function and β > 0. We take without loss of generality y = 0, R = 4 and η C 1 0(B 4 ). We denote B R = B(0, R). Now v is a legitimate test function and v = η u β 1 [(β 1) u + u] + η η(u β 1 u kβ ) Here we used the fact that u = u when the latter is nonzero. Later we will use also that u = u when the latter is nonzero. Because u is a subsolution we obtain (remember, b = c = d = f = g = 0 in our proof) η u β 1 [ (β 1) u + u ] dx C η η( u u β 1 u)dx. Here we used the fact that u β 1 u kβ 0 to drop the k β term in the right hand side. After hiding the term involving u from the right hand side in the left hand side, we have (new constant C!) η u β 1 [ (β 1) u + u ] dx C We let now w = u β 1 u. The inequality above implies η w dx C(β + 1) η w dx η u u β 1 dx In view of the fact that (ηw) = η w + w η and the estimate above we have, using the Sobolev embedding, ( ) n (ηw) n n n dx C(β + 1) η w dx if n >. (If n = we replace n in the left hand side by any q < ). We n choose appropriately now the test function η so that, for balls B r B R we obtain ) n ( ) w (Br n n n 1 dx C(β + 1) u β+1 dx R r B R Writing q = β + 1, we have (Br u nq ) n n dx n ( ) 1 Cq u q dx R r B R 3

4 for any q > 1 and r < R. Letting and then k 0 we obtain ) 1 u + nq q q (C L n u + (B r) (R r) L q (B R ) Denoting the amplification factor by a = 0, 1,..., q i = pa i, r i = + 1 i, we obtain and so and thus n n u + L q i+1 (Bri+1 ) C i a i u + L q i(bri ) u + L q i(bri ) C P i a i u + L p (B 4 ) u + L (B ) C u + L p (B 4 ) > 1, and choosing for i = proving the theorem for subsolutions. The same idea of proof works for supersolutions. Now we prove a lemma, a weak Harnack inequality. Theorem. If u is a supersolution of (1) that is nonnegative in a ball B(y, 4R) then ( ) R n p u L p (B(y,R) C inf u + k(r) B(y,R) holds for 1 p < n with C = C(n, Λ, q, p, RΓ). n λ The idea of proof is similar to the one for L bounds except we are going to use negative powers, and a result of John-Nirenberg, of independent interest. We start by assuming without loss of generality that y = 0, R = 1, and so on. Also, we may assume u is bounded (either by the previous result, or by a trimming procedure like in the proof above). We set v = η ū β with η C 1 0(B 4 ) a nonnegative cutoff function, ū = u + k with k > 0 and β 0. Using the fact that u is a supersolution we have, after hiding one term, η ū β 1 u dx C( β ) η ū β+1 dx 4

5 The constant C( β ) is bounded when β > ɛ > 0. We set w = ū β+1 if β 1 and w = log ū if β = 1. Letting γ = β + 1 we have η w C( β )γ η w dx if β 1, and η w C η dx ( ) if β = 1. We will refer to this inequality at the end of the proof. Thus, for n >, from the Sobolev inequality we obtain the inequality Choosing η like before, we have with a = ηw L n n C (1 + γ ) η w L w L a (B r1 ) C(1 + γ )(r r 1 ) 1 w L (B r ) n as before, and r n > r 1. Denote We have the inequalities Φ(p, r) = ( ( ) C(1+ γ ) γ r r 1 B r ū p dx ) 1 p Φ(aγ, r 1 ) Φ(γ, r ), if γ > 0 ( ) Φ(γ, r ) C(1+ γ ) γ r r 1 Φ(aγ, r 1 ), if γ < 0. For β < 0, we take any 0 < p 0 < p < a and we have, with r m = 1 + m, m = 0,..., Φ( p 0, 3) CΦ(, 1) We know (from homework!) that Φ(, r) = inf Br ū. We also can prove one step Φ(p, ) CΦ(p 0, 3) We would be therefore done if we could find some p 0 > 0 such that Φ(p 0, 3) CΦ( p 0, 3) This follows from a result of F. John and Nirenberg. 5

6 Theorem 3. (F. John-L. Nirenberg) Let u W 1,1 (U) where U is convex. Suppose that there exists a constant K so that U B r u dx Kr n 1 holds for all balls B r. Then there exists σ 0 > 0 and C depending only on n such that ( σ ) exp K u u U dx C(diam U) n holds with σ = σ 0 U (diam U) n and u U = U u. Let us note that, from the inequality ( ), we obtain, via Schwartz Br w dx Cr n ( B r w dx ) 1 Cr n 1 Therefore, by the theorem of John-Nirenberg, there exists a constant p 0 > 0 such that exp (p 0 w w 3 )dx C B 3 where w 3 = B 3 wdx. Therefore ( B 3 e p 0w dx)( B 3 e p 0w dx) Ce pow 3 e p 0w 3 = C This concludes the proof of the weak Harnack inequality, modulo the John- Nirenberg result. Putting together Theorem 1 and Theorem we have the full Harnack inequality for weak solutions: Theorem 4. If u is a weak W 1, () solution with u 0 then there exists a constant C so that sup u C inf u B(y,R) B(y,R) holds for any y so that B(y, 4R). oreover, for any U there exists a constant, depending on U and so that sup u C inf u U U 6

7 Hölder continuity The weak Harnack inequality is sufficient to prove the Hölder continuity of weak solutions. Theorem 5. Let u be a weak W 1, () solution of (1). Then u is locally Hölder continuous in. For every ball B 0 = B(y, R 0 ) there exists a constant C = C(n, Λ, Γ, q, R λ 0) such that osc B(y,R) u CR α (R α 0 sup B 0 u + k) where α = α(n, Λ, ΓR λ 0, q) > 0 and k = λ 1 ( f L q + g q L ). oreover, for every U, there exists α = α(n, Λ, Γd) > 0 where λ d = dist(u, ), so that u C α (U) C( u L () + k) We provide the proof for the case b = c = d = f = g = 0. Let 0 = sup B0 u, 4 = sup B(y,4R) u, m 4 = inf B(y,4R) u, 1 = sup B(y,R) u, m 1 = inf B(y,R) u. We have L( 4 u) = 0, L(u m 4 ) = 0 so, we can apply the weak Harnack inequality of Theorem with p = 1. We obtain R n ( 4 u)dx C( 4 1 ) B(y,R) and R n (u m 4 ) C(m 1 m 4 ) Adding, we obtain and so with γ = C 1 C B(y,R) ( 4 m 4 ) C[ 4 m 4 ( 1 m 1 )] 1 m 1 γ( 4 m 4 ) < 1. We have thus, for ω(r) = osc B(y,R)u It follows by iteration (exercise!) that ω(r) γω(4r) ω(r) CR α ω(r 0 ) 7

8 3 Riesz potentials and the John-Nirenberg inequality We use the notation of Gilbarg and Trudinger (V µ f)(x) = x y n(µ 1) f(y)dy with µ (0, 1]. This is the same as the Riesz potential of order nµ of fχ, where is a bounded open set and χ its characteristic (indicator) function. Note that V µ 1 µ 1 ωn 1 n µ Indeed, choosing R so that = ω n R n, we have (V µ 1)(x) = x y n(µ 1) dy B(x,R) x y n(µ 1) dy because \B(x, R) = B(x, R)\ and the points in \B(x, R) are farther away and hence have smaller potential the points in B(x, R) \. Lemma 1. The operator V µ maps L p () continuously into L q (), 1 q oreover V µ f L q 0 δ = δ(p, q) = p 1 q 1 < µ The proof: choose r so that ( ) 1 δ 1 δ ωn 1 µ µ δ f L p µ δ r 1 = 1 + q 1 p 1 = 1 δ Then h(x y) = x y n(µ 1) is in L r () and, using the same trick as above h L r ( ) 1 δ 1 δ ωn 1 µ µ δ µ δ Now we write h f = h r q h r(1 p 1) f p q f pδ 8

9 and using Hölder we get [ hr (x y) f(y) p dy ] 1 q [ V µ (x) hr (x y)dy ] 1 p 1 f pδ L p But sup x [ hr (x y)dy ] 1 r is finite and raising the inequality above to power q and integrating we obtain the desired result. Lemma. Let be convex and u W 1,1 (). Let S be any measurable set. Then u(x) u S dn x y 1 n u(y) dy n S a.e. in, where and d = diam. u S = 1 S S udy It is enough to establish the inequality for C 1 () functions. Then u(x) u(y) = x y 0 r u(x + rω)dr where ω = y x x y and r = ω. Integrating in y over S: We define and thus we have S (u(x) u S ) = V (x) = u(x) u S 1 = 1 S S dy x y { r u(x), if x 0, if x / S 0 ω =1 0 r u(x + rω)dr x y d dy 0 V (x + rω)dr dω d dr V (x + 0 ω =1 0 rω)ρn 1 dρ V (x + rω)drdω x y 1 n r u(y) dy = dn n S = dn n S We introduce now orrey spaces: We say that f p () if there exists a constant K so that f dx Kr n(1 1 p ) B r 9

10 holds for all balls B r = B(x 0, r). The norm f p () is the smallest such constant K. Lemma 3. Let f p (), and δ = p 1 < µ. Then Then, V µ f(x) 1 δ µ δ (diam )n(µ δ) f p () Proof. We extend f by zero outside and denote m(r) = B(x,r) f dy V µ f(x) rn(µ 1) f(y) dy, r = x y, = d 0 rn(µ 1) m (r)dr, d = diam = d n(µ 1) m(d) + n(1 µ) d 0 rn(µ 1) 1 m(r)dr C 1 δ µ δ dn(µ δ) We note here a generalization of orrey s inequality Proposition 1. Let u W 1,1 () and assume that there exist K > 0 and 0 < α 1 so that u dx Kr n 1+α B r for all balls B r. Then u C α () and osc Br u CKr α The proof is a direct application of Lemma with S = = B r Lemma 3 with = B r. and Lemma 4. Let f p () with p > 1 and let g = V µ f with µ = p 1. Then there exist constants c 1, c depending only on n and p so that ( ) g exp dx c (diam ) n c 1 K where K = f p (). 10

11 Proof: we write, for q 1: x y n(µ 1) = x y ( µ q 1) n q x y n(1 1 q )( µ q +µ 1) and by Hölder By Lemma 3 and by Lemma 1 g(x) ) 1 (V ( ) µq f q 1 1 V µ+ µ f q q (1 µ)q V µ+ µ f d n pq q µ K, d = diam (p 1)qd n pq K V µq f dx pqω1 1 pq n 1 pq f L 1 pqω n Kd n(1 1 p + 1 pq ) Therefore g q dx p(p 1) q 1 ω n q q d n K q where p = p ω n {(p 1)qK} q d n p p 1. Choosing c 1 > e(p 1) and summing, we have g m m!(c 1 dx p ω K) m n d ( n p 1 c 1 c d n ) m m m Combining Lemma and Lemma 4 we proved Theorem 3. m! 11