Elliptic PDEs of 2nd Order, Gilbarg and Trudinger

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1 Elliptic PDEs of 2nd Order, Gilbarg and Trudinger Chapter 2 Laplace Equation Yung-Hsiang Huang Mimic the proof for Theorem Proof. I think we should assume u C 2 (Ω Γ). Let W be an open ball containing some interior part Γ of the given portion Γ. Extend u(x) on W to be 0 if x W \Ω. This extended function, still call it u, is in C 2 (W ) due to the zero Dirichlet and Newmann boundary condition. We further consider a smaller portion Γ Γ and a smaller ball W W containing Γ Given x Γ and then given open ball B W with center at x, then B u dx = B u n ds = u B Ω n ds = u (B Ω) n ds = u dx = 0. B Ω This implies u = 0 on Γ and hence implies u = 0 on W. By analyticity of u in W, and the vansishing property on w \ Ω which contains an open set by the regularity of boundary portion Γ, we see that u 0 on W and hence on W Ω. By analyticity again, u 0 on Ω. 3. Proof. 4. Proof. It s clear that U is continuous on Ω := Ω + T Ω and harmonic on Ω. Given (x 0, 0) T, let B be a ball centered at x 0 and compactly contained in Ω. By translating and dilating the coordinates, we may assume that x 0 = 0 and B is the unit ball. Since u is continous on B, we can find a harmonic function v C 2 (B) C(B) such that v = u on B. By the Poisson formula and the fact u(x, t) = u(x, t), v(x, t) = v(x, t). In particular v = 0 on Department of Math., National Taiwan University. d042200@ntu.edu.tw

2 T and hence v = u on B + T and B T. Apply the maximum principle twice, we see v = u on B and hence u is harmonic on B. Remark. If we assume u C 2 (Ω + ), we can show U = 0 in Ω directly. 5. Proof. By dilating and translating the coordinates, we may consider Ω = B (0) \ B λ (0) for some 0 < λ <. We also assume d 3 first. Let Γ(r) = c d r 2 d (r = x ) be the fundamental solution of Laplace operator with singularity at the origin, x = respect to B and x = λ2 x x 2. x x 2 be the reflection of x with Then u = Γ( x y ) Γ( x x y ) is x-harmonic in Ω \ {y}, equals to zero if y B and equals to u 2 = Γ( x λ x y ) Γ(λ λ 2 x y ) if y B λ. Note that u u 2 is x-harmonic in Ω \ {y}, equals to zero if y B λ and equals to u 3 = Γ(λ λ 2 x y ) + Γ(λ x λ 2 x y ) if y B. Then we consider u u 2 + u 3 which is x-harmonic in Ω \ {y}, equals to zero if y B and equals to u 4 = Γ( x λ 2 λ 2 x y ) Γ(λ 2 λ 4 x y ) if y B λ. Fix y Ω. Define the function u(x) on Ω \ {y} by Γ( x y )+ Γ(λ k λ 2k x y )+Γ(λ k λ 2k x y ) Γ(λ k x λ 2k x y )+Γ(λ k x λ 2k x y ). k=0 Note that each term converges uniformly in x on Ω \ {y} by M-test (comparing to geometric series). By the previous argument, we know u is harmonic on Ω \ {y}, u = 0 for all y B since u 2k = Γ(λ k λ 2k+2 x y ) Γ(λ k x λ 2k+2 x y ) 0, and u = 0 for all y B λ since u 2k = Γ( x λ k λ 2k 2 x y ) Γ(λ k λ 2k x y ) 0. For planar case (d = 2), the boundary behavior is used the same argument. But I don t see the pointwise convergence for the series, c 2 u(x) = log x y + since the tails does NOT tend to zero. k=0 ( λ k λ 2k x y log λ k x λ 2k x y λ k λ 2k x y λ k x λ 2k x y ), Two-Dimensional Case. incomplete! 6. Estimate the denominator in the integrand and then apply mean value theorem. 7. Proof. If u is subharmonic, then we derive the local sub mean-value inequality by taking the harmonic function h on B δ with h = u on B δ. 2

3 On the other hand, if u satisfies the local sub mean-value inequality, then given B Ω and any harmonic function h in B with u h 0 on B. Since u h satisfies the sub mean-value inequality in B, by the same proof of maximum principle, u h in B. 8. Proof. Let φ ɛ be the standard mollifiers. Then (φ ɛ u) 0 on Ω ɛ := {x Ω : d(x, Ω) < ɛ}. For 0 < ɛ <, since φ ɛ u C 2 (Ω ɛ ), we see for each B r (x) Ω ɛ Ω, (φ ɛ u)(x) φ nω n r n ɛ u ds. B r Since u is continuous in Ω, φ ɛ u u locally uniform, and hence u(x) = lim(φ ɛ u)(x) lim φ ɛ 0 ɛ 0 nω n r n ɛ u ds = u ds. B r nω n r n B r So u is subharmonic in Ω by Exercise 2.7. Remark 2. This is the special case of Weyl s lemma, or so-called elliptic regularity theory. 9. Proof. Obviously, (i) implies (iii). (iii) implies (ii) by Exercise 2.8. Now we assume u is subharmonic in Ω. Suppose u < 0 at some point x, then u < 0 in some open ball B r (x) since u C 2. But this will contradict to the sub-mean inequality we obtained in Exercise 2.7 by the same argument as the proof of Theorem Proof. This problem is done if we know the sequence of the harmonic lifting V n is monotone increasing. But this is proved if v n is monotone increasing, for example, we can modify the definition of v n to be max{v, v 2,, v n, inf φ}, which is similar to the construction of Problem Proof. 2. This exercise is the special case of Exercise 6.3, and sometimes called the Zaremba s criterion. One of the counterparts of exterior cone condition for parabolic equations is the exterior tusk condition. See Lieberman [2, Exercise 3.] and Lorenz [3, Section 3..4]. Proof. Since the Laplace equation is invariant under translation and rotation, we may assume ξ = 0 and the exterior cone C is a right cone, that is, C = {x : xn x =: cos θ > cos α} for some α (0, π 2 ). Let V = B 2(0) C c. By truncating the cone properly and scaling, we may assume Ω B 2 (0) B 2 (0) C c = V. Consider the Dirichlet problem on V with continuous boundary value g(x) = x, then we have a harmonic function u C 2 (V ) C 0 (V \{0}) such that u = g on V \ {0}. Since g is subharmonic in V, and with the same boundary value as u, u(x) x > 0 in V. (In this proof, the geometry property of cone condition is only used here?) 3

4 If we can prove that u(x) 0 as x 0, then u is a local barrier at 0 and hence 0 is regular. Let v(x) := u(2x) defined on the set V 2 \ {0}, where V 2 := { x : x V }. Since the constant 2 function 2 and 2 are superfunction and subfunction to g on V respectively, 0 < u < 2 on V by strong maximum principle and hence on V 2 B (0). Since v = 2 on V 2 B (0) and both u and v are continuous on V 2 B (0), there is some 2 < α < such that u < αv on V 2 B (0). On the remainder of V 2 \ {0}, u = 2 v. Therefore, u < αv on V 2 \ {0}. Since u and v are continuous on V 2 \ {0}, for y V 2 \ {0}, lim z y,z V2 αv(z) u(z) 0. Consider the function αv u + ɛγ, where Γ(x) = x 2 n for n 3 and Γ(x) = log x for n = 2. Then for each ɛ > 0, lim inf x y,x V2 αv(x) u(x) + ɛγ(x) 0 if y V 2 \ {0} and lim inf x 0,x V2 αv(x) u(x) + ɛγ(x) =. Then weak minimum principle implies that αv u + ɛγ 0 on V 2. For each x V 2, since 0 αv(x) u(x) + ɛγ(x) αv(x) u(x) as ɛ 0. Therefore, 0 c := lim sup x 0,x V 2 u(x) α lim sup x 0,x V 2 u(2x) = αc. Since α <, 0 = c = lim sup x 0,x V2 u(x) lim inf x 0,x V2 u(x) 0. Remark 3. Counterexamples of Lebesgue and Osgood s theorem on R 2 Second Proof. incomplete! We try to follow the hint given there. WLOG, we may assume ξ = 0 and the exterior cone C is a right cone. Consider the function w(r, θ) = r λ f(θ) 3. Proof. If sup u =, then it s done. Otherwise, consider v = u u(x 0 ) and apply the meanvalue theorem for D i v in ball B := B dx0 (x 0 ), we have D i u(x 0 ) = D i v(x 0 ) = D w n d n i v(y) dy = 0 B w n d n 0 B v(y)ν i dy n d 0 sup v. If we take the square and sum up the index i, then we have the first inequality. For the second one, D i u(x 0 ) = w n d n 0 B D i u(y) dy = u(y)ν w n d n i dy 0 B w n d n 0 B u(y) dy = n d 0 u(x 0 ). 4. Proof. (a) Apply Exercise 2.3 with d 0. (c) Consider v = max{ x 2 d, }. (b) We may assume max B u = 0 on by adding some constant. Since u is bounded above, there is a sequence of radii {M n } as n such that u(x) < log x for all x M n n. For each n, apply the Maximum principle to u(x) log x on B n M n (0) \ B (0), we see that u 0 on R d \ B (0). Then we see that, for each r >, max Br u attains at the interior since 4

5 max B u = 0. Hence the strong maximum principle implies u is a constant function on B r for each r >, that is, on R d. Remark 4. First, note the unboundedness at infinity of two-dimensional fundamental solution for Laplace equation. Second, in the above proof, we see that the assumption can be weaken to assume that lim sup x u(x) log x Du 2 = div(udu) u u dx and then use the boundary condition and Young s inequality. 6. Proof. Given B ρ (x) Ω, there exists a dense subset {x k } of B ρ (x). Note that for each k, there is {v j k } S ϕ such that v j k (x k) u(x k ) as j. Define v j = max{v, j v j 2 v j j }, then for each k, u(x k ) v j (x k ) v j k (x k) and hence v j (x k ) u(x k ) for all k. Let V j := max{v j, inf Ω ϕ}, then inf Ω ϕ V j sup Ω ϕ by weak maximum principle. Consider the harmonic lifing Ṽ j of V j on B ρ (x), then by weak maximum principle v j V j Ṽ j, and hence Ṽ j converges to u(x k ) for each k. Moreover, by Theorem 2.9 (Harnack convergence theorem, see Problem 2.0) or Theorem 2. (interior gradient estimate and Arzela-Ascoli), we see that for each 0 < r < ρ we can find a subsequence of {Ṽ j } converges uniformly on B r (x) to v r which is hence harmonic in B r (x) and we can define v(y) = v r (y) if y B r (x) and hence v is harmonic in B ρ and u(x k ) = v(x k ) for each k. Given x 0 B r (x), we can choose a subsequence x kj and lower semi-continuity of u, we see of x k converging to x 0. By continuity of v v(x 0 ) = lim j v(x kj ) = lim inf j u(x k j ) u(x 0 ). If v(y) > u(y) for some y B r (x), then there is some Ṽ j S ϕ such that Ṽ j (y) > u(y) which contradicts to the definition of u. Hence u is harmonic in B ρ (x) since u = v. Since B ρ (x) Ω is arbitrary chosen, u is harmonic in Ω. 7. Proof. 8. Proof. Since the Laplace equation is isotropic everywhere, we may assume x 0 = 0. Note that for each s, w S n, cs bw = bs cw. By this fact, a = b2, Fubini s theorem and Poisson c integral formula, we have 5

6 Sn Sn u(aw)u(cw) dw = u(cw) bn b 2 a 2 u(bs) ds dw S nw n n b bs aw n Sn Sn = u(bs) cn c 2 b 2 u(cw) dw ds nw n c cs bw n Sn Sn = u(bs) cn c 2 b 2 u(cw) dw ds nw n c bs cw n = u(bs)u(bs) ds. S n If u is constant C (we may assume C = 0) on some open subset of Ω, then the open subset W := ({u = 0}) o is non-empty. If we can prove that W is closed in Ω, then by the connectedness of Ω, Ω = W. Let x 0 Ω \ W be a limit point of W, then we can find a radius c such that B(x 0, 2c) Ω and q W B(x 0, c/2). Let a be the largest radius such that B(q, a) W, then a c/2 and B(q, c) B(x 0, 2c) Ω. According to the integral identity we just proved, for each 0 < ɛ < a, S n u(q + (a ɛ)cw) 2 dw = 0. By weak maximum and minimum principle, we conclude that u = 0 on B(q, (a ɛ)c). But this contradicts to the definition of a since (a ɛ)c > a if ɛ is small. Therefore, if x 0 Ω is a limit point of W, then x 0 W. This is equivalent that W = W Ω and thus W is closed in Ω. Another interesting result is the following generalization of Liouville s theorem: Theorem 5. [, Theorem 6] u is harmonic in R n and vanishes at the origin. If a constant A exists such that for every ρ > 0 S n u + (ρw) dw A, then u 0. (This includes the case that u is bounded above.) Proof. By mean value theorem and continuity at the origin, for each ɛ > 0, we can find a radius ρ such that u(ρw) ɛ for all w S n and hence u(w) 2 dw = u(ρw)u(ρ w) = [u(ρw) + ɛ]u(ρ w) dw [u(ρw) + ɛ]u + (ρ w) dw S n S n S n S n 2ɛ u + (ρ w) dw 2ɛµ. S n 6

7 So u = 0 on S n and hence on B (0) by weak maximum and minimum principle. Therefore, u 0 by the exercise we just proved. Remark 6. The usual way to prove the uniqueness theorem for analytic functions are through power series expansion to show the closed subset K = {x Ω : D α f(x) = 0, α} of domain Ω is open. we also show the following theorem as an application of uniqueness theorem which distinguishes how different the real analyticity and complex holomorphy are. Theorem 7 (Open Mapping Theorem). Let Ω be a domain in C n and f : Ω C is holomorphic and non-constant, then for every open subset V of Ω, f(v ) is an open subset of C. Note that this is not true for g(x) = x 2 defined on R since g((, )) = [0, ). Proof. We may assume n > by assuming the one-dimensional open mapping theorem in complex analysis is known. Given open subset V and a V. By uniqueness theorem, there is some b V such that f(a) f(b). Let D = {s C : a + s(b a) V }, D is open in C since V is open. Let g(s) = f(a + s(b a)), then g is holomorphic on D and non-constant since g(0) = f(a) f(b) = g(). Then g(d) is an open set containing g(0) = f(a). Therefore f(v ) is open. Note that this Open Mapping Theorem implies the strong maximum principle easily. Acknowledgement The author would like to thank Jeaheang Bang for his/her reminder that the original proofs for problem 3 and 6 are wrong. ( ) References [] William Gustin. A bilinear integral identity for harmonic functions. American Journal of Mathematics, 70():22 220, 948. [2] Gary M Lieberman. Second Order Parabolic Differential Equations. World scientific, revised edition, [3] Thomas Lorenz. Mutational analysis: a joint framework for cauchy problems in and beyond vector spaces. Springer,