Elliptic PDEs of 2nd Order, Gilbarg and Trudinger

Elliptic PDEs of 2nd Order, Gilbarg and Trudinger Chapter 2 Laplace Equation Yung-Hsiang Huang 207.07.07. Mimic the proof for Theorem 3.. 2. 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. Email: d042200@ntu.edu.tw

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

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 2.. 0. 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 2.6.. 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

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

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 0. 5. 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

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

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. (207.07.08) 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, 2005. [3] Thomas Lorenz. Mutational analysis: a joint framework for cauchy problems in and beyond vector spaces. Springer, 200. 7