MATH 8.52 COUSE NOTES - CLASS MEETING # 9 8.52 Introduction to PDEs, Fall 20 Professor: Jared Speck Class Meeting # 9: Poisson s Formula, Harnack s Inequality, and Liouville s Theorem. epresentation Formula for Solutions to Poisson s Equation We now derive our main representation formula for solution s to Poisson s equation on a domain Ω. Theorem. (epresentation formula for solutions to the boundary value Poisson equation). Let Ω be a domain with a smooth boundary, and assume that f C 2 (Ω) and g C( Ω). Then the unique solution u C 2 (Ω) C(Ω) to (.0.) u(x) = f(x), x Ω n, u(x) = g(x), x Ω. can be represented as (.0.2) where G(x, y) is the Green function for Ω. u(x) = f(y)g(x, y) d n y + Ω g(σ) ˆN(σ) G(x, σ) dσ, Ω Poisson kernel Proof. Applying the epresentation formula for u Proposition, we have that (.0.3) u(x) = Ω Φ(x y)f(y) d n y Ω Φ(x σ) ˆN(σ) u(σ) dσ + Ω g(σ) ˆN(σ) Φ(x σ) dσ. ecall also that (.0.4) G(x, y) = Φ(x y) φ(x, y), where (.0.5) y φ(x, y) = 0, x Ω, and (.0.6) G(x, σ) = 0 when x Ω and σ Ω. The expression (.0.3) is not very useful since don t know the value of ˆN(σ) u(σ) along Ω. To fix this, we will use Green s identity. Applying Green s identity to the functions u(y) and φ(x, y), and recalling that y φ(x, y) = 0 for each fixed x Ω, we have that
2 MATH 8.52 COUSE NOTES - CLASS MEETING # 9 (.0.7) u(y) Φ(x σ) u(σ) 0 = φ(x, y) f(y) d n y Ω φ(x, σ) ˆNu(σ) dσ + Ω g(σ) ˆNφ(x, σ) dσ. Ω Subtracting (.0.7) from (.0.3), and using (.0.4), we deduce the formula (.0.2). 2. Poisson s Formula Let s compute the Green function G(x, y) and Poisson kernel P (x, σ) def = ˆNG(x, σ) from (.0.2) in the case that Ω def = B (0) 3 is a ball of radius centered at the origin. We ll use a technique called the method of images that works for special domains. Warning 2.0.. Brace yourself for a bunch of tedious computations that at the end of the day will lead to a very nice expression. The basic idea is to hope that φ(x, y) from the decomposition G(x, y) = Φ(x y) φ(x, y), where φ(x, y) is viewed as a function of x that depends on the parameter y, is equal to the Newtonian potential generated by some imaginary charge q placed at a point x B c (0). To ensure that G(x, σ) = 0 when σ B (0), q and x have to be chosen so that along the boundary {y 3 y = }, φ(x, y) = 4π x y. In a nutshell, we guess that (2.0.8) G(x, y) = 4π x y + q, 4π x y φ(x,y)? and we try to solve for q and x so that G(x, y) vanishes when y =. q emark 2.0.. Note that y 4π x y = 0, which is one of the conditions necessary for constructing G(x, y). By the definition of G(x, y), we must have G(x, y) = 0 when y =, which implies that (2.0.9) Simple algebra then leads to 4π x y = q 4π x y. (2.0.0) When y =, we use (2.0.0) to compute that (2.0.) x y 2 = q 2 x y 2. x 2 2x y + 2 = x y 2 = q 2 x y 2 = q 2 ( x 2 2x y + 2 ), where denotes the Euclidean dot product. Then performing simple algebra, it follows from (2.0.) that (2.0.2) x 2 + 2 q 2 ( 2 + x 2 ) = 2y (x q 2 x). Now since the left-hand side of (2.0.2) does not depend on y, it must be the case that the right-hand side is always 0. This implies that x = q 2 x, and also leads to the equation
MATH 8.52 COUSE NOTES - CLASS MEETING # 9 3 (2.0.3) Solving (2.0.3) for q, we finally have that q 4 x 2 q 2 ( 2 + x 2 ) + 2 = 0. (2.0.4) (2.0.5) Therefore, q = x, x = 2 x 2 x. (2.0.6) (2.0.7) φ(x, y) = 4π φ(0, y) = 4π, x 2 x x y, 2 where we took a limit as x 0 in (2.0.6) to derive (2.0.7). Next, using (2.0.8), we have (2.0.8) (2.0.9) G(x, y) = 4π x y + 4π G(0, y) = 4π y + 4π. For future use, we also compute that (2.0.20) x 2 x x y, x 0, 2 y G(x, y) = x y 4π x y + x y 3 4π x x y. 3 Now when σ B (0), (2.0.0) and (2.0.4) imply that (2.0.2) x σ = x σ. x Therefore, using (2.0.20) and (2.0.2), we compute that (2.0.22) σ G(x, σ) = x σ 4π x σ + x 2 3 4π 2 σ x 2 = ( 4π x σ 3 ). 2 Using (2.0.22) and the fact that ˆN(σ) = σ, we deduce x σ x σ = x σ 3 4π x σ + 3 4π x 2 2 2 x 2 x σ x σ 3 (2.0.23) ˆN(σ) G(x, σ) def = σ G(x, σ) ˆN(σ) 2 x 2 = 4π x σ. 3
4 MATH 8.52 COUSE NOTES - CLASS MEETING # 9 emark 2.0.2. If the ball were centered at the point p 3 instead of the origin, then the formula (2.0.23) would be replaced with (2.0.24) ˆN(σ) G(x, σ) def = σ G(x, σ) ˆN(σ) 2 x p 2 = 4π x σ. 3 Let s summarize this by stating a lemma. Lemma 2.0.. The Green function for a ball B (p) 3 is (2.0.25a) (2.0.25b) G(x, y) = 4π x y + 4π G(p, y) = 4π y p + 4π. Furthermore, if x B (p) and σ B (p), then (2.0.25c) x p 2 x p (x p) (y p), x p, 2 ˆN(σ) G(x, σ) = 2 x p 2 4π x σ 3. We can now easily derive a representation formula for solutions to the Laplace equation on a ball. Theorem 2. (Poisson s formula). Let B (p) 3 be a ball of radius centered at p = (p, p 2, p 3 ), and let x = (x, x 2, x 3 ) denote a point in 3. Let g C( B (p)). Then the unique solution u C 2 (B (p)) C(B (p)) of the PDE (2.0.26) can be represented using the Poisson formula: (2.0.27) { u(x) = 0, x B (p), u(x) = g(x), x B (p), u(x) = 2 x p 2 4π B (p) g(σ) x σ 3 dσ. emark 2.0.3. In n dimensions, the formula (2.0.27) gets replaced with (2.0.28) u(x) = 2 x p 2 ω n B (p) where as usual, ω n is the surface area of the unit ball in n. g(σ) x σ n dσ, Proof. The identity (2.0.27) follows immediately from Theorem. and Lemma 2.0.. 3. Harnack s inequality We will now use some of our tools to prove a famous inequality for Harmonic functions. The theorem provides some estimates that place limitations on how slow/fast harmonic functions are allowed to grow.
MATH 8.52 COUSE NOTES - CLASS MEETING # 9 5 Theorem 3. (Harnack s inequality). Let B (0) n be the ball of radius centered at the origin, and let u C 2 (B (0)) C(B (0)) be the unique solution to (2.0.26). Assume that u is non-negative on B (0). Then for any x B (0), we have that (3.0.29) n 2 ( x ) ( + x ) u(0) u(x) n 2 ( + x ) u(0). n ( x ) n Proof. We ll do the proof for n = 3. The basic idea is to combine the Poisson representation formula with simple inequalities and the mean value property. By Theorem 2., we have that (3.0.30) u(x) = 2 x 2 4π g(σ) B (0) x σ dσ. 3 By the triangle inequality, for σ B (0) (i.e. σ = ), we have that x x σ x +. Applying the first inequality to (3.0.30), and using the non-negativity of g, we deduce that (3.0.3) u(x) + x 2 x 2 4π g(σ) dσ. B (0) Now recall that by the mean value property, we have that (3.0.32) u(0) = Thus, combining (3.0.3) and (3.0.32), we have that (3.0.33) u(x) 4π 2 g(σ) dσ. B (0) ( + x ) ( x ) 2 u(0), which implies one of the inequalities in (3.0.29). The other one can be proved similarly using the remaining triangle inequality. We now prove a famous consequence of Harnack s inequality. The statement is also often proved in introductory courses in complex analysis, and it plays a central role in some proofs of the fundamental theorem of algebra. Corollary 3.0.2 (Liouville s theorem). Suppose that u C 2 ( n ) is harmonic on n. Assume that there exists a constant M such that u(x) M for all x n, or such that u(x) M for all x n. Then u is a constant-valued function. Proof. We first consider the case that u(x) M. Let v def = u + M. Observe that v 0 is harmonic and verifies the hypotheses of Theorem 3.. Thus, by (3.0.29), if x n and is sufficiently large, we have that (3.0.34) n 2 ( x ) ( + x ) v(0) v(x) n 2 ( + x ) v(0). n ( x ) n Allowing in (3.0.34), we conclude that v(x) = v(0). Thus, v is a constant-valued function (and therefore u is too).
6 MATH 8.52 COUSE NOTES - CLASS MEETING # 9 To handle the case u(x) M, we simply consider the function w(x) def = u(x) + M in place of v(x), and we argue as above.