Class Meeting # 9: Poisson s Formula, Harnack s Inequality, and Liouville s Theorem

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1 MATH 85 COUSE NOTES - CLASS MEETING # 9 85 Introduction to PDEs, Fall 0 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 ( Ω) and g C( Ω ) Then the unique solution u C (Ω) C( Ω) to (0) u( x) = f( x ), x Ω n, can be represented as u( x) = g( x ), x Ω (0) u(x) = f(y)g(x, y Ω where G(x, y) is the Green function for Ω ) d n y + g( ˆN( G(x, dσ, Ω Poisson kernel Proof Applying the epresentation formula for u Proposition, we have that (0) u(x) = Φ(x y)f(y) d n y Φ(x ˆN( u( dσ + g( ˆN ( Φ( x dσ ecall also that Ω (04) G(x, y) = Φ(x y) φ(x, y), where (05) y φ(x, y) = 0, x Ω, and (06) G( x, = 0 when x Ω and σ Ω Ω The expression (0) is not very useful since don t know the value of ˆN ( u( 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 Ω

2 MATH 85 COUSE NOTES - CLASS MEETING # 9 u(y) Φ(x u( (07) 0 = φ(x, y) f(y) d n y φ(x, ˆNu( dσ + g( ˆNφ(x, dσ Ω Ω Ω Subtracting (07) from (0), and using (04), we deduce the formula (0) Poisson s Formula Let s compute the Green function G( x, y) and Poisson kernel P ( x, = ˆNG(x, from (0) in the case that Ω = B ( 0) 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 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, = 0 when σ B ( 0 ), q and x have to be chosen so that along the boundary { y y = }, φ( x, y) = In a nutshell, we guess that (08) 4π x y G(x, y) = 4π x y + q 4 π x y, φ( x,y)? and we try to solve for q and x so that G( x, y) vanishes when y = q emark 0 Note that y 4π x y = 0, which is one of the conditions necessary for constructing G( x, y) By the inition of G(x, y), we must have G(x, y) = 0 when y =, which implies that (09) Simple algebra then leads to 4π x y = q 4π x y (00) x y = q x y When y =, we use (00) to compute that (0) x x y + = x y = q x y = q ( x x y + ), where denotes the Euclidean dot product Then performing simple algebra, it follows from (0) that (0) x + q ( + x ) = y ( x q x ) Now since the left-hand side of (0) does not depend on y, it must be the case that the right-hand side is always 0 This implies that x = q x, and also leads to the equation

3 MATH 85 COUSE NOTES - CLASS MEETING # 9 (0) q 4 x q ( + x ) + = 0 Solving (0) for q, we finally have that (04) q = x, (05) Therefore, x = x x (06) φ(x, y) = 4π x x x y (07) φ( 0, y) =,, where we took a limit as x 0 in (06) to derive (07) Next, using (08), we have (08) G(x, y) = 4π x y +, 4π x x x y x 0, (09) G( 0, y ) = + 4π y For future use, we also compute that (00) y G(x, y) = x y 4π x y Now when σ B ( 0 ), (00) and (04) imply that x y + 4π x x y (0) x σ = x σ x Therefore, using (00) and (0), we compute that ( ) = x σ x x σ x σ x x x σ (0) σg x, σ + = 4π x σ 4π x σ 4π x σ + 4π x σ σ x = ( ) 4π x σ Using (0) and the fact that Nˆ ( σ ) = σ, w e deduce x (0) ˆ ˆN ( G( x, = σg( x, N( = x σ

4 4 MATH 85 COUSE NOTES - CLASS MEETING # 9 emark 0 If the ball were centered at the point p instead of the origin, then the formula (0) would be replaced with ˆ x p (04) ˆN( G( x, = σg( x, N( σ ) = x σ Let s summarize this by stating a lemma Lemma 0 The Green function for a ball B ( ) is (05a) (05b) p ( ) = + G x, y, x p, 4π x y 4π x p (x p) (y p) G( p, y ) = 4π y p + Furthermore, if x B (p) and σ B (p), then ( ) = x p (05c) ˆN( G x, σ x σ We can now easily derive a representation formula for solutions to the Laplace equation on a ball Theorem (Poisson s formula) Let B (p) be a ball of radius centered at p = ( p, p, p ), and let x = ( x x ) ( ( )) ( ( )) (,, (x denote a point in Let g C B p Then the unique solution u C B p C B p)) of the PDE { u(x) = 0, x B ( p), (06) u( x) = g( x ), x B (p), can be represented using the Poisson formula: ( ) = x p (07) u x x p B (p) g( dσ x σ emark 0 In n dimensions, the formula (07) gets replaced with x p g( (08) u(x) = ω n B ( p) x σ where as usual, ωn is the surface area of the unit ball in Proof The identity (07) follows immediately from Theorem and Lemma 0 n n dσ, 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

5 MATH 85 COUSE NOTES - CLASS MEETING # 9 5 Theorem (Harnack s inequality) Let B ( 0) n be the ball of radius centered at the origin, and let u C ( B( 0)) C( B ( 0)) be the unique solution to (06) Assume that u is non-negative on B ( 0 ) Then for any x B ( 0 ), we have that (09) n ( x ) + x u( 0) u(x) n ( ) u ( 0 ) ( + x ) n ( x ) n Proof We ll do the proof for n = The basic idea is to combine the Poisson representation formula with simple inequalities and the mean value property By Theorem, we have that (00) ( ) = x u x B (0) g( x σ By the triangle inequality, for σ B ( 0) (ie σ = ), we have that x x σ x + Applying the first inequality to (00), and using the non-negativity of g, we deduce that (0) u(x) + x x B (0) Now recall that by the mean value property, we have that dσ g( dσ (0) u( 0) = g σ dσ ( ) (0) B Thus, combining (0) and (0), we have that ( + x ) (0) u(x) u(0), ( x ) which implies one of the inequalities in (09) 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 0 (Liouville s theorem) Suppose that u C ( 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 = u + M Observe that v 0 is harmonic and verifies the hypotheses of Theorem Thus, by (09), if x n and is sufficiently large, we have that n ( x ) n ( + x ) (04) ( + ) v(0) v(x) v(0 ) x n ( x ) n Allowing in (04), we conclude that v( x) = v( 0) Thus, v is a constant-valued function (and therefore u is too)

6 6 MATH 85 COUSE NOTES - CLASS MEETING # 9 To handle the case u(x) M, we simply consider the function w(x) = u(x) + M in place of v( x ), and we argue as above

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Class Meeting # 9: Poisson s Formula, Harnack s Inequality, and Liouville s Theorem

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