THE GREEN FUNCTION. Contents

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1 THE GREEN FUNCTION CRISTIAN E. GUTIÉRREZ NOVEMBER 5, 203 Contents. Third Green s formula 2. The Green function 2.. Estimates of the Green function near the pole Symmetry of the Green function The Green function for the ball Application Application 2 5 References 6. Third Green s formula Let n 3 and Γ(x) = ω n (2 n) x 2 n, where ω n is the surface area of the unit sphere in R n. The third Green formula reads ( Γ(x y) u(y) = u(x) Γ(x y) u ) η(x) η (x) dσ(x) + Γ(x y) u(x) dx, where u C 2 () C ( ), and u L () and the domain is sufficiently regular. This formula follows from the 2nd Green formula () applied in \ B ɛ (y) and letting ɛ 0. In obtaining the formula the average on the ball B ɛ (y) appears and this is the reason of the value of the constant in the definition of Γ. 2. The Green function Given y, define for x y, φ y (x) = Γ(x y). The function φ y is continuous in. Consider the Dirichlet problem h = 0 in and h = φ y on, and suppose that this problem has a solution h y (x). From the maximum principle this solution is unique.

2 2 C. E. GUTIÉRREZ NOVEMBER 5, 203 The Green function for the domain and with pole at the point y is defined by G(x, y) = h y (x) + Γ(x y). With the aid of G we will represent any solution of the Dirichlet problem u = F in with u = f on. For this we recall the 2nd Green formula: ( () (u(x) v(x) v(x) u(x)) dx = u(x) v ) (x) v(x) u η η (x) dσ(x). Applying this formula with v = h y and since h y (x) = Γ(x y) for x we get ( h y (x) u(x) dx = u(x) h ) y (x) Γ(x y) u η η (x) dσ(x) ( Γ(x y) = u(x) dσ(x) + u(x) (x) Γ(x y) u ) η η η (x) dσ(x) = u(x) η dσ(x) + u(y) Γ(x y) u(x) dx, from the third Green formula. We therefore obtain the following important representation formula valid for y and u C 2 () C ( ), with u L (): (2) u(y) = u(x) dσ(x) + G(x, y) u(x) dx. η(x) 2.. Estimates of the Green function near the pole. Since h y (x) is harmonic in, it follows from the maximum principle that (3) m y = min x h y (x) Γ(x y) G(x, y) M y = max h y (x), for x, x y. x Since G(x, y) = 0 for x, we have M y = max x Γ(x y) and m y = min x Γ(x y). From the definition of Γ, a simple calculation shows that M y = ω n (2 n) ( maxx x y ), m n 2 y = Then we obtain from (3) that We have other hand, M y Γ(x y) m y Γ(x y) = G(x, y) Γ(x y) M y Γ(x y) = m y ω n (2 n). dist(y, ) Γ(x y), for x, x y. x y, for x y dist(y, ). On the dist(y, ) x y for some x 0. Given 0 < α <, x 0 y

3 THE GREEN FUNCTION November 5, x y M y if x y α dist(y, ), then α and we obtain x 0 y Γ(x y) αn 2. Consequently, we get the following estimates for G (since Γ 0): (4) (5) G(x, y) Γ(x y), for 0 < x y < dist(y, ), and G(x, y) ( α n 2 ) Γ(x y), for 0 < x y α dist(y, ). Since G(x, y) = 0 for x, then from (4) and the maximum principle applied to G(, y) Γ( y) in the domain \ B dist(y, ) (y) yields that G(x, y) Γ(x y), for all x, x y. In addition, from (5) we get that G(x, y) 0 in B αdist(y, ) (y) and since G(x, y) = 0 on, then once again by the maximum principle G(x, y) 0, for all x, x y Symmetry of the Green function. We have the following G(x, y) = G(y, x), for all x, y, x y. Pick ɛ > 0 such that B ɛ (x), B ɛ (y) and B ɛ (x) B ɛ (y) =. Applying the second Green formula in the domain \(B ɛ (x) B ɛ (y)) with u(z) = G(z, x) and v(z) = G(z, y), and letting ɛ 0, the symmetry follows The Green function for the ball. Let B R (x 0 ) be the Euclidean ball in R n. Let y B R (x 0 ) and let y on the line joining x 0 and y such that (6) y x 0 y x 0 = R 2. We have y x 0 = λ(y x 0 ) and inserting in (6) yields λ = We claim that (7) y x y x y = x 0 + = R 2 y x 0 2 (y x 0). R y x 0, for all x x 0 = R. R 2 and so y x 0 2 Writing x = x 0 + Rw with w = it follows expanding the inner products and using (6) that y x 2 y x = R 2 2 y x 0 2

4 4 C. E. GUTIÉRREZ NOVEMBER 5, 203 and the claim follows. We then let for 0 < x y R R G(x, y) = Γ(x y) Γ(x y ). y x 0 We have that G(, y) is harmonic in B R (x 0 ) \ {y}, and G(x, y) = 0 for x y = R. We also have by calculation D x G(x, y) = ( ) 2 x y n ω R n y x 0 (x x 0 ) and since the normal η = x x 0 we obtain x x 0 G(x, y) = η ω n x y n x x 0 and for x x 0 = R we get the Poisson kernel (8) ( y x0 R R 2 y x 0 2 G(x, y) = P(x, y) =. η ω n R x y n Since the function one is harmonic in and equals one on, then from the uniqueness of the solution to the Dirichlet problem and (2) we get that dσ(x) =, η(x) in particular, P(x, y) dσ(x) =. x x 0 =R We now solve the Dirichlet problem in the ball B R (x 0 ). Let f C( B R (x 0 )) and let (9) u(y) = f (x) P(x, y) dσ(x), for y x 0 < R, x x 0 =R and u(y) = f (y) for y B R (x 0 ). Since y is in the interior of B R (x 0 ) and x x 0 = R, the function P(x, y) is C in a neighborhood of the point y and all its derivatives are bounded in this neighborhood uniformly for x x 0 = R. We can then infinitely differentiate under the integral sign in (9). By direct calculation, y P(x, y) = 0 for each y R n \ {x} and only when x x 0 = R. We then obtain that u = 0 in B R (x 0 ). It remains to show that u(z) f (y) when z y, z B R (x 0 ) for each y B R (x 0 ). Indeed, we write ( ) u(z) f (y) = f (x) f (y) P(x, z) dσ(x) = + = I+II. x x 0 =R x x 0 =R, x y δ x x 0 =R, x y >δ ) 2,

5 THE GREEN FUNCTION November 5, We have I max x y δ f (x) f (y). If z y < δ/2, then x z x y y z δ δ/2 = δ/2. Hence on the region of integration in II, we get P(x, z) C n (2/δ) n ( R 2 z x 0 2) = C n (2/δ) n (R z x 0 )(R + z x 0 ) C n δ n 2R(R z x 0 ) for z y < δ/2. Then II C δ max BR (x 0 ) f (x) (R z x 0 ) for z y δ/2. If z y, then R z x 0 0 and so II Application. Since the Dirichlet problem is solvable in a ball, we obtain the following. Theorem. If u C() satisfies the mean value property on any ball B, then u is harmonic in. Proof. Fix a ball B and let v be harmonic in B such that v = u on B. Then v satisfies the mean value property on any ball B B and so v u also satisfies the same property. Since the mean value property is enough to prove the strong maximum principle, we obtain that v u = 0 in B and we are done Application 2. A function u C() is subharmonic in if u(y) u(x) dσ(x) B r (y) for each ball B R (y). Also u is superharmonic in if u is subharmonic in. Theorem 2. If u C 2 () and u 0 in, then u is subharmonic in. Conversely, if u C 2 () is subharmonic in, then u 0 in. Proof. Let G(x, y) be the Green function of the ball. From (2) u(x 0 ) = u(x) G(x, x 0) dσ(x) + G(x, x 0 ) u(x) dx. η(x) Also from (8) = u(x) G(x, x 0) η(x) ω n R n dσ(x) = u(x) dσ(x) = u(x)p(x, x 0 ) dσ(x) u(x) dσ(x). Then (0) u(x 0 ) u(x) dσ(x) = G(x, x 0 ) u(x) dx. Since G 0 in, the first part of the theorem follows.

6 6 C. E. GUTIÉRREZ NOVEMBER 5, 203 On the other hand, if u is subharmonic, then the left hand side of (0) is less than or equal to zero and so G(x, x 0 ) u(x) dx 0. Since G(, x 0 ) is integrable in, we then can write u(x 0 ) G(x, x 0 ) dx G(x, x 0 ) ( u(x 0 ) u(x)) dx. Since G 0, we obtain u(x 0 ) and we are done. G(x, x 0) ( u(x 0 ) u(x)) dx G(x, x 0) dx 0, as r 0, References Department of Mathematics, Temple University, Philadelphia, PA 922 address: gutierre@temple.edu