SOLUTION OF POISSON S EQUATION. Contents

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1 SOLUTION OF POISSON S EQUATION CRISTIAN E. GUTIÉRREZ OCTOBER 5, 2013 Contents 1. Differentiation under the integral sign 1 2. The Newtonian potential is C The Newtonian potential from the 3rd Green formula 4 4. Second derivatives 5 5. Solution in all space 9 6. The Newtonian potential is not necessarily 2nd order differentiable Helmholtz decomposition 12 References Differentiation under the integral sign A pretty general result that is often useful is the following. Theorem 1. Let kx, y) be a function defined for all x O R m open and for a.e. y such that kx, ) L 1 ) for all x O. Then a) The function ux) = kx, y) dy is well defined for all x O. b) Fix 1 i m. Assume that xi kx, y) exists for all x O and for almost all y, and there exists g L 1 ) such that xi kx, y) gy), x O, and for almost all y. Assume further that xi k, y) CO) for almost all y. 1

2 2 C. E. GUTIÉRREZ OCTOBER 5, 2013 Then the derivative xi u exists for all x O and xi ux) = xi kx, y) dy, for all x O. c) If the gradient D x k, y) CO) and there exists g L 1 ) such that D x kx, y) gy), x O, and for almost all y, then D x ux) = D x kx, y) dy, for all x O. d) If xj xi k, y) CO) for a.e. y, and there exists g L 1 ) such that then xj xi kx, y) gy), x O, and for almost all y, xj xi ux) = xj xi kx, y) dy. Proof. It follows immediately writing down the differential increment, using the mean value theorem, the continuity assumption and the Lebesgue dominated convergence theorem. 2. The Newtonian potential is C 1 Theorem 2. If ρ L 1 ) L ), then wx) = ρy)γx y) dy is well defined at each x and w C 1 ). Proof. Let ηt) be a smooth function such that ηt) = 0 for t 1, 0 η 1 everywhere, and ηt) = 1 for t 2. Let n 3, Γx) = 1 for n = 2, Γx) = log x ) x n 2 and consider the function Γ ɛ x) = Γx) η x /ɛ), where ɛ > 0. Step 1 If ρ L 1 ), then the function w ɛ x) = ρy)γ ɛ x y) dy is well defined for all x. This follows inmediately since Γ ɛ is bounded.

3 SOLUTION OF POISSON S EQUATION October 5, Step 2 If ρ L 1 ), then w ɛ C 1 ). First prove using the previous theorem that w ɛ has derivative at each point taking kx, y) = ρy)γ ɛ x y). Notice that the function Fx, y) = D x Γx y)η x y /ɛ )) Cɛ) for ɛ x y 2ɛ and Fx, y) = 0 otherwise. Therefore D x kx, y) Cɛ)ρy) for all x, y. Second we show that xi wx) is continuous in x for each ɛ fixed, i = 1,, n. Write j w ɛ x) j w ɛ z) = ρy) xj Γ ɛ x y) xj Γ ɛ z y) ) dy R n = ρy) D ) ) xj Γ ɛ tx + 1 t)z y) x z) dy. Differentiating we obtain as before that D ) xj Γ ɛ x) Cɛ) for all x, and therefore j w ɛ x) j w ɛ z) Cɛ) ρy) dy x z and the continuity of j w ɛ follows. Step 3 If ρ L 1 ) L ), then w ɛ x) wx) = ρy)γx y) dy uniformly for x, where wx) is well defined for all x. Clearly, wx) is well defined since Γ is integrable near the origin, Γ is bounded away from the origin, ρ is bounded and integrable. Write wx) w ɛ x) = ρy) Γx y) 1 η x y /ɛ) ) dy. So wx) w ɛ x) ρy) Γx y) dy C ρ ɛ 2. Step 4 If ρ L 1 ) L ), then Dw ɛ v uniformly in, where vx) = ρy)d x Γx y)) dy = v 1 x),, v n x)),

4 4 C. E. GUTIÉRREZ OCTOBER 5, 2013 where the last integral is well defined. It is clear that v is well defined. Write j w ɛ x) v j x) = ρy) xj Γ ɛ x y) xj Γx y) ) dy R n = ρy) ) xj Γx y)η x y /ɛ) + Γx y) xj η x y /ɛ) xj Γx y) ) dy R n = ρy) xj Γx y) η x y /ɛ) 1 ) ) + Γx y) ) xj η x y /ɛ) dy. So j w ɛ x) v j x) ρy) C n ρ xj Γx y) + Γx y) 1 ) ɛ η x y /ɛ) dy ) 1 x y dy = C ρ n 1 x y n 2 ɛ. ɛ Step 5 As a consequence of steps 2-4, the functions w and v are continous when ρ L 1 ) L ). Step 6 If ρ L 1 ) L ), then w C 1 ), and Dwx) = vx). By the fundamental theorem of calculus, w ɛ x 1, x 2,, x n ) = x1 and from the uniform convergence y 1 x1 w ɛ t, x 2,, x n ) dt + w ɛ y 1, x 2,, x n ), wx 1, x 2,, x n ) = x1 y 1 v 1 t, x 2,, x n ) dt + wy 1, x 2,, x n ). 3. The Newtonian potential from the 3rd Green formula 1 Let n 3 and Γx) = ω n 1 2 n) x 2 n, where ω n 1 is the surface area of the unit sphere in. The third Green formula reads Γx y) uy) = ux) Γx y) u ) ηx) η x) dσx) + Γx y) ux) dx, where u C 2 ) C 1 ), and u L 1 ) and the domain is sufficiently regular. Suppose u C 2 ) solves the Poisson equation u = ρ

5 SOLUTION OF POISSON S EQUATION October 5, in all space, with ρ L 1 ) L ). Applying the third Green formula in the ball B R y) yields Γx y) uy) = ux) Γx y) u ) B R y) ηx) η x) dσx) + Γx y) ρx) dx. B R y) Suppose in addition that ux) = O1/ x ), Dux) = O1/ x 2 ), when x. Then, when R, the surface integral tends to zero and we obtain uy) = Γx y) ρx) dx, for each y. 4. Second derivatives We assume f : R is locally Hölder continuous, that is, there exists 0 < α 1 such that for each K compact there is a constant C K > 0 such that f x) f y) C K x y α, x, y K. We also assume f is bounded in. The goal is to prove the representation formula 1) D ij wx) = D ij Γ)x y) f y) f x)) dy f x) 0 D i Γx y)ν j y) dσy), 0 for the second derivatives of the Newtonian potential, where 0 is any domain for which the divergence theorem holds, with 0 ; and f x) = f x) for x and f x) = 0 in 0 \. Assume n 3, let Γx) = C n x 2 n be the fundamental solution, η : R R is smooth, ηt) = 0 for t 1, 0 η 1, ηt) = 1 for t 2 and 0 η 2. Define for x ux) = D ij Γ)x y) f y) f x)) dy f x) D i Γx y)ν j y) dσy). 0 0 Since D ij Γx) C n x n, and f is bounded and locally Hölder continuous, then the function ux) is well defined for all x. Let ɛ > 0 and define v ɛ x) = D i Γx y)ηx y)/ɛ) f y) dy,

6 6 C. E. GUTIÉRREZ OCTOBER 5, 2013 and let vx) = D i wx), where wx) = Γx y) f y) dy. If f is integrable in, then, as in Theorem 2, v ɛ C 1 ), and if in addition f is bounded, w C 1 ). We have D j v ɛ x) = D xj Di Γx y)ηx y)/ɛ) ) f y) dy = D xj Di Γx y)ηx y)/ɛ) ) f y) dy 0 = D xj Di Γx y)ηx y)/ɛ) ) f y) f x)) dy 0 + f x) D xj Di Γx y)ηx y)/ɛ) ) dy 0 = D xj Di Γx y)ηx y)/ɛ) ) f y) f x)) dy 0 f x) D yj Di Γx y)ηx y)/ɛ) ) dy 0 = D xj Di Γx y)ηx y)/ɛ) ) f y) f x)) dy 0 f x) D i Γx y)ηx y)/ɛ)ν j y) dσy) 0 from the divergence theorem. Since x, if we take ɛ distx, 0 )/2, then x y 2ɛ for y 0 and so ηx y)/ɛ) = 1. So D j v ɛ x) = D xj Di Γx y)ηx y)/ɛ) ) f y) f x)) dy 0 f x) D i Γx y)ν j y) dσy). 0

7 SOLUTION OF POISSON S EQUATION October 5, Then subtracting we get ux) D j v ɛ x) = Dij Γx y) D xj Di Γx y)ηx y)/ɛ) )) f y) f x)) dy 0 [ ) = D xj 1 ηx y)/ɛ) Di Γx y) ] f y) f x)) dy 0 [ ) = D xj 1 ηx y)/ɛ) Di Γx y) ] f y) f x)) dy [ = D ij Γx y) 1 ηx y)/ɛ) ) D i Γx y) 1 ɛ η x y /ɛ) x ] j y j f x y y) f x)) dy. Let K be compact, and K = {y : disty, K) distk, )/2}. We have K is compact, K, and if ɛ < distk, )/2, then B ɛ x) K for all x K. Therefore estimating the last integral we obtain for x K [ DijΓx ux) D j v ɛ x) y) Di + Γx y) ] C f y) f x) dy ɛ C K [ C x y + C C n x y n 1 ɛ ] x y α dy C K ɛ α. Therefore D j v ɛ u uniformly on compact subsets of as ɛ 0, so u is continuous in. In addition, v ɛ v uniformly in this follows directly by subtracting). This implies that w C 2 ) and ux) = D ij wx) for x by the fundamental theorem of calculus, because we write v ɛ x 1,, x j 1, y j, x j+1,, x n ) = yj x j D j v ɛ x 1,, x j 1, t, x j+1,, x n ) dt + v ɛ x), and pass to the limit as ɛ 0. So we have proved the following representation formula for the second derivatives of the Newtonian potential w for x : 2) D ij wx) = D ij Γ)x y) f y) f x)) dy f x) D i Γx y)ν j y) dσy), 0 0 under the assumption that f is bounded in and locally Hölder continuous for some 0 < α 1. We next prove that w = f in. Let x, take R sufficiently large such that B R x), and apply 2) with 0 = B R x), then D ii wx) = D ij Γ)x y) f y) f x)) dy f x) x y <R D i Γx y)ν j y) dσy), x y =R

8 8 C. E. GUTIÉRREZ OCTOBER 5, 2013 adding over 1 i n and using that Γx) = 0 for x 0, we obtain wx) = f x) DΓx y) νy) dσy). x y =R We have DΓx) = C n 2 n) x y x and νy) = x n x y. Then wx) = f x)c n2 n)ω n, 1, we are 2 n)ω n where ω n is the surface area of the unit sphere in. Since C n = done. We also have that wx) = 0 for x, which follows differentiating twice under the integral sign the justification follows from Theorem 1). So we have proved the following theorem. Theorem 3. Let be a bounded open set, ρ be a bounded function in. Then 1) w C 1 ); 2) w C 2 \ ); 3) w = 0 in \ ; 4) if in addition ρ is Hölder continuous, then w C 2 ) and w = ρ in. 5) the second derivatives of w might not exist in. Actually, the last item was not proved and we leave it as an exercise. For that consider in R 3 the ball B0, R). Assume that the ball is homogeneous with density λ. The Newtonian potential due to this ball is given by 1 Ux) = G λ B0,R) y x dy. a) Show that Gλ 4 1 Ux) = 3 πr3 x, for x > R Gλ 2 3 π3r2 x 2 ), for 0 x R. HINT: to simplify the calculation notice that UOx) = Ux) for any rotation O around the origin, that is, U is a radial function. b) Prove that the first order derivatives of U exist and are continuous everywhere. c) Prove that all the second derivatives of U exist and are continuous in R 3 \ {x : x = R}. Prove that they do not exist on x = R.

9 SOLUTION OF POISSON S EQUATION October 5, d) Prove that Gλ4π, for 0 x < R Ux) = div DUx) = 0, for x > R. 5. Solution in all space Let wx) = Γx y)ρy) dy. We shall prove that under some conditions on ρ, we have that w is C 2 everywhere and wx) = ρx) for all x. Fix R > 0 and let φ C ) be such that φy) = 1 for y 2R, φy) = 0 for y 3R and 0 φ 1. Write wx) = Γx y)ρy)φy) dy + Γx y)ρy)1 φy)) dy We have := w 1 x) + w 2 x). w 1 x) = Γx y)ρy)φy) dy B 3R 0) and from what we proved before we have that if ρ is locally Hölder continuous in, then w 1 C 2 B 3R 0)) and w 1 x) = ρx)φx) for all x B 3R 0). Therefore w 1 x) = ρx) for x B 2R 0). We shall prove that w 2 x) = 0 for all x B R 0). If ρ L 1 ) L ), then from Theorem 2 we have that w 2 C 1 ) and 3) Dw 2 x) = D x Γx y)ρy)1 φy)) dy. Let x 0 B R 0), then B R/2 x 0 ) B 3R/2 0). Notice that the integrand in w 2 vanishes for y 2R, and therefore the domain of integration reduces to the set y 2R. We want to differentiate under the integral sign in 3), and for that we need to verify the hypotheses in Theorem 1. Indeed, we have that 2 C x j x j Γx y) x y, n for x y, and since x y R/2 for x B R x 0 ) and y 2R xj x j Γx y) 2n C there. So if we let gy) = 2n C ρy)1 φy)), then we obtain that Rn hx, y) := xj x j Γx y)ρy)1 φy)) gy)

10 10 C. E. GUTIÉRREZ OCTOBER 5, 2013 for all x B R/2 x 0 ) and for all y. In addition, h, y) CB R/2 x 0 )) for all y. Therefore, we can differentiate under the integral sign and we obtain xj x j w 2 x) = xj x j Γx y)ρy)1 φy)) dy, for all x B R/2 x 0 ) and so w 2 x) = Γx y)ρy)1 φy)) dy = 0, for all x B R/2 x 0 ), for each x 0 B R 0). Since R is arbitrary, we therefore have proved the following theorem: Theorem 4. If ρ defined in is locally Hölder continuous of order α, for some 0 < α 1, and ρ L 1 ) L ), then the Newtonian potential wx) = Γx y)ρy) dy is C 2 in all space, and satisfies the Poisson equation wx) = ρx) in. 6. The Newtonian potential is not necessarily 2nd order differentiable We show that there exists f continuous in B 2 0) such that wx) = Γx y) f y) dy B 2 0) is not second order differentiable at the origin. The example is from [GT83, Exercise 4.9]. Assume n 3. Let c k be a sequence such that c k 0 and k and c k = +. Let η C 0 B 20)) with ηx) = 1 for x < 1, and let P be a harmonic polynomial of degree two in with D α P 0 for some α = 2, e.g., P = x 1 x 2. Let f x) = c k ηp)2 k x). Given x 0 in B 2 0) there exists a unique integer N 0 such that 2 N x < 2 1 N. Notice that ηp)x) = 0 for x > 2 and for x < 1. We have f x) = = c N ηp)2 N x). {k:2 k x <2 1 k }

11 SOLUTION OF POISSON S EQUATION October 5, This function is clearly continuous in B 2 0). We write wx) = = Γx y) f y) dy 2 k y <2 1 k c k Γx y) ηp)2 2 k y) dy = k y <2 1 k c k I k x). Changing variables we have I k x) = 2 2k 1 z <2 Γ2 k x z) ηp)z) dz = 2 2k Γ2 k x z) ηp)z) dz and from the 3rd Green formula since ηp has compact support we get as in Section 3 that Γ2 k x z) ηp)z) dz = ηp)2 k x). Therefore wx) = c k 2 2k ηp)2 k x). Given x < 2 there exists a unique N 0 integer such that 2 N x < 2 1 N. So 2 k x 2 k N 2 for k N + 1, and so ηp)2 k x) = 0 for k N + 1. On the other hand, 2 k x < 2 k 2 1 N 1 if k N 1 and hence η2 k x) = 1 for k N 1. Therefore Thus N 1 wx) = c k 2 2k P2 k x) + c N 2 2N ηp)2 N x), for 2 N x < 2 1 N. N 1 D α wx) = c k D α P2 k x) + c N D α ηp)2 N x), α = 2, for 2 N x < 2 1 N. If P = x 1 x 2, then N 1 D 12 wx) = c k + c N D 12 ηp)2 N x), for 2 N x < 2 1 N. If x 0, then N and therefore D 12 wx) +.

12 12 C. E. GUTIÉRREZ OCTOBER 5, Helmholtz decomposition Let F be a vector field in R 3 we shall prove that under appropriate hypotheses F = R + S with div S = 0 and curl R = 0. This means that F = Du + curl H for some function u and some field H. Recall that if G is a vector field with div G = 0 in R 3, then there exists a field H such that G = curl H. Also, if G is a field with curl G = 0, then there exists a function u such that G = Du. In addition, if G = curl H 1 = curl H 2, then H 1 H 2 = Dφ for some function φ. Assuming the desired decomposition, we have div F = div S + div R = div R. Since curl R = 0, there exists u with R = Du, and so div F = div Du = u. Therefore, if div F satisfies the hypotheses of Theorem 4 and let u be the Newtonian potential of div F, then u = div F. So if we let R = Du, and S = F R, then curl R = 0, and div S = div F div R = div F div Du = 0. So we have: Theorem 5 Helmholtz decomposition). Let F : R 3 R 3 be a field such that div F is locally Hölder continuous in R 3, uniformly bounded and integrable over all space. If wx) = Γx y) div Fy) dy, R 3 then Fx) = Dwx) + curl Hx) for some field H. The field H is determined up to the gradient of an arbitrary function φ. If H is a field we define the Laplacian of F by H = Ddiv H) curl curl H). If H = a 1 x)i + a 2 x)j + a 3 x)k, then is easy to see calculating the components of H that H = a 1 x)i + a 2 x)j + a 3 x)k. If we assume div H = 0, then H = curl curl H) = curl F by the Helmholtz decomposition. But the equation H = curl F is a Poisson equation now between

13 SOLUTION OF POISSON S EQUATION October 5, vectors, that is, three scalar Poisson equations a i = curl F) i, i = 1, 2, 3. Solving each one as before with the Newtonian potential we get that Hx) = Γx y) curl Fy) dy, R 3 that is, H is the Newtonian potential now a vector) of curl F. This means that if for a given field F we know div F and curl F, then F is automatically determined by the formula ) ) Fx) = D Γx y) div Fy) dy + curl Γx y) curl Fy) dy. R 3 R 3 References [GT83] D. Gilbarg and N. S. Trudinger. Elliptic Partial Differential Equations of Second Order. Springer-Verlag, New York, 2nd edition, Department of Mathematics, Temple University, Philadelphia, PA address: gutierre@temple.edu