Fundamental Solution and Newtonian Potential 1. The Fundamental Solution. • We shall derive deterministic formulas involving various types of potentials, constructed using a special function, called the fundamental solution of the Laplace operator. • As we did for the diffusion equation, let us look at the invariance properties characterizing the operator ∆: the invariance by translations and rotations. (i) Invariance by translations means that the function v(x)=u(x − y), for each fixed y, is also harmonic, as is immediate to check. (ii) Invariance by rotations means that, given a rotation in Rn, represented by an orthogonal matrix M (MT = M−1), also v(x)=u(Mx) is harmonic in Rn. – To check it, observe that, if we denote by D2u the Hessian of u, we have ∆u =TrD2u = trace of the Hessian of u. Since D2v(x)=MT D2u(Mx)M and M is orthogonal, we have ∆v(x)=Tr[MT D2u(Mx)M]=TrD2u(Mx)=∆u(Mx)=0. and therefore v is harmonic. • Now, a typical rotation invariant quantity is the distance function from a point, for instance from the origin, that is, r = |x|. – Thus, let us look for radially symmetric harmonic functions u = u(r). (I) Consider first n = 2; using polar coordinates, we find ∂2u 1 ∂u + =0 ∂r2 r ∂r so that u(r)=C log r + C1. (II) In dimension n = 3, using spherical coordinates (r,ψ,θ), r>0, 0 <ψ<π, 0 <θ<2π, the operator ∆ has the following expression: ∂2 2 ∂ 1 1 ∂2 ∂2 ∂ ∆= + + + + cot θ . ∂r2 r ∂r r2 (sin ψ)2 ∂θ2 ∂ψ2 ∂ψ | radial{z part } spherical| part (Laplace-Beltrami{z operator)} The Laplace equation for u = u(r) becomes ∂2u 2 ∂u + =0 ∂r2 r ∂r Typeset by AMS-TEX 1 2 whose general integral is C u(r)= 1 + C ,C,C : constants. r 2 1 2 (III) In dimension n>3, the Laplace equation for u = u(r) becomes ∂2u n − 1 ∂u + =0, ∂r2 r ∂r whose general integral is C u(r)= 1 + C ,C,C : constants. rn−2−ξ 2 1 2 1 1 • Choose C2 = 0 and C1 = if n ≥ 3, C1 = − if n = 2. The function (n−2)ωn 2π − 1 | | 2π log x , if n =2, 1 Φ(x)= 4π|x| , if n =3, 1 1 n−2 , if n>3. (n−2)ωn |x| is called the fundamental solution for the Laplace operator ∆. – The above choice of the constant C is made in order to have ∆Φ(x)=−δ(x) where δ(x) denotes the Dirac measure at x; in other words, Z ∞ n ∆Φ(x)v(x)dx = −v(0), ∀v ∈ C0 (R ). Rn • Clearly, if the origin is replaced by a point y, the corresponding potential is Φ(x − y) and ∆xΦ(x − y)=−δ(x − y). By symmetry, we also have ∆yΦ(x − y)=−δ(x − y). • The physical meaning of Φ is remarkable: if n = 3, in standard units, a single 3 unit charge placed at a point ξ ∈ R is known to induce an electric field Eξ(x)= x−ξ 1 − |x−ξ|3 , and thus uξ(x)= |x−ξ| =4πΦ(x ξ) is a potential function for E, (i.e. E = −∇u), which represents the electrostatic potential that a unit charge of the same sign would have if brought from infinity to the point x. 3 The Newtonian Potential • Suppose that f(x) is the density of a charge located inside a compact set in R3. Then Φ(x − y)f(y)dy represents the potential at x due to the charge f(y)dy inside a small region of volume dy around y. – The full potential is given by the sum of all the contributions; we obtain 1 f(y) ( R n n−2 dy,n≥ 3 (1) u(x)=Z Φ(x − y)f(y)dy = (n−2)ωn R |x−y| − Rn 1 | − | 2π RR2 f(y) log x y dy,n=2, which is the convolution between f and Φ and is called the Newtonian poten- tial of f if n ≥ 3, and is called the logarithmic potential if n =2. – Formally, we have Z Z (2) ∆u(x)= ∆xΦ(x − y)f(y)dy = − δ(x − y)f(y)dy = −f(x). Rn Rn Under suitable hypotheses on f, (2) is indeed true. • Clearly, u is not the only solution of ∆v = −f, since u + c, c constant, is a solution as well. – However, the Newtonian potential is the only solution vanishing at infinity. Theorem 1. Let f ∈ C2(Rn) with compact support. Let u be the Newtonian potential of f, defined by (1). Then u is a solution in Rn of (3) ∆u = −f If n ≥ 3, u is the only solution which belongs to C2(Rn) and vanishes at infinity. Proof. (I) The uniqueness part follows from Liouville’s Theorem. – Let v ∈ C2(R3) be another solution to (3) vanishing at infinity. Then u − v is a bounded harmonic function in all R3 and therefore is constant. Since it vanishes at infinity, it must be zero; thus u = v. (II) To show that (1) belongs to C2(R3) and satisfies (3), observe that we can write (1) in the alternate form 1 f(x−y) ( R n n−2 dy n ≥ 3, u(x)=Z Φ(y)f(x − y)dy = (n−2)ωn R |y| Rn − 1 − | | 2π RR2 f(x y) log( y )dy,n=2. 1 (II.i) Since |y|n−2 is integrable near the origin and f is zero outside a compact set, we can take first and second order derivatives under the integral sign to obtain Z (4) uxj xk (x)= Φ(y)fxj xk (x − y)dy. Rn n Since fxj xk ∈ C(R ), formula (4) shows that also uxixj is continuous and there- fore u ∈ C2(R3). 4 (II.ii) It remains to prove (3). Since ∆xf(x − y)=∆yf(x − y), from (4) we have Z Z ∆u(x)= Φ(y)∆xf(x − y)dy = Φ(y)∆yf(x − y)dy. Rn Rn – We want to integrate by parts, but since Φ has a singularity at y = 0, we have first to isolate the origin, by choosing a small ball Br = Br(0) and writing Z Z (5) ∆u(x)= ···dy + ···dy ≡ Ir + Jr. 3 Br (0) R \Br (0) We have, using spherical coordinates, max |∆f| 1 max |∆f| r max |∆f| 2 n−2 dy = ρdρ= r ,n≥ 3 ( (n−2)ωn RBr (0) |y| n−2 R0 2(n−2) |Ir|≤ − max |∆f| log(|y|) dy = (max |∆f|)(r| log r|),n=2, 2π RBr (0) so that Ir → 0, if r → 0. – Keeping in mind that f vanishies outside a compact set, we can integrate Jr by parts (twice), obtaining Z Z Jr = Φ(σ)∇σ f(x − σ) · νσ dσ − ∇Φ(y) ·∇yf(x − y) dy n ∂Br (0) R \Br (0) Z Z = Φ(σ)∂νσ f(x − σ) dσ − f(x − σ)∂νσ Φ dσ, (∵ ∆Φ =0) R3\B (0) ∂Br (0) ∂Br (0) r =Lr + Kr. We have −1 Z (n − 2) r max |∇f|→0,n≥ 3 |Lr|≤max |∇f| Φ dσ ≤ r| log r| max |∇f| n =2 ∂Br Hence Lr → 0asr → 0. 1 y On the other hand, ∇Φ(y)=− n and the outward pointing unit normal nωn |y| − y on ∂Br is νy = r , so that 1 Z − ∇ · Z − → → Kr = f(x σ) Φ(σ) νσ dσ = n−1 f(x σ)dσ f(x), as r 0. ∂Br(0) ωnr ∂Br (0) Thus Jr →−f(x)asr → 0. – Passing to the limit as r → 0 in (5), we get ∆u(x)=−f(x). Remark. Theorem 1 actually holds under much less restrictive hypotheses on f. For instance, it is enough that f ∈ C1(Rn) and |f(x)|≤C|x|n−ε, ε>0. Remark. The logarithmic potential does not vanish at infinity; its asymptotic behavior is M 1 (6) u(x)=− log |x| + O as |x|→∞, 2π |x| where M = Z f(y)dy. R2 The logarithmic potential is the only solution of ∆u = −f in R2 satisfying (6)..