6.3. Fundamental solutions in Rd 151

6.3 Fundamental solutions in Rd

Since Lapalce equation is invariant under translations, and rotations (see Exercise 6.4), welook for solutions to Laplace equation having such symmetric properties.

Let us fix ξ ∈ Rd , and look for solutions to ∆u = 0 having the form vξ (x) = ψ(r ),where

r = ∥x − ξ ∥=√√√√ d∑

i=1

(xi − ξi )2.

Substituting the formula for vξ in the equation, we get

∆vξ (x) =ψ′′(r )+ d − 1

rψ′(r ) = 0.

Solving the last equation, we get

ψ′(r ) =C r 1−d (6.18)

Integrating the last equation we get

ψ(r ) =

C ln r if d = 2,C

2−d r 2−d if d ≥ 3.

Definition 6.9. The fundamental solution of Laplacian operator is defined is defined as thefunction K : (Rd ×Rd ) \Diagonal→R by

K(x ,ξ ) =

1

2π ln∥x − ξ ∥ if d = 2,1

ωd (2−d )∥x − ξ ∥2−d if d ≥ 3,(6.19)

where Diagonal stands for the set (x ,ξ ) ∈Rd ×Rd : x = ξ .Remark 6.10. For each fixed ξ ∈ Rd , the function (x 7→ K(x ,ξ ) satisfies ∆K(x ,ξ ) = 0for every x = ξ . Thus K is a solution to Laplace equation. Since the family of these speicalsolutions K(.,ξ ) (indexed by ξ ∈ Rd ) generate all solutions to ∆u = f , K is the calledfundamental solution, which is proved in Theorem 6.13 (for d = 2), and Theorem 6.15(for d = 3).

Theorem 6.11. Let K(x ,ξ ) be as defined by (6.19). Let Ω be an open and bounded domainin Rd .

(i) Let ξ be a point of Ω. For u ∈C 2(Ω) the following identity holds

u(ξ ) =∫Ω

K(x ,ξ )∆u d x −∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ . (6.20)

(ii) If u ∈C 2(Ω) and harmonic in Ω (i.e.,∆u = 0 in Ω), then for ξ ∈Ω we get

u(ξ ) =−∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ . (6.21)

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152 Chapter 6. Laplace equation

(iii) The following equality holds in the sense of distributions on Rd :

∆K(x,ξ ) = δξ .

i.e., for every φ ∈C∞0 (Rd ) the following equality holds.

φ(ξ ) =∫Rd

K(x ,ξ )∆φ(x)d x . (6.22)

Proof. Step 1: Proof of (i):

Let u ∈ C 2(Ω) and ξ be a point of Ω. Note that we cannot apply Green’s identityII (6.6) with v(x) = K(x ,ξ ) since K(x ,ξ ) is singular at x = ξ . Thus we cut out a ballB(ξ ,ρ) fromΩ along with its boundary, and then apply Green’s identity II. Note that thedomain Ωρ;= Ω \ B[ξ ,ρ] is bounded by ∂ Ω and S(ξ ,ρ). Since ∆K(x ,ξ ) = 0 in Ωρ, wehave∫Ωρ

K(x ,ξ )∆u d x =∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ +∫

S(ξ ,ρ)(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ .

(6.23)

Let us now compute the second term on RHS of the equation (6.23).∫S(ξ ,ρ)

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ =∫

S(ξ ,ρ)K(x ,ξ )∂n u dσ −

∫S(ξ ,ρ)

u∂nK(x ,ξ )dσ .

(6.24)

Note that for x on the circle S(ξ ,ρ), we have K(x ,ξ ) =ψ(ρ). Using this information,and divergence theorem, we get∫

S(ξ ,ρ)K(x ,ξ )∂n u dσ =ψ(ρ)

∫S(ξ ,ρ)

∂n u dσ =−ψ(ρ)∫

B(ξ ,ρ)∆u d x . (6.25)

Note that the outward unit normal n on the circle S(ξ ,ρ) points towards its center ξ .Also ∂nK(x ,ξ ) =−ψ′(ρ) holds at points on the circle S(ξ ,ρ). Thus we get

∫S(ξ ,ρ)

u∂nK(x ,ξ )dσ =−ρ1−d

ωd

∫S(ξ ,ρ)

u dσ . (6.26)

Since both u and ∆u are continuous at ξ , we have the following convergences asρ→ 0:

ψ(ρ)∫

B(ξ ,ρ)∆u d x→ 0, ρ1−d

∫S(ξ ,ρ)

u dσ→ωd u(ξ ), (6.27)

whereωd denotes the surface area of the unit sphere in Rd .Now passing to the limit as ρ→ 0 in the equation (6.23), we get∫

Ω

K(x ,ξ )∆u d x =∫∂ Ω

(K(x ,ξ )∂n u − u∂nK(x ,ξ )) dσ + u(ξ ) (6.28)

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6.3. Fundamental solutions in Rd 153

in view of (6.25), (6.26), (6.27). Note that the equation (6.28) is the required (6.20).

Step 2: Proof of (ii): Equation (6.21) is an immediate consequence of (6.20).

Step 3: Proof of (iii): In the equation (6.20) if we take u = φ ∈ C∞0 (Ω), then we get(6.22).

Remark 6.12 (On the formula (6.20)). (i) If Dirichlet problem (6.8) has a solution,then by Theorem 6.8 Dirichlet problem has exactly one solution. Let the solutionto Dirichlet problem be denoted by u. The formula (6.20) gives a representation ofthe solution in terms of the boundary values of u and its normal derivative ∂n u on∂ Ω.

(ii) Note however that in Dirichlet problem, only the values of u are prescribed on ∂ Ω,and ∂n u is an unknown function on ∂ Ω, and thus the formula (6.20) is not usefulfor computing the solution.

(iii) Note that boundary values of u already determine a solution to Dirichlet problem,and thus the quantity ∂n u is already determined.

(iv) A related question concerning Laplace equation is whether Cauchy problem is well-posed for Laplace equation. The answer to this question is negative, and is madeprecise in Theorem 6.36.

Theorem 6.13 (Logarithmic potential). Let f ∈C 2(R2) having compact support. Definethe Logarithmic potential on R2 by

u(ξ ) =1

2π

∫R2

ln∥x − ξ ∥ f (x)d x . (6.29)

Then

(i) Logarithmic potential satisfies∆u = f .(ii) u(ξ )→∞ as ∥ξ ∥ →∞ The Logarithmic potential has the following asymptotic be-

haviour as ∥ξ ∥→∞.

u(ξ ) =M2π

ln∥ξ ∥+O

1∥ξ ∥

, (6.30)

where M =∫R2 f (x)d x .

(iii) Logarithmic potential is the only solution to∆u = f satisfying (6.30).

Proof. Proof of (i): Note that the integral in (6.29) is meaningful since logarithm hasan integrable singularity in d = 2, and the integral is on a compact set as f has compactsupport. We will now check that∆u = f holds. By a change of variable, we get

u(ξ ) =1

2π

∫R2

ln∥x∥ f (x + ξ )d x . (6.31)

Since f ∈C 2(R2), on applying laplacian∆ξ w.r.t. ξ on both sides of (6.31) yields

∆ξ u(ξ ) =1

2π

∫R2

ln∥x∥∆ξ f (x + ξ )d x

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154 Chapter 6. Laplace equation

=1

2π

∫R2

ln∥x∥∆x f (x + ξ )d x

=1

2π

∫Bρ(0)

ln∥x∥∆x f (x + ξ )d x +1

2π

∫R2\Bρ(0)

ln∥x∥∆x f (x + ξ )d x

where ρ> 0. We will show that the first and second terms on the RHS of the last equality,tend to 0 and f (ξ ) respectively as ρ→ 0, and thus proving (i).

(a) We have the following estimate of the first term: 12π∫

Bρ(0)ln∥x∥∆x f (x + ξ )d x

≤ maxx∈Supp f

|∆ f (x)|2π

∫Bρ(0)|ln∥x∥| d x

≤ maxx∈Supp f

|∆ f (x)|∫ ρ

0r | ln |r | | d r

Since the integrand r | ln |r | | is bounded on the interval [0,ρ], as ρ→ 0 the integraltends to zero. Thus the first term tends to zero as ρ→ 0.

(b) Performing integration by parts in the second integral, we get

12π

∫R2\Bρ(0)

ln∥x∥∆x f (x + ξ )d x

=− 12π

∫R2\Bρ(0)

∇ (ln∥x∥) .∇ f (x + ξ )d x +lnρ2π

∫S(0,ρ)

∂n f (x + ξ )dσ(x).

Integrating by parts in the first term on RHS, in view of the fact that ln∥x∥ is har-monic in any domain not containing the origin, we get

12π

∫R2\Bρ(0)

ln∥x∥∆x f (x + ξ )d x

=− 12π

∫S(0,ρ)

∂n (ln∥x∥) f (x + ξ )dσ(x)+lnρ2π

∫S(0,ρ)

∂n f (x + ξ )dσ(x).

Note that n denotes the unit outward normal (w.r.t. the domainR2\Bρ(0)) at pointsof the circle S(0,ρ). We have

∇ (ln∥x∥) = x∥x∥2 ,

n(x) =− xρ

for x ∈ S(0,ρ),

∂n (ln∥x∥) =− 1ρ

for x ∈ S(0,ρ).

Thus we get

12π

∫R2\Bρ(0)

ln∥x∥∆x f (x + ξ )d x

=1

2πρ

∫S(0,ρ)

f (x + ξ )dσ(x)+lnρ2π

∫S(0,ρ)

∂n f (x + ξ )dσ(x).

(6.32)

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6.3. Fundamental solutions in Rd 155

The first term on the RHS of (6.32) tends to f (ξ ) as ρ→ 0, and the second termtends to zero as the integrand is a bounded function and ρ lnρ→ 0 as ρ→ 0.

Thus we have shown that the logarithmic potential u defined by (6.29) solves∆u = f .Proof of (ii): The equation (6.29) may be re-written as

u(ξ ) =1

2π

∫R2

ln∥ξ ∥ f (x)d x +1

2π

∫R2

ln∥x − ξ ∥∥ξ ∥

f (x)d x . (6.33)

Note that the first term on RHS of (6.33) coincides with that of (6.30). Thus (6.30) followsif the second term on RHS of (6.33) has the following asymptotic behaviour

12π

∫R2

ln∥x − ξ ∥∥ξ ∥

f (x)d x =O

1∥ξ ∥

as ∥ξ ∥→∞. (6.34)

Since f is of compact support, the integral in (6.34) is on a compact set, and thus (6.34)follows from the estimate ln

∥x − ξ ∥∥ξ ∥ ≤ R∥ξ ∥ +

4R2

∥ξ ∥2 ≤C∥ξ ∥ . (6.35)

for some C > 0 and ∥ξ ∥ large enough. We now turn to the proof of the estimate (6.35).Let R> 0 be such that the support of f is contained in B(0, R). Then the inequality

1− R∥ξ ∥ ≤ 1− ∥x∥∥ξ ∥ ≤

∥x − ξ ∥∥ξ ∥ ≤ 1+

∥x∥∥ξ ∥ ≤ 1+

R∥ξ ∥ (6.36)

holds for every x belonging to the support of f , which follows from triangle inequality.Since logarithm function is an increasing function, we get for ∥ξ ∥> 2R

ln

1− R∥ξ ∥≤ ln∥x − ξ ∥∥ξ ∥≤ ln

1+R∥ξ ∥

. (6.37)

By Taylor’s theorem applied to the logarithm function ln(1+ s ) about the point s = 0,there exists − R

∥ξ ∥ < η1 < 0 and 0< η2 <R∥ξ ∥ such that

ln

1− R∥ξ ∥=− R∥ξ ∥ −

1(1+η1)2

R2

∥ξ ∥2 ,

ln

1+R∥ξ ∥=

R∥ξ ∥ −

1(1+η2)2

R2

∥ξ ∥2 .

Thus for every x ∈ Supp f , and for every ξ ∈ R2 such that ∥ξ ∥ > 2R, the inequalities(6.37) yield

− R∥ξ ∥ −

1(1+η1)2

R2

∥ξ ∥2 ≤ ln∥x − ξ ∥∥ξ ∥≤ R∥ξ ∥ −

1(1+η2)2

R2

∥ξ ∥2 . (6.38)

Since ∥ξ ∥> 2R, we get− 12 < η1 < 0, and 0< η2 <

12 . Thus for every x ∈ Supp f , and for

every ξ ∈R2 such that ∥ξ ∥> 2R, the inequalities (6.38) yield

− R∥ξ ∥ −

4R2

∥ξ ∥2 ≤ ln∥x − ξ ∥∥ξ ∥≤ R∥ξ ∥ . (6.39)

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156 Chapter 6. Laplace equation

From the inequalities (6.39), the following estimate holds for every x ∈ Supp f , and forevery ξ ∈R2 such that ∥ξ ∥> 2R: ln

∥x − ξ ∥∥ξ ∥ ≤ R∥ξ ∥ +

4R2

∥ξ ∥2 (6.40)

≤ R∥ξ ∥

1+4R∥ξ ∥≤ 3R∥ξ ∥ . (6.41)

This finishes the proof of (6.35), and thus of (i i ).Proof of (iii): If u, v are solutions to ∆u = f , and both satisfy (6.30), then u − v

is a harmonic function on R2, and vanishes at infinity. By Liouville’s theorem u − v isa constant function, and the constant has to be zero due to vanishing at infinity. Thisshows that Logarithmic potential is the only solution to ∆u = f having the asymptoticbehaviour (6.30).

Remark 6.14 (Interpretation of potential). A function u that satisfies ∆u = F is saidto be the potential due to the charge F , in the context of electrostatics.

Theorem 6.15 (Newtonian potential). Let f ∈ C 2(R3) having compact support. Definethe Newtonian potential on R3 by

u(ξ ) =− 14π

∫R3

1∥x − ξ ∥ f (x)d x . (6.42)

Then

(i) Newtonian potential satisfies∆u = f .(ii) u(ξ )→ 0 as ∥ξ ∥→∞.

(iii) Newtonian potential is the only solution to ∆u = f that is in C 2(R3) and vanishes atinfinity.

Proof. Proof of (i) is similar to the case of Logarithmic potential. Statement (ii) is provedby comparing 1

∥x−ξ ∥ with 1∥ξ ∥−∥x∥ for ξ such that ∥ξ ∥ > 2R, where R is such that the

support of f is contained in the ball of radius R with center at the origin. Statement (iii)follows by applying Liouville’s theorem. Writing down the rigorous proofs is left to thereader as an exercise.

6.4 A representative formula for the solution to Dirichlet BVPusing Green’s functionGreen’s function is defined as a solution to the BVP

∆G(x ,ξ ) = δξ in Ω, (6.43a)G(x ,ξ ) = 0 on ∂ Ω. (6.43b)

In this section we will find explicitly Green’s function corresponding to two domains: R3+,

and a ball. Using Green’s function for a ball, we obtain a representation of the solutionto Dirichlet BVP for Poisson equation in terms of Green’s function.

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6.4. A representative formula for the solution to Dirichlet BVP using Green’s function 157

Since the fundamental solution K(x ,ξ ) satisfies equation (6.43a), we look for G(x ,ξ )having the form

G(x ,ξ ) =K(x ,ξ )−φ(x ,ξ ), (6.44)

where φ solves the following boundary value problem

∆φ(x ,ξ ) = 0 in Ω, (6.45a)φ(x ,ξ ) =K(x ,ξ ) on ∂ Ω. (6.45b)

We employ the method of images (also called, method of electrostatic images) to ob-tain Green’s functions. The method prescribes that φ(x ,ξ ) is the potential (denoted byψ(x ,ξ ∗)) due to an imaginary charge q placed at a point ξ ∗ with ξ ∗ /∈Ω, and such that atpoints x on the boundary ∂ Ω, the value of φ(x ,ξ ) equals the potential K(x ,ξ ) createdby the unit charge at ξ That is,

φ(x ,ξ ) =K(x ,ξ ). (6.46)

The potential due to a charge q at a point ξ ∗ /∈Ω is given by − 14π

q∥x−ξ ∗∥ . Thus we take

φ(x ,ξ ) =− 14π

q∥x − ξ ∗∥ . (6.47)

In order to determine Green’s function, we need to find q and ξ ∗ so that (6.46) is satisfied.

Green’s function for Ω=R3+

Consider Ω=R3+, for which boundary is given by the set

(x1, x2, x3) ∈R3 : x3 = 0.For ξ = (ξ1,ξ2,ξ3) ∈ R3

+, define ξ ∗ = (ξ1,ξ2,−ξ3). Clearly ξ ∗ /∈ R3+. Letting q = 1,

the condition (6.46) is satisfied, since for x ∈ ∂ R3+ we have

∥x − ξ ∥=Æ(x1− ξ1)2+(x2− ξ2)2+ ξ 23 = ∥x − ξ ∗∥.

Thus Green’s function for the domain R3+ is given by

G(x ,ξ ) =− 14π

1∥x − ξ ∥ +

14π

1∥x − ξ ∗∥ . (6.48)

Green’s function for Ω= B(0, R)

We need to find q and ξ ∗ such that the following equality is satisfied, for each x with∥x∥= R:

− 14π

q∥x − ξ ∗∥ =−

14π

1∥x − ξ ∥ . (6.49)

The equation (6.49), on simplification, reduces to

q∥x − ξ ∥= ∥x − ξ ∗∥. (6.50)

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158 Chapter 6. Laplace equation

On squaring both sides of the last equation, and re-arranging terms yields

R2+ ∥ξ ∗∥2− q2(R2+ ∥ξ ∥2) = 2x .(ξ ∗− q2ξ ) (6.51)

Since the LHS of the equation (6.51) is independent of x , and RHS depends on x , bothLHS and RHS must be equal to zero. That is,

x .(ξ ∗− q2ξ ) = 0, (6.52a)R2+ ∥ξ ∗∥2− q2(R2+ ∥ξ ∥2) == 0. (6.52b)

Since the equation (6.52a) holds for every x ∈ S(0, R), we get ξ ∗− q2ξ = 0. Thus q andξ ∗ satisfy the conditions

ξ ∗ = q2ξ , (6.53a)R2+ ∥ξ ∗∥2− q2(R2+ ∥ξ ∥2) = 0. (6.53b)

Substituting the value of ξ ∗ from (6.53a) into (6.53b) gives

R2+ q4∥ξ ∥2− q2(R2+ ∥ξ ∥2) = 0. (6.54)

If q = 1, then ξ ∗ = ξ and thus ξ ∗ ∈ B(0, R). Since ξ ∗ must be in the complement ofB(0, R), q = 1. Thus for ξ = 0, equation (6.54) yields

q =R∥ξ ∥ . (6.55)

Thus Green’s function G(x ,ξ ) for ξ = 0 is given by

G(x ,ξ ) =− 14π

1∥x − ξ ∥ +

14π

R∥ξ ∥

1∥x − ξ ∗∥ . (6.56)

Since

∥x − ξ ∗∥=p

R4− 2R2ξ .x + ∥x∥2∥ξ ∥2∥ξ ∥ , (6.57)

we get

∥ξ ∥∥x − ξ ∗∥=ÆR4− 2R2ξ .x + ∥x∥2∥ξ ∥2 ξ→0−→ R2. (6.58)

Thus we may define G(x ,0) by

G(x ,0) =− 14π∥x∥ +

14πR

. (6.59)

As noted in Remark 6.12, the formula (6.20) gives a representation of the solutionin terms of the boundary values of u and its normal derivative ∂n u on ∂ Ω. Since thesaid formula features ∂n u which is an unknown function on ∂ Ω, the formula (6.20) isnot useful in determining a solution to Dirichlet problem (6.8). However we can use thisformula and obtain an expression for the solution by eliminating the term involving ∂n u,as illustrated below. Recall that the formula (6.20) takes the form

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6.5. Poisson’s formula and its applications 159

u(ξ ) =∫Ω

K(x ,ξ ) f (x)d x −∫∂ Ω

K(x ,ξ )∂n u dσ +∫∂ Ω

g (x)∂nK(x ,ξ )dσ . (6.60)

Applying Green’s identity II (6.6) with u = u and v = φ(x ,ξ ) yields

0=−∫Ω

φ(x ,ξ ) f (x)d x +∫∂ Ω

φ(x ,ξ )∂n u dσ −∫∂ Ω

g (x)∂nφ(x ,ξ )dσ . (6.61)

Adding the two equations (6.60) and (6.61), and using the definition of G(x ,ξ ) (see (6.44)),we get

u(ξ ) =∫Ω

G(x ,ξ ) f (x)d x +∫∂ Ω

g (x)∂nG(x ,ξ )dσ . (6.62)

as K(x ,ξ ) = φ(x ,ξ ) for x ∈ ∂ Ω (see (6.43b)).In the special case, when u is a harmonic function in Ω, i.e., f = 0, then the above

formula (6.62) reduces to

u(ξ ) =∫∂ Ω

g (x)∂nG(x ,ξ )dσ . (6.63)

For Ω= B(0, R), the quantity ∂nG(x ,ξ ) for x ∈ S(0, R) and ξ = 0 is given by

∂nG(x ,ξ ) =1

4π∂n

R∥ξ ∥

1∥x − ξ ∗∥ −

1∥x − ξ ∥

=1

4π∇x

R∥ξ ∥

1∥x − ξ ∗∥ −

1∥x − ξ ∥

.xR

=1

4π

− R∥ξ ∥

x − ξ ∗∥x − ξ ∗∥3 +

x − ξ∥x − ξ ∥3

.xR

In view of the relations (6.50) and (6.55), the expression for ∂nG(x ,ξ ) reduces to

∂nG(x ,ξ ) =R2−∥ξ ∥2

4πR1

∥x − ξ ∥3 (6.64)

Substituting the value of ∂nG(x ,ξ ) from the last equation into equation (6.63) gives

u(ξ ) =R2−∥ξ ∥2

4πR

∫S(0,R)

g (x)∥x − ξ ∥3 dσ . (6.65)

The last expression is known as Poisson’s formula, and we derive the same formula byfollowing the method of separation of variables in Section 6.5.

6.5 Poisson’s formula and its applicationsIn this section, we consider Dirichlet boundary value problem for Laplace equation on thedomain Ω= B(0, 1) = (x1, x2) : x2

1 + x22 < 1 in R2. For a given function f ∈C (S(0, 1)),

we consider the boundary value problem

∆u = 0 in B(0, 1), (6.66a)u = f on S(0, 1). (6.66b)

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