6.3 Fundamental solutions in d

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1 6.3. Fundamental solutions in d Fundamental solutions in d Since Lapalce equation is invariant under translations, and rotations (see Exercise 6.4), we look for solutions to Laplace equation having such symmetric properties. Let us fix ξ d, and look for solutions to u = 0 having the form v ξ (x) = ψ(r ), where r = x ξ = d (x i ξ i ). i= Substituting the formula for v ξ in the equation, we get Solving the last equation, we get Integrating the last equation we get v ξ (x) = ψ (r ) + d ψ (r ) = 0. r ψ (r ) = C r d (6.8) C ln r if d =, ψ(r ) = C d r d if d 3. Definition 6.9. The fundamental solution of Laplacian operator is defined is defined as the function K : ( d d ) \ Diagonal by K(x,ξ ) = ln x ξ if d =, ω d ( d) x ξ d if d 3, where Diagonal stands for the set {(x,ξ ) d d : x = ξ }. (6.9) emark 6.0. For each fixed ξ d, the function (x K(x,ξ ) satisfies K(x,ξ ) = 0 for every x ξ. Thus K is a solution to Laplace equation. Since the family of these speical solutions K(.,ξ ) (indexed by ξ d ) generate all solutions to u = f, K is the called fundamental solution, which is proved in Theorem 6.3 (for d = ), and Theorem 6.5 (for d = 3). Theorem 6.. Let K(x,ξ ) be as defined by (6.9). Let be an open and bounded domain in d. (i) Let ξ be a point of. For u C () the following identity holds u(ξ ) = K(x,ξ ) u d x (K(x,ξ ) n u u n K(x,ξ )) dσ. (6.0) (ii) If u C () and harmonic in (i.e., u = 0 in ), then for ξ we get u(ξ ) = (K(x,ξ ) n u u n K(x,ξ )) dσ. (6.) October, 05

2 5 Chapter 6. Laplace equation (iii) The following equality holds in the sense of distributions on d : K(x,ξ ) = δ ξ. i.e., for every φ C 0 (d ) the following equality holds. φ(ξ ) = K(x,ξ ) φ(x) d x. (6.) d Proof. Step : Proof of (i): Let u C () and ξ be a point of. Note that we cannot apply Green s identity II (6.6) with v(x) = K(x,ξ ) since K(x,ξ ) is singular at x = ξ. Thus we cut out a ball B(ξ,ρ) from along with its boundary, and then apply Green s identity II. Note that the domain ρ ;= \ B[ξ,ρ] is bounded by and. Since K(x,ξ ) = 0 in ρ, we have ρ K(x,ξ ) u d x = (K(x,ξ ) n u u n K(x,ξ )) dσ + Let us now compute the second term on HS of the equation (6.3). (K(x,ξ ) n u u n K(x,ξ )) dσ = K(x,ξ ) n u dσ (K(x,ξ ) n u u n K(x,ξ )) dσ. u n K(x,ξ ) dσ. (6.4) Note that for x on the circle, we have K(x,ξ ) = ψ(ρ). Using this information, and divergence theorem, we get K(x,ξ ) n u dσ = ψ(ρ) n u dσ = ψ(ρ) u d x. (6.5) B(ξ,ρ) Note that the outward unit normal n on the circle points towards its center ξ. Also n K(x,ξ ) = ψ (ρ) holds at points on the circle. Thus we get u n K(x,ξ ) dσ = ρ d ω d u dσ. (6.6) Since both u and u are continuous at ξ, we have the following convergences as ρ 0: ψ(ρ) u d x 0, ρ d u dσ ω d u(ξ ), (6.7) B(ξ,ρ) where ω d denotes the surface area of the unit sphere in d. Now passing to the limit as ρ 0 in the equation (6.3), we get K(x,ξ ) u d x = (K(x,ξ ) n u u n K(x,ξ )) dσ + u(ξ ) (6.8) (6.3) IIT Bombay

3 6.3. Fundamental solutions in d 53 in view of (6.5), (6.6), (6.7). Note that the equation (6.8) is the required (6.0). Step : Proof of (ii): Equation (6.) is an immediate consequence of (6.0). Step 3: Proof of (iii): In the equation (6.0) if we take u = φ C0 (6.). (), then we get emark 6. (On the formula (6.0)). (i) If Dirichlet problem (6.8) has a solution, then by Theorem 6.8 Dirichlet problem has exactly one solution. Let the solution to Dirichlet problem be denoted by u. The formula (6.0) gives a representation of the solution in terms of the boundary values of u and its normal derivative n u on. (ii) Note however that in Dirichlet problem, only the values of u are prescribed on, and n u is an unknown function on, and thus the formula (6.0) is not useful for 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 wellposed for Laplace equation. The answer to this question is negative, and is made precise in Theorem Theorem 6.3 (Logarithmic potential). Let f C ( ) having compact support. Define the Logarithmic potential on by u(ξ ) = ln x ξ f (x) d x. (6.9) Then (i) Logarithmic potential satisfies u = f. (ii) u(ξ ) as The Logarithmic potential has the following asymptotic behaviour as. u(ξ ) = M ln + O, (6.30) where M = 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 (6.9) is meaningful since logarithm has an integrable singularity in d =, and the integral is on a compact set as f has compact support. We will now check that u = f holds. By a change of variable, we get u(ξ ) = ln x f (x + ξ ) d x. (6.3) Since f C ( ), on applying laplacian ξ w.r.t. ξ on both sides of (6.3) yields ξ u(ξ ) = ln x ξ f (x + ξ ) d x October, 05

4 54 Chapter 6. Laplace equation = ln x x f (x + ξ ) d x = B ρ (0) ln x x f (x + ξ ) d x + \B ρ (0) ln x x f (x + ξ ) d x where ρ > 0. We will show that the first and second terms on the HS 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: B ρ (0) max f (x) ln x x f (x + ξ ) d x x Supp f max x Supp f f (x) B ρ (0) ρ 0 ln x d x r ln r d r Since the integrand r ln r is bounded on the interval [0,ρ], as ρ 0 the integral tends to zero. Thus the first term tends to zero as ρ 0. (b) Performing integration by parts in the second integral, we get \B ρ (0) ln x x f (x + ξ ) d x = \B ρ (0) (ln x ). f (x + ξ ) d x + lnρ n f (x + ξ ) dσ(x). S(0,ρ) Integrating by parts in the first term on HS, in view of the fact that ln x is harmonic in any domain not containing the origin, we get ln x x f (x + ξ ) d x \B ρ (0) = S(0,ρ) n (ln x ) f (x + ξ ) dσ(x) + lnρ n f (x + ξ ) dσ(x). S(0,ρ) Note that n denotes the unit outward normal (w.r.t. the domain \B ρ (0)) at points of the circle S(0,ρ). We have (ln x ) = x x, n(x) = x for x S(0,ρ), ρ n (ln x ) = for x S(0,ρ). ρ Thus we get ln x x f (x + ξ ) d x \B ρ (0) = f (x + ξ ) dσ(x) + lnρ ρ S(0,ρ) n f (x + ξ ) dσ(x). S(0,ρ) (6.3) IIT Bombay

5 6.3. Fundamental solutions in d 55 The first term on the HS of (6.3) tends to f (ξ ) as ρ 0, and the second term tends to zero as the integrand is a bounded function and ρlnρ 0 as ρ 0. Thus we have shown that the logarithmic potential u defined by (6.9) solves u = f. Proof of (ii): The equation (6.9) may be re-written as u(ξ ) = ln f (x) d x + x ξ ln f (x) d x. (6.33) Note that the first term on HS of (6.33) coincides with that of (6.30). Thus (6.30) follows if the second term on HS of (6.33) has the following asymptotic behaviour x ξ ln f (x) d x = O as. (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 x ξ ln + 4 C. (6.35) for some C > 0 and large enough. We now turn to the proof of the estimate (6.35). Let > 0 be such that the support of f is contained in B(0, ). Then the inequality x x ξ + x + (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 > ln x ξ ln ln +. (6.37) By Taylor s theorem applied to the logarithm function ln( + s) about the point s = 0, there exists < η < 0 and 0 < η < such that ln = ( + η ), ln + = ( + η ). Thus for every x Supp f, and for every ξ such that >, the inequalities (6.37) yield x ξ ( + η ) ln ( + η ). (6.38) Since >, we get < η < 0, and 0 < η <. Thus for every x Supp f, and for every ξ such that >, the inequalities (6.38) yield 4 ln x ξ. (6.39) October, 05

6 56 Chapter 6. Laplace equation From the inequalities (6.39), the following estimate holds for every x Supp f, and for every ξ such that > : x ξ ln + 4 (6.40) (6.4) This finishes the proof of (6.35), and thus of (ii). Proof of (iii): If u, v are solutions to u = f, and both satisfy (6.30), then u v is a harmonic function on, and vanishes at infinity. By Liouville s theorem u v is a constant function, and the constant has to be zero due to vanishing at infinity. This shows that Logarithmic potential is the only solution to u = f having the asymptotic behaviour (6.30). emark 6.4 (Interpretation of potential). A function u that satisfies u = F is said to be the potential due to the charge F, in the context of electrostatics. Theorem 6.5 (Newtonian potential). Let f C ( 3 ) having compact support. Define the Newtonian potential on 3 by u(ξ ) = f (x) d x. (6.4) x ξ 3 Then (i) Newtonian potential satisfies u = f. (ii) u(ξ ) 0 as. (iii) Newtonian potential is the only solution to u = f that is in C ( 3 ) and vanishes at infinity. Proof. Proof of (i) is similar to the case of Logarithmic potential. Statement (ii) is proved by comparing x ξ with x for ξ such that >, where is such that the support of f is contained in the ball of radius with center at the origin. Statement (iii) follows by applying Liouville s theorem. Writing down the rigorous proofs is left to the reader as an exercise. 6.4 A representative formula for the solution to Dirichlet BVP using Green s function Green s function is defined as a solution to the BVP G(x,ξ ) = δ ξ in, (6.43a) G(x,ξ ) = 0 on. (6.43b) In this section we will find explicitly Green s function corresponding to two domains: 3 +, and a ball. Using Green s function for a ball, we obtain a representation of the solution to Dirichlet BVP for Poisson equation in terms of Green s function. IIT Bombay

7 6.4. A representative formula for the solution to Dirichlet BVP using Green s function 57 Since the fundamental solution K(x,ξ ) satisfies equation (6.43a), we look for G(x,ξ ) having the form where φ solves the following boundary value problem G(x,ξ ) = K(x,ξ ) φ(x,ξ ), (6.44) φ(x,ξ ) = 0 in, (6.45a) φ(x,ξ ) = K(x,ξ ) on. (6.45b) We employ the method of images (also called, method of electrostatic images) to obtain Green s functions. The method prescribes that φ(x, ξ ) is the potential (denoted by ψ(x,ξ )) due to an imaginary charge q placed at a point ξ with ξ /, and such that at points x on the boundary, the value of φ(x,ξ ) equals the potential K(x,ξ ) created by 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 / is given by q x. Thus we take ξ φ(x,ξ ) = 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 = 3 + Consider = 3 +, for which boundary is given by the set {(x, x, x 3 ) 3 : x 3 = 0}. For ξ = (ξ,ξ,ξ 3 ) 3 +, define ξ = (ξ,ξ, ξ 3 ). Clearly ξ / 3 +. Letting q =, the condition (6.46) is satisfied, since for x 3 + we have x ξ = (x ξ ) + (x ξ ) + ξ 3 = x ξ. Thus Green s function for the domain 3 + is given by Green s function for = B(0, ) G(x,ξ ) = x ξ + x ξ. (6.48) We need to find q and ξ such that the following equality is satisfied, for each x with x = : q x ξ = The equation (6.49), on simplification, reduces to x ξ. (6.49) q x ξ = x ξ. (6.50) October, 05

8 58 Chapter 6. Laplace equation On squaring both sides of the last equation, and re-arranging terms yields + ξ q ( + ) = x.(ξ q ξ ) (6.5) Since the LHS of the equation (6.5) is independent of x, and HS depends on x, both LHS and HS must be equal to zero. That is, x.(ξ q ξ ) = 0, + ξ q ( + ) == 0. (6.5a) (6.5b) Since the equation (6.5a) holds for every x S(0, ), we get ξ q ξ = 0. Thus q and ξ satisfy the conditions ξ = q ξ, + ξ q ( + ) = 0. (6.53a) (6.53b) Substituting the value of ξ from (6.53a) into (6.53b) gives + q 4 q ( + ) = 0. (6.54) If q =, then ξ = ξ and thus ξ B(0, ). Since ξ must be in the complement of B(0, ), q. Thus for ξ 0, equation (6.54) yields q = Thus Green s function G(x,ξ ) for ξ 0 is given by. (6.55) Since G(x,ξ ) = x ξ + x ξ. (6.56) we get x ξ 4 ξ.x + x =, (6.57) Thus we may define G(x,0) by x ξ = 4 ξ.x + x ξ 0. (6.58) G(x,0) = x +. (6.59) As noted in emark 6., the formula (6.0) gives a representation of the solution in terms of the boundary values of u and its normal derivative n u on. Since the said formula features n u which is an unknown function on, the formula (6.0) is not useful in determining a solution to Dirichlet problem (6.8). However we can use this formula and obtain an expression for the solution by eliminating the term involving n u, as illustrated below. ecall that the formula (6.0) takes the form IIT Bombay

9 6.5. Poisson s formula and its applications 59 u(ξ ) = K(x,ξ ) f (x) d x K(x,ξ ) n u dσ + g(x) n K(x,ξ ) dσ. (6.60) Applying Green s identity II (6.6) with u = u and v = φ(x, ξ ) yields 0 = φ(x,ξ ) f (x) d x + φ(x,ξ ) n u dσ g(x) n φ(x,ξ ) dσ. (6.6) Adding the two equations (6.60) and (6.6), and using the definition of G(x,ξ ) (see (6.44)), we get u(ξ ) = G(x,ξ ) f (x) d x + g(x) n G(x,ξ ) dσ. (6.6) as K(x,ξ ) = φ(x,ξ ) for x (see (6.43b)). In the special case, when u is a harmonic function in, i.e., f = 0, then the above formula (6.6) reduces to u(ξ ) = g(x) n G(x,ξ ) dσ. (6.63) For = B(0, ), the quantity n G(x,ξ ) for x S(0, ) and ξ 0 is given by n G(x,ξ ) = n x ξ x ξ = x x ξ. x x ξ = x ξ x ξ + x ξ. x 3 x ξ 3 In view of the relations (6.50) and (6.55), the expression for n G(x,ξ ) reduces to n G(x,ξ ) = x ξ 3 (6.64) Substituting the value of n G(x,ξ ) from the last equation into equation (6.63) gives u(ξ ) = g(x) dσ. (6.65) x ξ 3 S(0,) The last expression is known as Poisson s formula, and we derive the same formula by following the method of separation of variables in Section Poisson s formula and its applications In this section, we consider Dirichlet boundary value problem for Laplace equation on the domain = B(0,) = {(x, x ) : x + x < } in. For a given function f C (S(0,)), we consider the boundary value problem u = 0 in B(0,), (6.66a) u = f on S(0,). (6.66b) October, 05

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