Hints and answers

21.7 HINTS AND ANSWERS
in V and takes the specified form φ = f on S, the boundary of V . The
Green’s function, G(r, r ), to be used satisfies
∇2 G − m2 G = δ(r − r )
and vanishes when r is on S.
(b) When V is all space, G(r, r ) can be written as G(t) = g(t)/t, where t = |r − r |
and g(t) is bounded as t → ∞. Find the form of G(t).
(c) Find φ(r) in the half-space x > 0 if ρ(r) = δ(r − r1 ) and φ = 0 both on x = 0
and as r → ∞.
21.28
Consider the PDE Lu(r) = ρ(r), for which the differential operator L is given by
L = ∇ · [ p(r)∇ ] + q(r),
where p(r) and q(r) are functions of position. By proving the generalised form of
Green’s theorem,
0
(φLψ − ψLφ) dV = p(φ∇ψ − ψ∇φ) · n̂ dS,
V
S
show that the solution of the PDE is given by
0
∂G(r, r0 )
∂u(r)
dS(r),
u(r0 ) =
G(r, r0 )ρ(r) dV (r) + p(r) u(r)
− G(r, r0 )
∂n
∂n
V
S
where G(r, r0 ) is the Green’s function satisfying LG(r, r0 ) = δ(r − r0 ).
21.7 Hints and answers
21.1
21.3
21.5
21.7
21.9
21.11
21.13
21.15
21.17
21.19
21.21
21.23
(a) C exp[λ(x2 + 2y)]; (b) C(x2 y)λ .
u(x, y, t) = sin(nπx/a) sin(mπy/b)(A sin ωt + B cos ωt).
(a) 6u/r 2 , −6u/r2 , 0, = 2 (or −3), m = 0;
(b) 2u/r2 , (cot2 θ − 1)u/r2 ; −u/(r2 sin2 θ), = 1 (or −2), m = ±1.
Solutions of the form r give as −1, 1, 2, 4. Because of the asymptotic form of
ψ, an r 4 term cannot be present. The coefficients of the three remaining terms are
determined by the two boundary conditions u = 0 on the sphere and the form of
ψ for large r.
Express cos2 φ in terms of cos 2φ; T (ρ, φ) = A + B/2 + (Bρ2 /2a2 ) cos 2φ.
(A cos mx + B sin mx + C cosh mx + D sinh mx) cos(ωt + ), with m4 a4 = ω 2 .
A
En = 16ρA2 c2 /[(2n + 1)2 π 2 L]; E = 2ρc2 A2 /L = 0 [2T v/( 21 L)] dv.
Note that the boundary value function is a square wave that is symmetric in φ.
Since there is no heat flow at x = ±a, use a series of period 4a, u(x, 0) = 100 for
0 < x ≤ 2a, u(x, 0) = 0 for −2a ≤ x < 0.
∞
k(2n + 1)2 π 2 t
200 1
(2n + 1)πx
exp −
.
u(x, t) = 50 +
sin
2
π n=0 2n + 1
2a
4a s
Taking only the n = 0 term gives t ≈ 2300 s.
u(x, t) = [a/(a2 + 4κt)1/2 ] exp[−x2 /(a2 + 4κt)].
Fourier-transform Poisson’s equation to show that ρ̃(α) = 0 (α2 + q 2 )Ṽ (α).
Follow the worked example that includes result (21.95). For part of the explicit
integration, substitute ρ = z tan α.
Φ(0, 0, z) =
z(1 + z 2 )1/2 − z 2 + (1 + z 2 )1/2 − 1
.
z(1 + z 2 )1/2
773
PDES: SEPARATION OF VARIABLES AND OTHER METHODS
21.25
The terms in G(r, r0 ) that are additional to the fundamental solution are
∞
!
−1/2
1 (−1)n (x − x0 )2 + (y − y0 )2 + (z + (−1)n z0 − nc)2
4π n=2
−1/2 "
.
+ (x − x0 )2 + (y − y0 )2 + (z + (−1)n z0 + nc)2
21.27
(a) As given in equation (21.86), but with r0 replaced by r .
(b) Move the origin to r and integrate the defining Green’s equation to obtain
t
dG
2
4πt2
G(t ) 4πt dt = 1,
− m2
dt
0
leading to G(t) = [−1/(4πt)]e−mt .
(c) φ(r) = [−1/(4π)](p−1 e−mp − q −1 e−mq ), where p = |r − r1 | and q = |r − r2 | with
r1 = (x1 , y1 , z1 ) and r2 = (−x1 , y1 , z1 ).
774