MATH 617-010
Prof. D. A. Edwards
Introduction to Applied Mathematics II
Due: Apr. 2, 2013
Homework Set 5 Solutions
1. (6 points) By direct calculation, verify that the integral representation for J0 (z)
does indeed solve Bessel’s equation of order 0.
Solution. Substituting the integral representation
Z
1 π
cos(z sin φ) dφ
J0 (z) =
π 0
into Bessel’s equation, we obtain
Z π
Z
Z π
2
d 1 π
1
2 d
21
z
cos(z sin φ) dφ + z
cos(z sin φ) dφ + z
cos(z sin φ) dφ = 0
dz 2 π 0
dz π 0
π 0
π
d2
d
[cos(z sin φ)] + z [cos(z sin φ)] + z 2 cos(z sin φ) dφ = 0
2
dz
dz
0
Z π
(z 2 − z 2 sin2 φ) cos(z sin φ) − z sin φ[sin(z sin φ)] dφ = 0
0
Z π
d sin(z sin φ)
− z sin φ[sin(z sin φ)] dφ = 0
(z cos φ)
dφ
0
Z π
π
[(z cos φ) sin(z sin φ)]0 +
z sin φ[sin(z sin φ)] − z sin φ[sin(z sin φ)] dφ = 0
Z
z2
0
0 = 0.
2. (8 points) Use analogous arguments to those in Theorem 1 on page 308 of Guenther and Lee to prove uniqueness of the solution to the following problem:
∇2 u − k 2 u = 0,
x ∈ D ⊂ R3 ;
u(x) = f (x),
x ∈ ∂D.
(5.1)
Solution. Let u1 and u2 be two solutions of (5.1). Then φ = u1 − u2 satisfies
∇2 φ − k 2 φ = 0,
x ∈ D ⊂ R3 ;
φ(x) = 0,
x ∈ ∂D.
(A)
We consider the following integral:
ZZZ
ZZZ
2
2
J=
φ(∇ φ − k φ) dx =
∇ · (φ∇φ) − |∇φ|2 − k 2 φ2 dx
ZDZ
D
ZZZ
φ∇φ dS −
=
∂D
Copyright ©2013
D. A. Edwards
2
2 2
ZZZ
|∇φ| + k φ dx = −
D
|∇φ|2 + k 2 φ2 dx,
D
All Rights Reserved
M617S13Sol5.2
T (1, θ) = f (θ)
insulated
where we have used the boundary condition in (A). From the equation in (A), we have
that J = 0, so φ = 0 and u1 = u2 .
3. (10 points) A semicircle of radius 1 is insulated along the flat side and has an
imposed temperature of f (θ) at the outer boundary (see figure). Show that the
solution to Laplace’s equation in this geometry is given by
Z
1 − r2 π
f (φ)(1 + r2 − 2r cos θ cos φ)
T (r, θ) =
dφ
2 2
2
2
π
0 (1 + r ) − 4r(1 + r ) cos θ cos φ + 2r (cos 2θ + cos 2φ)
Solution. We note that if we extended the problem in an even way about θ = 0 by
letting f (−θ) = f (θ), we would obtain a solution where
∂T
(r, 0) = 0.
∂θ
We also note that such a solution would also be even about θ = π, since f (π − a) =
f (a − π) = f (2π + (a − π)) = f (π + a). Therefore, from the standard Poisson integral
formula we have
Z π
Z π
1
1 − r2
1
1 − r2
T (r, θ) =
f (φ)
dφ
+
f
(φ)
dφ.
2π 0
1 + r2 − 2r cos(θ + φ)
2π 0
1 + r2 − 2r cos(θ − φ)
1
T (r, θ) =
2π
Z
π
(1 − r2 )f (φ)×
0
2(1 + r2 ) − 2r[cos(θ + φ) + cos(θ − φ)]
dφ
(1 + r2 )2 − 2r(1 + r2 )[cos(θ + φ) + cos(θ − φ)] + 4r2 cos(θ + φ) cos(θ − φ)
Z π
1
T (r, θ) =
(1 − r2 )f (φ)×
2π 0
2(1 + r2 ) − 4r cos θ cos φ
dφ
(1 + r2 )2 − 4r(1 + r2 ) cos θ cos φ + 2r2 (cos 2θ + cos 2φ)
Z
1 − r2 π
f (φ)(1 + r2 − 2r cos θ cos φ)
T (r, θ) =
dφ.
2 2
2
2
π
0 (1 + r ) − 4r(1 + r ) cos θ cos φ + 2r (cos 2θ + cos 2φ)
M617S13Sol5.3
4. Consider the dimensionless wave equation for the function Ψ(r, θ, φ, t) in a sphere
of radius 1. In addition, suppose that on the surface Ψ oscillates in the following
manner:
Ψ(1, θ, φ, t) = eiωt f (θ, φ).
(5.2)
(a) (2 points) Make an appropriate variable substitution to obtain the following
governing equation for the function ψ:
1 ∂
∂ψ
1 ∂2ψ
1 ∂
2 ∂ψ
r
+
sin φ
+
+ k 2 ψ = 0, (5.3)
r2 ∂r
∂r
sin φ ∂φ
∂φ
sin2 φ ∂ 2 θ
and calculate the value of k. (This is called the Helmholtz equation.) Also
calculate the boundary condition at r = 1.
Solution. We note from (5.2) that we should let Ψ(r, θ, φ, t) = eiωt ψ(r, θ, φ). Then
substituting into the general form of the multidimensional wave equation, we have
∂ 2 (eiωt ψ)
∂t2
2
2 2
∇ ψ = −c ω ψ
∇2 (eiωt ψ) = c2
∇2 ψ + k 2 ψ = 0,
k 2 = c2 ω 2 .
By substituting the spherical form of the Laplacian into the above, we obtain (5.3). Clearly
we have that
ψ(1, θ, φ) = f (θ, φ).
(b) (9 points) Show that the appropriate radial eigenfunctions are given by
Rn (r) =
Jn+1/2 (kr)
,
r1/2
n > 0.
(5.4)
Solution. Separating variables by letting ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ), we have
1
d(r2 R0 )
RΘ d(sin φΦ0 ) RΦΘ00
ΘΦ
+
+
+ k 2 RΘΦ = 0
r2
dr
sin φ
dφ
sin2 φ
sin2 φ d(r2 R0 ) sin φ d(sin φΦ0 ) Θ00
+
+
+ k 2 r2 sin2 φ = 0.
R
dr
Φ
dφ
Θ
Again we see that Θ00 /Θ must be a constant, and the same arguments as in class imply
that
Θ00
Θ(θ) = sin mθ, cos mθ
=⇒
= −m2 .
(B.1)
Θ
Substituting this result into our equation, we have
1 d(r2 R0 )
1 d(sin φΦ0 )
m2
2 2
+k r =−
−
.
R dr
sin φΦ
dφ
sin2 φ
M617S13Sol5.4
The right-hand side is the same is in class, so we have that
Φ(φ) = Pnm (cos φ),
(B.2)
and the right-hand side is n(n + 1), so we have
d(r2 R0 )
+ k 2 r2 R = n(n + 1)R.
dr
Letting
R(r) =
Z(z)
,
r−1/2
z = kr
in the above, we have
Z
z2
n(n + 1)Z
d z2 d
k
+
Z=
2
1/2
1/2
dz k dz r
r
r1/2
Z
k 1/2
− 3/2
+ 1/2 [z 2 − n(n + 1)]Z = 0
2z
z
2
k
1/2
z 3/2 Z 00 +
d
dz
0
Z
z2
z 1/2
1
1
3z 1/2 0 z 1/2 0
Z −
Z − 1/2 Z + 1/2 [z 2 − n(n + 1)]Z = 0
2
2
4z
z
2
1
2 00
0
2
z Z + zZ + [z − n +
]Z = 0
2
Z(z) = Jn+1/2 (z)
Jn+1/2 (kr)
, n ≥ 0.
r1/2
= O(rn+1/2 ) as r → 0, so these are
Rn (r) =
Here the restriction on n is due to the fact that Jn+1/2
the only ones that are bounded at the origin.
(c) (5 points) Write the solution Ψ to the problem.
Solution. Using (5.4) and (B) in our sum, we have
∞ X
n
X
Jn+1/2 (kr) m
Pn (cos φ)[amn sin mθ + bmn cos mθ].
ψ(r, θ, φ) =
r1/2
n=0 m=0
Examining the boundary condition, we have
∞ X
n
X
ψ(1, θ, φ) =
Jn+1/2 (k)Pnm (cos φ)[amn sin mθ + bmn cos mθ]
f (θ, φ) =
n=0 m=0
∞ X
n
X
Jn+1/2 (k)Pnm (cos φ)[amn sin mθ + bmn cos mθ].
n=0 m=0
To determine the constants, we realize that since the θ- and φ-functions are the same as
in class, with the only difference being the additional factor Jn+1/2 (k). Thus we have that
Z 2π Z π
1
f (θ, φ)Pnm (cos φ) sin mθ sin φ dφ dθ,
amn =
m
2
πJn+1/2 (k)||Pn (cos φ)|| 0
0
Z 2π Z π
1
bmn =
f (θ, φ)Pnm (cos φ) cos mθ sin φ dφ dθ.
m
2
πJn+1/2 (k)||Pn (cos φ)|| 0
0