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PH 116C, Solution set 5
1. Read Boas, Ch. 13, Sec. 1-4
2. Boas, p. 637, problem 13.4-3 or 13.4-4.
Here is 13.4-4; 13.4-3 is very similar.
y0 (x) =



4h
x,
l
2h − 4hl x,
−4h + 4hl x,
0 < x < 4l ,
l
< x < 34 l,
4
3
l < x < l;
4
(1)
The wave equation is
∂ 2y
1 ∂ 2y
=
;
∂ 2x
v2 ∂ 2t
separating the variables y(x, t) = X(x)T (t), the solution is
sin kx
sin ωt
y(x, t) =
×
,
ω = kv.
cos kx
cos ωt
(2)
(3)
Now we must apply the given boundary and initial conditions. The string is fixed at
its ends so that cos kx does not contribute; because y(l, t) = 0, we have kn = nπ
. The
l
solution can be written as:
X
y(x, t) =
sin kn x (An cos ωn t + Bn sin ωn t)
(4)
n
Because the string is initially not moving we have ∂y
(x, 0) = 0, implying Bn = 0.
∂t
Finally, we find the coefficients An from the initial configuration of the string:
X
An sin kn x = y0 (x);
y(x, 0) =
(5)
n
Z
Z
2 l/4
4h
nπ
2 3l/4
4h
nπ
y0 (x) sin kn x dx =
dx x sin
x+
dx (−x + 2l ) sin
x+
l 0
l
l
l l/4
l
l
0
Z
2 l
4h
nπ
+
dx (x − l) sin
x=
(6)
l 3l/4
l
l
("
l/4
l/4 #)
8h
l
nπ l2
nπ −x
cos
x
+ 2 2 sin
x
(7)
= 2
l
nπ
l 0
nπ
l 0
("
3l/4
3l/4 #
2
8h
l
nπ
nπ
l
+ 2
−(−x + 2l )
cos
x
− 2 2 sin
x
l
nπ
l l/4
nπ
l l/4
"
l
l #)
nπ l2
nπ 32h
nπ
nπ
l
+ −(x − l)
cos
x
+ 2 2 sin
x
= 2 2 cos
sin
(8)
nπ
l 3l/4 n π
l 3l/4
nπ
2
4
2
An =
l
Z
l
1
For odd n, cos nπ/2 = 0, so that only the even n’s contribute. The solution to the
wave equation with the given initial condition is
y=
2nπx
2nπvt
nπ
8h X 1
sin
cos
sin
2
2
π n n
2
l
l
(9)
3. Boas, p. 638, problem 13.4-7 or 13.4-8.
Here is 13.4-7. A string of length l is initially stretched straight; its ends are fixed for
all t. At the time t = 0 its points are given a velocity V (x). Determine the shape of
the string at a time t.
3
h l x,
0 < x < 3l ,
V (x) =
(10)
l
h 2l3 (l − x)
< x < l;
3
The problem is the same as in the last exercise, except that we have different initial
conditions: instead of the initial position y(x, 0) we have the initial velocity ∂y
(x, 0).
∂t
As before the solutions to the equation (2) are given by (3). Because the ends are fixed,
cos kx does not contribute and because y(l, t) = 0, we have kn = nπ
. The solution can
l
be written as:
X
sin kn x (An cos ωn t + Bn sin ωn t)
(11)
y(x, t) =
n
This time, because y(x, 0) = 0, we have An = 0; we find the Bn ’s applying the initial
condition:
X nπv
∂y
Bn
(x, 0) =
sin kn x = V (x);
(12)
∂t
l
n
Z
Z
Z
nπv
2 l
2 l/3 3h
nπx 2 l
3h
nπx
Bn =
V (x) sin kn x dx =
dx x sin
+
dx (l − x) sin (13)
l
l 0
l 0
l
l
l l/3 2l
l
#)
("
l/3
l/3
l2
6h
nπx nπx l
+ 2 2 sin
(14)
= 2
− x cos
l
nπ
l 0
nπ
l 0
( "
l
l #)
6h 1
l
nπx nπx l2
sin
+ 2
− (l − x) cos
−
l
2
nπ
l l/3 n2 π 2
l l/3
=
nπ
9h
sin
n2 π 2
3
Bn =
9hl
nπ
sin
n3 π 3 v
3
The solution with these initial conditions is then
9hl X 1
nπ
nπx
nπvt
y= 3
sin
sin
sin
.
π v n n3
3
l
l
(15)
(16)
The only difference between 13.4-7 and 13.4-8 is the initial velocity. As above, the
solution is of the form
X
nπx
nπvt
sin
,
y(x, t) =
Bn sin
l
l
n
2
since y(0, t) = y(l, t) = 0 and y(x, 0) = 0. The initial velocity it
(
X nπv
nπx
∂y sin 2πx/l, 0 < x < l/2
sin
= V (x) ≡
.
Bn
=
∂t t=0
l
l
0,
l/2
<
x
<
l
n
Multiplying through by sin mπx
(with m ∈ Z) and integrating from x = 0 to l gives
l
Z
0
l
Z
mπx X Bn nπv l
nπx
Bm mπv
mπx
dx V (x) sin
=
sin
=
dx sin
l
l
l
l
2
0
n
Z l/2
mπx
2πx
dx sin
=
sin
.
l
l
0
This can be solved using the product-to-sum identity
sin θ sin φ =
1
[cos(θ − φ) − cos(θ + φ)] .
2
Applying this gives
Z l/2
1
(m − 2)πx
(m + 2)πx
Bm =
dx cos
− cos
mπv 0
l
l
l
(m − 2)π
l
(m + 2)π
1
sin
−
sin
=
.
mπv (m − 2)π
2
(m + 2)π
2
For m 6= 2, this simplifies to
1
l
l
mπ
Bm =
−
+
sin
mπv
(m − 2)π (m + 2)π
2
(
m−1
4l
2 ,
m odd
2
2 (−1)
.
= mπ v(4−m )
0,
m even
This doesn’t work for m = 2, since the expression goes to infinity. This case must be
handled separately. From above,
1
B2 =
πv
Z
l/2
dx sin2
0
2πx
l
=
,
l
4πv
where we integrated by parts. Therefore the final answer is
y(x, t) =
l
2πx
2πvt
4l X (−1)(m−1)/2
mπx
mπvt
sin
sin
+ 2
sin
sin
. 2
4πv
l
l
π v m odd m(4 − m )
l
l
4. Boas, p. 638, problem 13.4-11: just solve problem [3 or 4], for the end pinned at x = 0
and free at x = l.
5. You are stranded on an elliptical desert island. In the center is a superhot volcano.
The shore is nice and cool but has tidal waves and sharks with lasers. You need to
3
find the right place to be on the island that’s far from shore but not too hot. For that
you need to solve the Laplace equation in coordinates given by:
x = a cosh u cos v,
y = a sinh u sin v.
Fortunately, you remember that in general, you can write the Laplacian as:
∂
1 X ∂ √
g (gij )−1
U (~x),
∇2 U (~x) = √
g i,j ∂xi
∂xj
where xi = (u, v) for i = (1, 2), g = det gij , (gij )−1 is the matrix inverse of gij , and gij
is given by:
X
dl2 =
gij dxi dxj .
i,j
The latter defines the gij in term of the usual Euclidean distance
dl2 = dx2 + dy 2
once you find the differentials dxi in terms of dx and dy.
(a) Show that lines of constant u are ellipses and lines of constant v are hyperbolae.
Solution:
To show that curves of constant u = u0 are ellipses, we must find constants A
and B such that
y2
x2
+
= 1.
A2 B 2
Substituting in for x and y gives
y2
a2 cosh2 u0
a2 sinh2 u0
x2
2
+
=
cos v +
sin2 v.
A2 B 2
A2
B2
By taking A = a cosh u0 and B = a sinh u0 , this becomes cos2 v + sin2 v = 1.
Thus curves of constant u are ellipses with axes a cosh u0 and a sinh u0 .
Similarly, to show that curves of constant v = v0 are hyperbolae, we must find
constants C and D such that
x2
y2
−
= 1.
C 2 D2
Substituting in for x and y gives
x2
y2
a2 cos2 v0
a2 sin2 v0
2
−
=
cosh
u
−
sin2 u.
C 2 D2
A2
B2
By taking C = a cos v0 and D = a sin v0 , this becomes cosh2 u − sinh2 u = 1.
Thus curves of constant u are hyperbolae with axes a cos v0 and a sin v0 .
4
(b) Find the expression for the Laplacian ∇2 U (~x) in the u − v coordinates using this
method.
In terms of du and dv, the differentials of the Cartesian coordinates are
∂x
∂x
du +
dv
∂u
∂v
= a sinh u cos vdu − a cosh u sin vdv
∂y
∂y
dy =
du +
dv
∂u
∂v
= a cosh u sin vdu + a sinh u cos vdv.
dx =
The line element is therefore
dl2 = dx2 + dy 2
= a2 sinh2 u cos2 vdu2 − 2 cosh u sinh u cos v sin vdudv + cosh2 u sin2 vdv 2
+ a2 cosh2 u sin2 vdu2 + 2 cosh u sinh u cos v sin vdudv + sinh2 u cos2 vdv 2
= a2 sinh2 u cos2 v + cosh2 u sin2 v (du2 + dv 2 ).
We can find the metric gij in elliptical coordinates by comparing the above expression with the definition of the metric:
dl2 = guu du2 + guv dudv + gvu dvdu + gvv dv 2 = guu du2 + 2guv dudv + gvv dv 2
=⇒ guu = gvv = a2 sinh2 u cos2 v + cosh2 u sin2 v .
In matrix form, this is
gij = a
2
1 0
sinh u cos v + cosh u sin v
.
0 1
2
2
2
2
This implies that
2
g ≡ det gij = a4 sinh2 u cos2 v + cosh2 u sin2 v
√
=⇒ g = a2 sinh2 u cos2 v + cosh2 u sin2 v ,
−1 1 0
2
2
−1
−2
2
2
gij = a
sinh u cos v + cosh u sin v
0 1
1 1 0
=√
.
g 0 1
Therefore
√
5
ggij−1 = δij ,
so
1 X ∂ √ −1 ∂
ggij
U (x)
∇2 U (x) = √
g i,j ∂xi
∂xj
1 X ∂
∂
=√
δij
U (x)
g i,j ∂xi ∂xj
1 X ∂2
=√
U (x)
g i ∂x2i
=
a2
2
1
∂
∂2
U (x). +
sinh2 u cos2 v + cosh2 u sin2 v ∂u2 ∂v 2
6. Boas, p. 643, problem 13.5-3(a)
Find the steady-state temperature distribution in a solid cylinder of height 10 and
radius 1 if the top and curved surfaces are held at 0◦ and the base at 100◦ .
We have to solve the Laplace equation
~ 2 u(r, θ, z) = 0
∇
subject to the boundary conditions
u(r, θ, 10) = u(1, θ, z) = 0 ,
u(r, θ, 0) = 100 .
(17)
In cylindrical coordinates, Laplace’s equation readsm[cf. eq. (5.3) on p. 639 of Boas]:
1 ∂
∂u
1 ∂ 2u ∂ 2u
(18)
r
+ 2 2 + 2 = 0.
r ∂r
∂r
r ∂θ
∂z
Employing the separation of variables technique, we write u(r, θ, z) = R(r)Θ(θ)Z(z)
and manipulate eq. (18) into the following form,
1 d2 Z
11 d
dR
1 1 d2 Θ
=
−
r
−
= K2 ,
(19)
Z dz 2
R r dr
dr
Θ r2 dθ2
where K 2 is the separation constant. By convention, we take K to be non-negative.
The Z equation subject to the boundary condition Z(10) = 0 [which is a consequence
of the first boundary condition in eq. (17)] is
Z(z) = sinh K(10 − z) .
Next we use eq. (19) to separate the variables r and θ. This yields
1 d2 Θ
r d
dR
−
=
r
+ K 2 r 2 = n2 ,
Θ dθ2
R dr
dr
6
(20)
where n2 is a second separation constant. The solution to Θ, subject to the condition
that Θ(θ + 2π) = Θ(θ) [since in cylindrical coordinates, (r, θ, z) and (r, θ + 2π, z)
represent the same point], is
(
einθ
Θ(θ) =
for n = 0, 1, 2, 3, . . . .
e−inθ
Note that there is only one allowed solution corresponding to n = 0, since the second
solution, Θ(θ) = θ does not satisfy the periodicity requirement. Finally, eq. (20) yields
the radial equation,
r2 R00 + rR + (K 2 r2 − n2 )R = 0 ,
where R0 ≡ dR/dr, etc. We recognize this as the Bessel equation, whose solutions are
(
Jn (Kr)
R(r) =
Nn (Kr)
Since the problem asks us to solve for the temperature inside the cylinder (which
includes r = 0), we must reject Nn (Kr) as a possible solution since this diverges as
r → 0. Hence, the general solution to Laplace’s equation prior to imposing the second
an third boundary condition in eq. (17) is
u(r, θ, z) = Jn (Kr) sinh K(10 − z)e±inθ ,
for n = 0, 1, 2, 3, . . . .
Noting that the boundary conditions are independent of θ, it follows that only solutions
with n = 0 are allowed. Hence, the allowed solutions are of the form
u(r, θ, z) = J0 (Kr) sinh K(10 − z) .
If we now impose the boundary condition u(1, θ, z) = 0, it follows that J0 (K) = 0.
That is K must be a zero of the Bessel function J0 . We label these zeros by km
where 0 < k1 < k2 < · · · . Hence, the most general solution prior to imposing the last
boundary condition of eq. (17) is
u(r, θ, z) =
∞
X
cm J0 (km r) sinh[km (10 − z)] .
m=1
The coefficients cm will be determined by imposing the boundary condition u(r, θ, 0) =
100.
∞
X
u(r, θ, 0) = 100 =
cm J0 (km r) sinh(10km ) .
(21)
m=1
This is a Fourier-Bessel series. Recall that the set of functions {J0 (km r)} are an
orthogonal set (with weight function r) on the interval 0 ≤ r ≤ 1. In particular,
eq. (19.10) on p. 602 of Boas is
Z 1
rJ0 (km r)J0 (kn r) dr = 21 [J1 (kn )]2 δmn .
0
7
Thus, if we multiply eq. (21) by rJ0 (kn r) and integrate from 0 to 1, we obtain
Z 1
Z 1
∞
X
rJ0 (km r)J0 (kn r) dr
rJ0 (kn r) dr =
cm sinh(10km )
100
0
0
m=1
=
∞
X
cm sinh(10km ) 21 [J1 (kn )]2 δmn
m=1
= cn sinh(10kn ) 21 [J1 (kn )]2 .
Hence,
200
cn =
sinh(10kn )[J1 (kn )]2
Z
1
rJ0 (kn r) dr .
(22)
0
The integral above is obtained by using the recursion relation given in eq. (15.1) on
p. 592 of Boas,
d p
x Jp (x) = xp Jp−1 (x) .
dx
Setting x = kn r and p = 1 and integrating both sides of the above equation yields,
1
Z 1
2
kn
r J0 (kn r) dr = kn r J1 (kn r) = kn J1 (kn ) .
0
0
Using this result in eq. (22), we end up with
cn =
200
.
kn J1 (kn ) sinh(10kn )
Inserting this result back into eq. (21) yields
∞
X
J0 (km r) sinh[km (10 − z)]
.
u(r, θ, z) = 200
km J1 (km ) sinh(10km )
m=1
7. Boas, p. 643, problem 13.5-4
Since the system’s boundary conditions are azimuthally symmetric, the final solution
will not have any angular dependence, and the heat equation we must solve is
∇2 u(r, t) =
1 ∂u(r, t)
.
α2 ∂t
We assume that u(r, t) is a linear combination of separable solutions of the form
R(r)T (t). Substituting this into the equation and dividing through by R(r)T (t) gives
1 2
1 dT
∇ R= 2
.
R
α T dt
Taking the separation constant to by −k 2 gives an equation that can easily be solved
for T :
1 dT
2 2
= −k 2 =⇒ T (t) = e−k α t .
2
α T dt
8
There is only one solution for T since the heat equation is first order in time. We
now see why we chose the separation constant to be −k 2 : it causes the temperature
fluctuations to die off at t → ∞, in accordance with our physical intuition.
The R equation becomes a Bessel equation:
dR
d2 R 1 dR
1 d
2
r
= 2 +
= −k 2 R
∇ R=
r dr
dr
dr
r dr
d2 R
dR
=⇒ r2 2 + r
+ k2 r2 R = 0
dr
dr
J0 (kr)
=⇒ R(r) =
→ J0 (kr).
N0 (kr)
The Neumann function is discarded since it diverges at r = 0, which is in the domain
relevant to the problem. Since u(a, t) = 0 for t ≥ 0, we require that J0 (ka) = 0,
which means k = km /a, where km is a zero of J0 (x). Our solution is therefore a linear
combination of the eigenfunctions we have found:
∞
X
km r −km
2 α2 t/a2
cm J0
u(r, t) =
e
.
a
m=1
Finally, we must determine the coefficients cm using the initial conditions provided:
∞
X
km r
cm J0
.
u(r, 0) = 100 =
a
m=1
Letting w ≡ r/a, we use orthogonality to find cm :
100 =
∞
X
cm J0 (km w)
m=1
Z
100
1
dw wJ0 (kn w) =
0
X
Z
cm
0
m=1
1
1
dw J0 (kn w)J0 (km w) = cn J12 (km ).
2
Since
d
[wJ1 (kn w)] = kn wJ0 (kn w),
dw
the LHS integral becomes
Z
Z
1
100 1
100 1
d
100
2
cn J1 (km ) =
dw kn wJ0 (kn w) =
dw
[wJ1 (kn w)] =
J1 (kn )
2
kn 0
kn 0
dw
kn
200
=⇒ cn =
.
kn J1 (kn )
Therefore
∞
X
200
u(r, t) =
J0
k J (k )
m=1 m 1 m
8. Boas, p. 643, problem 13.5-14
9. Boas, p. 646, problem 13.6-3
9
km r
a
2
e−km α
2 t/a2
.