Solution 37.19
u
xx
+ u
yy
= f(x, y), 0 < x < a, 0 < y < b,
u(0, y) = u(a, y) = 0, u
y
(x, 0) = u
y
(x, b) = 0
We will solve this problem with an eigenfunction expansion in x. To determine a suitable set of eigenfunctions, we
substitute the separation of variables u(x, y) = X(x)Y (y) into the homogeneous partial differential equation.
u
xx
+ u
yy
= 0
(XY )
xx
+ (XY )
yy
= 0
X

X
= −
Y

Y

= −λ
2
With the boundary conditions at x = 0, a, we have the regular Sturm-Liouville problem,
X

= −λ
2
X, X(0) = X(a) = 0,
which has the solutions,
λ
n
=
nπ
a
, X
n
= sin

nπx
a

, n ∈ Z
+
.
We expand u(x, y) in a series of the eigenfunctions.
u(x, y) =
∞

n=1
u

n
(y) sin

nπx
a

We substitute this series into the partial differential equation and boundary conditions at y = 0, b.
∞

n=1

−

nπ
a

2
u
n
(y) sin

nπx
a

+ u

n
(y) sin

nπx

a


= f(x)
∞

n=1
u

n
(0) sin

nπx
a

=
∞

n=1
u

n
(b) sin

nπx
a

= 0
1774
We expand f(x, y) in a Fourier sine series.

f(x, y) =
∞

n=1
f
n
(y) sin

nπx
a

f
n
(y) =
2
a

a
0
f(x, y) sin

nπx
a

dx
We obtain the ordinary differential equations for the coefficients in the expansion.
u

n
(y) −


nπ
a

2
u
n
(y) = f
n
(y), u

n
(0) = u

n
(b) = 0, n ∈ Z
+
.
We will solve these ordinary differential equations with Green functions.
Consider the Green function problem,
g

n
(y; η) −

nπ
a

2
g

n
(y; η) = δ(y −η), g

n
(0; η) = g

n
(b; η) = 0.
The homogeneous solutions
cosh

nπy
a

and cosh

nπ(y −b)
a

satisfy the left and right boundary conditions, respectively. We compute the Wronskian of these two solutions.
W (y) =




cosh(nπy/a) cosh(nπ(y −b)/a)
nπ
a
sinh(nπy/a)
nπ

a
sinh(nπ(y −b)/a)




=
nπ
a

cosh

nπy
a

sinh

nπ(y −b)
a

− sinh

nπy
a

cosh

nπ(y −b)
a


= −
nπ
a
sinh

nπb
a

The Green function is
g
n
(y; η) = −
a cosh(nπy
<
/a) cosh(nπ(y
>
− b)/a)
nπ sinh(nπb/a)
.
1775
The solutions for the coefficients in the expansion are
u
n
(y) =

b
0
g
n
(y; η)f

n
(η) dη.
Solution 37.20
u
tt
+ a
2
u
xxxx
= 0, 0 < x < L, t > 0,
u(x, 0) = f(x), u
t
(x, 0) = g(x),
u(0, t) = u
xx
(0, t) = 0, u(L, t) = u
xx
(L, t) = 0,
We will solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differential
equation and the homogeneous boundary conditions. We will use the initial conditions to determine the coefficients in
the expansion. We substitute the separation of variables, u(x, t) = X(x)T(t) into the partial differential equation.
(XT)
tt
+ a
2
(XT)
xxxx
= 0
T


a
2
T
= −
X

X
= −λ
4
Here we make the assumption that 0 ≤ arg(λ) < π/2, i.e., λ lies in the first quadrant of the complex plane. Note that
λ
4
covers the entire complex plane. We have the ordinary differential equation,
T

= −a
2
λ
4
T,
and with the boundary conditions at x = 0, L, the eigenvalu e problem,
X

= λ
4
X, X(0) = X

(0) = X(L) = X

(L) = 0.

For λ = 0, the general solution of the differential equation is
X = c
1
+ c
2
x + c
3
x
2
+ c
4
x
3
.
1776
Only the trivial solution satisfies the boundary conditions. λ = 0 is not an eigenvalue. For λ = 0, a set of linearly
independent solutions is
{
e
λx
,
e
ıλx
,
e
−λx
,
e
−ıλx
}.

Another linearly independent set, (which wil l be more useful for this problem), is
{cos(λx), sin(λx), cosh(λx), sinh(λx)}.
Both sin(λx) and sinh(λx) satisfy the left boundary conditions. Consider the linear combination c
1
cos(λx)+c
2
cosh(λx).
The left boundary conditions impose the two constraints c
1
+ c
2
= 0, c
1
− c
2
= 0. Only the trivial linear combin ation
of cos(λx) and cosh(λx) can satisfy the lef t boundary condition. Thus the solution has the form,
X = c
1
sin(λx) + c
2
sinh(λx).
The right boundary conditions impose the constraints,

c
1
sin(λL) + c
2
sinh(λL) = 0,
−c

1
λ
2
sin(λL) + c
2
λ
2
sinh(λL) = 0

c
1
sin(λL) + c
2
sinh(λL) = 0,
−c
1
sin(λL) + c
2
sinh(λL) = 0
This set of equations has a nontrivial solution if and only if the determinant is zero,




sin(λL) sinh(λL)
−sin(λL) sinh(λL)





= 2 sin(λL) sinh(λL) = 0.
Since sinh(z) is nonzero in 0 ≤ arg(z) < π/2, z = 0, and sin(z) has the zeros z = nπ, n ∈ N in this domain, the
eigenvalues and eigenfunctions are,
λ
n
=
nπ
L
, X
n
= sin

nπx
L

, n ∈ N.
1777
The differential equation for T becomes,
T

= −a
2

nπ
L

4
T,
which has the solutions,


cos

a

nπ
L

2
t

, sin

a

nπ
L

2
t

.
The eigen-solutions of the partial differential equation are,
u
(1)
n
= sin

nπx
L


cos

a

nπ
L

2
t

, u
(2)
n
= sin

nπx
L

sin

a

nπ
L

2
t

, n ∈ N.
We expand the solution of the partial differential equation in a series of the eigen-solutions.

u(x, t) =
∞

n=1
sin

nπx
L


c
n
cos

a

nπ
L

2
t

+ d
n
sin

a

nπ
L


2
t

The initial condition for u(x, t) and u
t
(x, t) allow us to determine the coefficien ts in the expansion.
u(x, 0) =
∞

n=1
c
n
sin

nπx
L

= f(x)
u
t
(x, 0) =
∞

n=1
d
n
a

nπ

L

2
sin

nπx
L

= g(x)
c
n
and d
n
are coefficients in Fourier sine series.
c
n
=
2
L

L
0
f(x) sin

nπx
L

dx
d
n

=
2L
aπ
2
n
2

L
0
g(x) sin

nπx
L

dx
1778
Solution 37.21
u
t
= κu
xx
+ I
2
αu, 0 < x < L, t > 0,
u(0, t) = u(L, t) = 0, u(x, 0) = g(x).
We will solve this problem with an expansion in eigen-solutions of the partial differential equation. We substitute the
separation of variables u(x, t) = X(x)T(t) into the partial differential equation.
(XT)
t
= κ(XT)

xx
+ I
2
αXT
T

κT
−
I
2
α
κ
=
X

X
= −λ
2
Now we have an ordinary differential equation for T and a Sturm-Liouville eigenvalue problem for X. (Note that we
have followed the rule of thumb that the problem will be easier if we move all the parameters out of the eigenvalue
problem.)
T

= −

κλ
2
− I
2
α


T
X

= −λ
2
X, X(0) = X(L) = 0
The eigenvalues and eigenfunctions for X are
λ
n
=
nπ
L
, X
n
= sin

nπx
L

, n ∈ N.
The differential equation for T becomes,
T

n
= −

κ

nπ

L

2
− I
2
α

T
n
,
which has the solution,
T
n
= c exp

−

κ

nπ
L

2
− I
2
α

t

.

1779
From this solution, we see that the critical current is
I
CR
=

κ
α
π
L
.
If I is greater that this, then the eigen-solution for n = 1 will be exponentially growing. This would make the whole
solution exponentially growing. For I < I
CR
, each of the T
n
is exponentially decaying. The eigen-solutions of the
partial differential equation are,
u
n
= exp

−

κ

nπ
L

2

− I
2
α

t

sin

nπx
L

, n ∈ N.
We expand u(x, t) in its eigen-solutions, u
n
.
u(x, t) =
∞

n=1
a
n
exp

−

κ

nπ
L


2
− I
2
α

t

sin

nπx
L

We determine the coefficients a
n
from the initial condition.
u(x, 0) =
∞

n=1
a
n
sin

nπx
L

= g(x)
a
n
=

2
L

L
0
g(x) sin

nπx
L

dx.
If α < 0, then the solution is exponentially decaying regardless of current. Thus there is no critical current.
Solution 37.22
1780
a) The problem is
u
t
(x, y, z, t) = κ∆u(x, y, z, t), −∞ < x < ∞, −∞ < y < ∞, 0 < z < a, t > 0,
u(x, y, z, 0) = T, u(x, y, 0, t) = u(x, y, a, t) = 0.
Because of symmetry, the partial differential equation in four variables is reduced to a problem in two variables,
u
t
(z, t) = κu
zz
(z, t), 0 < z < a, t > 0,
u(z, 0) = T, u(0, t) = u(a, t) = 0.
We will solve this problem with an expansion in eigen-solutions of the partial differential equation that satisfy the
homogeneous boun dary conditions. We substitute the separation of variables u(z, t) = Z(z)T (t) into the partial
differential equation.
ZT


= κZ

T
T

κT
=
Z

Z
= −λ
2
With the boundary conditions at z = 0, a we have the Sturm-Liouville eigenvalue problem,
Z

= −λ
2
Z, Z(0) = Z(a) = 0,
which has the solutions,
λ
n
=
nπ
a
, Z
n
= sin

nπz

a

, n ∈ N.
The problem for T becomes,
T

n
= −κ

nπ
a

2
T
n
,
with the solution,
T
n
= exp

−κ

nπ
a

2
t

.

1781
The eigen-solutions are
u
n
(z, t) = sin

nπz
a

exp

−κ

nπ
a

2
t

.
The solution for u is a linear combination of the eigen-solutions. The slowest decaying eigen-solution is
u
1
(z, t) = sin

πz
a

exp


−κ

π
a

2
t

.
Thus the e-folding time is
∆
e
=
a
2
κπ
2
.
b) The problem is
u
t
(r, θ, z, t) = κ∆u(r, θ, z, t), 0 < r < a, 0 < θ < 2π, −∞ < z < ∞, t > 0,
u(r, θ, z, 0) = T, u(0, θ, z, t) is bounded, u(a, θ, z, t) = 0.
The Laplacian in cylindrical coordinates is
∆u = u
rr
+
1
r
u

r
+
1
r
2
u
θθ
+ u
zz
.
Because of symmetry, the solution does not depend on θ or z.
u
t
(r, t) = κ

u
rr
(r, t) +
1
r
u
r
(r, t)

, 0 < r < a, t > 0,
u(r, 0) = T, u(0, t) is bounded, u(a, t) = 0.
We will solve this problem with an expansion in eigen-solutions of the partial differential equation that satisfy
the homogeneous boundary conditions at r = 0 and r = a. We substitute the separation of variables u(r, t) =
1782
R(r)T (t) into the partial differential equation.

RT

= κ

R

T +
1
r
R

T

T

κT
=
R

R
+
R

rR
= −λ
2
We have the eigenvalue problem,
R

+

1
r
R

+ λ
2
R = 0, R(0) is bounded, R(a) = 0.
Recall that the Bessel equation,
y

+
1
x
y

+

λ
2
−
ν
2
x
2

y = 0,
has the general solution y = c
1
J
ν

(λx) + c
2
Y
ν
(λx). We discard the Bessel function of the second kind, Y
ν
, as it
is unbounded at the origin. The solution for R(r) is
R(r) = J
0
(λr).
Applying the boundary condition at r = a, we see that the eigenvalues and eigenfunctions are
λ
n
=
β
n
a
, R
n
= J
0

β
n
r
a

, n ∈ N,
where {β

n
} are the positive roots of the Bessel function J
0
.
The differential equation for T becomes,
T

n
= −κ

β
n
a

2
T
n
,
which has the solutions,
T
n
= exp

−κ

β
n
a

2

t

.
1783
The eigen-solutions of the partial differential equation for u(r, t) are,
u
n
(r, t) = J
0

β
n
r
a

exp

−κ

β
n
a

2
t

.
The solution u(r, t) is a linear combination of the eigen- solutions, u
n
. The slowest decaying eigenfunction is,

u
1
(r, t) = J
0

β
1
r
a

exp

−κ

β
1
a

2
t

.
Thus the e-folding time is
∆
e
=
a
2
κβ
2

1
.
c) The problem is
u
t
(r, θ, φ, t) = κ∆u(r, θ, φ, t), 0 < r < a, 0 < θ < 2π, 0 < φ < π, t > 0,
u(r, θ, φ, 0) = T, u(0, θ, φ, t) is bounded, u(a, θ, φ, t) = 0.
The Laplacian in spherical coordinates is,
∆u = u
rr
+
2
r
u
r
+
1
r
2
u
θθ
+
cos θ
r
2
sin θ
u
θ
+
1

r
2
sin
2
θ
u
φφ
.
Because of symmetry, the solution does not depend on θ or φ.
u
t
(r, t) = κ

u
rr
(r, t) +
2
r
u
r
(r, t)

, 0 < r < a, t > 0,
u(r, 0) = T, u(0, t) is bounded, u(a, t) = 0
1784
We will solve this problem with an expansion in eigen-solutions of the partial differential equation that satisfy
the homogeneous boundary conditions at r = 0 and r = a. We substitute the separation of variables u(r, t) =
R(r)T (t) into the partial differential equation.
RT


= κ

R

T +
2
r
R

T

T

κT
=
R

R
+
2
r
R

R
= −λ
2
We have the eigenvalue problem,
R

+

2
r
R

+ λ
2
R = 0, R(0) is bounded, R(a) = 0.
Recall that the equation,
y

+
2
x
y

+

λ
2
−
ν(ν + 1)
x
2

y = 0,
has the general solution y = c
1
j
ν
(λx) + c

2
y
ν
(λx), where j
ν
and y
ν
are the spherical Bessel functions of the first
and second kind. We discard y
ν
as it is unbounded at the origin. (The spherical Bessel functions are related to
the Bessel functions by
j
ν
(x) =

π
2x
J
ν+1/2
(x).)
The solution for R(r) is
R
n
= j
0
(λr).
Applying the boundary condition at r = a, we see that the eigenvalues and eigenfunctions are
λ
n

=
γ
n
a
, R
n
= j
0

γ
n
r
a

, n ∈ N.
The problem for T becomes
T

n
= −κ

γ
n
a

2
T
n
,
1785

which has the solutions,
T
n
= exp

−κ

γ
n
a

2
t

.
The eigen-solutions of the partial differential equation are,
u
n
(r, t) = j
0

γ
n
r
a

exp

−κ


γ
n
a

2
t

.
The slowest decaying eigen-solution is,
u
1
(r, t) = j
0

γ
1
r
a

exp

−κ

γ
1
a

2
t


.
Thus the e-folding time is
∆
e
=
a
2
κγ
2
1
.
d) If the edges are perfectly insulated, then n o heat escapes through the boundary. The temperature i s constant for
all time. There is no e-folding time.
Solution 37.23
We will solve this problem with an eigenfun ction expansion. Since the partial differential equation is homogeneous, we
will find eigenfunctions in both x and y. We substitute the separation of variables u(x, y, t) = X(x)Y (y)T (t) into the
partial differential equation.
XY T

= κ(t) (X

Y T + XY

T )
T

κ(t)T
=
X


X
+
Y

Y
= −λ
2
X

X
= −
Y

Y
− λ
2
= −µ
2
1786
First we have a Sturm-Liouville eigenvalue problem for X,
X

= µ
2
X, X

(0) = X

(a) = 0,
which has the solutions,

µ
m
=
mπ
a
, X
m
= cos

mπx
a

, m = 0, 1, 2, . . . .
Now we have a Sturm-Liouville eigenvalue problem for Y ,
Y

= −

λ
2
−

mπ
a

2

Y, Y (0) = Y (b) = 0,
which has the solutions,
λ

mn
=


mπ
a

2
+

nπ
b

2
, Y
n
= sin

nπy
b

, m = 0, 1, 2, . . . , n = 1, 2, 3, . . . .
A few of the eigenfunctions, cos

mπx
a

sin

nπy

b

, are shown in Figure 37.3.
The differential equation for T becomes,
T

mn
= −


mπ
a

2
+

nπ
b

2

κ(t)T
mn
,
which has the solutions,
T
mn
= exp

−



mπ
a

2
+

nπ
b

2


t
0
κ(τ) dτ

.
The eigen-solutions of the partial differential equation are,
u
mn
= cos

mπx
a

sin

nπy

b

exp

−


mπ
a

2
+

nπ
b

2


t
0
κ(τ) dτ

.
The solution of the partial differential equation is,
u(x, y, t) =
∞

m=0
∞


n=1
c
mn
cos

mπx
a

sin

nπy
b

exp

−


mπ
a

2
+

nπ
b

2



t
0
κ(τ) dτ

.
1787
We determine the coefficients from the initial condition.
u(x, y, 0) =
∞

m=0
∞

n=1
c
mn
cos

mπx
a

sin

nπy
b

= f(x, y)
c
0n

=
2
ab

a
0

b
0
f(x, y) sin

nπ
b

dy dx
c
mn
=
4
ab

a
0

b
0
f(x, y) cos

mπ
a


sin

nπ
b

dy dx
Solution 37.24
The steady state temperature satisfies Laplace’s equation, ∆u = 0. The Laplacian in cyli ndric al coordinates is,
∆u(r, θ, z) = u
rr
+
1
r
u
r
+
1
r
2
u
θθ
+ u
zz
.
Because of the homogeneity in the z direction, we reduce the partial differential equation to,
u
rr
+
1

r
u
r
+
1
r
2
u
θθ
= 0, 0 < r < 1, 0 < θ < π.
The boundary conditions are,
u(r, 0) = u(r, π) = 0, u(0, θ) = 0, u(1, θ) = 1.
We will solve this problem with an eigenfunction expansion. We substitute the separation of variables u(r, θ) = R(r)T (θ)
into the partial differential equation.
R

T +
1
r
R

T +
1
r
2
RT

= 0
r
2

R

R
+ r
R

R
= −
T

T
= λ
2
1788
We have the regular Sturm-Liouville eigenvalue problem,
T

= −λ
2
T, T(0) = T (π) = 0,
which has the solutions,
λ
n
= n, T
n
= sin(nθ), n ∈ N.
The problem for R becomes,
r
2
R


+ rR

− n
2
R = 0, R(0) = 0.
This is an Euler equation. We substitute R = r
α
into the differential equation to obtain,
α(α − 1) + α − n
2
= 0,
α = ±n.
The general solution of the differential equation for R is
R
n
= c
1
r
n
+ c
2
r
−n
.
The solution that vanishes at r = 0 is
R
n
= cr
n

.
The eigen-solutions of the differential equation are,
u
n
= r
n
sin(nθ).
The solution of the partial differential equation is
u(r, θ) =
∞

n=1
a
n
r
n
sin(nθ).
We determine the coefficients from the boundary condition at r = 1.
u(1, θ) =
∞

n=1
a
n
sin(nθ) = 1
a
n
=
2
π


π
0
sin(nθ) dθ =
2
πn
(1 − (−1)
n
)
1789
The solution of the partial differential equation is
u(r, θ) =
4
π
∞

n=1
odd n
r
n
sin(nθ).
Solution 37.25
The problem is
u
xx
+ u
yy
= 0, 0 < x, 0 < y < 1,
u(x, 0) = u(x, 1) = 0, u(0, y) = f(y).
We substitute the separation of variables u(x, y) = X(x)Y (y) into the partial differential equation.

X

Y + XY

= 0
X

X
= −
Y

Y
= λ
2
We have the regular Sturm-Liouville problem,
Y

= −λ
2
Y, Y (0) = Y (1) = 0,
which has the solutions,
λ
n
= nπ, Y
n
= sin(nπy), n ∈ N.
The problem for X becomes,
X

n

= (nπ)
2
X,
which has the general solution,
X
n
= c
1
e
nπx
+c
2
e
−nπx
.
The solution that is bounded as x → ∞ is,
X
n
= c
e
−nπx
.
1790
The eigen-solutions of the partial differential equation are,
u
n
=
e
−nπx
sin(nπy), n ∈ N.

The solution of the partial differential equation is,
u(x, y) =
∞

n=1
a
n
e
−nπx
sin(nπy).
We find the coefficients from the boundary condition at x = 0.
u(0, y) =
∞

n=1
a
n
sin(nπy) = f(y)
a
n
= 2

1
0
f(y) sin(nπy) dy
Solution 37.26
The Laplacian in polar coordinates is
∆u ≡ u
rr
+

1
r
u
r
+
1
r
2
u
θθ
.
Since we have homogeneous boundary conditions at θ = 0 and θ = α, we will solve this problem with an e igenfun ction
expansion. We substitute the separation of variables u(r, θ) = R(r)Θ(θ) into Laplace’s equation.
R

Θ +
1
r
R

Θ +
1
r
2
RΘ

= 0
r
2
R


R
+ r
R

R
= −
Θ

Θ
= λ
2
.
1791
We have a regular Sturm-Liouville eigenvalue problem for Θ.
Θ

= −λ
2
Θ, Θ(0) = Θ(α) = 0
λ
n
=
nπ
α
, Θ
n
= sin

nπθ

α

, n ∈ Z
+
.
We have Euler equations for R
n
. We solve them with the substitution R = r
β
.
r
2
R

n
+ rR

n
−

nπ
α

2
R
n
= 0, R
n
(a) = 0
β(β − 1) + β −


nπ
α

2
= 0
β = ±
nπ
α
R
n
= c
1
r
nπ/α
+ c
2
r
−nπ/α
.
The solution, (up to a multiplicative constant), that vanishes at r = a is
R
n
= r
nπ/α
− a
2nπ/α
r
−nπ/α
.

Thus the series expansion of our solution is,
u(r, θ) =
∞

n=1
u
n

r
nπ/α
− a
2nπ/α
r
−nπ/α

sin

nπθ
α

.
We determine the coefficients from the boundary condition at r = b.
u(b, θ) =
∞

n=1
u
n

b

nπ/α
− a
2nπ/α
b
−nπ/α

sin

nπθ
α

= f(θ)
u
n
=
2
α (b
nπ/α
− a
2nπ/α
b
−nπ/α
)

α
0
f(θ) sin

nπθ
α


dθ
1792
Solution 37.27
a) The mathematical statement of the problem is
u
tt
= c
2
u
xx
, 0 < x < L, t > 0,
u(0, t) = u(L, t) = 0,
u(x, 0) = 0, u
t
(x, 0) =

v for |x − ξ| < d
0 for |x − ξ| > d.
Because we are interest in the harmonics of the m otion, we will solve this problem with an eigenfunction expansion in
x. We substitute the separation of variables u(x, t) = X(x)T (t) into the wave equation.
XT

= c
2
X

T
T


c
2
T
=
X

X
= −λ
2
The eigenvalue problem for X is,
X

= −λ
2
X, X(0) = X(L) = 0,
which has the solutions,
λ
n
=
nπ
L
, X
n
= sin

nπx
L

, n ∈ N.
The ordinary differential equation for the T

n
are,
T

n
= −

nπc
L

2
T
n
,
which have the linearly independent solutions,
cos

nπct
L

, sin

nπct
L

.
1793
The solution for u(x, t) is a linear combination of the eigen-solutions.
u(x, t) =
∞


n=1
sin

nπx
L


a
n
cos

nπct
L

+ b
n
sin

nπct
L

Since the string initially has zero displacemen t, each of the a
n
are zero.
u(x, t) =
∞

n=1
b

n
sin

nπx
L

sin

nπct
L

Now we use the initial velocity to determine the coefficients in the expansion. Because the position is a continuous
function of x, and there is a jump discontinuity in the velocity as a function of x, the coefficients in the expansion will
decay as 1/n
2
.
u
t
(x, 0) =
∞

n=1
nπc
L
b
n
sin

nπx
L


=

v for |x − ξ| < d
0 for |x − ξ| > d.
nπc
L
b
n
=
2
L

L
0
u
t
(x, 0) sin

nπx
L

dx
b
n
=
2
nπc

ξ+d

ξ−d
v sin

nπx
L

dx
=
4Lv
n
2
π
2
c
sin

nπd
L

sin

nπξ
L

The solution for u(x, t) is,
u(x, t) =
4Lv
π
2
c

∞

n=1
1
n
2
sin

nπd
L

sin

nπξ
L

sin

nπx
L

sin

nπct
L

.
1794
b) The form of the solution is again,
u(x, t) =

∞

n=1
b
n
sin

nπx
L

sin

nπct
L

We determine the coefficients in the expansion from the initial velocity.
u
t
(x, 0) =
∞

n=1
nπc
L
b
n
sin

nπx
L


=

v cos

π(x−ξ)
2d

for |x − ξ| < d
0 for |x − ξ| > d.
nπc
L
b
n
=
2
L

L
0
u
t
(x, 0) sin

nπx
L

dx
b
n

=
2
nπc

ξ+d
ξ−d
v cos

π(x − ξ)
2d

sin

nπx
L

dx
b
n
=

8dL
2
v
nπ
2
c(L
2
−4d
2

n
2
)
cos

nπd
L

sin

nπξ
L

for d =
L
2n
,
v
n
2
π
2
c

2nπd + L sin

2nπd
L

sin


nπξ
L

for d =
L
2n
The solution for u(x, t) is,
u(x, t) =
8dL
2
v
π
2
c
∞

n=1
1
n(L
2
− 4d
2
n
2
)
cos

nπd
L


sin

nπξ
L

sin

nπx
L

sin

nπct
L

for d =
L
2n
,
u(x, t) =
v
π
2
c
∞

n=1
1
n

2

2nπd + L sin

2nπd
L

sin

nπξ
L

sin

nπx
L

sin

nπct
L

for d =
L
2n
.
c) The kinetic energy of the string is
E =
1
2


L
0
ρ (u
t
(x, t))
2
dx,
1795
where ρ is the density of the string per unit length.
Flat Hammer. The n
th
harmonic is
u
n
=
4Lv
n
2
π
2
c
sin

nπd
L

sin

nπξ

L

sin

nπx
L

sin

nπct
L

.
The kinetic energy of the n
th
harmonic is
E
n
=
ρ
2

L
0

∂u
n
∂t

2

dx =
4Lv
2
n
2
π
2
sin
2

nπd
L

sin
2

nπξ
L

cos
2

nπct
L

.
This will be maximized if
sin
2


nπξ
L

= 1,
nπξ
L
=
π(2m − 1)
2
, m = 1, . . . , n,
ξ =
(2m − 1)L
2n
, m = 1, . . . , n
We note that the kinetic energies of the n
th
harmonic decay as 1/n
2
.
Curved Hammer. We assume that d =
L
2n
. The n
th
harmonic is
u
n
=
8dL
2

v
nπ
2
c(L
2
− 4d
2
n
2
)
cos

nπd
L

sin

nπξ
L

sin

nπx
L

sin

nπct
L


.
The kinetic energy of the n
th
harmonic is
E
n
=
ρ
2

L
0

∂u
n
∂t

2
dx =
16d
2
L
3
v
2
π
2
(L
2
− 4d

2
n
2
)
2
cos
2

nπd
L

sin
2

nπξ
L

cos
2

nπct
L

.
1796
This will be maximized if
sin
2

nπξ

L

= 1,
ξ =
(2m − 1)L
2n
, m = 1, . . . , n
We note that the kinetic energies of the n
th
harmonic decay as 1/n
4
.
Solution 37.28
In mathematical notation, the problem is
u
tt
− c
2
u
xx
= s(x, t), 0 < x < L, t > 0,
u(0, t) = u(L, t) = 0,
u(x, 0) = u
t
(x, 0) = 0.
Since this is an inhomogeneous partial differential equation, we will expand the solution in a series of eigenfunctions in
x for which the coefficients are functions of t. The solution for u has the form,
u(x, t) =
∞


n=1
u
n
(t) sin

nπx
L

.
Substituting this expression in to the inhomogeneous partial differential equation will give us ordinary differential equa-
tions for each of the u
n
.
∞

n=1

u

n
+ c
2

nπ
L

2
u
n


sin

nπx
L

= s(x, t).
We expand the right side in a series of the eigenfunctions.
s(x, t) =
∞

n=1
s
n
(t) sin

nπx
L

.
1797