Homework #7 Solutions
Math 182, Spring 2014
Instructor: Dr. Doreen De Leon
1
Exercises 4.1: 1(b), 2(c), 5(d)
1(b) Solve the heat equation ut = 2uxx for a rod of length L with both ends held at 0◦ , if L = 1, u(x, 0) =
x. (See Example 2, Section 3.6.)
Solution: We are solving the initial-boundary-value problem
ut = 2uxx ,
0 < x < 1, t > 0,
t ≥ 0,
u(0, t) = u(1, t) = 0,
u(x, 0) = x,
0 ≤ x ≤ 1.
As before, we use separation of variables to solve, letting u(x, t) = X(x)T (t). Then, we obtain
dT
d2 X
X=2 2T
dt
dx
1 dT
1 d2 X
=⇒
=
= −λ.
2T dt
X dx2
We also must separate the boundary conditions:
u(0, t) = 0 =⇒ X(0)T (t) = 0 =⇒ X(0) = 0,
u(1, t) = 0 =⇒ X(1)T (t) = 0 =⇒ X(1) = 0.
Thus, we obtain the following equations:
d2 X
+ λX = 0,
dx2
dT
+ 2λT = 0.
dt
X(0) = X(1) = 0,
From our previous work, we know that the solution to the eigenvalue problem is
λn = (nπ)2 and Xn (x) = sin nπx, n = 1, 2, . . . ,
and
2
Tn (t) = cn e−2(nπ) t .
Then, for each n, un (x, t) = Xn (x)Tn (t), so
2
un (x, t) = cn e−2(nπ) t sin nπx.
1
By superposition,
u(x, t) =
∞
X
2
cn e−2(nπ) t sin nπx.
n=1
We need to determine the coefficients cn using the initial condition. Since
x = u(x, 0) =
∞
X
cn sin nπx, n = 1, 2, . . . ,
n=1
we see that the right-hand side must be the Fourier sine series of x. By Example 2 in Section 3.6,
then, we have that
2(−1)n+1
.
cn =
nπ
Therefore,
∞
X
2(−1)n+1 −2(nπ)2 t
u(x, t) =
e
sin nπx .
nπ
n=1
2(c) Solve the heat equation ut = 4uxx for a rod of length L with both ends insulated, if L = 2,
(
10, if 0 ≤ x ≤ 1,
u(x, 0) =
.
0, if 1 < x ≤ 2.
Solution: We are solving the initial-boundary-value problem
ut = 4uxx ,
0 < x < 1, t > 0,
u(0, t) = u(2, t) = 0, t > 0,
(
10, if 0 ≤ x ≤ 1,
u(x, 0) =
0, if 1 < x ≤ 2.
As before, we use separation of variables to solve, letting u(x, t) = X(x)T (t). Then, we obtain
dT
d2 X
X=4 2T
dt
dx
1 dT
1 d2 X
=⇒
=
= −λ.
4T dt
X dx2
We also must separate the boundary conditions:
dX
dX
(0)T (t) = 0 =⇒
(0) = 0,
dx
dx
dX
dX
ux (2, t) = 0 =⇒
(2)T (t) = 0 =⇒
(2) = 0.
dx
dx
ux (0, t) = 0 =⇒
Thus, we obtain the following equations:
d2 X
+ λX = 0,
dx2
dT
+ 4λT = 0.
dt
dX
dX
(0) =
(2) = 0,
dx
dx
2
From our previous work, we know that the solution to the eigenvalue problem is
λ0 = 0, X0 (x) = 1,
nπ 2
nπx
, Xn (x) = cos
,
λn =
2
2
and
T0 (t) = c0 ,
Tn (t) = cn e−4(
n = 1, 2, . . . ,
nπ 2
t
2
) , n = 1, 2, . . . .
Then, for each n,
u0 (x, t) = c0
un (x, t) = cn e−4(
) cos nπx , n = 1, 2, · · · .
2
nπ 2
t
2
Therefore, by superposition,
u(x, t) = c0 +
∞
X
cn e−4(
n=1
) cos nπx .
2
nπ 2
t
2
Next, we determine the coefficients using the initial condition. Since
u(x, 0) = c0 +
∞
X
cn cos
n=1
nπx
,
2
we know that the right-hand side is the Fourier cosine series of u(x, 0), which we now must determine.
Z
2 2
a0 2c0 =
u(x, 0) dx since c0 =
2 0
2
Z 1
Z 2
=
10 dx +
0 dx
0
1
= 10
=⇒ c0 = 5.
cn =
2
2
Z
2
Z
u(x, 0) cos
0
nπx
dx
2
1
nπx
dx + 0
2
0
20
nπx 1
=
sin
nπ
2 0
20
nπ
=
sin
.
nπ
2
=
10 cos
Therefore, the solution is
u(x, t) = 5 +
∞
X
20
nπ −4( nπ )2 t
nπx
2
sin
e
cos
.
nπ
2
2
n=1
3
5(d) Solve the general heat equation ut = α2 uxx subject to initial conditions u(x, 0) = f (x) and to the
boundary conditions ux (0, t) = u(L, t) = 0.
Solution: We are solving the initial-boundary-value problem
ut = α2 uxx ,
0 < x < L, t > 0,
ux (0, t) = u(L, t) = 0,
u(x, 0) = f (x),
t > 0,
0 < x < L.
As before, we use separation of variables to solve, letting u(x, t) = X(x)T (t). Then, we obtain
d2 X
dT
X = α2 2 T
dt
dx
1 dT
1 d2 X
= −λ.
=⇒ 2
=
α T dt
X dx2
We also must separate the boundary conditions:
dX
dX
(0)T (t) = 0 =⇒
(0) = 0,
dx
dx
u(L, t) = 0 =⇒ X(L)T (t) = 0 =⇒ X(L) = 0.
ux (0, t) = 0 =⇒
From our previous work, we know that the solution to the eigenvalue problem is
(2n − 1)π 2
(2n − 1)πx
λn =
, Xn (x) = cos
, n = 1, 2, . . . ,
2L
2L
and
−α2
(2n−1)π
2L
2
2
(2n − 1)πx
.
2L
Tn (t) = cn e
t
.
Then, for each n,
un (x, t) = cn e
By superposition,
u(x, t) =
∞
X
−α2
−α2
(2n−1)π
2L
cn e
(2n−1)π
2L
t
cos
2
t
n=1
cos
(2n − 1)πx
.
2L
Finally, we need to determine the coefficients cn using the initial condition u(x, 0) = f (x). We have
f (x) = u(x, 0) =
∞
X
n=1
cn cos
(2n − 1)πx
.
2L
From
the discussion
in Section 3.7 and from Exercise 1 in that section, we know that the functions
(2n − 1)πx
cos
form a complete orthogonal set on 0 ≤ x ≤ L, and therefore, we can do the
2L
(2m − 1)πx
following: multiply both sides of the equation by cos
, integrate both sides with respect
2L
to x, and solve. If we do this, we will obtain
Z
2 L
(2n − 1)πx
cn =
f (x) cos
dx.
L 0
2L
Therefore, the solution is
u(x, t) =
∞
X
n=1
cn e
−α2
(2n−1)π
2L
2
t
(2n − 1)πx
2
, cn =
cos
2L
L
4
Z
L
f (x) cos
0
(2n − 1)πx
dx.
2L
2
Exercises 4.2: 2(b), 3(a), 5(a)
2(b) Solve the wave equation utt = 4uxx for a string of length π, subject to the boundary conditions
ux (0, t) = ux (π, t) = 0 and to the initial conditions u(x, 0) = sin x, ut (x, 0) = 0. (See Exercise 3,
Section 3.6.)
Solution: We are solving the initial-boundary-value problem
utt = 4uxx ,
0 < x < π, t > 0,
ux (0, t) = ux (π, t) = 0,
u(x, 0) = sin x,
ut (x, 0) = 0,
t > 0,
0 ≤ x ≤ π,
0 ≤ x ≤ π.
As before, we use separation of variables to solve, letting u(x, t) = X(x)T (t). Then, we obtain
d2 X
d2 T
X
=
4
T
dt2
dx2
1 d2 T
1 d2 X
=⇒
=
= −λ.
4T dt2
X dx2
We also must separate the boundary conditions:
dX
dX
(0)T (t) = 0 =⇒
(0) = 0,
dx
dx
dX
dX
ux (π, t) = 0 =⇒
(π)T (t) = 0 =⇒
(π) = 0.
dx
dx
ux (0, t) = 0 =⇒
Thus, we obtain the following equations:
d2 X
+ λX = 0,
dx2
d2 T
+ 4λT = 0.
dt2
dX
dX
(0) =
(π) = 0,
dx
dx
From our previous work, we know that the solution to the eigenvalue problem is
λ0 = 0,
λn = n2 ,
X0 (x) = 1,
Xn (x) = cos nx,
n = 1, 2, . . . ,
Then, we need to solve the equation in T for both cases of λ.
λ = 0 =⇒ T = c1 + c2 t,
λn = n2 =⇒ T 00 + 4n2 T = 0
=⇒ Tn (t) = c1 cos 2nt + c2 sin 2nt.
So, for each n, un (x, t) = Xn (x)Tn (t), so
u0 (x, t) = c0 + d0 t and un (x, t) = cn cos 2nt cos nx + dn sin 2nt cos nx.
By superposition,
u(x, t) = c0 + d0 t +
∞
X
cn cos 2nt cos nx +
n=1
∞
X
n=1
5
dn sin 2nt cos nx.
We now apply the initial conditions. First,
sin x = u(x, 0) = c0 +
∞
X
cn cos nx.
n=1
Therefore, the right-hand side must be the Fourier cosine series of f (x) = sin x, which we have
computed in Exercise 3.6, problem 3.
∞
Fc (x) =
2 X 2 (1 + (−1)n )
+
·
cos nx.
π
π
n2 − 1
n=1
So, we see that
2
,
nπ
2 (1 + (−1)n )
cn = ·
.
π
n2 − 1
c0 =
We now have
∞
u(x, t) =
∞
X 2 (1 + (−1)n )
X
2
+ d0 t +
·
cos
2nt
cos
nx
+
dn sin 2nt cos nx
nπ
π
n2 − 1
n=1
=⇒ ut (x, t) = d0 +
=⇒ ut (x, 0) = d0 +
∞
X
n=1
∞
X
−
n=1
(−1)n )
4n (1 +
·
π
n2 − 1
sin 2nt cos nx +
∞
X
2ndn cos 2nt cos nx
n=1
2ndn cos nx.
n=1
But, since ut (x, 0) = 0, we have
d0 = 0,
dn = 0, n = 1, 2, . . . .
Therefore, the solution is
∞
u(x, t) =
X 2 (1 + (−1)n )
2
+
·
cos 2nt cos nx .
nπ
π
n2 − 1
n=1
3(a) Solve the wave equation utt = uxx subject to u(0, t) = ux (1, t), u(x, 0) = 0, ut (x, 0) = 1.
Solution: We are solving the initial-boundary-value problem
utt = uxx ,
0 < x < 1, t > 0,
u(0, t) = ux (1, t) = 0,
t > 0,
u(x, 0) = 0,
0 ≤ x ≤ 1,
ut (x, 0) = 1,
0 < x < 1.
As before, we use separation of variables to solve, letting u(x, t) = X(x)T (t). Then, we obtain
d2 X
d2 T
X
=
T
dt2
dx2
1 d2 T
1 d2 X
=⇒
=
= −λ.
T dt2
X dx2
6
We also must separate the boundary conditions:
u(0, t) = 0 =⇒ X(0)T (t) = 0 =⇒ X(0) = 0,
dX
dX
(1)T (t) = 0 =⇒
(1) = 0.
ux (1, t) = 0 =⇒
dx
dx
Thus, we obtain the following equations:
d2 X
+ λX = 0,
dx2
d2 T
+ λT = 0.
dt2
X(0) =
dX
(1) = 0,
dx
From previous work (Exercises 1.7, problem 16(b)), we know that the solution to the eigenvalue
problem, since L = 1, is
(2n − 1)πx
(2n − 1)π 2
, n = 1, 2, . . . ,
λn =
, Xn (x) = sin
2
2
and therefore,
(2n − 1)πt
(2n − 1)πt
+ dn sin
.
2
2
Then, for each n, un (x, t) = Xn (x)Tn (t), so
Tn (t) = cn cos
un (x, t) = cn cos
(2n − 1)πx
(2n − 1)πt
(2n − 1)πx
(2n − 1)πt
sin
+ dn sin
sin
.
2
2
2
2
By superposition,
u(x, t) =
∞
X
n=1
∞
(2n − 1)πt
(2n − 1)πx X
(2n − 1)πt
(2n − 1)πx
cn cos
sin
+
dn sin
sin
.
2
2
2
2
n=1
We now apply the initial conditions. First,
0 = u(x, 0) =
∞
X
cn sin
n=1
(2n − 1)πx
=⇒ cn = 0 for all n.
2
We, thus, have
u(x, t) =
=⇒ ut (x, t) =
=⇒ 1 = ut (x, 0) =
∞
X
n=1
∞
X
n=1
∞
X
n=1
dn sin
(2n − 1)πx
(2n − 1)πt
sin
2
2
(2n − 1)π
(2n − 1)πt
(2n − 1)πx
dn cos
sin
2
2
2
(2n − 1)π
(2n − 1)πx
dn sin
.
2
2
(2n − 1)πx
(2m − 1)πx
is a complete orthogonal set, we multiply both sides by sin
2
2
and integrate with respect to x to find
Z 1
Z 1
∞
X
(2n − 1)π
(2m − 1)πx
(2m − 1)πx
(2n − 1)πx
dx =
dn
sin
sin
dx,
sin
2
2
2
2
0
0
Since
sin
n=1
7
and
(2m − 1)πx
2
(2m − 1)πx 1
sin
dx = −
cos
,
2
(2m − 1)π
2
0
0
(
Z 1
0, if n 6= m,
(2n − 1)πx
(2m − 1)πx
sin
sin
dx = 1
2
2
0
2 , if n = m.
Z
1
=⇒
(2m − 1)π
2
=
dm
(2m − 1)π
4
8
.
=⇒ dm =
(2m − 1)2 π 2
Therefore, we obtain the solution
∞
X
u(x, t) =
n=1
(2n − 1)πt
(2n − 1)πx
8
sin
sin
.
(2n − 1)2 π 2
2
2
5(a) Solve the initial-boundary-value problem
utt = uxx − 4ut ,
u(x, 0) = 1,
ut (x, 0) = 0,
u(0, t) = u(π, t) = 0.
Solution: We will use separation of variables to solve this problem. Let u(x, t) = X(x)T (t). Then
ut = X(x)
d2 T
d2 X
dT
, utt = X(x) 2 , uxx = T (t) 2 .
dt
dt
dx
Substituting into the PDE gives
d2 T
d2 X
dT
=
T
(t)
− 4X(x)
2
2
dt
dx
dt
d2 T
dT
d2 X
X(x) 2 + 4X(x)
= T (t) 2 .
dt
dt
dx
X(x)
Divide both sides by X(x)T (t) to obtain
1 d2 T
dT
1 d2 X
+
4
=
= −λ.
T dt2
dt
X dx2
We also separate the boundary conditions.
u(0, t) = 0 =⇒ X(0)T (t) = 0 =⇒ X(0) = 0
u(π, t) = 0 =⇒ X(π)T (t) = 0 =⇒ X(π) = 0.
Thus, we obtain the following equations:
X 00 + λX = 0,
d2 T
dt2
+4
X(0) = X(π) = 0,
dT
+ λT = 0.
dt
8
From our previous work, we know that the solution to the eigenvalue problem is
λ n = n2 ,
Xn (x) = sin nx, n = 1, 2, . . . .
Then, we must solve the equation in T :
T ”0 + 4T 0 + n2 T = 0
Characteristic equation:
r2 + 4r + n2 = 0
√
16 − 4n2
p2
= −2 ± 4 − n2
r1,2 =
−4 ±
If n = 1, then
r1,2 = −2 ±
√
−2t
=⇒ T1 (t) = c1 e
3
cosh
√
√
3t + d1 e−2t sinh 3t.
If n = 2, then
r1 = r2 = −2
=⇒ T2 (t) = c2 e−2t + d2 te−2t .
If n > 2, then
p
r1,2 = −2 ± i n2 − 4
p
p
=⇒ Tn (t) = cn e−2t cos(t n2 − 4) + dn e−2t sin(t n2 − 4).
This gives the following, then, for un (x, t)
√
√
u1 (x, t) = c1 e−2t cosh 3t sin x + d1 e−2t sinh 3t sin x,
u2 (x, t) = c2 e−2t sin 2x + d2 te−2t sin 2x,
p
p
un (x, t) = cn e−2t cos(t n2 − 4) sin nx + dn e−2t sin(t n2 − 4) sin nx, n = 3, 4, . . . .
By superposition, we have
√
√
u(x, t) = c1 e−2t cosh 3t sin x + d1 e−2t sinh 3t sin x + c2 e−2t sin 2x + d2 te−2t sin 2x
∞
∞
p
p
X
X
cn e−2t cos(t n2 − 4) sin nx +
dn e−2t sin(t n2 − 4) sin nx.
+
n=3
n=3
We now apply the initial conditions. First,
1 = u(x, 0) = c1 sin x + c2 sin 2x +
∞
X
n=3
=
∞
X
cn sin nx.
n=1
9
cn sin nx
Therefore, the right-hand side is the Fourier sine series of u(x, 0) = 1, and so
Z
2 π
sin nx dx
cn = bn =
π 0
2
=−
cos nx|π0
nπ
2
(1 − cos nπ)
=
nπ
2
=
(1 − (−1)n ).
nπ
So, we have
√
√
4
u(x, t) = e−2t cosh 3t sin x + d1 e−2t sinh 3t sin x + d2 te−2t sin 2x
π
∞
∞
p
p
X
X
2
+
(1 − (−1)n )e−2t cos(t n2 − 4) sin nx +
dn e−2t (sin t n2 − 4) sin nx
nπ
n=3
n=3
√
√
√
√
8 −2t
4 3 −2t
=⇒ ut (x, t) = − e cosh 3t sin x +
e sinh 3t sin x − 2d1 e−2t sinh 3t sin x
π √
π
√
−2t
+ 3d1 e cosh 3t sin x + d2 e−2t sin 2x − 2d2 te−2t sin 2x
√
∞
∞
p
p
X
X
−4 −2t
2 n2 − 4
2
e cos(t n − 4) sin nx +
(1 − (−1)n ) sin(t n2 − 4) sin nx
+
nπ
nπ
+
n=3
∞
X
n=3
∞
X
p
−2dn e−2t sin(t n2 − 4) sin nx +
−dn
p
p
n2 − 4e−2t cos(t n2 − 4) sin nx.
n=3
n=3
Therefore,
∞
∞
n=3
n=3
p
X 4
X
√
8
0 = ut (x, 0) = − sin x + 3d1 sin x + d2 sin 2x +
−
sin nx +
−dn n2 − 4 sin nx.
π
nπ
Re-arranging gives
∞
∞
n=3
n=3
p
X 4
X
√
8
=⇒
sin x +
sin nx = 3d1 sin x + d2 sin 2x +
−dn n2 − 4 sin nx.
π
nπ
Equating coefficients, we see that:
√
8
8
=⇒ d1 = √
π
3π
d2 = 0
p
4
− n2 − 4 dn =
, n ≥ 3,
nπ
4
=⇒ dn = − √
, n ≥ 3.
πn n2 − 4
3d1 =
Therefore, the solution is
u(x, t) =
∞
p
X
√
√
4 −2t
8
2
e cosh 3t sin x + √ e−2t sinh 3t sin x +
(1 − (−1)n )e−2t cos(t n2 − 4) sin nx
π
nπ
3π
n=3
+
∞
X
n=3
−
p
4
√
e−2t sin(t n2 − 4) sin nx.
πn n2 − 4
10
Since lim e−2t = 0, | cos at| ≤ 1, | sin at| ≤ 1, and
t→∞
3
√
3 < 2, we have lim u(x, t) = 0.
t→∞
Exercises 4.3: 3, 7, 12(c)
For problems 3 and 7, solve Laplace’s eequation subject to the given boundary conditions. Work each
problem out completely, rather than referring to the solutions in this section.
3. u(x, 0) = u(x, π) = 0, u(0, y) = y, u(1, y) = 0
Solution: We are solving the initial-boundary-value problem
4u = 0,
0 < x < 1, 0 < y < π,
u(x, 0) = u(x, π) = 0,
0 < x < 1,
u(0, y) = y,
0 < y < π,
u(1, y) = 0,
0 < y < π.
We will solve this problem using separation of variables. Let u(x, y) = X(x)Y (y). Then, we obtain
d2 Y
d2 X
Y
(y)
+
X(x) = 0.
dx2
dy 2
Divide by XY to obtain
1 d2 X
1 d2 Y
=
−
= λ.
X dx2
Y dy 2
We need to obtain the boundary conditions, as well:
0 = u(x, 0) = X(x)Y (0) =⇒ Y (0) = 0,
0 = u(x, π) = X(x)Y (π) =⇒ Y (π) = 0.
We thus obtain the following equations:
Y 00 + λY = 0,
Y (0) = Y (π) = 0,
00
X − λX = 0.
The eigenvalue problem is similar to one we have seen before, so we know that
λ n = n2 ,
Yn (y) = sin ny, n = 1, 2, . . . .
We then solve the equation for X:
X 00 − n2 X = 0
The characteristic equation is
r 2 − n2 = 0
=⇒ r1,2 = ±n
=⇒ Xn (x) = cn cosh nx + dn sinh nx.
11
For each n, un (x, y) = Xn (x)Yn (y) is given by
un (x, y) = cn cosh nx sin ny + dn sinh nx sin ny.
By superposition,
u(x, y) =
∞
X
cn cosh nx sin ny +
n=1
∞
X
dn sinh nx sin ny.
n=1
We now may apply the remaining boundary conditions to determine cn and dn .
y = u(0, y) =
∞
X
cn sin ny.
n=1
Therefore, the right-hand side represents the Fourier sine series of u(0, y) = y. This tells us
Z
2 π
y sin ny
cn =
π 0
2
= − cos nπ
n
2
= − (−1)n
n
2
= (−1)n+1 .
n
So, we have
u(x, y) =
=⇒ 0 = u(1, y) =
=⇒
∞
X
dn sinh n sin ny =
n=1
∞
∞
X
X
2
(−1)n+1 cosh nx sin ny +
dn sinh nx sin ny
n
n=1
∞
X
n=1
∞
X
n=1
2
(−1)n+1 cosh n sin ny +
n
n=1
∞
X
dn sinh n sin ny
n=1
2
(−1)n+1 cosh n sin ny.
n
Equating coefficients requires
2
(−1)n+1 cosh n
n
2
=⇒ dn = (−1)n+1 coth n.
n
Therefore, the solution to our problem is
dn sin h =
u(x, y) =
∞
∞
X
X
2
2
(−1)n+1 cosh nx sin ny +
(−1)n+1 coth n sinh nx sin ny .
n
n
n=1
n=1
7. uy (x, 0) = 0, u(x, 1) = x, u(0, y) = u(1, y) = 0
Solution: We are solving the initial-boundary-value problem
4u = 0,
0 < x < 1, 0 < y < 1,
uy (x, 0) = 0,
0 < x < 1,
uy (x, 1) = x,
0 < x < 1,
u(0, y) = u(1, y) = 0,
12
0 < y < 1.
We will solve this problem using separation of variables. Let u(x, y) = X(x)Y (y). Then, we obtain
d2 Y
d2 X
Y
(y)
+
X(x) = 0.
dx2
dy 2
Divide by XY to obtain
1 d2 Y
1 d2 X
=
−
= −λ.
X dx2
Y dy 2
We need to obtain the boundary conditions, as well:
0 = u(0, y) = X(0)Y (y) =⇒ X(0) = 0,
0 = u(1, y) = X(1)Y (y) =⇒ X(1) = 0.
We thus obtain the following equations:
X 00 + λX = 0,
X(0) = X(1) = 0,
00
Y − λY = 0.
The eigenvalue problem is similar to one we have seen before, so we know that
λn = (nπ)2 ,
Xn (x) = sin nπx, n = 1, 2, . . . .
We then solve the equation for X:
Y 00 − (nπ)2 Y = 0
The characteristic equation is
r2 − (nπ)2 = 0
=⇒ r1,2 = ±nπ
=⇒ Yn (y) = cn cosh nπy + dn sinh nπy.
For each n, un (x, y) = Xn (x)Yn (y) is given by
un (x, y) = cn cosh nπy sin nπx + dn sinh nπy sin nπx.
By superposition,
u(x, y) =
∞
X
cn cosh nπy sin nπx +
n=1
∞
X
dn sinh nπy sin nπx.
n=1
We now may apply the remaining boundary conditions to determine cn and dn .
0 = uy (x, 0) =
∞
X
dn nπ cos nπx,
n=1
=⇒ dn = 0, n = 1, 2, . . . .
∞
X
x = u(x, 1) =
cn cosh nπ sin nπx
n=1
13
Therefore, the right-hand side must represent the Fourier sine series for u(x, 1) = x.
Fs (x) =
∞
X
2(−1)n+1
n=1
nπ
sin nπx
2(−1)n+1
nπ
2
=⇒ cn =
(−1)n+1 .
nπ cosh nπ
=⇒ cn cosh nπ =
Therefore, the solution to the problem is
u(x, y) =
∞
X
n=1
2
(−1)n+1 cosh nπy sin nπx .
nπ cosh nπ
12(c) What is the solution of the Neumann problem
uxx + uyy = 0,
uy (x, 0) = f1 (x),
uy (x, b) = f2 (x),
ux (0, y) = g1 (y),
ux (a, y) = g2 (y)?
Solution: First, write u(x, y) = u1 (x, y) + u2 (x, y), where u1 solves
u1xx + u1yy = 0,
u1y (x, 0) = 0,
u1y (x, b) = 0,
u1x (0, y) = g1 (y),
u1x (a, y) = g2 (y),
and u2 solves
u2xx + u2yy = 0,
u2y (x, 0) = f1 (x),
u2y (x, b) = f2 (x),
u2x (0, y) = 0,
u2x (a, y) = 0.
First, solve for u1 (x, y). Write u1 (x, y) = X1 (x)Y1 (y). Then, we have
Y1 (y)
d2 Y1
d2 X1
+
X
(x)
= 0.
1
dx2
dy 2
Divide both sides by X1 Y1 to obtain
1 d2 X1
1 d2 Y1
=
−
= λ.
X1 (x) dx2
Y1 (x) dy 2
14
We need to obtain the boundary conditions, as well.
0 = u1y (x, 0) = X1 (x)Y1y (0) =⇒ Y1y (0) = 0,
0 = u1y (x, b) = X1 (x)Y1y (b) =⇒ Y1y (b) = 0.
We thus obtain the following equations:
Y100 + λY1 = 0,
Y10 (0) = Y10 (b) = 0,
X100 − λX1 = 0.
The eigenvalue problem is similar to one we have seen before, so we know that
λ0 = 0, Y10 (y) = 1,
nπ 2
nπy
, Y1n (y) = cos
, n = 1, 2, . . . .
λn =
b
b
Then, solving for X1 gives us
X10 (x) = c0 + d0 x,
nπx
nπx
X1n (x) = cn cosh
+ dn sinh
.
b
b
For each n, then u1n (x, y) is given by
u10 (x, y) = c0 + d0 x,
nπy
nπx
nπy
nπx
cos
+ dn sinh
cos
, n = 1, 2, . . . .
u1n (x, y) = cn cosh
b
b
b
b
By superposition, then
u1 (x, y) = c0 + d0 x +
∞
X
∞
cn cosh
nπy X
nπx
nπy
nπx
cos
+
dn sinh
cos
.
b
b
b
b
n=1
n=1
We must apply the boundary conditions to solve for cn and dn .
g1 (y) = u1x (0, y) = c0 +
∞
X
cos
n=1
nπy
.
b
Therefore, the series on the right-hand side must be the Fourier cosine series of g1 (y), and
Z
Z
1 b
2 b
2c0 =
g1 (y) dy =⇒ c0 =
g1 (y) dy,
b 0
b 0
Z
2 b
nπy
cn =
g1 (y) cos
dy.
b 0
b
∞
∞
X
nπa
nπa
nπy X
nπy
g2 (y) = u1x (a, y) = c0 + d0 a +
cn cosh
cos
+
dn sinh
cos
b
b
b
b
=⇒ g2 (y) = c0 + d0 a +
n=1
∞ X
n=1
15
n=1
cn cosh
nπa
nπa nπy
+ dn sinh
cos
.
b
b
b
The right-hand side must represent the Fourier cosine series of g2 (y), and so
Z
Z
1 1 b
2 b
g2 (y) dy =⇒ d0 =
g2 (y) dy − c0
2(c0 + d0 a) =
b 0
a b 0
Z
nπa
nπy
nπa
2 b
cn cosh
g2 (y) cos
+ dn sinh
=
dy
b
b
b 0
b
Z b
1
2
nπy
nπa
=⇒ dn =
g2 (y) cos
.
dy − cn cosh
sinh nπa
b 0
b
b
b
Therefore, the solution is given by
u1 (x, y) = c0 + d0 x +
∞
X
n=1
∞
nπx
nπy X
nπx
nπy
cn cosh
cos
+
dn sinh
cos
,
b
b
b
b
n=1
where cn and dn , n = 0, 1, . . . are defined as above. Next, we solve for u2 (x, y) using separation of
variables. Let u2 (x, y) = X2 (x)Y2 (y). Then,
Y2 (y)
d2 Y2
d2 X2
+
X
(x)
= 0.
2
dx2
dy 2
Divide both sides by X2 Y2 to obtain
1 d2 X2
1 d2 Y2
=−
= −λ.
2
X2 (x) dx
Y2 (x) dy 2
We need to obtain the boundary conditions, as well.
0 = u1y (0, y) = X2x (0)Y2 (y) =⇒ X2x (0) = 0,
0 = u1y (a, y) = X2x (a)Y2 (y) =⇒ X2x (a) = 0.
We thus obtain the following equations:
X200 + λX2 = 0 X20 (0) = X20 (a) = 0,
Y200 − λY2 = 0.
The eigenvalue problem is similar to one we have seen before, so we know that
λ0 = 0, X20 (x) = 1,
nπ 2
nπx
, X2n (y) = cos
, n = 1, 2, . . . .
λn =
a
a
Then, solving for Y2 gives us
Y20 (y) = p0 + q0 y,
nπy
nπy
Y2n (y) = pn cosh
+ qn sinh
.
a
a
For each n, then u2n (x, y) is given by
u20 (x, y) = p0 + q0 y,
nπy
nπx
nπy
nπx
u2n (x, y) = pn cosh
cos
+ qn sinh
cos
,
a
a
a
a
16
and by superposition,
u2 (x, y) = p0 + q0 y +
∞
X
n=1
∞
nπx X
nπx
nπy
nπy
cos
+
cos
.
qn sinh
pn cosh
a
a
a
a
n=1
We must apply the boundary conditions to solve for pn and qn .
f1 (x) = p0 +
∞
X
nπx
a
pn cos
n=1
Therefore, the right-hand side is the Fourier cosine series of f1 (x). So,
Z
Z
1 a
2 a
f1 (x) dx =⇒ p0 =
f1 (x) dx
2p0 =
a 0
a 0
Z
2 a
nπx
dx.
pn =
f1 (x) cos
a 0
a
∞
∞
X
nπb
nπx X
nπb
nπx
f2 (x) = u2x (x, b) = p0 + q0 b +
pn cosh
cos
+
qn sinh
cos
a
a
a
a
n=1
n=1
∞ X
nπb
nπx
nπb
+ qn sinh
cos
.
=⇒ g2 (y) = p0 + q0 b +
pn cosh
a
a
a
n=1
The right-hand side must represent the Fourier cosine series of f2 (x), and so
Z
Z
2 a
1 1 a
2(p0 + q0 b) =
f2 (x) dx =⇒ q0 =
f2 (x) dx − p0
a 0
b a 0
Z a
nπb
nπb
2
nπx
pn cosh
+ qn sinh
=
f2 (x) cos
dx
a
a
a 0
a
Z a
2
nπx
nπb
1
f2 (x) cos
dx − pn cosh
.
=⇒ qn =
a 0
a
a
sinh nπb
a
Therefore, the solution is given by
u2 (x, y) = p0 + q0 y +
∞
X
n=1
∞
nπy
nπx X
nπx
nπy
pn cosh
cos
+
cos
,
qn sinh
a
a
a
a
where pn and qn , n = 0, 1, . . . are defined as above. Then,
u(x, y) = u1 (x, y) + u2 (x, y).
17
n=1