Homework
§12.1
1. Solve the PDEs(for u = u(x, y)).
(a) uxx = 4y 2 u
(b) uyy = 4xuy
1
(c) uxy = ux
(d) uyy + 10uy + 25u = e−5y
Sol. (a) If u = u(x), then u(x) = Ae2yx + Be−2yx . Thus the solution of the PDE is
u(x, y) = A(y)e2yx + B(y)e−2yx ,
where A(y) and B(y) are arbitrary functions.
(b) Set uy = p. We have py = 4xp ⇒ p = c(x)e4xy and
c(x) 4xy
e + B(x) = A(x)e4xy + B(x),
4x
where A(x) and B(x) are arbitrary functions.
u(x, y) =
(c) Set ux = p. We have py = p ⇒ p = A(x)ey and
Z
u(x, y) = c(x)dxey + B(y) = A(x)ey + B(y),
where A(x) and B(y) are arbitrary functions.
(d) If u = u(y), then u00 + 10u0 + 25u = e−5y . Then uh = (A + By)e−5y .
Let up = Cy 2 e−5y . u0p = C(2y − 5y 2 )e−5y and u00p = C(2 − 20y + 25y 2 )e−5y .
u00p + 10u0p + 25up = C(2 − 20y + 25y 2 )e−5y + 10C(2y − 5y 2 )e−5y + 25Cy 2 e−5y
= 2Ce−5y = e−5y
⇒
C = 1/2.
Thus u = uh + up = (A + By + 12 y 2 )e−5y , and so the solution of the PDE is
1
u(x, y) = (A(x) + B(x)y + y 2 )e−5y ,
2
where A(x) and B(x) are arbitrary functions.
2. Solve (for u = u(x, y)) (a) uxx = 0, uxy = 0
(b) uxx = 0, uyy = 0 .
Sol. (a) Set ux = p. We have
px = 0, py = 0 ⇒ p = A = constant
and
u(x, y) = Ax + B(y),
where B(y) is an arbitrary function.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
(b)
2
Consider uxx = 0, we have
u(x, y) = A(y) + B(y)x
and
uyy = A00 (y) + B 00 (y)x = 0 ⇒ A00 (y) = B 00 (y) = 0.
Hence A(y) = ay + b, B(y) = cy + d and so
u(x, y) = ay + b + (cy + d)x = b + ay + dx + cxy,
where a, b, c, d are arbitrary constants.
§12.3
1. Find u(x, t) for the string of length L = 1 and c2 = 1 when the initial velocity is
zero and the initial deflection f (x) is as follows.
(a) f (x) = 0.01(sin πx − 31 sin 3πx)
½
0.2x
if 0 ≤ x ≤ 0.5
(b) f (x) =
.
−0.2x + 0.2
if 0.5 ≤ x ≤ 1
Sol. We want to find a solution of the wave equation
∂ 2u ∂ 2u
= 2
∂t2
∂x
satisfying the conditions
u(0, t) = 0, u(1, t) = 0 for all t ≥ 0,
u(x, 0) = f (x), ut (x, 0) = 0 for all 0 ≤ x ≤ 1.
By the method of separating variables, we have
∞
X
u(x, t) =
(Bn cos nπt + Bn∗ sin nπt) sin nπx.
n=1
(Give the details of your answer.)
From the Initial Conditions
u(x, 0) =
∞
X
Bn sin nπx = f (x),
n=1
∞
X
∂u ¯¯
Bn∗ nπ sin nπx = 0.
¯ =
∂t t=0 n=1
Using the Fourier Sine Series, Bn∗ = 0 and
Z 1
Bn = 2
f (x) sin nπxdx, n = 1, 2, · · · .
0
Computational Science & Engineering (CSE)
C. K. Ko
Homework
3
1
1
(a) B1 = 100
, B3 = − 300
and B2 = B4 = B5 = · · · = 0.
The solution u is
1
1
u(x, t) =
cos πt sin πx −
cos 3πt sin 3πx.
100
300
(b)
Z 1/2
Z 1
Bn = 2[
0.2x sin nπxdx +
0.2(1 − x) sin nπxdx]
0
=
1/2
0.8 sin nπ
2
.
2
(nπ)
The solution u is
∞
X
0.8 sin nπ
2
u(x, t) =
cos nπt sin nπx
2
(nπ)
n=1
=
4
1
1
[cos
πt
sin
πx
−
cos
3πt
sin
3πx
+
cos 5πt sin 5πx + · · · ].
5π 2
32
52
2. What happens to the frequency of the fundamental mode of the vibrating string
if we double all the length of the string and the mass per length and the tension?
Sol. The solution of the wave equation
2
∂ 2u
2∂ u
=
c
∂t2
∂x2
(c2 =
T
)
ρ
satisfying u(0, t) = 0, u(L, t) = 0 for all t is
∞
X
nπx
u(x, t) =
(Bn cos λn t + Bn∗ sin λn t) sin
.
L
n=1
Here ρ is a mass per length, T is the tension, L is the length and λn =
The frequency of the fundamental mode of the vibrating string is
√
λn
cn
Tn
=
= √ .
2π
2L 2 ρL
cnπ
L .
Hence if we double all the length of the string and the mass per length and the
tension, the frequency became a half.
3. Find the deflection u(x, t) of the string of length L = π and c2 = 1 for zero initial
displacement and the initial velocity ut (x, 0) = 0.01x if 0 ≤ x ≤ 12 π, ut (x, 0) =
0.01(π − x) if 21 π ≤ x ≤ π.
Computational Science & Engineering (CSE)
C. K. Ko
Homework
4
Sol. We want to find a solution of the wave equation
∂ 2u ∂ 2u
= 2
∂t2
∂x
satisfying the conditions
u(0, t) = 0, u(π, t) = 0 for all t,
u(x, 0) = 0,
½
0.01x
ut (x, 0) = g(x) =
0.01(π − x)
if
if
0 ≤ x ≤ 12 π
.
1
2π ≤ x ≤ π
By the method of separating variables, we have
∞
X
u(x, t) =
(Bn cos nt + Bn∗ sin nt) sin nx.
n=1
(Give the details of your answer.)
From the Initial Conditions
u(x, 0) =
∞
X
Bn sin nx = 0,
n=1
∞
X
∂u ¯¯
nBn∗ sin nx = g(x).
¯ =
∂t t=0 n=1
Using the Fourier Sine Series, Bn = 0 and
Z
2 π
∗
g(x) sin nxdx
nBn =
π 0
Z
Z π
2 π/2
= [
0.01x sin nxdx +
0.01(π − x) sin nxdx]
π 0
π/2
0.04 sin nπ
0.04 sin nπ
∗
2
2
=
⇒
Bn =
.
n2 π
n3 π
The solution u is
u(x, t) =
∞
X
0.04 sin nπ
2
n=1
=
n3 π
sin nt sin nx
0.04
1
1
[sin t sin x − 3 sin 3t sin 3x + 3 sin 5t sin 5x + · · · ].
π
3
5
Computational Science & Engineering (CSE)
C. K. Ko
Homework
5
4. Find the deflection u(x, t) of the string of length L = 1 and c2 = 1 for the initial
displacement u(x, 0) = 100 sin 2πx and the initial velocity ut (x, 0) = 2x(1 − x).
Sol. We want to find a solution of the wave equation
∂ 2u ∂ 2u
= 2
∂t2
∂x
satisfying the conditions
u(0, t) = 0, u(1, t) = 0 for all t,
u(x, 0) = 100 sin 2πx,
ut (x, 0) = 2x(1 − x).
By the method of separating variables, we have
∞
X
u(x, t) =
(Bn cos nπt + Bn∗ sin nπt) sin nπx.
n=1
(Give the details of your answer.)
From the Initial Conditions
∞
X
u(x, 0) =
Bn sin nπx = 100 sin 2πx,
n=1
∞
X
∂u ¯¯
nπBn∗ sin nπx = 2x(1 − x).
¯ =
∂t t=0 n=1
Using the Fourier Sine Series,
B2 = 100, B1 = B3 = B4 = · · · = 0
and
Z
1
8(1 − (−1)n )
= 2
2x(1 − x) sin nπxdx =
n3 π 3
0
8(1 − (−1)n )
.
⇒
Bn∗ =
n4 π 4
nπBn∗
The solution u is
u(x, t) = 100 cos 2πt sin 2πx +
∞
X
8(1 − (−1)n )
n=1
n4 π 4
sin nπt sin nπx
= 100 cos 2πt sin 2πx
16
1
1
+ 4 [sin πt sin πx + 4 sin 3πt sin 3πx + 4 sin 5πt sin 5πx + · · · ].
π
3
5
Computational Science & Engineering (CSE)
C. K. Ko
Homework
6
§12.4
1. Using d’Alembert’s solution, find the deflection u(x, t) of a vibrating string
(length L = 1, ends fixed, c = 1) staring with initial velocity 0 and initial deflection
f (x) = 100 sin πx.
Sol. The d’Alembert’s solution of the wave equation
∂ 2u ∂ 2u
= 2
∂t2
∂x
satisfying u(x, 0) = f (x) = 100 sin πx, ut (x, 0) = 0 is
1
u(x, t) = [f (x + t) + f (x − t)]
2
= 50{sin[π(x + t)] + sin[π(x − t)]} = 100 sin πx cos πt.
(Give the details of your answer.)
2. Show that the Tricomi equation yuxx + uyy = 0 is of mixed type.
Sol. Note that AC − B 2 = y. The type of the Tricomi equation is elliptic over the
region y > 0, and hyperbolic over the region y < 0. Thus the Tricomi equation is a
mixed type.
3. Find the type, transform to normal form, and solve.
(a) uxx + uxy − 2uyy = 0 (b) uxx + 6uxy + 9uyy = 0 (b) uxx + 2uxy + 5uyy = 0.
Sol. (a) discriminant AC − B 2 = −9/4 < 0 =⇒ type : hyperbolic,
characteristic equation : y 02 − y 0 − 2 = 0, y 0 = −1, 2, y = −x + c, y = 2x + c0 ,
new variables : v = x + y, w = 2x − y
normal form : uvw = 0 (Show the details of your work.)
uw =R k(w),
u = k(w)dw + g(v) = f (w) + g(v) = f (2x − y) + g(x + y).
Thus the solution is u = f (2x − y) + g(x + y).
(b) discriminant AC − B 2 = 9 − 9 = 0 −→ type : parabolic,
characteristic equation : y 02 − 6y 0 + 9 = 0, y 0 = 3, y = 3x + c,
new variables : v = Φ = x, w = Ψ = 3x − y
normal form : uvv = 0 (Show the details of your work.)
uv = f (w),
u = vf (w) + g(w) = xf (3x − y) + g(3x − y).
Thus the solution is u = xf (3x − y) + g(3x − y).
Computational Science & Engineering (CSE)
C. K. Ko
Homework
7
(c) discriminant AC − B 2 = 5 − 1 = 4 > 0 −→ type : elliptic,
characteristic equation : y 02 − 2y 0 + 5 = 0,
y 0 = 1 − 2i, 1 + 2i, y = (1 − 2i)x + c, y = (1 + 2i)x + c0 ,
new variables : v = [(y − (1 − 2i)x) + (y − (1 + 2i)x)]/2 = y − x,
w = [(y − (1 − 2i)x) − (y − (1 + 2i)x)]/(2i) = 2x
normal form : uvv + uww = 0 (Show the details of your work.)
(Note)
new variables : v = y − (1 − 2i)x + y, w = y − (1 + 2i)x
normal form : uvw = 0 (Show the details of your work.)
Ruw = k(w),
u = k(w)dw + g(v) = f (w) + g(v) = f (y − (1 + 2i)x) + g(y − (1 − 2i)x).
Thus the solution is u = f (y − (1 + 2i)x) + g(y − (1 − 2i)x).
§12.5
1. A laterally insulated bar of length 10cm and constant cross-sectional area 1cm2 ,
of density 10.6gm/cm3 , thermal conductivity 1.04cal/(cm sec ◦ C), and specific heat
0.056 cal/(gm ◦ C)(this corresponds to silver, a good heat conductor) has initial temperature f (x) and is kept at 0◦ C at the ends x = 0 and x = 10. Find the temperature
u(x, t) at later times.
(a) f (x) = sin 0.1πx + 21 sin 0.2πx
(b) f (x) = 1 − 0.2|x − 5|
K
1.04
Sol. Note c2 = σρ
= 0.056×10.6
≈ 1.75, c ≈ 1.32.
By the method of separating variables, the heat equation
2
∂u
2∂ u
=c
∂t
∂x2
satisfying u(0, t) = 0, u(10, t) = 0 for all t, u(x, 0) = f (x), 0 ≤ x ≤ 10 has a
solution
∞
X
nπx −λ2n t
u(x, t) =
Bn sin
e
,
10
n=1
2 2 2
c n π
2
2 2
where λ2n = ( cnπ
10 ) = 100 = 0.0175π n . (Give the details of your answer.)
From the Initial Conditions
∞
X
nπx
u(x, 0) =
Bn sin
= f (x),
10
n=1
and using the Fourier sine series
R 10
Bn = 15 0 f (x) sin nπx
10 dx, n = 1, 2, · · · .
Computational Science & Engineering (CSE)
C. K. Ko
Homework
8
(a) B1 = 1, B2 = 12 , B3 = B4 = · · · = 0.
The solution u(x, t) is
2
2
u(x, t) = sin(0.1πx)e−λ1 t + 0.5 sin(0.2πx)e−λ2 t
2
2
= sin(0.1πx)e−0.0175π t + 0.5 sin(0.2πx)e−0.07π t .
(b)
Bn
Z
1 10
nπx
=
(1 − 0.2|x − 5|) sin
dx
5 0
10
Z 5
Z 10
nπx
nπx
8
1
nπ
=
[
x sin
dx +
(10 − x) sin
dx] =
sin
.
25 0
10
10
(nπ)2
2
5
The solution is
u(x, t) =
∞
X
n=1
=
8
nπ
nπx −0.0175π2 n2 t
sin
sin
e
(nπ)2
2
10
8
1
−0.0175π 2 t
−0.1575π 2 t
[sin(0.1πx)e
−
sin(0.3πx)e
π2
32
1
2
+ 2 sin(0.5πx)e−0.4375π t + · · · ].
5
2. (a) For the completely insulated bar, ux (0, t) = 0, ux (π, t) = 0, u(x, 0) = π 2 − x2 ,
find the temperature with c = 1. What is the temperature (u, t) in the bar after a
long time(theoretically, as t → ∞?
(b) Find the temperature of the bar in (a) if the left end is kept at 0◦ C, the right
end is insulated, and the initial temperature is u0 = const.
Sol. (a) By the method of separating variables, the heat equation
∂u ∂ 2 u
= 2
∂t
∂x
satisfying ux (0, t) = 0, ux (π, t) = 0 for all t and u(x, 0) = f (x) = π 2 −x2 , 0 ≤ x ≤ π
has a solution
∞
X
2
u(x, t) = A0 +
An cos nxe−n t .
n=1
(Give the details of your answer.)
From the Initial Conditions
u(x, 0) = A0 +
∞
X
An cos nx = f (x) = π 2 − x2 ,
n=1
Computational Science & Engineering (CSE)
C. K. Ko
Homework
9
using the Fourier cosine series
Z
1 π 2
2π 2
2
A0 =
(π − x )dx =
,
π 0
3
Z
2 π 2
4 cos nπ
4(−1)n+1
2
An =
(π − x ) cos nxdx = −
=
.
π 0
n2
n2
The solution is
∞
2π 2 X 4(−1)n+1
−n2 t
u(x, t) =
+
cos
nxe
3
n2
n=1
=
1
2π 2
1
+ 4[cos xe−t − 2 cos 2xe−4t + 2 cos 3xe−9t + · · · ].
3
2
3
After a long time(theoretically, as t → ∞, the temperature becomes
2π 2
3 .
(b) Consider
∂u ∂ 2 u
= 2
∂t
∂x
u(0, t) = 0, ux (π, t) = 0 for all t and u(x, 0) = u0 = const, 0 ≤ x ≤ π .
By the method of separating variables, set u(x, t) = F (x)G(t). Then
00
Ġ F
=
= k,
G
F
and we have two ODEs
00
F − kF = 0, Ġ − kG = 0.
From the Boundary Conditions
u(0, t) = F (0)G(t) = 0 for all t,
ux (π, t) = F 0 (π)G(t) = 0 for all t.
If G ≡ 0, then u ≡ 0, which is of no interest. Hence, G 6= 0 and F (0) = F 0 (π) = 0.
Case k > 0, k = µ2 . F 00 − µ2 F = 0 and F (x) = Aeµx + Be−µx .
Since F (0) = F 0 (π) = 0, F ≡ 0 and u ≡ 0, which is of no interest.
Case k = 0. F 00 = 0 ⇒ F (x) = A + Bx, F (0) = A = 0, F 0 (π) = B = 0.
F ≡ 0 and u ≡ 0, which is of no interest.
Case k < 0, k = −p2 (p > 0). F 00 + p2 F = 0 ⇒ F (x) = A cos px + B sin px.
F (0) = A = 0, and F 0 (π) = Bp cos(pπ) = 0.
p = 2n−1
2 , n = 1, 2, · · · and setting B = 1, we obtain
F (x) = Fn (x) = sin
2n − 1
x, n = 1, 2, · · · .
2
Computational Science & Engineering (CSE)
C. K. Ko
Homework
The other ODE Ġ + p2 G = 0, p =
2n−1
2 ,
Gn (t) = Bn e−(
10
has a solution
2n−1 2
2 ) t
, n = 1, 2, · · · .
Hence, solutions of the heat equation are
un (x, t) = Fn (x)Gn (t) = Bn sin
2n − 1 −( 2n−1 )2 t
xe 2 , n = 1, 2, · · · .
2
∞
X
Since the heat equation is linear and homogeneous,
un (x, t) is a solution.
n=1
u(x, t) =
∞
X
un (x, t) =
n=1
∞
X
Bn sin
n=1
2n − 1 −( 2n−1 )2 t
xe 2 .
2
From the Initial Conditions
u(x, 0) =
∞
X
n=1
Bn sin
2n − 1
x = u0
2
and using the Fourier sine series
Bn =
4u0
(2n−1)π (1
− cos (2n−1)π
)=
2
4u0
(2n−1)π .
The solution is
u(x, t) =
∞
X
n=1
=
4u0
2n − 1 −( 2n−1 )2 t
sin
xe 2
(2n − 1)π
2
1
4u0
x 1
3x 3 2
1
5x 5 2
[sin e− 22 t + sin e−( 2 ) t + sin e−( 2 ) t + · · · ].
π
2
3
2
5
2
3. Consider the two dimensional heat equation
ut = c2 (uxx + uyy ).
Solve the steady state solutions(temperatures) in the square plate(with the side’s
length 2) satisfying the conditions u(x, 2) = sin πx and u(x, 0) = u(0, y) = u(2, y) =
0 on the other sides.
Sol. The steady state solution is the solution of the Laplace equation uxx + uyy = 0.
By the method of separating variables, the Laplace equation
∂ 2u ∂ 2u
+
=0
∂x2 ∂y 2
Computational Science & Engineering (CSE)
C. K. Ko
Homework
11
satisfying u(x, 0) = u(0, y) = u(2, y) = 0 for all x, y ∈ [0, 2] has a solution
u(x, y) =
∞
X
An sin
n=1
nπx
nπy
sinh
.
2
2
(Give the details of your answer.)
From the boundary conditions
u(x, 2) =
∞
X
An sin
n=1
nπx
sinh nπ = sin πx
2
we have
A1 = A3 = A4 = A5 = · · · = 0, A2 =
The solution is
u(x, y) =
1
sinh 2π
1
sin πx sinh πy.
sinh 2π
Computational Science & Engineering (CSE)
C. K. Ko