# Solution Set 7 1. Consider waves in a resistant medium that satisfy ```Solution Set 7
1. Consider waves in a resistant medium that satisfy the problem
utt = c2 uxx − rut for 0 &lt; x &lt; L
u(0, t) = u(L, t) = 0
u(x, 0) = φ(x)
ut (x, 0) = ψ(x)
where r is a constant, 0 &lt; r &lt; 2πc/L. Write down the series expansion of the
solution.
Separation gives
T 00 + rT 0
X 00
=
= −λ.
c2 T
X
X is a solution of the standard Dirichlet BVP. Therefore, Xn(x) = C sin βn x
with βn = nπ/l, n = 1, 2, ... is a solution.
The T equation becomes
T 00 + rT + c2 βn3 T = 0.
The characteristic equation ofpthis ODE is s2 + rs + c2 βn2 = 0. It has
two distinct roots s = (−r &plusmn; r2 − 4c2 βn2 )/2 = −r/2 &plusmn; iωn where ωn =
p
(2nπc/l)2 − r2 /2 is a real positive number since 0 &lt; r &lt; 2πc/l and
n ≥ 1. Therefore, Tn = e−rt/2 (A cos ωn t + B sin ωn t).
The series solution is
∞
X
−rt/2
u(x, t) = e
(An cos ωn t + Bn sin ωn t) sin βn x.
n=1
with
φ(x) =
∞
X
An sin βn x,
ψ(x) =
n=1
∞
X
r
(− An + Bn ωn ) sin βn x.
2
n=1
2. Use the method of subtraction to solve utt = 9uxx for 0 &lt; x &lt; 1 with u(0, t) =
h, u(1, t) = k, where h and k are given constants, and u(x, 0) = 0, ut (x, 0) = 0.
1
Let U(x, t) = (1 − x)h + xk = h + (k − h)x. Let v = u − U. Then, v
satisfies the equation vtt = 9vxx with the Dirichlet BC v(0, t) = v(1, t) = 0
and initial conditions v(x, 0) = u(x, 0) − U(x, 0) = −h − (k − h)x and
vt (x, 0) = ut (x, 0) − Ut (x, 0) = 0.
The series solution for this Dirichlet BC is
∞
X
v(x, t) =
(An cos 3nπt + Bn sin 3nπt) sin nπx.
n=1
First, note that vt (x, 0) = 0 ⇒ Bn = 0.
1
1
sin2 nπxdx = , the remaining coefficients can be computed as
2
0
Z 1
h
1
An = −2
[h+(k−h)x] sin nπxdx = −2
[1 − (−1)n ] − (k − h) (−1)n
nπ
nπ
0
2
= − [h + k(−1)n+1 ].
nπ
As
Z
Therefore,
∞
2X
u(x, t) = h + (k − h)x −
[h + k(−1)n+1 ] cos 3nπt sin nπx.
π n=1
3. Use the method of descent to derive d’Alembert’s formula for the one-dimensional
Cauchy problem from Poisson’s formula for two space dimensions.
Poisson’s formula for two space dimensions is
&quot;Z Z
#
1 ∂
φ(ξ, η)
p
u(x, y, t) =
dξdη
2πc ∂t
(ct)2 − (ξ − x)2 − (η − y)2
Dt
1
+
2πc
ZZ
Dt
ψ(ξ, η)
p
(ct)2 − (ξ − x)2 − (η − y)2
dξdη
where Dt is a disk centered at (x, y) with radius ct. To make a onedimensional problem, let’s consider points on (x, 0). Furthermore, u, φ
and ψ are made independent of y (or η inside the double integral). The
formula can be written as
2
&quot;Z Z
#
1 ∂
φ(ξ)
p
dξdη
2πc ∂t
(ct)2 − (ξ − x)2 − η 2
Dt
ZZ
1
ψ(ξ)
p
+
dξdη.
2πc Dt (ct)2 − (ξ − x)2 − η 2
As an example, consider the how the double integral for φ can be handled.
The integral inside the rectangular bracket can be written as
!
Z x+ct
Z +√(ct)2−(ξ−x)2
1
p
φ(ξ)
dη dξ.
√ 2
(ct)2 − (ξ − x)2 − η 2
x−ct
− (ct) −(ξ−x)2
p
Let a = (ct)2 − (ξ − x)2 and η = a sin θ. The integal insider the large
bracket can be evaluated as
Z +π/2
1
p
a cos θdθ = π.
2
2
2
−π/2
a − a sin
Therefore,
ZZ
Z x+ct
φ(ξ)
p
dξdη = π
φ(ξ)dξ.
(ct)2 − (ξ − x)2 − η 2
Dt
x−ct
u(x, t) =
This gives the solution of CP as
Z x+ct
Z
1 ∂
1 x+t
u(x, t) =
φ(ξ) dξ +
ψ(ξ) dξ
2c ∂t x−ct
2c x−t
Z
1
1 x+ct
= (φ(x + t) + φ(x − t)) +
ψ(ξ) dξ,
2
2c x−ct
d’Alembert’s solution.
4. Solve the problem:
utt − c2 (uxx + uyy + uzz ) = 0
ut=0 = x2 + yz,
ut t=0 = 0.
Kirchhoff-Poisson’s solution gives:
Z 2πZ π
1 ∂
u=
t
(x + ct sin θ cos φ)2 + (y + ct sin θ sin φ)(z + ct cos θ) sin θdθdφ
4π ∂t 0 0
∂
1
=
x2 t + c2 t3 + yzt = x2 + c2 t2 + yz
∂t
3
3
```