HOMEWORK 4 SOLUTIONS
Section 3.1
1.
According to formula (6) in page 59, we have
Z ∞
Z ∞
(x−y)2
(x+y)2
1
1
− 4kt
−y
e
e− 4kt e−y dy = I1 − I2 .
e dy − √
u(x, t) = √
4πkt 0
4πkt 0
For the first part of the integral, I1 , we have
Z ∞
(x−y)2
1
I1 = √
e− 4kt −y dy
4πkt Z0
∞
−x2 +2xy−y 2 −4kty
1
4kt
dy
=√
e
4πkt Z0
∞
−x2 +2(x−2kt)y−y 2
1
4kt
=√
e
dy
4πkt Z0
∞
−x2 +(x−2kt)2 −(x−2kt)2 +2(x−2kt)y−y 2
1
4kt
dy
e
=√
4πkt Z0
∞
−x2 +(x−2kt)2 −(x−2kt−y)2
1
4kt
=√
e
dy
4πkt Z0
∞
−4xkt+4k2 t2 −(x−2kt−y)2
1
4kt
dy
e
=√
4πkt 0
Z
e−x+kt ∞ −(x−2kt−y)2
4kt
=√
e
dy
4πkt 0
Now make a change of variable
p=
(x − 2kt − y)
√
,
4kt
then
Z
e−x+kt −∞ −p2
e dp
I1 = − √
x−2kt
π
√
4kt
x−2kt
−x+kt Z √4kt
e
2
= √
e−p dp
π −∞
Z
Z x−2kt
√
4kt
e−x+kt 0 −p2
e−x+kt
2
= √
e dp + √
e−p dp
π −∞
π 0
e−x+kt e−x+kt
x − 2kt
=
+
Erf( √
).
2
2
4kt
1
For the second part of the integral, I2 , we have
Z ∞
(x+y)2
1
e− 4kt −y dy
I2 = √
4πkt Z0
∞
−x2 −2xy−y 2 −4kty
1
4kt
e
dy
=√
4πkt Z0
∞
−x2 −2(x+2kt)y−y 2
1
4kt
e
dy
=√
4πkt Z0
∞
−x2 +(x+2kt)2 −(x+2kt)2 −2(x+2kt)y−y 2
1
4kt
=√
e
dy
4πkt Z0
∞
−x2 +(x+2kt)2 −(x+2kt+y)2
1
4kt
e
=√
dy
4πkt Z0
∞
+4xkt+4k2 t2 −(x+2kt+y)2
1
4kt
=√
e
dy
4πkt 0
Z ∞
−(x+2kt+y)2
ex+kt
4kt
e
dy
=√
4πkt 0
Now make a change of variable
p=
(x + 2kt + y)
√
,
4kt
then
ex+kt
I2 = √
π
∞
Z
2
e−p dp
x+2kt
√
4kt
Z
Z x+2kt
√
4kt
ex+kt ∞ −p2
ex+kt
2
= √
e dp − √
e−p dp
π 0
π 0
x+kt
x+kt
e
x + 2kt
e
).
−
Erf( √
=
2
2
4kt
The solution can then be written as
u(x, t) = I1 − I2
e−x+kt e−x+kt
x − 2kt
ex+kt ex+kt
x + 2kt
+
Erf( √
)−
+
Erf( √
)
2
2
2
2
4kt
4kt
ekt −x
x − 2kt
x + 2kt
kt
x
= e sinh x +
e Erf( √
) + e Erf( √
)
2
4kt
4kt
=
2.
Consider a function v(x, t), which is a solution to
vt = kvxx ;
v(x, 0) = −1;
2
v(0, t) = 0
on the half line 0 < x < ∞. According to formula (6) in page 59, we have
Z ∞
Z ∞
(x−y)2
(x+y)2
1
1
− 4kt
v(x, t) = − √
e
e− 4kt dy.
dy + √
4πkt 0
4πkt 0
Now we let
u(x, t) = v(x, t) + 1,
then u(x, t) solves our original PDE problem, with u(x, 0) = v(x, 0) + 1 = 0 and
u(0, t) = v(0, t)+1 = 1. We still need to evaluate the two integrals for v(x, t). Make
the change of variable
x−y
,
p= √
4kt
the first integral can be written as
Z ∞
(x−y)2
1
√
e− 4kt dy
4πkt 0
Z x/√4kt
1
2
=√
e−p dp
π −∞
1 1
x
= + Erf( √ )
2 2
4kt
Make the change of variable
x+y
q= √
,
4kt
the first integral can be written as
Z ∞
(x+y)2
1
√
e− 4kt dy
4πkt
Z ∞0
1
2
e−p dp
=√
√
π x/ 4kt
1 1
x
= − Erf( √ )
2 2
4kt
Hence
v(x, t) = −
Thus,
x
x
1 1
1 1
x
+ Erf( √ ) + − Erf( √ ) = −Erf( √ ).
2 2
2 2
4kt
4kt
4kt
x
u(x, t) = v(x, t) + 1 = 1 − Erf( √ ).
4kt
3.
For the half line Neumann problem
wt − kwxx = 0;
wx (0, t) = 0;
3
w(x, 0) = φ(x)
on the half line 0 < x < ∞, we first extend the initial condition φ(x) to an even
function ψ defined on the whole real line, such that ψ(x) = φ(x) when x ≥ 0 and
ψ(x) = φ(−x) when x < 0. Now we consider a new PDE
ut − kuxx = 0;
u(x, 0) = ψ(x),
with −∞ < x < ∞. According to formula (8) in page 49, the solution to this PDE
is
Z ∞
−(x−y)2
1
u(x, t) = √
e 4kt ψ(y)dy.
4πkt −∞
Since the initial condition ψ(x) is an even function, the solution u(x, t) is always
an even function in x at any time t, hence ux (0, t) = 0 for any t. That means,
w(x, t) = u(x, t) for x > 0 is the solution to our original Neumann problem. Now
we are to calculate w(x, t).
Z ∞
−(x−y)2
1
e 4kt ψ(y)dy
w(x, t) = √
4πkt −∞
Z ∞
Z 0
−(x−y)2
−(x−y)2
1
1
e 4kt φ(y)dy + √
e 4kt φ(−y)dy
=√
4πkt 0
4πkt −∞
Let z = −y, then
1
√
4πkt
Z
0
e
−∞
−(x−y)2
4kt
1
φ(−y)dy = √
4πkt
Z
∞
e
−(x+z)2
4kt
φ(z)dz
0
Hence
Z ∞
Z ∞
−(x−y)2
−(x+z)2
1
1
e 4kt φ(y)dy + √
e 4kt φ(z)dz
w(x, t) = √
4πkt Z0
4πkt Z0
∞
∞
−(x−y)2
−(x+y)2
1
1
=√
e 4kt φ(y)dy + √
e 4kt φ(y)dy
4πkt 0 4πkt
0
Z ∞
−(x−y)2
−(x+y)2
1
=√
e 4kt + e 4kt
φ(y)dy
4πkt 0
4
Section 3.2
1.
For the Neumann problem
utt − c2 uxx = 0,
u(x, 0) = φ(x),
ut (x, 0) = ψ(x),
ux (0, t) = 0
with 0 < x < ∞, we extend φ and ψ to even functions h(x) = φ(x) for x ≥ 0,
h(x) = φ(−x) for x < 0, and g(x) = ψ(x) for x ≥ 0, g(x) = ψ(−x) for x < 0.
Consider the wave equation
vtt − c2 vxx = 0,
v(x, 0) = h(x),
vt (x, 0) = g(x)
with −∞ < x < ∞. According to formula (8) in page 36, the solution to this wave
equation is given by
Z
1 x+ct
1
g(s)ds.
v(x, t) = (h(x + ct) + h(x − ct)) +
2
2c x−ct
Now let u(x, t) = v(x, t) for x > 0, then u(x, t) solves the original Neumann problem. The question is now how to compute u(x, t). When x > c|t|,
Z
1 x+ct
1
ψ(s)ds.
u(x, t) = (φ(x + ct) + φ(x − ct)) +
2
2c x−ct
When 0 < x < c|t|,
u(x, t) =
=
=
=
Z
1 x+ct
1
(h(x + ct) + h(x − ct)) +
g(s)ds
2
2c x−ct
Z
Z
1
1 x+ct
1 0
ψ(−s)ds +
ψ(s)ds
(φ(x + ct) + φ(ct − x)) +
2
2c x−ct
2c 0
Z
Z
1
1 ct−x
1 x+ct
(φ(x + ct) + φ(ct − x)) +
ψ(s)ds +
ψ(s)ds
2
2c 0
2c 0
Z
Z
1 x+ct
1
1 ct−x
(φ(x + ct) + φ(ct − x)) +
ψ(s)ds +
ψ(s)ds
2
c 0
2c ct−x
5.
According to formulae (2) and (3) in page 62, the solution to this Dirichlet
problem is
Z
1
1 x+2t
u(x, t) = (φ(x + 2t) + φ(x − 2t)) +
ψ(y)dy = 1
2
4 x−2t
for x > 2|t|, and
1
1
u(x, t) = (φ(x + 2t) − φ(2t − x)) +
2
4
Z
x+2t
ψ(y)dy = 0
2t−x
for x < 2|t|. Hence the solution is not continuous and has a singularity at x = 2|t|.
5
10.
We are to solve utt = 9uxx in 0 < x < π/2, u(x, 0) = cos x, ut (x, 0) = 0,
ux (0, t) = 0, u(π/2, t) = 0. Notice that at x = 0 we have Neumann boundary
condition, so we need an even extension at x = 0; Notice that at x = π/2, we have
Dirichlet boundary condition, so we need an odd extension at x = π/2. It turns out
that φ(x) = cos x itself is an even extension at x = 0 because cos x = cos(−x); and
φ(x) = cos x itself is an odd extension at x = π/2 because cos x = − cos(π − x). So
we just need to solve the problem
vtt − 9vxx = 0,
v(x, 0) = cos x,
vt (x, 0) = 0
over the whole real line −∞ < x < ∞. According to formula (8) in page 36, the
solution to that is
v(x, t) =
1
(cos (x + ct) + cos (x − ct)) = cos x cos (ct)
2
6