HOMEWORK 5 SOLUTIONS
Section 3.3
1
We’ll solve the inhomogeneous diffusion equation on the half-line with Dirichlet boundary conditions using the reflection method. The setup of the problem is as follows:
u− kuxx = f (x, t) 0 < x < ∞, 0 < t < ∞
u(0, t) = 0
u(x, 0) = φ(x).
Because we have a Dirichlet boundary condition, we’ll use odd extensions of f and φ. Now we must
solve the problem
vt − kvxx = fodd (x, t) − ∞ < x < ∞, 0 < t < ∞
v(0, t) = 0
v(x, 0) = φodd (x).
Now we use the formula for the solution on the whole line:
v(x, t) = ∫
∞
−∞
∞
=∫
0
+∫
−∞
∞
0
=∫
0
∞
t
S(x − y, t)φ(y)dy + ∫
t
∫
∫
∞
−∞
∞
0
0
0
=∫
S(x − y, t)φodd (y)dy + ∫
S(x − y, t − s)f (y, s)dyds
0
S(x − y, t)(−φ(−y))dy + ∫
t
0
S(x − y, t − s)fodd (y, s)dyds
∫
0
−∞
S(x − y, t − s)(−f (−y, s))dyds
[S(x − y, t)φ(y) − S(x + y, t)φ(y)] dy + ∫
[S(x − y, t) − S(x + y, t)] φ(y)dy + ∫
0
t
∫
t
∫
0
0
∞
0
∞
[S(x − y, t − s)f (y, s) − S(x + y, t − s)f (y, s)] dyds
[S(x − y, t − s) − S(x + y, t − s)] f (y, s)dyds
3
Let W (x, t) = w(x, t) − xh(t). Then we have
Wt − kWxx = wt − kwxx − xh′ (t) = −xh′ (t)
Wx (0, t) = wx (0, t) − h(t) = h(t) − h(t) = 0
W (x, 0) = w(x, 0) − xh(0) = φ(x) − xh(0).
Now we solve the inhomogeneous Neumann problem on the whole line using even extentions.
Let γ(x) = φ(x) − xh(0) and f (x, t) = −xh′ (t) for 0 < x < ∞. Then we have
1
2
HOMEWORK 5 SOLUTIONS
∞
W (x, t) = ∫
−∞
∞
=∫
0
t
+∫
0
=∫
∞
t
0
=∫
0
=∫
∞
0
∞
0
t
0
∫
∞
−∞
S(x − y, t − s)feven (y, s)dyds
S(x − y, t)γ(−y)dy
S(x − y, t − s)f (y, s)dyds + ∫
t
∞
0
∞
∫
0
−∞
0
0
S(x − y, t − s)f (−y, s)dyds
S(x + y, t)γ(y)dy
S(x − y, t − s)f (y, s)dyds − ∫
t
0
[S(x − y, t) + S(x + y, t)]γ(y)dy + ∫
∫
0
∞
t
0
∫
S(x + y, t − s)f (y, s)dyds
∞
0
[S(x − y, t − s) + S(x + y, t − s)]f (y, s)dyds
[S(x − y, t) + S(x + y, t)][φ(y) − yh(0)]dy
t
0
0
−∞
S(x − y, t)γ(y)dy − ∫
∫
∞
= +∫
S(x − y, t)γ(y)dy + ∫
∫
0
+∫
S(x − y, t)γeven (y)dy + ∫
∫
∞
0
[S(x − y, t − s) + S(x + y, t − s)][−yh′ (s)]dyds.
Finally, we use the fact that W (x, t) = w(x, t) − xh(t) so we have
w(x, t) = ∫
∞
0
[S(x−y, t)+S(x+y, t)][φ(y)−yh(0)]dy−∫
t
0
∫
∞
0
[S(x−y, t−s)+S(x+y, t−s)][yh′ (s)]dyds+xh(t)
3.4
1
We use Theorem 1 with f (x, t) = xt, φ(x) = 0 and ψ(x) = 0. This gives
u(x, t) =
=
=
=
=
1
∫∫ f
2c
△
t
x+c(t−s)
1
ysdyds
∫ ∫
2c 0 x−c(t−s)
t
1
1 2 x+c(t−s)
∫ s [ y ∣x−c(t−s) ] ds
2c 0
2
t
1
1
s((x2 + 2xc(t − s) + c2 (t − s)2 ) − (x2 − 2xc(t − s) + c2 (t − s)2 ))ds
∫
2c 0 2
t
1
∫ s(4xc(t − s))ds
4c 0
=∫
0
t
xts − xs2 ds
t
1
1
= xts2 − xs3 ∣
0
2
3
1 3 1 t
= xt − xt
2
3
1 3
= xt .
6
HOMEWORK 5 SOLUTIONS
3
3
We use Theorem 1 with f (x, t) = cos(x), φ(x) = sin(x) and ψ(x) = 1 + x. This gives
x+ct
1
1
1
u(x, t) = [φ(x + ct) + φ(x − ct)] + ∫
ψ(s)ds + ∫∫ f
2
2c x−ct
2c
△
x+ct
t
x+c(t−s)
1
1
1
= [sin(x + ct) + sin(x − ct)] + ∫
(1 + s)ds + ∫ ∫
cos(y)dyds
2
2c x−ct
2c 0 x−c(t−s)
We’ll evaluate each term separately. The first term simplifies to sin(x) cos(ct) by the angle sum
formula (although you are not required to perform this simplification).
The second term is:
x+ct
x+ct
1
1
1
ψ(s)ds = (s + s2 )∣
∫
x−ct
2c x−ct
2c
2
1
= (2ct + 2xct)
2c
= t + xt.
The third term is:
t
x+c(t−s)
t
x+c(t−s)
1
1
cos(y)dyds =
∫ ∫
∫ sin(y)∣x−c(t−s) ds
2c 0 x−c(t−s)
2c 0
t
1
=
∫ 2 cos(x) sin(c(t − s))ds
2c 0
t
1
= 2 cos(x) cos(c(t − s))∣
0
c
1
= 2 cos(x)(1 − cos(ct)).
c
Putting it all together we have
1
u(x, t) = sin(x) cos(ct) + t + xt + 2 cos(x)(1 − cos(ct)).
c
5
This problem isn’t pretty, but here it is worked out in excruciating detail. We’ll make extensive
use of Theorem 3 from Appendix A.3, which says the following:
b(t)
b(t) d
d
f (x, t)dx = ∫
f (x, t)dx − f (a(t), t)a′ (t) + f (b(t), t)b′ (t).
∫
dt a(t)
a(t) dt
First we’ll verify the initial conditions:
0
x+c(0−s)
1
f (y, s)dyds
∫ ∫
2c 0 x−c(0−s)
=0
u(x, 0) =
since we’re integrating from 0 to 0.
t
x+c(t−s)
d 1
f (y, s)dyds
∫ ∫
dt 2c 0 x−c(t−s)
t d
x+c(t−s)
x+c(t−t)
1
=
(∫
[∫
f (y, s)dy] ds + ∫
f (y, t)dy) .
2c 0 dt x−c(t−s)
x−c(t−t)
ut (x, t) =
When we set t = 0 the first term disappears because we’re integrating from 0 to 0 and the second
term disappears because we’re integrating from x to x. Thus, ut (x, 0) = 0 as required.
4
HOMEWORK 5 SOLUTIONS
Now we proceed to verify that utt = c2 uxx + f . Continuing the computation from above we have
ut (x, t) =
t d
x+c(t−s)
x+c(t−t)
1
(∫
[∫
f (y, s)dy] ds + ∫
f (y, t)dy)
2c 0 dt x−c(t−s)
x−c(t−t)
=
t d
x+c(t−s)
1
[∫
f (y, s)dy] ds
∫
2c 0 dt x−c(t−s)
=
t
x+c(t−s) d
1
f (y, s)dy − f (x − c(t − s), s)(−c) + f (x + c(t − s), s)(c)) ds
∫ (∫
2c 0
x−c(t−s) dt
Since neither y nor s depend on t, we have
d
dt f (y, s)
= 0 so this becomes
t
1
∫ cf (x − c(t − s), s) + cf (x + c(t − s), s)ds.
2c 0
Now we differentiate with respect to t again:
t
d 1
utt (x, t) =
∫ cf (x − c(t − s), s) + cf (x + c(t − s), s)ds
dt 2c 0
t d
1
(∫
[cf (x − c(t − s), s) + cf (x + c(t − s), s)] ds + cf (x − c(t − t), t) + cf (x + c(t − t), t))
=
2c 0 dt
t d
1
(∫
[cf (x − c(t − s), s) + cf (x + c(t − s), s)] ds + 2cf (x, t))
=
2c 0 dt
t d
1
=
(∫
[cf (x − c(t − s), s) + cf (x + c(t − s), s)] ds) + f (x, t)
2c 0 dt
In order to differential the terms inside the integral, we’ll need to apply the chain rule. To make it
easier to read, let’s let a = x − c(t − s) and b = x + c(t − s). Then we have
t d
1
=
(∫
[cf (a, s) + cf (b, s)] ds) + f (x, t)
2c 0 dt
t
1
∂f ∂b ∂f ∂s
∂f ∂a ∂f ∂s
=
(∫ c [
+
] + c[
+
]) ds + f (x, t)
2c 0
∂a ∂t ∂s ∂t
∂b ∂t ∂s ∂t
t ∂f ∂a
1
∂f ∂s
∂f ∂b ∂f ∂s
= (∫ [
+
]+[
+
]) ds + f (x, t)
2 0 ∂a ∂t ∂s ∂t
∂b ∂t ∂s ∂t
Since s doesn’t depend on t,
finally get a formula for utt :
∂s
∂t
= 0. We also have
utt (x, t) =
∂a
∂t
= −c and
∂b
∂t
= c. Putting this together we
t
c
∫ [−fa (a, s) + fb (b, s)]ds + f (x, t).
2 0
Now we’ll compute uxx :
t
x+c(t−s)
d d 1
f (y, s)dyds
∫ ∫
dx dx 2c 0 x−c(t−s)
t d
x+c(t−s)
d 1
=
[
f (y, s)dy] ds
∫
∫
dx 2c 0 dx x−c(t−s)
uxx (x, t) =
=
t
x+c(t−s) d
d 1
(
f (y, s)dy − f (x − c(t − s), s) + f (x + c(t − s), s)) ds
∫
∫
dx 2c 0
x−c(t−s) dx
As before, since f (y, s) doesn’t depend on x, the
t
d
dx f (y, s)
term drops out and we have
d 1
∫ (−f (x − c(t − s), s) + f (x + c(t − s), s)) ds
dx 2c 0
t d
1
=
[−f (x − c(t − s), s) + f (x + c(t − s), s)] ds
∫
2c 0 dx
=
HOMEWORK 5 SOLUTIONS
5
As before, let’s let a = x − c(t − s) and b = x + c(t − s). Then we have
t d
1
=
[−f (a, s) + f (b, s)] ds
∫
2c 0 dx
t
1
∂f ∂a ∂f ∂s
∂f ∂b ∂f ∂s
=
+
]+[
+
] ds
∫ −[
2c 0
∂a ∂x ∂s ∂x
∂b ∂x ∂s ∂x
t
1
=
∫ −fa (a, s) + fb (b, s)ds
2c 0
t
1
=
∫ −fa (a, s) + fb (b, s)ds
2c 0
That was a lot of work, but now we have nice expressions for utt and uxx . They are as follows:
t
c
utt = ∫ −fa (a, s) + fb (b, s)ds + f (x, t)
2 0
t
1
uxx =
∫ −fa (a, s) + fb (b, s)ds
2c 0
It is now immediately clear that utt = c2 uxx + f (x, t) as desired.