Math 257/316 Assignment 7 Solutions
1. The concentration u(x, t) of a reactive chemical diffusing in one dimension satisfies

0 < x < 2, t > 0
 ut = uxx − u,
u(0, t) = −1, u(2, t) = 1
.

u(x, 0) = 0
where the ‘loss’ term represents a reaction which consumes the chemical. Find u(x, t)
and sketch the solution for 3 different values of t. Hint: first find the steady-state,.
You may find it helpful to know that (you can check this by integrating by parts) the
Fourier sine series coefficients for sinh(x − 1) on [0, 2] are:
0
n odd,
bn =
−4nπ
sinh(1)
n even.
4+n2 π 2
A steady state u(x, t) = v(x) should satisfy
v 00 = v, 0 < x < 2,
v(0) = −1, v(2) = 1.
The general solution of this ODE is a linear combination of ex and e−x , or sinh(x)
and cosh(x), OR, thinking ahead, since the BCs are odd with respect to x = 1,
v(x) = A cosh(x − 1) + B sinh(x − 1).
(Note: you do not have to do it this way – leaving things in terms of ex and e−x is
also fine – it just makes the formulas and computations a little simpler.) Then
−1 = v(0) = A cosh(1) − B sinh(1),
1 = v(2) = A cosh(1) + B sinh(1),
so (adding the two) A = 0, and B = 1/ sinh(1), and the steady-state is
v(x) =
sinh(x − 1)
.
sinh(1)
Now write u(x, t) = v(x) + w(x, t), so that w solves
wt = ut = uxx − u = v 00 − v + wxx − w = wxx − w,
w(0, t) = u(0, t) − v(0) = 0,
1
w(2, t) = u(2, t) − v(2) = 0.
Now do separation of variables:
T0
X 00
+1=
= −λ = const.
T
X
w(x, t) = X(x)T (t) =⇒
We already know that the X problem
X 00 (x) = −λX(x),
X(0) = 0 = X(2)
has solutions Xn (x) = sin(nπx/2), λn = n2 π 2 /4, n = 1, 2, 3, . . .. Then for each n,
the T problem
T 00 (t) = −(λn + 1)T (t) = −(n2 π 2 /4 + 1)T (t)
2 π 2 /4+1)t
has solution T (t) = (const)e−(n
w(x, t) =
∞
X
. So the general solution is
2 π 2 /4+1)t
cn sin(nπx/2)e−(n
.
n=1
The initial condition is
w(x, 0) =
∞
X
cn sin(nπx/2) = u(x, 0) − v(x) = −
n=1
sinh(x − 1)
sinh(1)
and so (using the hint)
1
cn = −
sinh(1)
Z
2
sinh(x − 1) sin(nπx/2)dx =
0
0
4nπ
4+n2 π 2
n odd
.
n even
Finally then,
u(x, t) =
sinh(x − 1)
+ 4π
sinh(1)
∞
X
n=1,n
even
n
2 2
sin(nπx/2)e−(n π /4+1)t .
2
2
4+n π
2. There is a gas leak at the end x = 1 of a corridor 0 ≤ x ≤ 1. The concentration of
gas satisfies
ut = uxx ,
0<x<1
ux (0, t) = 0,
ux (1, t) = 1
u(x, 0) = 0
(a) Find the solution for u(x, t) by first finding a particular solution of the form
v(x, t) = ax2 + bx + ct that satisfies the PDE and BCs; then write u(x, t) =
v(x, t) + w(x, t), and find and solve the homogeneous problem for w.
2
Plugging v(x, t) = ax2 + bx + ct into the heat equation gives c = 2a. So v =
ax2 + bx + 2at, vx = 2ax + b, and the boundary conditions imply
0 = vx (0) = b,
1 = vx (1) = 2a + b = 2a
=⇒ a = 1/2,
so v(x, t) = x2 /2 + t. Now write u(x, t) = v(x, t) + w(x, t) and note that w solves
the homogeneous problem
wt = wxx ,
w(x, 0) = u(x, 0) − v(x, 0) = −x2 /2,
w(0, t) = 0 = w(1, t),
so
∞
w(x, t) =
a0 X
2 2
+
an cos(nπx)e−n π t
2
n=1
with
Z
a0 = 2
1
(−x2 /2)dx = −
0
1
3
and for n = 1, 2, 3, . . .,
Z 1
Z 1
2
2
2
2
1
an = 2
sin(nπx)(−x /2)|0 −
cos(nπx)(−x /2)dx =
sin(nπx)(−x)dx
nπ
nπ 0
0
Z 1
2
2
2
= − 2 2 cos(nπx)(x)|10 + 2 2
cos(nπx)dx = − 2 2 (−1)n .
n π
n π 0
n π
Thus
∞
1 2
1
2 X (−1)n
2 2
u(x, t) = x + t − − 2
cos(nπx)e−n π t .
2
2
6 π
n
n=1
(b) An alarm in the middle of the corridor (x = 1/2) is triggered when the gas
concentration reaches 1. By considering the size of the different terms in your
solution, deduce that the alarm goes off at approximately t ≈ 25/24.
Now
∞
1
1 2 X (−1)n
1
1
2 2
2
u(1/2, t) = + t − − 2
cos(nπ/2)e−n π t = t − + 2 e−4π t + · · ·
8
6 π
n2
24 2π
n=1
where the omitted terms get rapidly smaller for t > 0. We seek the t for which
u(1/2, t) = 1 so we estimate
1 = u(1/2, t) ≈ t −
2
1
24
=⇒
t≈
25
24
(and note that by this time that e−4π t is much smaller than 1, justifying the
omission of that term (and all the other, even smaller, terms) in our approximation).
3
3. Solve the following non-homogeneous problem for the heat equation.
ut = uxx + sin(πx) t e−π
u(0, t) = 0,
2t
0 < x < 1, t > 0
u(1, t) = 5
u(x, 0) = 5x.
To take care of the BCs, we seek a steady-state solution v(x) of the problem without
the source term:
v 00 (x) = 0,
v(0) = 0, v(1) = 5
=⇒ v(x) = 5x,
and so writing u(x, t) = 5x + w(x, t), we find that w solves

−π 2 t

 wt = wxx + sin(πx)te
w(0, t) = 0 = w(1, t)


w(x, 0) = 5x − 5x = 0
For zero BCs, our ‘eigenfunctions’ are sin(kπx) (k = 1, 2, 3, . . .), so we seek w(x, t)
as an eigenfunction expansion
w(x, t) =
∞
X
ck (t) sin(kπx),
k=1
which when we plug it into the PDE for w reads
∞
X
0
2
ck (t) + π 2 k 2 ck (t) sin(kπx) = te−π t sin(πx).
k=1
Note we do not have to expand the source term here, as it is already in the form of
a (one-term) eigenfunction expansion, and we can just match coefficients:
0
k 6= 1
0
2 2
ck + π k ck =
.
2t
−π
te
k=1
Solving these ODEs yields, for k 6= 1,
ck (t) = bk e−k
2 π2 t
,
2
and for k = 1, using the integrating factor eπ t ,
h 2
i0
eπ t c1 (t) = t
=⇒
1
2
eπ t c1 (t) = t2 + b1
2
4
1
2
2
=⇒ c1 (t) = t2 e−π t + b1 e−π t .
2
So the general solution for w is
∞
X
1
2
2 2
w(x, t) = t2 e−π t sin(πx) +
bk e−k π t sin(kπx).
2
k=1
But the initial conditions read
0 = w(x, 0) =
∞
X
bk sin(kπx),
k=1
2
so we see b1 = b2 = b3 = · · · = 0, and so w(x, t) = 21 t2 e−π t sin(πx), and thus
1
2
u(x, t) = 5x + t2 e−π t sin(πx).
2
5