Math 257/316 Assignment 10 Solutions
1. Find the eigenvalues and eigenfunctions of the following Sturm-Liouville problem on
[0, 1]
X 00 (x) + λX(x) = 0, X 0 (0) = 0, X(1) = 0,
and use them to solve the following heat equation with mixed BCs and a source term:
ut = uxx + cos(5πx/2)t,
ux (0, t) = 0,
0 < x < 1, t > 0,
u(1, t) = 0
u(x, 0) = cos(7πx/2)
There are three cases to consider:
• Case 1: λ := −µ2 < 0. Then the general solution is X(x) = Aeµx + Be−µx .
The first BC is 0 = X 0 (0) = µ(A − B) so B = A, and then the second BC is
0 = X(1) = A(eµ + e−µ ) which implies A = 0. So no non-zero solutions.
• Case 2: λ = 0. Then the general solution is X(x) = Ax + B. The first BC is
0 = X 0 (0) = A so A = 0, and then the second BC is 0 = X(1) = B. So, again,
no non-zero solutions.
• Case 3: λ := µ2 > 0. Then X(x) = A sin(µx) + B cos(µx). The first BC is
0 = X 0 (0) = µA, so A = 0, and then the second BC is 0 = X(1) = B cos(µ), so
µ = (1/2 + k)π, k = 0, 1, 2, 3, . . ..
So the eigenvalues and eigenfunctions are:
λk = (1/2 + k)2 π 2 ,
Xk (x) = cos((1/2 + k)πx),
k = 0, 1, 2, 3, . . .
To solve the heat problem, we use an eigenfunction expansion
u(x, t) =
∞
X
ck (t)Xk (x) =
k=0
∞
X
k=0
1
ck (t) cos((k + 1/2)πx)
(which has the correct BCs built in). Plugging into the equation, we find
0 = ut − uxx − t cos(5πx/2)
∞
∞
X
X
=
c0k (t) cos((k + 1/2)πx) −
ck (t)[−(k + 1/2)2 π 2 ] cos((k + 1/2)πx) − t cos(5πx/2)
k=0
=
=
∞
X
k=0
c0k (t) + (k + 1/2)2 π 2 ck (t) cos((k + 1/2)πx) − t cos(5πx/2)
k=0
∞
X
0
ck (t) + (k + 1/2)2 π 2 ck (t) cos((k + 1/2)πx)
k=0,k6=2
+ c02 (t)
+ (5/2)2 π 2 c2 (t) − t cos(5πx/2)
So for k 6= 2, we require
c0k (t) = −(k + 1/2)2 π 2 ck (t)
=⇒
ck (t) = bk e−(k+1/2)
2 π2 t
,
while for k = 2 we need
c02 (t) + (5π/2)2 c2 (t) = t.
h
i0
2
2
2
The integrating factor is e(5π/2) t giving e(5π/2) t c2 (t) = te(5π/2) t , and so integration by parts yields
2
2
2
e(5π/2) t c2 (t) = (5π/2)−2 e(5π/2) t (t−(5π/2)−2 )+b2 =⇒ c2 (t) = (5π/2)−2 (t−(5π/2)−2 )+b2 e−(5π/2) t .
It remains to find the constants bk , using the initial condition:
∞
X
cos(7πx/2) = u(x, 0) =
bk cos((k + 1/2)πx) + −(5π/2)−4 + b2 cos(5πx/2),
k=0,k6=2
and inspection shows that
b2 = (5π/2)−4 , b3 = 1, bk = 0 for k 6= 2, 3,
and so
h
i
2
2
u(x, t) = (5π/2)−2 t + (5π/2)−4 (e−(5π/2) t − 1) cos(5πx/2) + e−(7π/2) t cos(7πx/2).
2. Consider the problem of steady heat conduction in a semi-circular annulus
∆u =
1 ∂2u
∂ 2 u 1 ∂u
+
+ 2 2 = 0,
2
∂r
r ∂r
r ∂θ
2
1 < r < 2,
0 < θ < π,
u(1, θ) = u(2, θ) = 0,
u(r, 0) = 0,
u(r, π) = 1.
Separate variables, u(r, θ) = R(r)Θ(θ), and find the Euler equation to be solved for
R(r). Show that it can also be converted to the Sturm-Liouville form
1
(rR0 )0 + λ R = 0,
r
R(1) = R(2) = 0.
Find the eigenfunctions Rn and eigenvalues λn , and check the orthogonality condition
Z 2
1
0
m 6= n
Rn (r)Rm (r) dr =
,
C
m=n
r
1
where you should determine the constant C.
Hence, find the solution u(r, θ) by eigenfunction expansion.
Hint: remember to include the weighting function 1/r when calculating the generalized
Fourier coefficients.
Substituting u(r, θ) = R(r)Θ(θ) into the equation gives
r2
R0
Θ00
R00
+r
=−
= −λ.
R
R
Θ
The R problem is the Euler ODE
0 = r2 R00 (r) + rR0 (r) + λR(r) = 0,
R(1) = 0 = R(2),
(for convenience we have written the separation constant with a minus sign, since
it is the R problem which has the zero BCs at both ends), which we can re-write in
Sturm-Liouville form:
1
(rR0 )0 + λ R = 0,
r
R(1) = 0 = R(2).
Solutions of the Euler ODE are of the form rµ where µ solves the indicial equation
√
0 = µ2 + λ =⇒ µ = ±i λ,
and so the general solution is
√
√
R(r) = A sin( λ ln x) + B cos( λ ln x).
The BCs give
0 = R(1) = B
and
√
0 = R(2) = A sin( λ ln 2)
√
=⇒ R(r) = A sin( λ ln x)
√
=⇒
3
λ ln 2 = nπ, n = 1, 2, 3, . . . ,
so
λn =
nπ
n2 π 2
,
R
(r)
=
sin
ln
r
.
n
ln 2
ln2 (2)
Compute, by changing variables y = ln r/ ln 2 (so dy = (1/ ln 2)(dr/r)),
Z
1
2
1
Rn (r)Rm (r) dr =
r
2
Z
sin(nπ ln r/ ln 2) sin(mπ ln r/ ln 2)dr/r
Z 1
0 m 6= n
= ln 2
sin(nπy) sin(mπy)dy =
.
ln 2
m=n
0
2
1
Now the Θ problem,
Θ00 = λn Θ,
Θ(0) = 0
has solutions
Θn (θ) = sinh(nπθ/ ln 2),
so
u(r, θ) =
∞
X
cn sinh(nπθ/ ln 2) sin(nπ ln r/ ln 2).
n=1
It remains to satisfy the last BC:
1 = u(r, π) =
∞
X
cn sinh(nπ 2 / ln 2) sin(nπ ln r/ ln 2).
n=1
Using the eigenfunction expansion coefficient formula, the coefficients are given by
R2
Z 1
1
2
2
1 sin(nπ ln r/ ln 2) r dr
sinh(nπ / ln 2)cn = R 2 2
=
ln 2
sin(nπy)dy
1
ln 2
0
1 sin (nπ ln r/ ln 2) r dr
4
2
n odd
nπ
(1 − cos(nπ)) =
=
.
0 n even
nπ
So
4
u(r, θ) =
π
∞
X
n=1,n
odd
sinh(nπθ/ ln 2)
ln r
sin nπ
.
n sinh(nπ 2 / ln 2)
ln 2
4