Math 241
Practice problems for Midterm 2
1. (a) (8.2.4.) Solve the two dimensional heat equation ut = k(uxx + uyy ) subject to the
time-independent boundary conditions
u(0, y, t) = 0,
uy (x, 0, t) = 0
u(L, y, t) = 0,
u(x, H, t) = g(x)
and the initial condition
u(x, y, 0) = f (x, y).
Analyze the limit as t → ∞.
(b) Solve the two dimensional wave equation utt = c2 (uxx + uyy ) with the same boundary
conditions as above and the initial conditions
u(x, y, 0) = f (x, y)
ut (x, y, 0) = 0.
Solution 1:
(a) We look for an equilibrium solution. That is, uE doesn’t depend on t, and satisfies the
boundary conditions
uE (0, y) = 0, (uE )y (x, 0) = 0
uE (L, y) = 0,
uE (x, H) = g(x).
Since (uE )t = 0 and (uE )t = k((uE )xx + (uE )yy ), uE should satisfy (uE )xx + (uE )yy = 0,
i.e. the Laplace equation. This is a fairly straightforward problem; we separate variables
uE (x, y) = φ(x)h(y), and get the equations
φ00 + λφ = 0 φ(0) = φ(L) = 0
h00 − λh = 0 h0 (0) = 0.
nπ 2
Solving for φ gives
us φ(x) = sin nπ
; then solving for h, we have
L x with λ =
L
h(y) = cosh nπ
y
.
So
a
general
solution
will
look
like
L
uE (x, y) =
∞
X
Bn sin
n=1
nπ nπ x cosh
y .
L
L
We use the nonhomogeneous BC to solve for the coefficients Bn :
uE (x, H) =
∞
X
n=1
Bn cosh
nπ
nπ H sin
x = g(x),
L
L
which gives us
Bn cosh
Z
nπ 2 L
H =
g(x) sin
x dx.
L
L 0
L
nπ
Now that we have uE , let’s let v(x, y, t) = u(x, y, t) − uE (x, y). Plugging in u = v + uE
into the original PDE and ICs, we get vt = k(vxx + vyy ) with BCs
v(0, y, t) = 0,
vy (x, 0, t) = 0
v(L, y, t) = 0,
v(x, H, t) = 0
and IC v(x, y, 0) = f (x, y) − uE (x, y).
Now we separate variables to solve for v: v(x, y, t) = ψ(x, y)P (t); plugging in and
dividing by ψP gives
P0
∆ψ
=
= −µ,
ψ
kP
where µ is a constant. We get
∆ψ + µψ = 0 ψ(0, y) = ψ(L, y) = 0
ψy (x, 0) = ψ(x, H) = 0
0
P + µkP = 0.
Now we separate ψ: ψ(x, y) = F (x)G(y). Plugging in and dividing by F G gives
F 00
G00
=−
− µ = −ν,
F
G
which gives
F 00 + νF = 0 F (0) = F (L) = 0,
G00 + (µ − ν)G = 0 G0 (0) = G(H) = 0.
2
Solving for F , we get µ = mπ
, with F (x) = sin mπ
x . Solving for G gives µ − ν =
L
L
2
(2n−1)π
(2n−1)π
y
. So
,
with
G(y)
=
cos
2L
2L
µmn
ψmn
(2n − 1)π 2
=
+
L
2L
mπ (2n − 1)π
= sin
x cos
y .
L
2L
mπ 2
Solving for P gives P (t) = e−µmn kt , so our product solution is ψmn (x, y)e−µmn kt . Then
the general solution is
v(x, y, t) =
∞ X
∞
X
Amn ψmn (x, y)e−µmn kt ;
m=1 n=1
plugging in the ICs, we have
v(x, y, 0) =
∞ X
∞
X
Amn ψmn (x, y) = f (x, y) − uE (x, y),
m=1 n=1
so the coefficients Amn are given by
RR
ψmn (x, y)(f (x, y) − uE (x, y)) dx dy
RR
Amn =
.
ψmn (x, y)2 dx dy
And (finally) the solution is u(x, y, t) = v(x, y, t) + uE (x, y).
(b) Fortunately, we can reuse a lot of work from the previous part. Let uE be exactly as
before, and let v(x, y, t) = u(x, y, t) − v(x, y); then v satisfies the PDE vtt = c2 (vxx + vyy )
with BCs
v(0, y, t) = 0,
vy (x, 0, t) = 0
v(L, y, t) = 0,
v(x, H, t) = 0
and ICs v(x, y, 0) = f (x, y)−uE (x, y), vt (x, y, 0) = 0. Separating v(x, y, t) = ψ(x, y)Q(t),
we see that ψ is exactly as it was in the previous part, and Q00 + c2 µQ = 0. We note
that µmn > 0, so the solution is
√
√
Q(t) = c1 cos( µmn ct) + c2 sin( µmn ct).
Looking ahead to the ICs, we can see that the cos terms will contribute to u(x, y, 0) and
the sin terms to ut (x, y, 0). Since ut (x, y, 0) = 0, we can ignore the sines. So the solution
is given by
∞ X
∞
X
√
v(x, y, t) =
Amn ψmn (x, y) cos( µmn ct),
m=1 n=1
(where ψmn and µmn are as in the previous part). For the ICs, we have
v(x, y, 0) =
∞ X
∞
X
Amn ψmn (x, y) = g(x, y) − uE (x, y),
m=1 n=1
so the coefficients Amn are as in the previous problem too. And finally, we get u = v+uE .
2. Let u(x, t) = F (x − ct) + G(x + ct), where F and G are two functions.
(a) Show that u satisfies the wave equation utt = c2 uxx .
(b) Suppose that u(x, 0) = α(x) and ut (x, 0) = β(x). What are F and G? (You may get an
undetermined constant; this is fine.)
Solution 2:
(a) Differentiating with respect to x and t, we have:
ux = F 0 (x − ct) + G0 (x + ct)
uxx = F 00 (x − ct) + G00 (x + ct)
ut = −cF 0 (x − ct) + cG0 (x + ct)
utt = c2 F 00 (x − ct) + c2 G00 (x + ct),
so we clearly have utt = c2 uxx .
(b) We have
u(x, 0) = F (x) + G(x) = α(x)
ut (x, 0) = −cF 0 (x) + cG0 (x) = β(x),
so
F 0 (x) =
cα0 (x) − β(x)
,
2c
and so
cα0 (x) − β(x)
dx + C
2c
Z
α(x)
1
=
−
β(x) dx + C.
2
2c
Z
F (x) =
Here C is an arbitrary constant; we can set C = 0 (or whatever) if we want. Then
G(x) = α(x) − F (x)
Z
α(x)
1
=
+
β(x) dx − C.
2
2c
3. Consider the eigenvalue problem
xφ00 (x) + xφ0 (x) + λφ(x) = 0,
φ(1) = 0,
φ(2) = 0.
(a) Write this problem in standard Sturm-Liouville form.
(b) Suppose we know the eigenvalues λ1 , λ2 , λ3 , . . ., with eigenfunctions φ1 , φ2 , φ3 , . . .. What
orthogonality condition do the eigenfunctions φn satisfy? If f (x) is a function, and
f (x) =
∞
X
an φn (x),
n=1
write down a formula for an .
(c) Now consider the PDE
ut = xuxx + xux
with boundary conditions
u(1, t) = 0,
u(2, t) = 0
and initial conditions
u(x, 0) = f (x),
(where f (x) is as in part (b)). What is u(x, t)?
(d) Are all eigenvalues nonnegative (λ ≥ 0)? (Bonus: can we have λ = 0?)
(e) Estimate λ for large λ.
Solution 3:
(a) This almost looks like exercise 5.3.3; we divide by x to get
φ00 (x) + φ0 (x) +
λ
φ(x) = 0.
x
Now we use exercise 5.3.3, or we can do this directly. Multiply by H(x) and compare
with the standard Sturm-Liouville form:
H(x)φ00 + H(x)φ + λ
H(x)
φ=0
x
d
(p(x)φ0 ) + (λσ(x) + q(x))φ = 0;
dx
so p(x) = H(x) = p0 (x); we get H(x) = p(x) = ex . Then
ex φ00 + ex φ0 + λ
which we rewrite as
ex
φ = 0,
x
d x 0
ex
(e φ ) + λ φ = 0.
dx
x
Note that p(x) = ex , q(x) = 0, and σ(x) =
ex
x.
(b) The eigenfunctions are orthogonal with weight σ(x) =
2
Z
φm (x)φn (x)
1
ex
dx,
x
ex
x.
So we have
if m 6= n.
The coefficients are given by
R2
1
an =
x
f (x)φn (x) ex dx
R2
1
x
φn (x)2 ex dx
.
(c) Separating variables, we have u(x, t) = φ(x)g(t); plugging in and dividing by φg, we get
xφ00 + xφ0
g0
=
= −λ.
φ
g
So we get xφ00 + xφ0 + λφ = 0 and g 0 + λg = 0, with BCs φ(1) = φ(2) = 0. As in
the previous problems, we assume we know eigenvalues and eigenfunctions λn and φn ;
solving for g, we have g(t) = e−λn t . So the product solutions are φn (x)e−λn t , and the
solution is
∞
X
u(x, t) =
cn φn (x)e−λn t ,
n=1
for some coefficients cn . Using the ICs,
u(x, 0) =
∞
X
cn φn (x) = f (x),
n=1
but f (x) =
∞
X
an φn (x) (as in the previous part), so we just get cn = an . Thus the
n=1
solution is
∞
X
u(x, t) =
an φn (x)e−λn t .
n=1
(d) The Rayleigh quotient is
Rb
−pφφ0 |ba + a [p(φ0 )2 − qφ2 ] dx
λ=
.
Rb
2 σ dx
φ
a
Here a = 1, b = 2, p = ex , q = 0, σ =
simplifies to
ex
x.
Since φ(1) = φ(2) = 0 by the BCs, the formula
R2
ex (φ0 )2 dx
λ = 1R 2 ex
.
2
1 φ x dx
x
Since ex (φ0 )2 ≥ 0 and φ2 ex ≥ 0, we have λ ≥ 0.
For the second part, the only way to have λ = 0 is if ex (φ0 (x))2 = 0 for all x. But then
φ0 (x) = 0 for all x, so φ must be constant. But the BCs say that φ(1) = 0, so we must
have φ = 0. But we can’t have a zero eigenfunction, so we can’t have λ = 0.
(e) For large λ, we have
φ(x) ≈ A(x)(c1 sin ψ(x) + c2 cos ψ(x)),
where
√ Z
ψ(x) = λ
x
1
s
σ(x0 )
dx0
p(x0 )
r
√ Z x
1
dx0
= λ
x0
1
√
√
= λ · 2( x − 1),
and A(x) = (σp)−1/4 (in practice, √
the important facts are that A(x) is never 0, and A(x)
and A0 (x) are small compared to λ).
Now we use the BCs:
φ(1) = A(1)(c1 sin(0) + c2 cos(0)) = A(1) · c2 = 0.
Since A(1) 6= 0, we have c2 = 0. Then
√ √
φ(2) = A(2) · c1 sin(2 λ( 2 − 1)) = 0.
√ √
So 2 λ( 2 − 1) = nπ; solving for λ, we get
λ=
nπ
√
2( 2 − 1)
2
.
4. Consider the Laplace equation ∆u = 0 on the sphere ρ < a subject to the boundary condition
u(a, θ, φ) = sin(φ).
Find u. Write down formulas for the coefficients, but don’t evaluate any integrals. (Note:
You should probably refer to your notes to solve some of the equations. On the test, I’ll give
you any necessary formulas.) (Hint: The BCs don’t depend on θ.)
Solution 4:
Ordinarily, we would separate u(ρ, θ, φ) = f (ρ)q(θ)g(φ). But the BCs don’t depend on θ, so
the solution should also be independent of θ. So we have u(ρ, φ) = f (ρ)g(φ), and we have
∂2u
= 0. So writing out the equation in spherical coordinates, we have
∂θ2
1 ∂
ρ2 ∂ρ
2 ∂u
ρ
∂ρ
1
∂
+ 2
ρ sin φ ∂φ
∂u
sin φ
∂φ
Separating, we get (eventually):
d
dg
sin φ
+ (µ sin φ)g = 0 |g(0)| < ∞,
dφ
dφ
d
2 df
ρ
− µf = 0 |f (0)| < ∞.
dρ
dρ
= 0.
|g(π)| < ∞
The solution to the first equation is µ = n(n + 1), and g(φ) = Pn (cos φ) (Pn is also written
as Pn0 ). Then the second equation becomes
ρ2
df
d2 f
+ 2ρ − n(n + 1)f = 0;
2
dρ
dρ
this is a Cauchy-Euler equation, so we use f = ρs . Plugging this in, we get
s2 + s − n(n + 1) = 0,
which has solutions s = n, −n − 1. So f = c1 ρn + c2 ρ−n−1 . We want |f (0)| < ∞, so we need
c2 = 0. Thus the solution is f = c1 ρn .
So the product solution is ρn Pn (cos φ), and the general solution is
u(ρ, φ) =
∞
X
An ρn Pn (cos φ).
n=1
For the coefficients, we use the BCs:
u(a, φ) =
∞
X
An an Pn (cos φ) = sin φ.
n=1
We remember that the eigenfunctions Pn (cos φ) are orthogonal with weight sin φ, so we have
Ra
Pn (cos φ)(sin φ)2 dφ
n
An a = R0a
.
2
0 (Pn (cos φ)) sin φ dφ
5. Consider the heat equation ut = u(xx) with boundary conditions
ux (0, t) = 0,
ux (1, t) = 1
and initial conditions
u(x, 0) = 0.
(a) Does this problem have an equilibrium solution? If so, what is it? If not, why not?
(b) Solve the problem.
Solution 5:
(a) Total flux in is ux (1, t) − ux (0, 2) = 1 and total heat from sources is 0, so the total energy
is not constant. Thus there is no equilibrium.
(b) We need a reference temperature which satisfies rx (0, t) = 0 and rx (1, t) = 1. One
2
2
possibility is r(x, t) = x2 . Then let v(x, t) = u(x, t) − r(x, t). Plugging in u = v + x2
into the PDE, we have
x2
x2
(v + )t = (v + )xx ,
2
2
and we can rearrange to get
vt = vxx + 1
with BCs
vx (0, t) = vx (1, t) = 0,
x2
and IC v(x, 0) = − 2 .
The homogeneous equation vt = vxx will give the eigenvalue problem
φ00 + λφ = 0,
φ0 (0) = φ0 (1) = 0,
which has solutions λn = (nπ)2 , φn = cos(nπx) for n = 0, 1, 2, . . .. (If we wanted to start
at n = 1 instead, we should set λn = ((n − 1)π)2 and φn = cos((n − 1)πx).)
∞
X
We expand v in terms of the eigenfunctions φn : v(x, t) =
an (t)φn (x). Then vt =
n=0
P 0
P
P
an (t)φn (x), and vx x =
an (t)φ00n (x) = an (t)(−λn )φn (x).
Then
vt − vxx =
=
∞
X
a0n (t)φn (x) −
n=0
∞
X
∞
X
an (t)(−λn )φn (x)
n=0
(a0n (t) + λn an (t))φn (x).
n=0
We expand 1 into eigenfunctions:
1=
∞
X
qn (t)φn (x).
n=0
Now we remember that φn (x) = cos(nx), so
1=
∞
X
n=0
qn (t) cos(nx);
so q0 (t) = 1, and qn (t) = 0 if n ≥ 1. Then setting vt − vxx = 1, we have
a00 + λ0 a0 = 1,
and
a0n + λn an = 0.
Plugging in λn = (nπ)2 , we have
so a0 (t) = t + a0 (0), and
a00 (t) = 1,
a0n (t) + nπan (t) = 0.
so an (t) = an (0)e−nπt for n ≥ 1. To calculate an (0), we use the IC:
v(0, t) =
∞
X
an (0) cos(nπx) = −
n=0
so
Z
1
x2
−
2
a0 (0) =
0
and
Z
an (0) = 2
0
1
x2
−
2
x2
,
2
dx
cos(nx) dx
for n ≥ 1. (If you actually do the integrals, I think you get a0 (0) = − 61 and an (0) =
(−1)n+1 n22π2 .)
Then the solution is
u(x, t) = v(x, t) + r(x, t)
∞
X
x2
=
an (t) cos(nπx) + .
2
n=0