MATH 617-010
Prof. D. A. Edwards
Introduction to Applied Mathematics II
Due: Apr. 9, 2013
Homework Set 6 Solutions
1. (12 points) Consider the one-phase Stefan problem where the condition at x̃ = 0
is a flux condition:
∂ T̃
−k
(0, t̃) = Q,
(6.1)
∂ x̃
where Q is a constant. Show that this problem may be scaled in such a way so
that no dimensionless parameters appear. Why does this happen?
Solution. Rewriting the dimensional equations from class, we have
∂ 2 T̃
∂ T̃
=α 2,
∂ x̃
∂ t̃
0 < x̃ < s̃(t̃);
T̃ (s̃(t̃), t̃) = Tm ,
−k
∂ T̃
ds̃
(s̃(t̃), t̃) = ρL .
∂ x̃
dt̃
(A)
We leave each of the scalings arbitrary for now, except that we subtract off Tm to leave a
temperature of zero at the front:
T (x, t) =
T̃ (x̃, t̃) − Tm
,
Tc
x=
x̃
,
xc
s(t) =
s̃(t̃)
,
xc
t=
t̃
.
tc
Substituting these results into (A), we have
T (s(t), t) = 0,
Tc ∂T
αTc ∂ 2 T
= 2
,
tc ∂t
xc ∂x2
0 < x < s(t),
from which we have that tc = x2c /α. Substituting these results into (6.1) and (A), we have
−
kTc ∂T
α ds
(s(t), t) = ρLxc 2 ,
xc ∂x
xc dt
∂T
ρLα ds
−Q
(s(t), t) =
.
∂x
xc dt
−
kTc ∂T
(0, t) = Q
xc ∂x
Therefore, by letting
xc =
we have
Copyright ©2013
ρLα
,
Q
Tc =
ρLα
,
k
tc =
ρ2 L2 α
,
Q2
∂T
∂2T
=
, 0 < x < s(t),
∂t
∂x2
∂T
ds
∂T
T (s(t), t) = 0,
(s(t), t) = − ,
(0, t) = −1.
∂x
dt
∂x
D. A. Edwards
All Rights Reserved
M617S13Sol6.2
The elimination of the parameter occurs because there is no artificially imposed temperature scale, and hence we may scale both temperature and flux by the same quantity.
2. (10 points) Write the solution to
u00 = f (x),
0 < x < 1;
u0 (0) = u0 (1) = 0,
(6.2)
in terms of a (regular or modified) Green’s function. Be sure to state any restrictions on f and how they affect any arbitrariness in your solution.
Solution. The homogeneous equation
u00h = 0,
u0h (0) = u0h (1) = 0,
has the solution uh = A, where A is a constant. Therefore, (6.2) has a solution if and only
if
Z 1
f (x) dx = 0.
0
The normalized eigenfunction has
Z
1
A2 dx = 1
=⇒
A = 1.
0
Therefore, to obtain the modified Green’s function, we must solve
G00 = δ(x − ξ) − (1)(1),
G0 (0) = G0 (1) = 0.
Solving the above subject to the boundary condition, we obtain
G=




−
x2
+ B− , 0 < x < ξ,
2
2


 − x + x + B+ , ξ < x < 1.
2
Enforcing continuity, we obtain

x2


 − + ξ + B, 0 < x < ξ,
2
G=
2

x

 − + x + B, ξ < x < 1.
2
Note that we have a multiple of the eigenfunction in our solution, and the jump condition
is automatically satisfied, since
G0 (ξ + ) − G0 (ξ − ) = (−ξ + 1) − (−ξ) = 1.
M617S13Sol6.3
By the Fredholm alternative, we know that hf, 1i = 0, so we may take B = 0 without loss
of generality to write our solution as

x2


Z 1
 − + ξ, 0 < x < ξ,
2
f (x)G(x|ξ) dξ + C,
G(x|ξ) =
u(x) =
2

0

 − x + x, ξ < x < 1.
2
where the arbitrary constant C is a multiple of the eigenfunction to the homogeneous
problem, and hence can be added on without loss of generality.
3. (18 points) Write the solution to
(xu0 )0 = f (x),
1 < x < e;
eu0 (e) = u(e),
u(1) = 0,
(6.3)
in terms of a (regular or modified) Green’s function. You should find that part
of the Green’s function looks like
−
x(log x − 2) log ξ + e log(ξ/x)
.
e−2
Be sure to state any restrictions on f and how they affect any arbitrariness in
your solution.
Solution. Solving the homogeneous equation, we have
xu0h = A
uh = A log x + B
uh (1) = 0
eu0 (e) = e
=⇒
B=0
A
= A = A log e = u(e),
e
so the homogeneous equation has the nontrivial solution uh = A log x, where A is a constant. Therefore, (6.3) has a solution if and only if
Z
e
f (x) log x dx = 0.
1
The normalized eigenfunction has
Z
u = log x
=⇒
e
A2 log2 x dx = 1
1
Z 1
Z 1
2 u 1
2
2
u
2
u
A
u (e du) = A
u e 0−
2ue du = 1
0
0
1
A2 e − 2 [ueu ]0 + 2
Z
0
1
n
o
1
u
e du = A2 −e + 2 [eu ]0 = 1
M617S13Sol6.4
A2 = (e − 2)−1 .
Therefore, to obtain the modified Green’s function, we must solve
(xG0 )0 = δ(x − ξ) −
log x log ξ
,
e−2
G(1) = 0,
eG0 (e) = G(e).
We begin by finding Gp , which is proportional to the particular solution:
(xG0p )0 = log x
xG0p = x(log x − 1)
G0p
(B)
= log x − 1
Gp = x(log x − 1) − x = x(log x − 2),
where we don’t have to worry about constants of integration since they will come into play
with the homogeneous conditions. Therefore, our modified Green’s function is given by

x(log x − 2) log ξ


+ A− log x + B− , 1 < x < ξ,
−
e−2
G=

x(log x − 2) log ξ

−
+ A+ log x + B+ , ξ < x < e.
e−2
Enforcing the boundary conditions, we have
2 log ξ
2 log ξ
+ B− = 0
=⇒
B− = −
e−2
e−2
e(1 − 2) log ξ
A+
= G(e) = −
+ A+ + B +
=⇒
eG0 (e) = e
e
e−2
G(1) =
B+ = −
e log ξ
e−2
Therefore, we have

[x(log x − 2) + 2] log ξ


+ A− log x,
−
e−2
G=

[x(log x − 2) + e] log ξ

−
+ A+ log x,
e−2
1 < x < ξ,
ξ < x < e.
Enforcing continuity, we obtain

x(log x − 2) log ξ + 2(log ξ − log x)


+ A log x,
−
e−2
G=

x(log x − 2) log ξ + e(log ξ − log x)

−
+ A log x,
e−2
1 < x < ξ,
ξ < x < e.
In the above we have tried to impose some symmetry on the problem. However, since
in doing so we added a multiple of the eigenfunction (which we can do without loss of
generality), other forms are possible.
M617S13Sol6.5
Note that we still have a multiple of the eigenfunction in our solution, and the jump
condition is automatically satisfied, since
2ξ − e−2
eξ + −
=
ξ G (ξ ) − ξ G (ξ ) =
= 1.
(e − 2)x x=ξ+
(e − 2)x x=ξ−
e−2
+
0
+
−
0
−
By the Fredholm alternative, we know that hf, log xi = 0 for there to be a solution, so we
may take A = 0 without loss of generality to write our Green’s function as

x(log x − 2) log ξ + 2 log(ξ/x)


, 1 < x < ξ,
−
e−2
G(x|ξ) =

x(log x − 2) log ξ + e log(ξ/x)

−
, ξ < x < e.
e−2
Hence the solution may be written as solution as
Z
u(x) =
e
f (x)G(x|ξ) dξ + C log x,
1
where a multiple of the eigenfunction can be added to the solution without loss of generality.