It would be a smart thing to have the solutions of this equation on your “cheat sheet,” in case here is how
one solves it. For
n
= 0 the equation is the same as
rR
′′
+
R
′
= 0; setting
v
=
R
′
one has
rv
′
+
v
= 0. This
separates as
dv/v
=

dr/r
, hence integrating (and ignoring the constantln
v
=

ln
r
+
C
= ln(
C
1
/r
), thus
R
′
=
v
=
C
1
/r
; integrating again,
R
=
C
1
ln
R
+
C
2
. We want differentiable solutions so the logarithm term
is out.
R
0
(
r
) = constant.
For
n
≥
0 we use that the equation has solutions of the form
r
a
; plugging in
R
(
r
) =
r
a
we get
a
(
a

1)
r
a
+
ar
a

n
2
r
a
= 0
,
i.e.
a
2

n
2
= 0
.
Thus
a
=
±
n
and the general solution is
R
(
r
) =
C
1
r
n
+
C
2
r
−
n
. Wanting only differentiable functions we
exclude the part that is singular at 0; have
R
n
(
r
) = constant
r
n
. We can now put everything together and
get the solution as the following superposition:
u
(
r, θ
) =
1
2
a
0
+
∞
∑
n
=1
(
a
n
cos
nθ
+
b
n
sin
nθ
)
r
n
.
For the boundary condition:
cos
θ
+ sin
θ
=
u
(1
, θ
) =
1
2
a
0
+
∞
∑
n
=1
(
a
n
cos
nθ
+
b
n
sin
nθ
)
.
We see that
a
1
= 1
, b
1
= 1 all other coeﬃcients are 0. Thus
u
(
r, θ
) = (cos
θ
+ sin
θ
)
r
. In cartesian coordinates
this is
u
(
x, y
) =
x
+
y
.
One can guess that this would be the solution because ∆(
x
+
y
) = 0 and clearly
x
+
y
equals
x
+
y
on the
boundary of the disk.
4.
(20 points)
Consider the ICBC problem
u
t
(
x, t
) = 4
u
xx
(
x, t
) +
γx,
0
< x <
2
, t >
0
,
u
x
(0
, t
) = 5
,
u
x
(2
, t
) = 3
, t >
0
,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
5
u
(
x,
0) =

1
2
x
2
+ 5
x,
0
< x <
2
,
Determine the value of
γ
for which there is an equilibrium solution (10 points) and find it (10 points).
Solution.
Note:
One of the boundary conditions was given as
u
x
(
L, t
) = 3. It should have been
u
x
(2
, t
) =
3. A student running into this condition should either realize that
L
= 2, or ask what was meant by
L
.
The equilibrium solution
u
∞
will satisfy
4
u
′′
∞
+
γx
= 0
,
u
′
∞
(0) = 5
,
u
′
∞
(2) = 0
.
Solving this:
u
′
∞
(
x
)
=

1
8
γx
2
+
C
1
,
u
∞
(
x
)
=

1
24
γx
3
+
C
1
x
+
C
2
.
From
u
′
∞
(0) = 5 we get
C
1
= 5, thus
u
′
∞
(
x
) =

1
8
x
2
+ 5; setting
x
= 2 we must have

1
2
γ
+ 5 = 3, thus
γ
= 4.
So far,
u
∞
(
x
) =

1
6
x
3
+ 5
x
+
C
2
.
We learned in class (and found a very imprecise explanation of the same thing in the textbook) that
when this type of boundary conditions allow for an equilibrium solution, then
∫
L
0
u
(
x, t
)
dx
is constant and
∫
L
0
u
∞
(
x
)
dx
=
∫
L
0
u
(
x,
0)
dx
. In our case
∫
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '13
 Schonbek
 Sin, Boundary value problem, J2

Click to edit the document details