n
= 1
,
2
,
3
, .. . .
Looking at the equation for
R
, with
λ
=
n
2
it has the form
r
2
R
′′
+
rR
′

n
2
R
= 0
.
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
,