(cos(
ξt
)) =

ξ
2
J
2
(
ξr
) cos(2
θ
) sin(
ξt
)
,
∂u
∂r
(
r, θ, t
)
=
ξJ
′
2
(
ξr
) cos(2
θ
) sin(
ξt
)
,
∂
2
u
∂r
2
(
r, θ, t
)
=
ξ
2
J
′′
2
(
ξr
) cos(2
θ
) sin(
ξt
)
,
∂
2
u
∂θ
2
(
r, θ, t
)
=

4
J
2
(
ξr
) cos(2
θ
) sin(
ξt
)
thus (from the last three equalities)
u
rr
(
r, θ, t
) +
1
r
u
r
(
r, θ, t
) +
1
r
2
u
θθ
(
r, θ, t
) =
(
ξ
2
J
′′
2
(
ξr
) +
ξ
r
J
′
2
(
rξ
)

4
r
2
J
2
(
ξr
)
)
cos(2
θ
) sin(
ξt
)
and proving that
u
tt
=
u
rr
+
r
−
1
u
r
+
r
−
2
u
θθ
reduces to showing that

ξ
2
J
2
(
ξr
) cos(2
θ
) sin(
ξt
) =
(
ξ
2
J
′′
2
(
ξr
) +
ξ
r
J
′
2
(
rξ
)

4
r
2
J
2
(
ξr
)
)
cos(2
θ
) sin(
ξt
)
.
Cancelling cos(2
θ
) sin(
ξt
), this is equivalent to showing that

ξ
2
J
2
(
ξr
) =
ξ
2
J
′′
2
(
ξr
) +
ξ
r
J
′
2
(
rξ
)

4
r
2
J
2
(
ξr
);
equivalently that
(
rξ
)
2
J
′′
2
(
ξr
) +
rξJ
′
2
(
rξ
) + ((
rξ
)
2

4)
J
2
(
ξr
) = 0
.
This last equation is just the defining ODE for
J
2
, evaluated at
z
=
rξ
; thus it holds. Finally,
u
(1
, θ, t
) =
J
2
(
ξ
) cos(2
θ
) sin(
ξt
) = 0
since
J
2
(
ξ
) = 0.
3.
(20 points)
Solve by separation of
polar
variables
∆
u
(
x, y
)
=
0
,
x
2
+
y
2
<
1
,
u
(
x, y
)
=
x
+
y,
x
2
+
y
2
= 1
.
This problem is not formulated in the way the text trained you. It is the same problem as
u
rr
+
1
r
u
r
+
1
r
2
u
θθ
=
0
,
0
≤
r <
1
,

π
≤
θ
≤
π,
u
(1
, θ
)
=
cos
θ
+ sin
θ,

π
≤
θ
≤
π
Solve the second problem (which is the same as the first). It is OK to leave the solution in the form
u
(
r, θ
) =
function of
r
and
θ
. But 5 bonus points will be given for returning it to cartesian coordinates. Will this be
an aha! moment?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
4
Solution.
We look for all nonzero solutions of the form
u
(
r, θ
) =
R
(
r
)Θ(
θ
). They will have to satisfy
R
′′
Θ +
1
r
R
′
Θ +
1
r
2
R
Θ
′′
= 0;
Multiplying by
r
2
/
(
R
Θ) and separating variables we get
r
2
R
′′
R
+
rR
′
R
=

Θ
′′
Θ
=
λ,
for constant
λ
. We have the equations
Θ
′′
+
λ
Θ
=
0
,
r
2
R
′′
+
rR
′

λR
=
0
.
For geometry reasons, the solutions of the first equation must be periodic of period 2
π
in
θ
; this forces
λ
=
n
2
,
n
= 0
,
1
,
2
,
3
, . . .
and the corresponding solutions are
Θ
0
(
θ
)
=
1
2
a
0
,
a
0
a constant
,
Θ
n
(
θ
)
=
a
n
cos
nθ
+
b
n
sin
nθ,
constants
a
n
, b
n
,
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
.
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