Solution.
As mentioned, this is a Calculus exercise. Suppose
u
(
r,θ,t
) =
J
2
(
ξr
) cos(2
θ
) sin(
ξt
). By a
celebrated calculus result that goes by the name of
chain rule
,
∂u
∂t
(
r,θ,t
)
=
∂
∂t
(
J
2
(
ξr
) cos(2
θ
) sin(
ξt
)) =
J
2
(
ξr
) cos(2
θ
)
∂
∂t
(sin(
ξt
)) =
J
2
(
ξr
) cos(2
θ
)
ξ
cos(
ξt
)
=
ξJ
2
(
ξr
) cos(2
θ
) cos(
ξt
)
,
∂
2
u
∂t
2
(
r,θ,t
)
=
ξJ
2
(
ξr
) cos(2
θ
)
∂
∂t
(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
)
²
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
)
²
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?