Our product solutions are of the form
u
nm
(
x, y, t
) =
(
a
nm
cos
√
n
2
+
m
2
πt
+
b
nm
sin
√
n
2
+
m
2
πt
)
sin
nπx
sin
nπy.
The general solution to the PDE and boundary conditions coming from the superposition of these solutions
is
u
(
x, y, t
) =
∞
∑
n,m
=1
(
a
nm
cos
√
n
2
+
m
2
πt
+
b
nm
sin
√
n
2
+
m
2
πt
)
sin
nπx
sin
nπy.
We have to satisfy the initial conditions. Setting
t
= 0 and equating to
u
(
x, y,
0), we get
sin(
πx
) sin(2
πy
) + 3 sin(5
πx
) sin(
πy
) =
∞
∑
n,m
=1
a
nm
sin
nπx
sin
nπ,
from which we get
a
12
= 1,
a
51
= 3, all other
a
nm
= 0. Differentiating with respect to
t
and equating to
u
t
(
x, y,
0),

sin(3
πx
) sin(3
πy
) =
∞
∑
n,m
=1
b
nm
√
n
2
+
m
2
π
sin
nπx
sin
nπ
from which
b
33
=

1
/
(
π
√
18), all other
b
nm
= 0. The solution is
u
(
x, t
) = cos(
√
5
πt
) sin(
πx
) sin(2
πy
) + 3 cos(
√
26
πt
) sin(3
πx
) sin(
πy
)

1
π
√
18
sin(
√
18
πt
) sin(3
πx
) sin(3
πy
).
Grading Note:
There being NO excuse for not doing this exercise correctly and completely, I graded: 50
points for doing it correctly (or almost correctly). That included showing how the separation of variables was
done, and showing all work.
30 points for doing something reasonable.
0 points for doing nothing or almost nothing.
A few grades fell between these parameters.
2.
(10 points)
Show that
u
(
r, θ, t
) =
J
2
(
ξr
) cos(2
θ
) sin(
ξt
), where
ξ
is one of the positive zeroes of the Bessel
function
J
2
, satisfies
u
tt
(
r, θ, t
) =
u
rr
(
r, θ, t
) +
1
r
u
r
(
r, θ, t
) +
1
r
2
u
θθ
(
r, θ, t
)
,
0
≤
r <
1
,

π
≤
θ
≤
π,
t >
0
,
u
(1
, θ, t
) = 0
,

π
≤
θ
≤
π,
t >
0
.
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3
This is really a calculus exercise. As a hint, recall that the equation satisfied by
J
2
(
z
) is
z
2
J
′′
2
(
z
) +
zJ
′
2
(
z
) +
(
z
2

4)
J
2
(
z
) = 0 so if by some miracle you happen to have somewhere a factor of the form
ξ
2
J
′′
2
(
ξr
) +
ξ
r
J
′
(
ξr
)

4
r
2
J
2
(
ξr
);
that is,
(
(
rξ
)
2
J
′′
2
(
ξr
) + (
rξ
)
J
′
(
ξr
)

4
J
2
(
ξr
)
)
1
r
2
,
it will equal (can be replaced by)

ξ
2
J
2
(
rξ
). Or viceversa.
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
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 Spring '13
 Schonbek
 Sin, Boundary value problem, J2

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