Math 311
Bessel function series and the heat
equation
Michael Anshelevich
December 4, 2009
O RTHOGONALITY
OF
B ESSEL
FUNCTIONS .
Let λ1, λ2, λ3, . . . be positive roots of Jn(x).
1.0
0.8
0.6
0.4
0.2
0.0
−20
−16
−12
−8
−4
−0.2
x
0
4
8
12
16
20
−0.4
y
−0.6
−0.8
−1.0
λ1
λ2
λ3
λ4
J0 2.405 5.520 8.654 11.792
cos x 1.571 4.712 7.854 10.995
1
Then
Z 1
0
x Jn(λk x) Jn(λmx) dx = 0.
In other words, for the inner product
hf, gi =
Z 1
0
xf (x)g(x) dx
we have
hJn(λk x), Jn(λmx)i = 0 for k 6= m.
Also,
2
kJn(λk x)k =
Z 1
0
x Jn(λk
x)2 dx
2
= 2
.
Jn+1(λk )
Different Jn(λk x) orthogonal but not orthonormal.
2
S ERIES
OF
B ESSEL
FUNCTIONS .
So if
∞
X
f (x) =
Ak Jn(λk x)
k=1
then
Ak =
hf, Jn(λk x)i
kJn(λk x)k2
Z
1
2
Ak = 2
x Jn(λk x) f (x) dx.
Jn+1(λk ) 0
3
H EAT
EQUATION IN A CIRCULAR PLATE ( OR CYLINDER ).
Faces insulated.
Rim kept at temperature 0.
Initial temperature (1) F (r). (2) F (r) = T0.
Find u(r, θ, t).
Clearly temperature circularly symmetric, so only u(r, t).
Heat equation in polar coordinates
µ
1
ut = a2 urr + ur
r
¶
1
(+ 2 uθθ ).
r
Boundary condition u(1, t) = 0.
Initial condition u(r, 0) = F (r).
4
Step I. Separation of variables.
u(r, t) = R(r)T (t),
1
RT 0 = a2(R00T + R0T ),
r
R00
1 R0
1 T0
2
=
+
=
−λ
a2 T
R
rR
So
T 0 = −a2λ2T
and
T (t) =
2 λ2 t
−a
e
.
Also
r2R00(r) + rR0(r) + λ2r2R(t) = 0.
After a substitution x = λr, get
x2R00 + xR0 + x2R = 0.
Bessel equation for n = 0.
5
So the first equation has solutions
R = a1J0(λr) + b1Y0(λr).
Step II. Boundary conditions.
R bounded at r = 0. So
R = a1J0(λr).
Also
R(1)T (t) = 0.
So
J0(λ) = 0
and
λ = λ1, λ2, λ3, . . . positive roots of J0.
Thus
R(r) = Ak J0(λk r)
and the product solutions are
2 2
u(r, t) = Ak e−a λk t J0(λk r).
6
Step III. Initial conditions.
From superposition
u(r, t) =
∞
X
2 2
Ak e−a λk t J0(λk r).
k=1
The initial condition is
u(r, 0) =
∞
X
Ak J0(λk r) = F (r).
k=1
So know
Z
1
2
Ak = 2
x J0(λk x) F (x) dx.
J1 (λk ) 0
7
If F (r) = T0 then
Z
1
2T0
Ak = 2
xJ0(λk x) dx
J1 (λk ) 0
2T
1
= 2 0
J1 (λk ) λ2
k
Recall
Z
Z λ
k
0
xJ0(x) dx.
xnJn−1(x) dx = xnJn(x) + C.
Then
2T0 1
2T0
λk
Ak = 2
[xJ1(x)]0 =
.
2
J
(λ
)λ
J1 (λk ) λk
1 k k
Thus finally,
∞
X
2T0
−a2 λ2
k t J0 (λk r).
e
u(r, t) =
J (λ )λ
k=1 1 k k
8