Hints to homework 6
7.7.1. We are trying to solve
utt = c2
1
1
(rur )r + 2 uθθ
r
r
with conditions u(a, θ, t) = 0, u(r, θ, 0) = 0 and ut (r, θ, 0) = α(r) sin 3θ.
Separate variables: u(r, θ, t) = R(r)Θ(θ)T (t) so
T 00
(rR0 )0
Θ00
=
+ 2 = −λ
2
c T
rR
r Θ
so T 00 + c2 λT = 0, and
r(rR0 )0
Θ00
+ λr2 = −
= µ.
R
Θ
Since Θ must be periodic with period 2π we have Θ = c1 cos nθ + c2 sin nθ and µ = n2 (for n =
0, 1, 2, . . .). The R equation becomes
r2 R00 + rR0 + (λr2 − n2 )R = 0,
which is Bessel’s equation. Since we must have R(a) = 0, we now know that λ > 0 and so
R(r) = Jn
z
nm r
a
and λ =
2
znm
.
a2
The initial condition u(r, θ, 0) = 0 tells us that we will only use the sine terms for T , so
√
cznm t
T = sin(c λt) = sin
.
a
So we have
u(r, θ, t) =
∞ X
∞
X
Jn
z
n=0 m=1
nm r
a
sin
cznm t
a
[anm cos nθ + bnm sin nθ].
To prepare for the initial condition ut (r, θ, 0) = α(r) sin 3θ, we see that we’ll have anm = 0 for all n
and m, and bnm = 0 unless n = 3. So
∞
z r
X
cz3m t
3m
u(r, θ, t) =
b3m J3
sin
sin nθ
a
a
m=1
and
ut (r, θ, 0) =
So we need
∞
X
cz3m b3m z3m r J3
sin 3θ.
a
a
m=1
∞
X
cz3m b3m z3m r α(r) =
J3
a
a
m=1
and so
Z
a
b3m
a
z r
3m
Z a
r dr
α(r)J3
z r
2
a
3m
0 Z
α(r)J
=
r dr
=
3
a
z r 2
acz3m J4 (z3m )2 0
a
3m
cz3m
J3 (
r dr
a
0
2
using the formula at the end of the disk notes.
7.7.2(d). We are trying to solve
2
utt = c
1
1
(rur )r + 2 uθθ
r
r
but this time ur (a, θ, t) = 0, u(r, θ, 0) = 0 and ut (r, θ, 0) = β(r, θ).
Start the same way as the previous problem. Separate variables: u(r, θ, t) = R(r)Θ(θ)T (t) so
T 00
(rR0 )0
Θ00
=
+ 2 = −λ
2
c T
rR
r Θ
√
so T 00 + c2 λT = 0 (we know that T will be sin(c λt) since u(r, θ, 0) = 0), and
r(rR0 )0
Θ00
+ λr2 = −
= µ.
R
Θ
Since Θ must be periodic with period 2π we have Θ = c1 cos nθ + c2 sin nθ and µ = n2 (for n =
0, 1, 2, . . .). The R equation becomes
r2 R00 + rR0 + (λr2 − n2 )R = 0,
but this time we need R0 (a) = 0. There are three cases in which this will be satisfied:
1. if λ = n = 0 and R(r) = 1, or
√
2. λ > 0 and n = 0 in which case Θ = 1, T = sin( λt) and
y r
0m
R = J0
a
where y0m is the mth positive zero of J00 (x) for m = 1, 2, . . ., or else
3. if n > 0,
2
ynm
a
a2
0
is the mth positive zero of Jn (x) for m = 1, 2, . . . .
R(r) = Jn
where ynm
y
nm r
and λ =
In the first case (λ = n = 0), we have Θ = cos 0θ = 1 and T = t so that T 00 = 0 and T (0) = 0.
So the product solution in this case is simply t.
So altogether we have
u(r, θ, t) = a00 t+
∞
X
a0m J0
m=1
y
0m r
a
sin
X
∞ X
∞
y r
cy0m t
cynm t
nm
+
Jn
sin
[anm cos nθ + bnm sin nθ] .
a
a
a
n=1 m=1
Since we must have
ut (r, θ, 0) = a00 +
∞
∞ ∞
X
cy0m a0m y0m r X X cynm ynm r J0
+
Jn
[anm cos nθ + bnm sin nθ] = β(r, θ),
a
a
a
a
m=1
n=1 m=1
3
the coefficients are given by
Z 2πZ a
Z 2πZ a
β(r, θ) r dr dθ
1
=
β(r, θ) r dr dθ
a00 = 0 Z0 2πZ a
πa2 0 0
r dr dθ
0
a0m
anm
bnm
0
Z 2πZ a
Z 2πZ a
y r
y r
0m
0m
β(r, θ)J0
a
β(r, θ)J0
r dr dθ
a
r dr dθ
a
a
0 0
0 0
Z
=
=
(m ≥ 1)
Z 2πZ a a
y r 2
y0m r 2
0m
2πcy
J
r
dr
J0
r dr dθ
cy0m
0m
0
a
a
0
0 0
Z 2πZ a
Z 2πZ a
y r
y r
nm
nm
β(r, θ)Jn
a
cos(nθ) r dr dθ
β(r, θ)Jn
cos(nθ) r dr dθ
a
a
a
0 0
0 0
Z
=
=
Z 2πZ a a
y r 2
y0m r 2
0m
r dr
πcynm
J0
cynm
J0
cos2 (nθ) r dr dθ
a
a
0
0 0
Z 2πZ a
Z 2πZ a
y r
y r
nm
nm
a
a
β(r, θ)Jn
sin(nθ) r dr dθ
sin(nθ) r dr dθ
β(r, θ)Jn
a
a
0 0
0 0
Z
=
=
Z 2πZ a a
y r 2
y0m r 2 2
0m
πcynm
J0
r dr
cynm
sin (nθ) r dr dθ
J0
a
a
0
0 0
7.7.10. In this problem, we are trying to solve
1
ut = k (rur )r
r
subject to u(a, t) = 0 and u(r, 0) = f (r). If u(r, t) = R(r)T (t), then
(rR0 )0
T0
=
= −λ
kT
rR
and λ > 0 since we need R(a) = 0. The R equation becomes
r2 R + rR + λr2 R = 0
which is Bessel’s equation of order zero, so
R(r) = J0
z
0m r
a
Therefore
T = e−
and we have
u(r, t) =
∞
X
2
z0m
.
a2
and λ =
2 λt
z0m
a2
2
2
am e−z0m kt/a J0
m=1
z
0m r
a
and the coefficients are
Z
am
a
z
0m r
f (r)J0
r dr
a
=
= 0Z a z0m r 2
J0
r dr
a
0
Z
2
0
a
f (r)J0
z
0m r
a
a2 J1 (z0m )2
r dr
(m ≥ 1 and n ≥ 1)
(m ≥ 1 and n ≥ 1)
4
Since all of the coefficients of t in the exponentials are negative, we have u(r, t) → 0 as t → ∞.
7.9.1(b). The differential equation is
1
1
(rur )r + 2 uθθ + uzz = 0
r
r
with boundary condition u(r, θ, 0) = α(r) sin 7θ on the bottom, and u = 0 on the top and side, i.e.,
u(r, θ, H) = 0 and u(a, θ, z) = 0. We’ll assume u(r, θ, z) = R(r)Θ(θ)Z(z) and then
(rR0 )0
Θ00
Z 00
+ 2 +
= 0.
rR
r Θ
Z
We separate out the Z first and get
Θ00
Z 00
(rR0 )0
+ 2 =−
= −λ
rR
r Θ
Z
since the boundary conditions for R are R(a) = 0. Thus Z 00 − λZ = 0 and we have
√
Z = sinh( λ(H − z))
because Z(H) = 0. The rest of the equation becomes
r(rR0 )0
Θ00
+ λr2 = −
=µ
R
Θ
after multiplying by r2 , and since Θ must be periodic with period 2π. Because of the form of the
boundary condition u(r, θ, 0) = α(r) sin 7θ, we have that
Θ = sin 7θ
and µ = 49.
So now the R equation becomes
r2 R00 + rR0 + (λr2 − 49)R = 0
which is Bessel’s equation of order 7, so
R = J7
z
7m r
and λ =
a
2
z7m
.
a2
Using the notation from the “Laplacian on the cylinder” notes, the solution to the problem is thus
∞
z r
X
z7m (H − z)
7m
u(r, θ, z) =
d7m J7
sinh
sin 7θ,
a
a
m=1
where
Z
2
a
rα(r)J7
z
7m r
0
d7m =
a2 J
n+1 (znm
)2 sinh
a
dr
znm H
a
7.9.2(b). The differential equation is the same,
1
1
(rur )r + 2 uθθ + uzz = 0,
r
r
for m ≥ 1
5
but the boundary conditions are that u is zero on the bottom anduz is zero on the top: u(r, θ, 0) = 0
and uz (r, θ, H) = 0, u = 0 on the flat side: u(r, 0, z) = u(r, π, z) = 0 and u(a, θ, z) = β(θ, z) on the
curved side. As usual, we’ll assume u(r, θ, z) = R(r)Θ(θ)Z(z) and then
Θ00
Z 00
(rR0 )0
+ 2 +
= 0.
rR
r Θ
Z
We separate out the Z first and get
(rR0 )0
Θ00
Z 00
+ 2 =−
=λ
rR
r Θ
Z
since Z(0) = 0 and Z 0 (H) = 0. We conclude that
Z = sin
(n + 21 )πz
H
(n + 12 )2 π 2
.
H2
and λ =
for n = 0, 1, 2, . . .. The rest of the equation becomes
r(rR0 )0
Θ00
− λr2 = −
=µ
R
Θ
after multiplying by r2 , and since Θ(0) = Θ(π) = 0. This gives us
Θ = sin mθ
and µ = m2
with m = 1, 2, . . .. Finally the R equation becomes
r2 R00 + rR0 − (λr2 + m2 )R = 0,
which is the modified Bessel equation of order m, and so
R = Im
(n + 21 )πr
H
.
Altogether, then,
u(r, θ, z) =
∞ X
∞
X
fmn Im
n=0 m=1
(n + 21 )πr
H
sin
(n + 21 )πz
H
sin mθ.,
where (since u(a, θ, z) = β(θ, z) and this is a “standard” (double) Fourier series for β,
Z πZ
4
fmn =
0
0
H
(n + 21 )πz
β(θ, z) sin
sin mθ dz dθ
H
(n + 12 )πa
πHIm
H
7.9.3(b)The differential equation is
ut = k
1
1
(rur )r + 2 uθθ + uzz
r
r
for n ≥ 0, m ≥ 1.
6
and the entire boundary is insulated, so ur (a, θ, z, t) = 0, uz (r, θ, 0, t) = 0, uz (r, θ, H, t) = 0,
uθ (r, 0, z, t) = 0 and uθ (r, π2 , z, t) = 0 and the initial conditions are
u(r, θ, z, 0) = f (r, θ, z).
Because the boundary is insulated, we expect the temperature to approach a constant equal to the
average initial temperature as t → ∞, i.e.,
lim u(r, θ, z, t) =
t→∞
4
πa2 H
Z HZ
Z a
π/2
f (r, θ, z) r dr dθ dz.
0
0
0
To solve the equation, we assume u(r, θ, z, t) = R(r)Θ(θ)Z(z)T (t) and the differential equation
becomes
T0
(rR0 )0
Θ00
Z 00
=
+ 2 +
= −λ
kT
rR
r Θ
Z
since we’re expecting the temperature to decay to the average (so λ ≥ 0). The T equation becomes
T 0 + kλT = 0, so T = e−kλt . Next, we can separate out the Z and get
(rR0 )0
Θ00
Z 00
+ 2 =−
−λ=µ
rR
r Θ
Z
so Z 00 + (λ + µ)Z = 0 with Z 0 (0) = Z 0 (H) = 0. Therefore
Z = cos
nπz H
and λ + µ =
n2 π 2
H2
for n = 0, 1, 2, . . .. Next, we’ll separate out the Θ, so
r(rR0 )0
Θ00
− µr2 = −
=ν
R
Θ
and since Θ0 (0) = Θ0 ( π2 ) = 0, we have
Θ = cos(2mθ)
and ν = 4m2
for ν = 0, 1, 2, . . .. Finally we have
r2 R00 + rR0 + (−µr2 − 4m2 )R = 0
which needs to become Bessel’s equation (not the modified Bessel equation) of order 2m since we
have R0 (a) = 0. We get
2
y
y(2m)`
(2m)` r
R = J2m
and µ = − 2
a
a
0
where y(2m)` is the `th positive zero of J2m
(x), for ` = 1, 2, . . . and m = 0, 1, 2, . . .. As in problem
7.7.2, we must also take into account the fact that we could have µ = m = 0, in which case we have
the constant solution R = 1. We conclude that
u(r, θ, z, t) = a000 +
∞ X
∞ X
∞
X
m=0 n=0 `=1
Problems from the notes:
amn` e−kλt J2m
y
(2m)` r
a
cos(2mθ) cos
nπz H
7
1. For formula (1), we have
∞
X
d
d n
(x Jn (x)) =
dx
dx
k=0
∞
X
=
(−1)k (n + k) x2n+2k−1
k!(k + n)! 2n−1+2k
k=0
= xn
!
(−1)k (2n + 2k)x2n+2k−1
k!(k + n)!
2n+2k
k=0
∞
X
=
(−1)k x2n+2k
k!(k + n)! 2n+2k
∞
X
k=0
xn+2k−1
(−1)k
k!(k + n − 1)! 2n−1+2k
= xn Jn−1 (x).
Similarly, for formula (2),
∞
X
d −n
d
(x Jn (x)) =
dx
dx
=
=
(−1)k
x2k
k!(k + n)! 2n+2k
k=0
∞
X
k=1
∞
X
k=1
(−1)k (2k)x2k−1
k!(k + n)! 2n+2k
(−1)k k x−n xn+2k−1
k!(k + n)! 2n−1+2k
= −x−n
= −x−n
∞
X
k=1
∞
X
k=0
= −x
!
−n
(−1)k−1
xn+1+2(k−1)
(k − 1)!(k − 1 + n + 1)! 2n+1+2(k−1)
(−1)k
xn+1+2k
k!(k + n + 1)! 2n+1+2k
Jn+1 (x).
From formula (1) we get
xn Jn0 (x) + nxn−1 Jn (x) = xn Jn−1 (x)
and dividing both sides by xn gives
Jn0 (x) +
n
Jn (x) = Jn−1 (x)
x
which is (3).
From formula (2) we get
x−n Jn0 (x) − nx−(n+1) Jn (x) = −x−n Jn+1 (x)
and multiplying both sides by xn we get
Jn0 (x) −
which is (4).
n
Jn (x) = −Jn+1 (x)
x
8
Add (3) and (4) together and get (5):
2Jn0 (x) + J(n − 1)(x) − Jn+1 (x)
because the (n/x)Jn (x) terms cancel.
Likewise, subtract (4) from (3) to get (6):
2n
Jn (x) = Jn−1 (x) + Jn+1 (x).
x
2. Recall that u(x) = Jn (αx) is a solution of the differential equation. Multiply the equation by
2x2 u0 and integrate from 0 to 1 and get:
Z
1
2x2 u0 u00 + 2x(u0 )2 + 2(α2 x2 − n2 )uu0 dx
0=
0
Z
1
(x2 (u0 )2 )0 + (α2 x2 − n2 )(u2 )0 dx
=
0
Z
1
(x2 (u0 )2 )0 + α(x2 u2 )0 − 2αxu2 − n2 (u2 )0 dx
=
0
1
Z
= [x (u ) + αx u − n u ] − 2α
0 2
2
2 2
1
2 2
xu2 dx
0
0
where the clever step comes after the third equals sign. Now we recall that u(x) = Jn (αx), so that
u0 (x) = αJn0 (αx), and move the integral over to the other side of the equation and get
1
Z
1
1
xJn (αx) dx = Jn0 (α)2 +
2
2
2
0
n2
1−
α
Jn (α)2 .
But if α = znm is a place where Jn is zero, then we have
Z
1
xJn (znm x)2 dx =
0
1 0
1
Jn (znm )2 = Jn+1 (znm )2 ,
2
2
using formula (4) (and the fact that Jn (znm ) = 0 again).
3. We start from the formula for u at the top of page 7 of the notes:
u(r, θ, t) =
∞ X
∞
X
p
p
Jn ( λnm r) cos( λnm ct) anm cos nθ + bnm sin nθ
n=0 m=1
∞ X
∞
X
+
p
p
Jn ( λnm r) sin( λnm ct) cnm cos nθ + dnm sin nθ
n=0 m=1
To calculate the coefficients cnm and dnm , we need to take the derivative of u with respect to t and
set t = 0 (which gives us the initial condition containing g):
g(r, θ) = ut (r, θ, 0) =
∞ X
∞ p
X
n=0 m=1
p
λnm cJn ( λnm r)(cnm cos nθ + dnm sin nθ)
9
If we view θ as the variable and r as constant for the moment, this becomes an ordinary Fourier
series for g(r, θ), so we have
∞ p
X
Z π
p
1
g(r, θ) dθ for n = 0,
λ0m c c0m J0 ( λ0m r) =
2π −π
m=1
Z
∞ p
X
p
1 π
λnm c cnm Jn ( λnm r) =
g(r, θ) cos mθ dθ for n ≥ 1,
π −π
m=1
Z
∞ p
X
p
1 π
g(r, θ) sin mθ dθ for n ≥ 1.
λnm c dnm Jn ( λnm r) =
π −π
m=1
But the left sides of these are Fourier-Bessel series, so using the results of the ”Orthogonality”
section of the notes we finally obtain the coefficients:
Z πZ a
p
1
rg(r, θ)J0 ( λ0m r) dr dθ
2π −π 0
Z a
c0m =
for n = 0, m ≥ 1
p
√
λnm c
rJ0 ( λ0m r)2 dr
0
Z Z
p
1 π a
rg(r, θ)Jn ( λnm r) cos nθ dr dθ
π −π 0
Z a
cnm =
for n ≥ 1, m ≥ 1
p
√
λnm c
rJn ( λnm r)2 dr
0
Z Z
p
1 π a
rg(r, θ)Jn ( λnm r) sin nθ dr dθ
π −π 0
Z a
for n ≥ 1, m ≥ 1
dnm =
p
√
λnm c
rJn ( λnm r)2 dr
0
and the denominators are given by
Z a
Z
p
2
rJn ( λnm r) dr =
0
a
rJn
z
0
nm
a
r
2
dr =
a2
Jn+1 (znm )2 .
2
4. Formula (1) from problem 1 is
d n
(x Jn (x)) = xn Jn−1 (x)
dx
for n ≥ 1
Replace n with n + 1 and get
d n+1
(x
Jn+1 (x)) = xn+1 Jn (x)
dx
for n ≥ 0
and turn it around into the integral formula
a
Z a
n+1
n+1
x
Jn (x) dx = [x
Jn+1 (x)] = an+1 Jn+1 (a).
0
0
Now to do the integral
Z
0
a
xn+1 Jn
αx a
dx
10
we let u = αx/a, so dx = (a/α) du and we get
a
Z
xn+1 Jn
0
Z
a α a n+1
u
Jn (u) du
α 0 α
a n+2 Z α
un+1 Jn (u) du
=
α
0
a n+2
an+2
=
αn+1 Jn+1 (α) =
Jn+1 (α)
α
α
αx dx =
a
which is the formula we needed to prove.
Now we’re ready to solve the PDE. You could either start from scratch and separate variables
(we’re looking for u(r, t) since there’s no dependence on θ, or else we can use the solution from the
“back to the wave equation” section of the notes, and just get rid of all the terms that have θ in
them, namely all the terms with n ≥ 1. So the solution is
u(r, t) =
∞
X
∞
X
p
p
p
p
c0m J0 ( λ0m r) sin(4 λ0m t)
a0m J0 ( λ0m r) cos(4 λ0m t) +
m=1
m=1
2
since c = 4, a = 1 and λ0m = z0m
(where z0m is the mth positive zero of the Bessel function J0 (x)).
Since we’re only using J0 in this problem, we’ll just write λm instead of λ0m , zm instead of z0m , and
am and cm instead of a0m and c0m .
Since the initial conditions are f (r) = 1 − r2 and g(r) = 1, we also know that
1
Z
am =
0
r(1 − r2 )J0 (zm r) dr
Z 1
rJ0 (zm r)2 dr
for m ≥ 1
0
and (from problem 3),
1
Z
rJ0 (zm r) dr
cm =
0
Z
4zm
for m ≥ 1
1
rJ0 (zm r)2 dr
0
and the integrals in the denominators evaluate to 21 J1 (zm )2 (from problem 2).
To evaluate the integral in the numerator of am , we first make the substitution x = zm r (so
dr = dx/zm and rewrite it as
Z
0
zm
x
zm
Z zm
x2
dx
1
2
1 − 2 J0 (x)
= 4
x(zm
− x2 )J0 (x) dx.
zm
zm
zm 0
2
Next, integrate by parts with u = zm
− x2 and dv = xJ0 (x) dx. Then from formula (1) in the notes
11
(with n = 1), v = xJ1 (x) so we get
Z
Z zm
1
2
x(zm
− x2 )J0 (x) dx
4
zm
0
zm
Z zm
1
1
2
= 4 [(zm
− x2 )xJ1 (x)] + 4
2x2 J1 (x) dx
zm
z
m 0
0
Z zm
2
2
= 4
x J1 (x) dx
zm 0
zm
2
2
= 4 x2 J2 (x) = 2 J2 (zm )
zm
z
m
0
1
r(1 − r2 )J0 (zm r) dr =
0
where we used (1) again, this time with n = 2. Therefore, we have that
Z
am =
0
1
2
J (z )
r(1 − r2 )J0 (zm r) dr
2 2 m
4J2 (zm )
zm
=
= 2
.
Z 1
1
zm J1 (zm )2
2
2
J
(z
)
1
m
rJ0 (zm r) dr
2
0
We can do a little better still — from equation (6) with n = 1, we can replace J2 (zm ) with
2J1 (zm )/zm , and write:
8
am = 3
,
zm J1 (zm )
and that’s about as simple as we can get it.
For cm , we need only use our substitution x = zm r and formula (1) to integrate
Z
1
Z
rJ0 (zm r) dr =
0
0
Therefore
Z
zm
x
J1 (zm )
dx
1
J0 (x)
= 2 zm J1 (zm ) =
zm
zm
zm
zm
1
J1 (zm )
1
zm
0
=
=
= 2
Z 1
1
2zm J1 (zm )
4zm J1 (zm )2
4zm
rJ0 (zm r)2 dr
2
0
rJ0 (zm r) dr
cm
Therefore, the solution is
∞
X
1
8
u(r, t) =
J0 (zm r) 3
cos(4zm t) + 2
sin(4zm t) .
zm J1 (zm )
2zm J1 (zm )
m=1
Whew!