Hints to homework 7
8.2.1(d). The problem is
ut = kuxx + k,
u(x, 0) = f (x),
u(0, t) = A,
u(L, t) = B.
It has a source term and inhomogeneous boundary conditions, but none of them depend on t. So
(as in example 3 of the notes) we should find the equilibrium temperature distribution ueq (x) and
subtract that from u. The function ueq (x) satisfies:
ku00eq + k = 0,
ueq (0) = A,
ueq (L) = B.,
so ueq = c1 + c2 x − 21 x2 . To have ueq (0) = A we need c1 = A and to have ueq (L) = B we need to
solve
L2
A + c2 L −
=B
2
for c2 . So
L B−A
c2 = +
2
L
and
x2
L B−A
+
x− .
ueq (x) = A +
2
L
2
Next, let v(x, t) = u(x, t) − ueq (x). Since k(ueq )xx = −k, we’ll have
vt = ut = kuxx + k = k(u − ueq )xx = kvxx ,
and since ueq (0) = A and ueq (L) = B, we’ll have
v(0, t) = 0
and v(L, t) = 0.
Finally, we have
v(x, 0) = u(x, 0) − ueq (x) = f (x) − A −
L B−A
−
2
L
x+
x2
.
2
This is a familiar heat equation initial/boundary-value problem and has solution
v(x, t) =
∞
X
cn e−kn
2
π 2 t/L2
sin
nπx n=1
where
cn =
2
L
Z
L
0
L
nπx L B−A
x2
f (x) − A −
−
x+
sin
dx.
2
L
2
L
Finally,
u(x, t) = A +
L B−A
+
2
L
∞
x−
nπx 2 2
2
x2 X
+
cn e−kn π t/L sin
2
L
n=1
with these same constants cn .
8.2.3. The problem is
ut =
k
(rur )r + Q(r),
r
u(r, 0) = f (r),
u(a, t) = T
2
together with the implicit condition that u(0, t) is finite. Once again, this is a problem with an
equilibrium solution ueq (r). We find it by solving
k
(ru0eq )0 + Q(r) = 0,
r
ueq (0) finite,
ueq (a) = T.
To solve the differential equation, move the Q(r) to the right and “unwind” the derivatives and
multiplications that are applied to ueq on the left as follows: First, multiply both sides by r/k and
integrate from 0 to r to get
Z
1 r
0
ρQ(ρ) dρ.
rueq (r) = −
k 0
Next, divide by r and integrate, this time from r to a (to avoid the 1/r singularity at r = 0 – be
careful to change the sign since the r is the lower limit, and add a constant to match the boundary
condition:
Z a
Z σ
1
ρQ(ρ) dρ dσ.
ueq (r) = T +
r kσ 0
Now, as usual, let v(r, t) = u(r, t) − ueq (r). Since (k/r)(ru0eq )0 = −Q(r), we’ll have
vt = ut =
k
k
k
(rur )r + Q(r) = (r(u − ueq )r )r = (rvr )r ,
r
r
r
and v(0, r) will be finite and v(a, r) = 0. The initial conditions for v will be
Z a
Z σ
1
v(r, 0) = u(r, 0) − ueq (r) = f (r) −
ρQ(ρ) dρ dσ.
r kσ 0
The v problem is a familiar heat equation problem on the disk and has solution
v(x, t) =
∞
X
2
2
cn e−kzn t/a J0
n=1
z r
n
,
a
where zn is the nth positive zero of the Bessel function J0 (x), and
Z a
Z σ
Z a
2
1
zn r r dr.
cn = 2
f
(r)
−
ρQ(ρ)
dρ
dσ
J0
2
a (J1 (zn )) 0
a
r kσ 0
Finally,
Z
u(r, t) =
r
a
1
kσ
σ
Z
ρQ(ρ) dρ dσ +
0
∞
X
2
2
cn e−kzn t/a J0
n=1
z r
n
a
with these same constants cn .
8.3.1(c). This time, the problem is
ur = kuxx + Q(x, t),
u(x, 0) = f (x),
u(0, t) = A(t),
ux (L, t) = 0.
No equilibrium solution this time, so we’ll just take something simple to get rid of the boundary
values:
ubd (x, t) = A(t),
so that u0bd is zero when x = L and ubd is equal to A(t) when x = 0. We have
(ubd )t = A0 (t)
and
(ubd )xx = 0,
3
so the function v(x, t) = u(x, t) − ubd (x, t) satisfies
vt = ut − (ubd )t
= kuxx + Q(x, t) − A0 (t)
= k(u − ubd )xx + Q(x, t) − A0 (t)
= kvxx + Q(x, t) − A0 (t)
together with the boundary conditions v(0, t) = 0 and vx (L, t) = 0 and the initial condition
v(x, 0) = f (x) − A(0).
Since the x part (actually, the non-t part) of the differential equation is simply vxx , we seek to expand
v(x, t) as a series of eigenfunctions of X 00 + λX = 0, X(0) = 0, X 0 (L) = 0 and with time-dependent
coefficients, as follows:
∞
X
(n + 21 )πx
v(x, t) =
an (t) sin
L
n=0
(and we know the eigenvalues are (n + 12 )2 π 2 /L2 ). We substitute this into the differential equation
for v to get
∞ X
k(n + 21 )2 π 2
(n + 12 )πx
a0n (t) +
a
(t)
sin
= Q(x, t) − A0 (t).
n
2
L
L
n=0
We have to expand the right side in a similar way (as a sinish series in x with t-dependent coefficients),
so let
Z
(n + 21 )πx
2 L
[Q(x, t) − A0 (t)] sin
dx.
rn (t) =
L 0
L
Then the equation becomes
X
∞ ∞
X
k(n + 12 )2 π 2
(n + 21 )πx
(n + 21 )πx
a0n (t) +
a
(t)
sin
=
r
(t)
sin
,
n
n
L2
L
L
n=0
n=0
and so we have to solve the differential equations
a0n +
k(n + 21 )2 π 2
an = rn (t)
L2
1 2
2
2
for an (t). The integrating factor for this first-order linear equation is ek(n+ 2 ) π t/L , so we have
Z t
2
1 2 2
−k(n+ 21 )2 π 2 t/L2
−k(n+ 12 )2 π 2 t/L2
an (t) = cn e
+e
ek(n+ 2 ) π τ /L rn (τ ) dτ.
0
We determine the constants cn using the initial conditions for v, since we have
∞
X
(n + 21 )πx
v(x, 0) =
cn sin
= f (x) − A0 (0).
L
n=1
Expand the right side as a Fourier series and learn that
Z
(n + 21 )πx
2 L
cn =
[f (x) − A0 (0)] sin
dx.
L 0
L
4
Put all of this together to get
u(x, t) = A(t) +
∞
X
an (t) sin
n=0
(n + 21 )πx
L
where
1 2
an (t) = cn e−k(n+ 2 )
π 2 t/L2
1 2
+ e−k(n+ 2 )
π 2 t/L2
Z
t
,
1 2
ek(n+ 2 )
π 2 τ /L2
rn (τ ) dτ
0
in which
2
rn (t) =
L
0
[Q(x, t) − A (t)] sin
0
and
2
cn =
L
L
Z
Z
(n + 21 )πx
L
[f (x) − A (0)] sin
(n + 21 )πx
L
u(x, 0) = f (x),
u(0, t) = 0,
L
0
0
dx
dx.
8.3.2. This time, the problem is
ur = kuxx + Q(x),
u( L, t) = 0.
Even though we expect an equilibrium solution, the text directs us to use eigenfunction expansion,
so we write
∞
nπx X
an (t) sin
u(x, t) =
L
n=1
(and we know the eigenvalues are n2 π 2 /L2 ). We substitute this into the differential equation for u
to get
∞ nπx X
kn2 π 2
0
an (t) +
a
(t)
sin
= Q(x).
n
L2
L
n=1
We expand the right side (as a sine series in x and let
Z
nπx 2 L
dx.
rn =
Q(x) sin
L 0
L
Then the equation becomes
∞ ∞
nπx X
nπx X
kn2 π 2
0
an (t) +
a
(t)
sin
=
r
sin
,
n
n
L2
L
L
n=1
n=1
and so we have to solve the differential equations
a0n +
kn2 π 2
an = rn
L2
for an (t). By integrating factors or undetermined coefficients, we get that the solution is
2 2
2
L2 rn
L2 rn
an (t) = cn − 2 2 e−kn π t/L + 2 2
kn π
kn π
(we write it this way so that an (0) = cn ). We determine the constants cn using the initial conditions
for u, since we have
∞
nπx X
= f (x).
u(x, 0) =
cn sin
L
n=1
5
Expand the right side as a Fourier series and learn that
Z
nπx 2 L
f (x) sin
dx.
cn =
L 0
L
Put all of this together to get
∞
X
u(x, t) =
an (t) sin
nπx L
n=1
where
2 2
2
L2 rn
L2 rn
an (t) = cn − 2 2 e−kn π t/L + 2 2
kn π
kn π
in which
2
rn =
L
Z
2
L
Z
Q(x) sin
0
and
cn =
L
L
f (x) sin
0
nπx dx
L
nπx L
dx.
As t → ∞, an (t) tends to the constant L2 rn /(kn2 π 2 ), so we have
lim u(x, t) =
t→∞
∞
X
nπx L2
r
sin
.
n
kn2 π 2
L
n=1
This function is clearly zero at both ends of the interval and its second derivative is
−
∞
nπx X
rn
1
sin
= − Q(x)
k
L
k
n=1
by the definition of rn , as we would expect.
8.3.6. The problem is
ut = uxx + e−2t sin 5x,
u(0, t) = 1,
u(π, t) = 0,
u(x, 0) = 0.
Let ubd (x) = 1 − x/π, and v(x, t) = u(x, t) − ubd (x). Since ubd is linear in x,
vt = ut = uxx + e−2t sin 5x = vxx + e−2t sin 5x,
and v(0, t) = v(π, t) = 0 and v(x, 0) = −ubd (x) = x/π − 1. Seek v(x, t) in the form
v(x, t) =
∞
X
an (t) sin nx.
n=1
The differential equation for v becomes
∞
X
(a0n + n2 an ) sin nx = e−2t sin 5x,
n=1
which tells us that a0n + n2 an = 0 unless n = 5, in which case a05 + 25a5 = e−2t . Therefore,
an (t) = cn e−n
2
t
if n 6= 5,
6
and
a5 (t) =
1
1
c5 −
e−25t + e−2t
23
23
(via the method of undetermined coefficients – we subtracted 1/23 from c5 so that a5 (0) = c5 , just
as an (0) = cn for all the other values of n). We determine the coefficients from the initial condition
v(x, 0) =
∞
X
an (0) sin nx =
n=1
∞
X
cn sin nx =
n=1
x
− 1.
π
Thus
cn =
2
π
Z
0
π
Z π
1
2
2 x cos nx π
+
cos
nx
dx
=− .
− 1 sin nx dx =
1−
π
π
π
n 0 nπ 0
nπ
x
We conclude that
v(x, t) =
and so
∞
X
1 −2t
2 −n2 t
(e
− e−25t ) sin 5x −
e
sin nx
23
nπ
n=1
∞
X
2 −n2 t
x
1 −2t
−25t
u(x, t) = 1 − + (e
−e
) sin 5x −
e
sin nx.
π 23
nπ
n=1
8.5.2(b). The problem is
utt = c2 uxx + g(x) cos ωt,
u(0, t) = u(L, t) = 0,
u(x, 0) = f (x),
ut (x, 0) = 0.
Since the boundary values are homogeneous, we launch into our eigenfunction expansion, letting
∞
X
u(x, t) =
an (t) sin
nπx L
n=1
,
where an (t) must satisfy
a00n +
c2 n2 π 2
an = rn cos ωt,
L2
where
2
rn =
L
Z
L
g(x) sin
nπx 0
L
dx
are the Fourier sine coefficients of g(x), and where
a(0) =
2
L
Z
L
f (x) sin
0
nπx L
and a0 (0) = 0
dx
to match the initial data.
We’ll use the method of undetermined coefficients to solve the differential equation for an (t). If
ω 6= cnπ/L then we guess an = A cos ωt and get that
A=
rn
c2 n2 π 2
−
L2
ω2
,
7
so
an (t) =
cn −
L2 rn
2
2
c n π 2 − L2 ω 2
cos
cnπt
L
cnπt
L
+ dn sin
+
L2 rn
cos ωt.
− L2 ω 2
c2 n2 π 2
If ω 6= cnπ/L for all n = 1, 2, 3, . . . then there is no resonance and we will have dn = 0 for all n and
cn = an (0), so can write the solution of the problem as
∞ nπx X
L2 rn
cnπt
L2 rn
u(x, t) =
cn − 2 2 2
cos
cos
ωt
sin
,
+
c n π − L2 ω 2
L
c2 n2 π 2 − L2 ω 2
L
n=1
where
2
rn =
L
Z
L
g(x) sin
0
nπx L
dx
2
and cn =
L
Z
L
f (x) sin
0
nπx L
dx.
If ω = cN π/L for some positive integer N , then we will have resonance, and we have to solve
separately for aN (t) from all the others (the other an ’s remain the same as calculated above). The
function aN (t) satisfies
c2 N 2 π 2
cN πt
a
=
r
cos
,
a00N +
N
N
L2
L
which we’ll write as
a00N + ω 2 aN = rN cos ωt
for simplicity, with rN , aN (0) and a0N (0) as before. For our undetermined coefficients this time,
we’ll have to guess aN (t) = At sin ωt, which gives us
a00N + ω 2 aN = (2Aω cos ωt − Aω 2 t sin ωt) + Aω 2 t sin ωt = 2Aω cos ωt = rN cos ωt,
so A = rN /(2ω) and so
rN t
sin ωt.
2ω
Since aN (0) = cN , we determine cN the same way as all the other cn ’s. And a0N (0) = ωdN is
supposed to be zero, so dN = 0 as well. Therefore, in the case of resonance where ω = cN π/L, we
have
X
∞ nπx rN t
N πx
L2 rn
cnπt
L2 rn
u(x, t) =
sin ωt sin
+
cn − 2 2 2
cos
+
cos
ωt
sin
,
2ω
L
c n π − L2 ω 2
L
c2 n2 π 2 − L2 ω 2
L
n=1
aN (t) = cN cos ωt + dN sin ωt +
where
rn =
2
L
Z
L
g(x) sin
0
nπx L
dx
and cn =
2
L
Z
L
f (x) sin
0
nπx L
dx.
The extra term with t sin ωt will make the solution oscillate more and more violently as t grows.
10.3.6. The Fourier transform of
f (x) =
0 if |x| > a
1 if |x| < a
is (assuming a > 0)
a
Z ∞
Z a
1
1
eiωx eiωa − e−iωa
sin ωa
iωx
iωx
F (ω) =
f (x)e dx =
e dx =
=
=
.
2π −∞
2π −a
2πiω −a
2πiω
πω
8
10.3.7. The inverse Fourier transform of F (ω) = e−|ω|α is (provided α > 0),
Z ∞
F (ω)e−iωx dω
f (x) =
−∞
0
Z
eαω e−iωx dω +
=
−∞
Z 0
=
e
(α−ix)ω
Z
Z
dω +
−∞
∞
e−αω e−iωx dω
0
∞
e(−α−ix)ω dx
0
=
ω=0
ω=∞
e(α−ix)ω e(−α−ix)ω +
α − ix ω=−∞
−α − ix ω=0
=
1
1
+
α − ix α + ix
=
α2
2α
+ x2
(the limiting values of the exponentials at ±∞ are zero because e−|a|x goes to zero as x goes to ±∞
and eiωx stays bounded).
10.3.8. The Fourier transform of xf (x) is
F[xf (x)](ω) =
1
2π
Z
∞
xf (x)e−iωx dx.
−∞
The derivative with respect to ω of the Fourier transform of f is
Z ∞
Z ∞
Z ∞
d
d
1
i
1
d
F[f (x)](ω) =
f (x)eiωx dx =
f (x) (eiωx ) dx =
xf (x)eiωx dx
dω
dω 2π −∞
2π −∞
dω
2π −∞
(we assume we can exchange the derivative and the integral — this will be true provided f (x) is
smooth and goes to zero quickly enough as x goes to ±∞). Comparing these two equations shows
that
d
F[f (x)] = iF[xf (x)]
dω
or, multiplying both sides by −i,
F[xf (x)] = −i
d
F[f (x)].
dω