Solving Problem Set 4 by Recipe
Here is a restructured presentation of selected solutions from Problem Set 4 in which the five-step recipe
introduced in Section 203 is prominently displayed.
2. Solve the wave equation on 0 < x < 1, t > 0, with boundary conditions u(0, t) = u(1, t) = 0 and initial
conditions
x,
0 ≤ x ≤ 1/2,
0 ≤ x ≤ 1,
(a) u(x, 0) =
ut (x, 0) = 0,
1 − x, 1/2 < x ≤ 1,
0 ≤ x ≤ 1.
(b) u(x, 0) = 0, ut (x, 0) = x,
1. Eigenfunctions. Try u(x, t) = X(x)T (t) in the wave equation and BC’s:
X 00 (x) + λX(x) = 0, 0 < x < 1;
X(0) = 0 = X(1).
Every nontrivial solution for X is a constant multiple of some function on this list:
Xk (x) = sin(kπx),
2. Declare. Let u(x, t) =
∞
X
k = 1, 2, 3, . . . .
Tk (t) sin(kπx) for functions Tk (t) to be determined.
k=1
3. Initialize. Plug the series into the IC’s and use FSS coefficient formulas. In part (a),
Z
∞
X
kπ
2 1
4
Tk (0) sin(kπx)
=⇒ Tk (0) =
u(x, 0) sin(kπx) dx = · · · = 2 2 sin
,
u(x, 0) =
1 0
k π
2
0 = ut (x, 0) =
k=1
∞
X
Tk0 (0) sin(kπx)
=⇒ Tk0 (0) = 0.
k=1
In part (b),
0 = u(x, 0) =
x = ut (x, 0) =
∞
X
k=1
∞
X
Tk (0) sin(kπx)
=⇒ Tk (0) = 0
Tk0 (0) sin(kπx)
Tk0 (0)
=⇒
k=1
2
=
1
Z
1
x sin(kπx) dx = · · · =
0
2(−1)k+1
.
kπ
4. Propagate. Plug the series form into the wave equation and rearrange.
0 = utt − c2 uxx =
∞
X
Tk00 (t) + k 2 π 2 c2 Tk (t) sin(kπx).
k=1
By FSS coefficient-matching, each Tk obeys
Tk00 (t) + k 2 π 2 c2 Tk (t) = 0,
t > 0,
so there must be constants Ak , Bk such that
Tk (t) = Ak cos(kπct) + Bk sin(kπct);
note
Tk (0) = Ak ,
Tk0 (0) = kπcBk .
Here Ak = Tk (0) and Bk = Tk0 (0)/(kπc) can be obtained from the results of Step 3.
∞
X
kπ
4
sin
cos(kπct) sin(kπx).
5. Summarize. For part (a), u(x, t) =
2
2
k π
2
k=1
∞
X
2(−1)k+1
cos(kπct) sin(kπx).
For part (b), u(x, t) =
k2 π2 c
k=1
File “hw04”, version of 2 Feb 2005, page 1.
Typeset at 16:45 February 2, 2005.
2
Recipe-Based approach to Problem Set 4
3. Consider the modified wave equation
1
utt + γ 2 u = uxx ,
0 < x < `,
t > 0,
c2
with boundary conditions
u(0, t) = 0 = u(`, t),
t>0
and initial conditions
u(x, 0) = f (x),
ut (x, 0) = 0,
0 < x < `.
Show by the method of separation of variables that the solution can be written as
!
r
∞
X
k2 π2
kπx
,
αk cos ct
+ γ 2 sin
u(x, t) =
`2
`
k=1
where
αk =
2
`
Z
`
f (x) sin
0
kπx
dx.
`
1. Eigenfunctions. Try u(x, t) = X(x)T (t) in the wave equation and BC’s:
X 00 (x) + λX(x) = 0, 0 < x < `;
X(0) = 0 = X(`).
Every nontrivial solution for X is a constant multiple of some function on this list:
kπx
,
k = 1, 2, 3, . . . .
Xk (x) = sin
`
2. Declare. Let u(x, t) =
∞
X
k=1
kπx
Tk (t) sin
`
for functions Tk (t) to be determined.
3. Initialize. Plug the series into the given IC’s: by FSS coefficient formulas,
Z
∞
X
kπx
kπx
2 `
Tk (0) sin
f (x) sin
=⇒ Tk (0) =
dx
f (x) = u(x, 0) =
`
` 0
`
k=1
∞
X
kπx
0
Tk (0) sin
=⇒ Tk0 (0) = 0.
0 = ut (x, 0) =
`
k=1
4. Propagate. Plug the series form into the PDE and rearrange.
# "
2
∞
X
kπ
kπx
1 00
1
2
2
.
T
(t)
+
γ
T
(t)
+
T
(t)
sin
0 = 2 utt + γ u − uxx =
k
k
c
c2 k
`
`
k=1
By FSS coefficient-matching, each Tk obeys
k2 π2
t > 0.
Tk (t) = 0,
Tk00 (t) + c2 γ 2 + 2
`
r
k2 π2
Write ωk = c γ 2 + 2 to express the general solution as
`
note
Tk (0) = αk , Tk0 (0) = ωk βk .
Tk (t) = αk cos(ωk t) + βk sin(ωk t);
Here αk = Tk (0) and βk = Tk0 (0)/ωk = 0 from the results of Step 3.
! r
∞
2 π2
X
kπx
k
αk cos c γ 2 + 2 t sin
, where
5. Summarize. u(x, t) =
`
`
k=1
Z
kπx
2 `
f (x) sin
dx,
k = 1, 2, 3, . . . .
αk =
` 0
`
File “hw04”, version of 2 Feb 2005, page 2.
Typeset at 16:45 February 2, 2005.
Recipe-Based approach to Problem Set 4
3
4. In one common application of the diffusion equation
∂2u
∂u
=
,
∂t
∂x2
0 < x < a,
u(x, t) represents the temperature at position x and time t along a straight bar of uniform cross-section. The
ends of the bar x = 0 and x = a are held at zero temperature, so that u(0, t) = u(a, t) = 0. Use the method
of separation of variables to obtain the solution satisfying the initial condition u(x, 0) = 1, 0 < x < a.
1. Eigenfunctions. Try u(x, t) = X(x)T (t) in the wave equation and BC’s:
X 00 (x) + λX(x) = 0, 0 < x < a;
X(0) = 0 = X(a).
Every nontrivial solution for X is a constant multiple of some function on this list:
kπx
,
Xk (x) = sin
a
2. Declare. Let u(x, t) =
∞
X
k=1
k = 1, 2, 3, . . . .
kπx
Tk (t) sin
a
for functions Tk (t) to be determined.
3. Initialize. Plug the series into the given IC’s: by FSS coefficient formulas,
1 = u(x, 0) =
∞
X
k=1
Z
kπx
kπx
2 a
2(1 − (−1)k )
.
Tk (0) sin
sin
=⇒ Tk (0) =
dx =
a
a 0
a
kπ
4. Propagate. Plug the series form into the PDE and rearrange.
0 = ut − uxx =
∞
X
"
Tk0 (t)
+
k=1
kπ
a
2
#
kπx
Tk (t) sin
.
a
By FSS coefficient-matching, each Tk obeys
Tk0 (t) +
Write αk =
k2 π2
Tk (t) = 0,
a2
t > 0.
k2 π2
to express the general solution as
a2
Tk (t) = Ak e−αk t;
Here Ak = Tk (0) is calculated in Step 3.
5. Summarize. u(x, t) =
∞
X
2(1 − (−1)k )
k=1
File “hw04”, version of 2 Feb 2005, page 3.
kπ
e
−k2 π 2 t/a2
note
Tk (0) = Ak .
kπx
sin
.
a
Typeset at 16:45 February 2, 2005.
4
Recipe-Based approach to Problem Set 4
5. A string is stretched between two infinitely long horizontal rods separated by a distance 2π. Its ends are
free to slide without friction along the rods, so the string’s displacement u(x, t) satisfies
0 < x < 2π, t > 0,
(PDE) utt = c2 uxx ,
(BC)
ux (0, t) = 0 = ux (2π, t),
t > 0.
Calculate and describe the motion of the string in two different situations:
(a) The string starts with u(x, 0) = 2 + 8 cos(4x) − cos(32x) and velocity ut (x, 0) = 0.
(b) The string starts with displacement u(x, 0) = 0 and velocity ut (x, 0) = sin(x/2).
1. Eigenfunctions. Try u(x, t) = X(x)T (t) in the wave equation and BC’s:
X 00 (x) + λX(x) = 0, 0 < x < 2π;
X 0 (0) = 0 = X 0 (2π).
Every nontrivial solution for X is a constant multiple of some function on this list:
kx
,
k = 0, 1, 2, 3, . . . .
Xk (x) = cos
2
This follows from Question 1(a).
2. Declare. Inspired by FCS, set u(x, t) = 12 T0 (t)+
∞
X
Tn (t) cos( 12 nx), for functions Tn (t) to be determined.
n=1
3. Initialize. See below.
4. Propagate. 0 = utt − uxx = 12 T000 (t) +
Tn00 (t) +
which leads to
Tn (t) =
∞ X
n 2 c2
Tn (t) cos( 12 nx) implies
Tn00 (t) +
4
n=1
n 2 c2
Tn (t) = 0,
4
n = 0, 1, 2, . . . ,
A0 + B0 t,
if n = 0,
An cos( 12 nct) + Bn sin( 12 nct), if n = 1, 2, . . ..
5. Summary. Without looking at the IC’s, we have the form
u(x, t) =
1
2
(A0 + B0 t) +
∞
X
An cos( 12 nct) + Bn sin( 12 nct) cos( 12 nx),
n=1
for constants An and Bn determined by the IC’s.
Problem part (a). The zero-velocity condition gives
0 = ut (x, 0) = 12 B0 +
∞
X
1
2 ncBn
cos( 12 nx),
n=1
which implies Bn = 0 for all n = 0, 1, 2, . . .. Then the displacement condition gives
2 + 8 cos(4x) − cos(32x) = u(x, 0) = 12 A0 +
∞
X
An cos( 12 nct) cos( 12 nx).
n=1
Coefficient-matching gives An = 0 for all n except these: A0 = 4, A8 = 8, A64 = −1. The solution is
u(x, t) = 2 + 8 cos(4ct) cos(4x) − cos(32ct) cos(32x).
File “hw04”, version of 2 Feb 2005, page 4.
Typeset at 16:45 February 2, 2005.
Recipe-Based approach to Problem Set 4
5
The string oscillates in a simple superposition of two modes about an average displacement of u = 2.
Problem part (b). The zero-displacement condition gives
0 = u(x, 0) = 12 A0 +
∞
X
An cos( 12 nct) cos( 12 nx),
n=1
so all An = 0, n = 0, 1, 2, . . .. The velocity condition gives
sin( 12 x) = ut (x, 0) = 12 B0 +
∞
X
1
1
2 ncBn cos( 2 nx).
n=1
The FCS extraction formula gives
[case n = 0]
[case n = 1]
B0 =
1
2 cB1
=
2
2π
2
2π
Z
2π
sin( 12 x) dx =
0
Z
0
4
,
π
2π
sin( 12 x) cos( 12 x) dx = 0,
while for n ≥ 2, famous trigonometric identities imply
1
2 ncBn
Z 2π
2
sin( 12 x) cos( 12 nx) dx
2π 0
Z
1−n
1+n
1 2π 1
x
+
sin
x
dx
sin
=
π 0 2
2
2
2π
1+n
1−n
2
2
2[(−1)n + 1]
1
cos
x +
cos
x
.
=−
=−
2π 1 + n
2
1−n
2
π(n2 − 1)
x=0
=
The solution is
u(x, t) =
∞
4 X (−1)n + 1
2
t−
sin( 12 nct) cos( 12 nx).
π
πc n=2 n(n2 − 1)
The series part describes a periodic oscillation of some shape, but the initial term indicates that the string has
a large-scale motion that causes its displacement to increase forever, because the initial condition provides
a nonzero average drift to the system.
File “hw04”, version of 2 Feb 2005, page 5.
Typeset at 16:45 February 2, 2005.