MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION D
1.
Which of the following is the solution u(x, t) (via Fourier’s method, not
d’Alembert’s) to the wave equation describing a vibrating string of length L = π/2
and wave velocity c = 1 with initial velocity g(x) = x cos x and initial position
f (x) = 0?
(a)
∞
8 X (−1)n+1
u(x, t) =
sin (2nx) sin (2nt)
π n=1 (4n2 − 1)2
(b)
u(x, t) =
∞
8X
1
sin (2nx) sin (2nt)
π n=1 (4n2 − 1)2
u(x, t) =
∞
8X
1
sin (2nx) cos (2nt)
π n=1 (4n2 − 1)2
(c)
(d)
u(x, t) =
1
1
(x − t) cos(x − t) + (x + t) cos(x + t)
2
2
(e)
1
u(x, t) =
2
Z
x+t
s cos(s) ds
x−t
(f)
u(x, t) = 0
(g)
u(x, t) = x cos x cos t
Solution: The Fourier formula for the solution of the wave equation looks only
like (a) or (b). So we have to figure out which. These two agree at the first term,
so we need to calculate b∗2 , the coefficient of the second term, where they disagree
in a minus sign, to see which one is correct.
Z L
2
2πx
∗
g(x) sin
dx
b2 =
2πc 0
L
Z
1 π/2
=
x cos(x) sin(4x) dx
π 0
It is easiest if we write
sin(4x) cos(x) =
1
1
sin(5x) + sin(3x).
2
2
Date: March 14, 2002.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION D
Then we calculate (integrate by parts) that
Z π/2
πa πa 1
π
+ 2 sin
.
x sin(ax) dx = − cos
2a
2
a
2
0
This gives
π/2
Z
x sin(5x) dx =
0
1
25
and
Z
π/2
1
9
0
so b∗2 must turn out negative, whatever it is, when we add these together. Therefore
the answer is (a).
x sin(3x) dx = −
2.
Consider again the same vibrating string as in the previous problem.
(a) Draw a rough graph of the function g(x).
(b) Examining the Fourier amplitudes in the solution, how many terms in a
partial sum do you think you would need to get a good picture of u(x, t)?
(c) Could you see this by just looking at the picture of g(x)?
(d) Draw what you expect the graph of u(x, t) looks like for time t = 0 and for
time t = 0.1.
Solution:
(a) See figure 1 on the facing page.
(b) Coefficients in the first few terms in the solution are
8
8
8
,−
,
,...
9π 225π 1225π
so that they get small very quickly. (In fact, to 2 decimals they are
.28, −.01, .00, . . . ).
So you only need one term.
(c) The picture looks just like a sine function, vanishing at both ends, so we
expect to be able to use a single sine function to approximate it.
(d) At time t = 0 it is flat. At time t = 0.1 it looks like g(x), except scaled to
be very small, so like a small rescaling of sin(2πx).
3.
Consider the heat equation
∂u
∂2u
= c2 2
∂t
∂x
on a wire of length L = 1 with c = 7 and with initial temperature u(x, 0) = x(1−x).
Suppose that the total heat at time t is given by
Z 1
Q=
u(x, t) dx
0
(a) What is the rate of change of the total heat at time t = 0? Hint: don’t
use any Fourier series: just differentiate Q with respect to t and bring the t
derivative under the integral sign. Then use the heat equation to turn time
derivatives into space derivatives.
(b) Is it cooling down or heating up?
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION D
0.5
0.4
0.3
0.2
0.1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x
Figure 1. The graph of g(x) = x cos x
Solution:
(a)
Z 1
dQ ∂ =
u(x, t) dx
dt t=0
0 ∂t t=0
Z 1
∂2 =
c2 2 u(x, t) dx
∂x t=0
0
Z 1
∂2
=
c2 2 x(1 − x) dx
∂x
0
Z 1
=
c2 (−2) dx
0
= −2c2
= −2(7)2
= −2(49)
= −98
so −98 degrees/second (or whatever the units are).
(b) Cooling down.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION D
4. Solve the heat equation for a wire with insulated ends, with initial temperature
u(x, 0) = 273o K.
Solution:
273o K.
It is at a constant temperature, so it always stays that way: u(x, t) =
5. Solve the heat equation for a wire of length L = 1 with ends held at u = 100
at x = 0 and u = 0 at x = L with diffusivity constant c = 1 and initial temperature
u = 100(1 − x) + 30 sin (πx) at time t = 0.
Solution: Subtract off the steady state, which is 100(1 − x), to get the problem
to have zero temperatures at the ends. Then apply Fourier’s formula and get:
u(x, t) = 100(1 − x) + 30 exp −π 2 t sin (πx)
6. Take a flat string of length L = 1 with constant c = 1. If you tap it at one end
(so you create an initial velocity g(x) in the string which is zero except very close
to x = 0) how long does it take for the tap to be noticed at the other end? Explain
using d’Alembert’s formula.
Solution:
Time L/c. You see this from
Z L+ct
1
g(s) ds
u(L, t) =
2c L−ct
so that to get a nonzero answer here, we need to have L − ct near 0, or (because
the picture is periodic with period 2L) L + ct near 2L. Both of these happen at
time t = L/c.