MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
1.
Suppose that a string in a still fluid is tied down at x = 0 and x = 1. The
force of resistance to motion of the string coming from the fluid is proportional to
the velocity of the string. Which of the following is the equation of motion of this
string?
(a)
∂u
∂u
∂2u
= −2k
+ c2 2
∂t
∂x
∂x
(b)
∂2u
∂u
∂2u
= −2k
+ c2 2
2
∂t
∂x
∂x
(c)
∂2u
∂u
∂2u
= −2k
+ c2 2
2
∂t
∂t
∂x
(d)
∂u
∂2u
= c2 2
∂t
∂x
(e)
∂2u
∂2u
= sinh u +
2
∂t
∂x2
Which two of the following are modes of the string?
(a)
cos (πnct) sin (πnx)
(b)
(cos (πnct) − kt) sin (πnx)
(c)
cos (πnct) cos (πnx)
(d)
2
exp − (πnc) t sin (πnx)
(e)
p
e−kt sin t π 2 c2 n2 − k 2 sin (πnx)
(f)
sin (πnct) sin (πnx)
(g)
2
exp − (πnc) t cos (πnx)
(h)
p
e−kt cos t π 2 c2 n2 − k 2 sin (πnx)
Date: March 22, 2002.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
(i)
sinh(πnt) sin(πnx)
Solution:
The equation of motion is (c), since we add in to the usual wave
equation a term proportional to velocity. The modes are (e) and (h), which we see
by taking derivatives. Intuitively clear, since the gradual loss of energy from the
friction with the fluid is expressed in the exponential decay over time.
2.
The total energy of a vibrating string of length L is
2
2 !
Z
∂u
∂u
1 L
2
H=
+c
dx.
2 0
∂t
∂x
(Warning: this is not what we called energy before, in talking about Fourier series.)
Calculate the total energy at time t of every mode
πnx sin
cos (λn t)
L
and
πnx sin
sin (λn t)
L
of the wave equation with ends of the string tied down (where λn = πnc/L). How
does the total energy of each mode vary in time?
Solution:
The total energy of each mode is
2
H=
(πcn)
.
4L
This is independent of time.
3.
Suppose that a wire of length L = π and diffusivity c = 1 with initial
temperature 100o is placed in an insulating tube. One end is kept at 100o with a
thermostat, while the other is kept at 0o .
(a) Find the temperature u(x, t). Your answer should be a sum of a steady state
(in this case a linear function of x) and a solution of a similar problem with
0o at both ends.
(b) Draw a picture of what the temperature looks like at time t = 0 and at
large time t when it has not quite reached the steady state.
Solution:
Depending on which end is which, either
∞
100
200 X (−1)n+1 sin (nx) −n2 t
(π − x) +
e
u(x, t) =
π
π n=1
n
(the left is 100o ) or
u(x, t) =
∞
100
200 X sin (nx) −n2 t
x+
e
π
π n=1
n
(the left is 0o ). In case the left is 0o , the steady state is 100
π x; just before the steady
state there is a small contribution from the first term, which looks like a small
multiple of sin x, so as in figure 1 on the facing page.
4.
Consider the wave equation with gravitational force
∂2u
∂2u
= c2 2 − g.
2
∂t
∂x
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
3
100
80
60
40
20
0
0.5
1
1.5
2
2.5
3
x
Figure 1. Near steady state
(a) Find the steady state (i.e. u = u(x) independent of time) which satisfies
u = u0 fixed (u0 some constant) at x = 0 and u = u1 (some other constant)
at x = L.
(b) Draw a picture of what it looks like for small positive g (for example, on
the moon) and for large positive g (for example, on Jupiter).
Solution:
(a) The equation of a steady state is
∂2u
g
= 2.
∂x2
c
This says that the second derivative is constant, so the first derivative is a
linear function, and the function u is quadratic. Working out the constants:
u(x) =
g 2 u1 − u0 − gL2 /2c2
x +
x + u0
2c2
L
(b) They are parabolas, with larger gravitational constant g giving larger constants on the x2 term. Obviously gravity pulls harder and they hang lower,
as in figure 2 on the next page.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
3
2.5
2
1.5
1
0
0.2
0.4
0.6
0.8
1
x
Figure 2. Steady states of the wave equation on the moon and Jupiter
5.
Consider d’Alembert’s solution
u(x, t) =
1
1
(f (x − ct) + f (x + ct)) +
2
2c
Z
x+ct
g(s) ds
x−ct
of the wave equation for a vibrating string, where f (x) is the odd function with
period 2L which on the interval 0 ≤ x ≤ L gives the initial position of the string,
and similarly g(x) gives the initial velocity on 0 ≤ x ≤ L and is odd and 2L periodic.
Suppose that the string has length L = π, and at time t = 0 has initial position
f (x) = sin x
and initial velocity
g(x) = 0 .
What are all of the times t at which the string will be flat? (By flat, I mean that
at that time t the height is u(x, t) = 0, for every point x.)
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
5
Solution:
The string height is
1
u(x, t) = (f (x − ct) + f (x + ct))
2
1
= (sin(x − ct) + sin(x + ct))
2
1
= (sin(x) cos(ct) − cos(x) sin(ct) + sin(x) cos(ct) + cos(x) sin(ct))
2
= sin(x) cos(ct)
The height will be zero then precisely when
cos(ct) = 0
which happens at
π
+ πn
2
π 1
t=
+n
c 2
ct =
for any integer n, or
for any integer n.
6.
A wire of length L = 1 with insulated ends and diffusivity c = 1 has
temperature f (x) = x at time t = 0. Find the temperature u(x, t) at time t and
position x.
Solution:
The solution is
∞
πnx X
πnc 2
u(x, t) =
An exp −
t cos
L
L
n=0
where
1
2
(
− π24n2
An =
0
A0 =
if n is odd
if n is even
The picture is in figure 3 on the following page.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
1
0.8
0.6
0.4
0.2
1
0
0
0.8
0.2
0.6
0.4
x
0.6
0.4
0.8
0.2
t
1
Figure 3. Heat in an insulated wire, starting at t = 0 with temperature u = x, and ending very nearly at u = 1/2, the steady
state