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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION E
- 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 ∂t
= − 2 k
∂u ∂x
∂^2 u ∂x^2 (b) ∂^2 u ∂t^2
= − 2 k
∂u ∂x
∂^2 u ∂x^2 (c) ∂^2 u ∂t^2
= − 2 k
∂u ∂t
∂^2 u ∂x^2 (d) ∂u ∂t
= c^2
∂^2 u ∂x^2 (e) ∂^2 u ∂t^2
= sinh u + ∂^2 u ∂x^2 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) exp
− (πnc)^2 t
sin (πnx) (e) e−kt^ sin
t
π^2 c^2 n^2 − k^2
sin (πnx) (f) sin (πnct) sin (πnx) (g) exp
− (πnc)^2 t
cos (πnx) (h) e−kt^ cos
t
π^2 c^2 n^2 − k^2
sin (πnx)
Date: March 22, 2002. 1
(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.
- The total energy of a vibrating string of length L is
H =
∫ L
0
∂u ∂t
∂u ∂x
dx.
(Warning: this is not what we called energy before, in talking about Fourier series.) Calculate the total energy at time t of every mode
sin
( (^) πnx L
cos (λnt)
and sin
( (^) πnx L
sin (λnt)
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
H =
(πcn)^2 4 L
This is independent of time.
- 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 0 o^ 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
u(x, t) =
π
(π − x) +
π
∑^ ∞
n=
(−1)n+1^ sin (nx) n
e−n
(^2) t
(the left is 100o) or
u(x, t) =
π
x +
π
∑^ ∞
n=
sin (nx) n
e−n
(^2) t
(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.
- Consider the wave equation with gravitational force
∂^2 u ∂t^2
= c^2
∂^2 u ∂x^2
− g.
1
2
3
0 0.2 0.4 0.6 0.8 1 x
Figure 2. Steady states of the wave equation on the moon and Jupiter
- Consider d’Alembert’s solution
u(x, t) =
(f (x − ct) + f (x + ct)) +
2 c
∫ (^) x+ct
x−ct
g(s) ds
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.)
Solution: The string height is
u(x, t) =
(f (x − ct) + f (x + ct))
=
(sin(x − ct) + sin(x + ct))
=
(sin(x) cos(ct) − cos(x) sin(ct) + sin(x) cos(ct) + cos(x) sin(ct))
= sin(x) cos(ct)
The height will be zero then precisely when
cos(ct) = 0
which happens at
ct =
π 2
for any integer n, or
t =
π c
for any integer n.
- 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
u(x, t) =
∑^ ∞
n=
An exp
( (^) πnc L
t
cos
( (^) πnx L
where
A 0 =
An =
− (^) π (^24) n 2 if n is odd 0 if n is even
The picture is in figure 3 on the following page.
