MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION A
1.
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 at initial time we have a flat string with nonzero velocity. At what
later time t will the string be completely flat again? Explain how you obtain your
answer.
Solution:
The string height is
Z
x+ct
u(x, t) =
g(s) ds
x−ct
and we need this height to be zero at some fixed time t, and every value of x. But
g(x) is 2L periodic and odd, so its integral over any interval of length 2L vanishes,
as in figure 1 on the following page. Therefore, if the interval
x − ct ≤ s ≤ x + ct
has length 2L, we will get zero height. This happens exactly when
(x + ct) − (x − ct) = 2L
or
t=
L
c
Frequent mistakes:
(1) Confusing g(x) (which is the velocity at time t = 0)
∂u g(x) =
∂t t=0
with the derivative f 0 (x) of the initial position. Since x is a space variable,
not a time variable, derivatives in x are not velocities.
2. Use separation of variables to find the general solution of the heat equation in
a rectangular plate with edges of lengths a and b, and with top and bottom edges
insulated, while right and left edges are kept at 0o .
Date: July 16, 2001.
1
2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION A
1
0.5
–3
–2
–1
1
x
2
3
–0.5
–1
Figure 1. An odd periodic function
Solution:
The equations to solve are
2
∂u
∂ u ∂2u
2
=c
+ 2
∂t
∂x2
∂y
u = 0 at x = 0 & x = a
∂u
= 0 at y = 0 & y = b
∂y
The last two equations use Fourier’s law to ensure that there is no heat flux through
the boundary. Separating variables:
u(x, y, t) = E(x)F (y)G(t)
(we just split u into a product of functions, one of each variable) we find the heat
equation becomes
EF G0 = c2 (E 00 F G + EF 00 G)
which, once we divide by EF G, is just
00
G0
E
F 00
= c2
+
.
G
E
F
The left side involves only t, and the right side only x and y. So there is a constant
γ so that
00
G0
E
F 00
= −γ = c2
+
.
G
E
F
Therefore
G(t) = e−γt
MATH 3150: PDE FOR ENGINEERS
while
−
or
MIDTERM TEST #2 VERSION A
3
γ
E 00
F 00
=
+
2
c
E
F
γ
F 00
E 00
= 2+
.
E
c
F
The left side depends only on x, and the right only on y, so that there is some
constant β so that
E 00
a
F 00
−
=β= 2 +
E
c
F
or
E 00
−
=β
E
F 00
γ
−
= 2 −β
F
c
We see that E(x) satisfies a harmonic oscillator type equation, and so does F (y).
Our conditions on the edges force
−
E(x) = 0 for x = 0 and x = a
and similarly
F 0 (y) = 0 for y = 0 and y = b .
Since the functions E and F 0 have to have two zeros, our analysis of harmonic
oscillator equations says that β and
γ
−β
c2
must both be positive. In particular, γ must be positive, so therefore G(t) is an
exponential decay, and up to rescaling
p E(x) = sin
βx
r
γ
F (y) = cos
−
βy
.
c2
But the vanishing of the functions E at x = a and F 0 at y = b gives
p
πm
β=
a
r
γ
πn
−β =
c2
b
for some integers m and n. In terms of m and n, everything looks like
πmx E(x) = sin
a πny
F (y) = cos
b
G(t) = e−γmn t
where
2 2
γmn = c π
n2
m2
+
a2
b2
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION A
So
2 2
u(x, y, t) = exp −c π
m2
n2
+
a2
b2
πmx πmy t sin
cos
a
b
and the general solution of the heat equation with top and bottom insulated and
0o at the sides is
2
∞ X
∞
πmx πmy X
n2
m
2 2
u(x, y, t) =
Amn exp −c π
+
t
sin
cos
a2
b2
a
b
m=1 n=0
Frequent mistakes:
(1) Incorrect boundary conditions: to have insulation at x = 0 you need
∂u =0
∂x x=0
while to have 0o at x = 0 you need
u|x=0 = 0 .
(2) Not paying attention to the boundary conditions in order to get conditions on E(x) and F (y) that give you whether they are sines or cosines or
whatever.
3.
Consider a Taylor expansion
u(z) =
∞
X
σk z k
k=0
in a complex variable z = x + iy with complex coefficients (constants) σk . Calculate
∇2 u .
(Hint: write z in polar coordinates as
z = reiθ
and use the polar coordinate expression for the Laplacian
∇2 u =
1 ∂2u
∂ 2 u 1 ∂u
+
+
.)
∂r2
r ∂r
r2 ∂θ2
Solution:
z k = reiθ
k
= rk eikθ
so that
∂ k
k
z = krk−1 eikθ = z k
∂r
r
∂
k
ikθ
= r ike = ikz k
∂θ
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION A
5
so that the second derivatives are
∂2 k
k(k − 1) k
z =
z
2
∂r
r2
∂2 k
ik 2 k
z =
z
∂r∂θ
r
∂2 k
z = −k 2 z k
∂θ2
so that
k
k2
k(k − 1)
2 k
∇ z =
+ 2 − 2 zk
r2
r
r
=0
Therefore
∇2 u =
X
σk ∇2 z k
k
=0
Frequent mistakes:
(1) Confusion about exponents.
z = reiθ
gives
k
z k = reiθ = rk ekiθ
(2) Summation signs often disappear for no reason in student’s answers in the
middle of calculations. Watch the summation signs.
4.
Draw a solution u(x, y) of the Laplace equation
0=
∂2u ∂2u
+ 2
∂x2
∂y
with nonzero second derivatives.
Solution: The second derivatives in x and in y will have to be negatives of each
other, so that one is negative and the other positive. This gives a saddle—curved
up in one direction and down in the other. The only exception is if both of these
second derivatives vanish. (Then it might not be curved at all in either direction.)
See figure 2 on the next page.
Frequent mistakes:
(1) Not drawing a picture.
(2) Drawing a function y(x) instead of u(x, y).
(3) Drawings which were not intelligible.
(4) Most common mistake: drawing the bump from the practice midterm. It is
not a steady state of the heat equation, as we saw in the practice midterm.
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION A
1
0.5
0
1
–0.5
0.5
–1
–1
0y
–0.5
x
0
–0.5
0.5
1
–1
Figure 2. A solution of the Laplace equation with nonvanishing
second derivatives