MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2
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
f (x) = sin πx
g(x) = 0
with L = 1 and c = 1. What is the first positive time t that the string returns to
its original shape?
Solution:
1
(sin π(x − t) + sin π(x + t))
2
and the hump of the first sine wave sits at
u(x, t) =
π(x − t) = π/2
which is
x = t + 1/2
while the other sine wave has hump sitting at
x = −t + 1/2
so that it goes back to its original shape when the two humps both align again, at
x = 1/2
up to multiples of 2 (the period). This first happens at time t = 2.
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 all edges insulated.
Solution:
The equations to solve are
2
∂u
∂ u ∂2u
= c2
+
∂t
∂x2
∂y 2
∂u
= 0 at x = 0 & x = a
∂x
∂u
= 0 at y = 0 & y = b
∂y
Date: November 12, 2000.
1
2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2
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
2
=c
+
.
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
2
= −γ = c
+
.
G
E
F
Therefore
G(t) = e−γt
while
E 00
F 00
γ
+
− 2 =
c
E
F
or
E 00
γ
F 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
γ
F 00
−
=β= 2 +
E
c
F
or
E 00
−
=β
E
γ
F 00
−
= 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 0 (x) = 0 for x = 0 and x = a
and similarly
F 0 (y) = 0 for y = 0 and y = b .
Since the derivatives E 0 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) = cos
βx
r
γ
− βy .
F (y) = cos
c2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2
3
But the vanishing of the derivatives at x = a and 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) = cos
a πny
F (y) = cos
b
G(t) = e−γmn t
where
γmn = c2 π 2
m2
n2
+ 2
2
a
b
So
πmx πmy m2
n2
+
t
cos
cos
u(x, y, t) = exp −c π
a2
b2
a
b
and the general solution of the heat equation with insulated boundary is
2
∞ X
∞
πmx πmy X
m
n2
u(x, y, t) =
Amn exp −c2 π 2
+
t cos
cos
2
2
a
b
a
b
m=0 n=0
3.
2 2
Under rescaling a region by a factor of ε in the x and y variables, by rescaling
x → εx
y → εy
the operators
∂
∂
and
∂x
∂y
get rescaled by
∂
1 ∂
→
∂x
ε ∂x
∂
1 ∂
→
∂y
ε ∂y
Calculate the rescaling that must occur to the time variable t in the heat equation
2
∂u
∂ u ∂2u
2
=c
+ 2
∂t
∂x2
∂y
in order that the constant c stay the same, and the heat equation still be satisfied
by the same function with the rescaled arguments. Recalling the basic idea that a
mode consists of an exponential decay in time multiplied by a function of x and y,
say
exp(−κt)f (x, y)
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2
how does this rescaling affect κ, the rate of decay of the mode? Given a replica of a
plate of 1/20 the scale of the original, made of the same material, how many times
faster will it cool or heat than the original?
Solution:
The heat equation tells us that the time variable has to be scaled like
t → ε2 t
so that the κ constant will be scaled like
κ
.
ε2
Therefore the rate of cooling is faster by 1/ε2 ; in particular a
κ→
ε = 1/20
scale replica cools at
1
1/20
2
= 400
times the rate of the original.
4.
Show that the function
un (r, θ) =
r n
R
(an cos(nθ) + bn sin(nθ))
satisfies the Laplace equation
∇2 u = 0
on the disk of radius R, where in polar coordinates
∇2 u =
∂ 2 u 1 ∂u
1 ∂2u
+
+ 2 2 .
2
∂r
r ∂r
r ∂θ
Now why does the function
u(r, θ) = a0 +
∞ n
X
r
n=1
R
(an cos (nθ) + bn sin (nθ))
satisfy the Laplace equation?
How should we pick the a0 and an and bn so that
u(R, θ) = f (θ)
is some specified temperature function f (θ) around the edge?
Solution:
The first derivatives are
∂
rn−1
un = n n (an cos nθ + bn sin nθ)
∂r
R
n
= un
r
r n
∂
un = n
(−an sin nθ + bn cos nθ)
∂θ
R
The second derivatives are
∂2
n(n − 1)
un =
un
2
∂r
r2
∂2
un = −n2 un
∂θ2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2
5
1
0.8
0.6
0.4
0.2
–1
–0.5
y0
0.5
1
1
0.5
0
x
–0.5
–1
Figure 1. A round peak
Putting these together, we find
2
1 ∂
1 ∂2
n(n − 1)
n
n2
∂
+
+
u
=
+
−
un = 0
n
∂r2
r ∂r r2 ∂θ2
r2
r2
r2
Adding up the un we find, by linearity of ∇2 , that u satisfies
∇2 u = 0 .
Moreover if we pick these a0 , an and bn according to
Z 2π
1
a0 =
f (θ) dθ
2π 0
Z 2π
1
an =
f (θ) cos(nθ) dθ
π 0
Z
1 2π
bn =
f (θ) sin(nθ) dθ
π 0
(as Fourier amplitudes of f (θ)) then
u(R, θ) = f (θ)
(plug in r = R) so that this is the steady state of the heat equation in the disk with
boundary held fixed at temperature u(R, θ) = f (θ).
5.
Looking at the wave equation
2
∂2u
∂ u ∂2u
2
=c
+ 2
∂t2
∂x2
∂y
and recalling second derivatives from calculus, explain why a round peak in the
graph of u (as in figure 1) will be accelerated downward.
Solution: The second derivatives in x and in y will be negative, so the right hand
side of the wave equation is negative. Therefore the left hand side, the acceleration,
must be negative too.