MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #3
1. A rectangular sheet is tied on all four sides. The top and bottom sides
have length a = 1 and the left and right sides have length b = 1. The
wave velocity is c = 1/π. Initially the sheet is flat, with initial velocity
g(x, y) = x(1 − y). Find the position u(x, y, t) of the sheet at time t.
Solution:
∞
X
u(x, y, t) =
Cmn sin (πmx) sin (πny) sin (ωmn t)
m,n=1
where
Cmn =
(−1)m+1 4
√
π 2 mn m2 + n2
and
ωmn =
p
m2 + n2 .
2. The steady state of heat in a rectangular plate of width a = 1 and length
b = 1 with left bottom corner at the origin of coordinates and right top
corner at (1, 1), with temperatures f1 (x) = x at bottom, f2 (x) = x + 1 at
top, g1 (y) = y at left side and g2 (y) = y + 1 at right side is:
(a)
x2 + y 2
(b)
∞
X
(−1)m+n
sin (πmx) sin (πny)
mn
n=0
(c)
r cos θ + ir sin θ
(d)
sin(πx) sin(πy)
(e)
x+y
(f )
∞
X
n=1
rn
1
(−1)m+n
cos(nθ) +
sin(nθ)
mn
mn
(g)
1−x−y
(h)
0
(i) y
Solution:
Looking at the edge temperatures, clearly
s(x, y) = x + y.
Date: April 22, 2002.
1
2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #3
For partial credit, you could calculate out that
s(x, y) =
∞
X
An sin (πnx) sinh (πn(1 − y))+Bn sin (πnx) sinh (πny)+Cn sinh (πn(1 − x)) sin (πny)+Dn sinh (πnx
n=1
with
(−1)n+1 2
sinh (πn) πn
1 + (−1)n+1 2
Bn = Dn = 2
.
sinh (πn) πn
An = Cn =
3. Suppose that the same rectangular plate as in the previous problem has
initial temperature f (x, y) = 1o and diffusivity constant c = 1/π. Find the
temperature U (x, y, t) as a function of time. Remember that U = s + u
where s is the steady state, and u = f − s at time t = 0.
Solution: First we write
U (x, y, t) = s(x, y) + u(x, y, t)
and we know the steady state is
s(x, y) = x + y.
Now we calculate u(x, y, t) which looks like
u(x, y, t) =
∞
X
amn exp (−βmn t) sin (πmx) sin (πny) .
m,n=1
These amn are given by
ZZ
amn = 4
(f (x, y) − s(x, y)) sin (πmx) sin (πny) dx dy
ZZ
=4
(1 − x − y) sin (πmx) sin (πny) dx dy
4 (−1)m+n+1 − 1
=
π 2 mn
So the solution is
∞ ∞ 8 X X 1, if m + n even
1
U (x, y, t) = x+y− 2
exp (−βmn t) sin (πmx) sin (πny)
π m=1 n=1 0, if m + n odd mn
with
βmn = m2 + n2 .
4. Take a disk of radius 1 and place thermostats on its edge so that the temperature is 1o on a quarter of the disk, and 0o on the other three quarters.
Align the disk so that the quarter heated to 1o is half above and half below
the x axis, with the center of the disk at the origin of the coordinates, as
in figure 1 on the facing page.
(a) Find the steady state temperature s(r, θ) at all points of the disk.
(b) Write down the first four terms of the expansion into modes of the
steady state.
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #3
3
Figure 1. The heated disk
1
0.8
0.6
0.4
0.2
0
1
0.5
0
–1
–0.5
–0.5
0
–1 1
0.5
Figure 2. The steady state inside the disk
(c) The steady state temperature at (x, y) = (0, 1/10) is : (you can use
the approximation π ∼ 3)
(a)
0.2466
(b)
0.2792
(c)
0.1900
(d)
0.25
(e)
1
(f ) 0
(g)
− 0.001
(h)
3.14159
4
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #3
Hint: recall that if we write the temperature of the edge of the disk as
f (θ), and write the Fourier amplitudes of f (θ) as
Z 2π
1
a0 =
f (θ) dθ
2π 0
Z
1 2π
am =
f (θ) cos(mθ) dθ
π 0
Z
1 2π
bm =
f (θ) sin(mθ) dθ
π 0
then the steady state temperature inside the plate, in polar coordinates, is
∞
X
s(r, θ) = a0 +
rn (an cos (nθ) + bn sin (nθ)) .
n=0
Solution:
(a) The edge temperature is
(
1o if − π/4 ≤ θ ≤ π/4
F (θ) =
0o otherwise
so that
a0 = 1/4
2 sin πn
4
an =
πn
bn = 0
and
s(r, θ) = 1/4 +
∞
πn X
2
sin
rn cos(nθ).
πn
4
n=1
(b)
s(r, θ) = a0 + a1 r cos θ + a2 r2 cos 2θ + a3 r3 cos 3θ + . . .
√
√
√
2 5
2 3
2
1
r cos 5θ + . . .
r cos 3θ + 0 · r4 −
= 1/4 +
r cos θ + r2 cos 2θ +
5π
3π
π
π
(c) In polar coordinates, (x, y) = (0, 1/10) is (r, θ) = (1/10, π/2), so
1 1
s = 1/4 + 0 −
+ 0 + 0 + ...
π 102
1
∼ 1/4 −
314
1
∼ 1/4 −
300
∼ 0.25 − 0.0033
∼ 0.2466.
For partial credit: without any calculation, since the temperature at
the center is the average of the temperatures from around the edge, it
must be 1/4. Then moving away from the center, up the y axis, we
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #3
5
are moving toward the cold part, so the temperature should go down
at least a little, so the answer must be (a) or (c).
5. Calculate ∇2 u where
1
.
u(x, y) = p
2
x + y2
Solution:
In polar coordinates
∇2 u =
1 ∂2u
∂ 2 u 1 ∂u
+
+ 2 2.
2
∂r
r ∂r
r ∂θ
But this u is
u=
1
= r−1 .
r
Therefore
∇2 u = (−1)(−2)r−3 + r−1 (−1)r−2 + 0 = r−3 .
In rectangular coordinates this is
∇2 u =
1
(x2
3/2
+ y2 )
.