112. Same as Problem 111 for
f () = cos(9) + cos(10)
This one is hard to do by hand, but is interesting because you get to see an example of
amplitude modulation, AM. Compare Figure 13 on page 31.
113
1
5
10
15
20
5
10
15
20
5
10
15
20
5
10
15
20
-1
1
-1
1
-1
1
-1
Figure 37: Top: What frequencies could be hidden in this function? Second: How about
now? The function has been modied a little to make
but somehow we do not know what the coefcients are yet. Dot with , for example.
You nd the number a2 by doing a dot product:
v = a1 + a2 + a3 k = 0 + a2 + 0 = a2
by the orthogonality and the fact that
20
15
10
5
0
-5
-10
-15
-20
0
10
20
30
40
50
Figure 38: A chaotic motion.
There are systems for which the study of linear equations does not prepare you.
Notice in Figure 38 that the function plotted
106. We need to check carefully that when u is a solution to the Laplace equation, then
so is 6u, or cu for any constant c. How can that be checked? Start with the rst order
derivative
u
r
cu
u
= c ?
At least that will be a solution to the Laplace equation (see Problem 108).
Then we have to choose the coefcients ck if possible to get the boundary
values. It will work to take c1 = 6 and c2 = 3. Ans
18.1 Laplace leads to Fourier
We are solving the boundary value problem
1
1
urr + ur + 2 u = 0
r
r
(0 < r < 1, 0 2)
with a specied value at the boundary circle
u(1, ) = f ()
and so far you have found
Figure 36: A membrane stretched, or the graph of a steady temperature.
So we are solving the boundary value problem
1
1
urr + ur + 2 u = 0
r
r
(0 < r < 1, 0 2)
with a specied value at the boundary cir
17.1 Sound
Here we start from the Euler equations, and imagine a small disturbance superimposed over an ambient stillness.
P RACTICE : Is it true that the ambient stillness u = 0, = (constant) is a so
From the differential equation we see this: suppose you have an interval like
[x1 , x2 ] where J0 is 0 at each end, and nonzero between. At any local min or
max where J0 = 0, the DE then tells us that
P RACTICE : Why does
fR fL
RL
tend to fx as R and L tend to x? Assume f differentiable.
We notice that we have two unknown functions, u and . Probably one equation
isnt going to be enough, so we look
goes from 0 to 2. One cycle in every
q2
T
x
1
seconds. The frequency is
T x1
2 .
For larger roots x2 etc you get shorter cycles, higher pitched sounds.
P RACTICE : When you tighten up the drum tensi
96. If you are not clear about the limit we found, try this. First tell what this limit is:
lim
h0
Then how about
lim
h0
f (x + h) f (x)
h
(x + h)g(x + h) xg(x)
?
h
Finally the one we need
lim
dr0
(r
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
0
5
10
15
20
25
Figure 34: Graph of Bessell function J0 (x) = 1
30
35
40
x2
x4
x6
+ 2 2 2 2 2 +
22
2 4
2 4 6
For that to be identically 0, we need all the coefcients to
(0,b)
T
T
(0,0)
(a,0)
T
Figure 33: A triangle of the drumhead. See Problem 93.
P ROBLEMS
93. Imagine the tension forces on a small triangle of material located anywhere in the
drumhead, as in Figure 3
rutt =
T
(rur )r or equivalently
utt =
T
1
urr + ur
r
We will nd some solutions to this having u(1, t) = 0 at the rim of the drum.
The solutions will be vibrations of our drum model, and hopefully wil
Figure 31: A wave traveling initially to the left is shown at six times, t increasing from top
to bottom in the picture. It is reected downward and reversed left-right when it bounces
off a wall at th
10
9
8
7
6
5
4
3
2
1
0
2
1
0
1
2
3
4
5
6
7
8
Figure 30: The wave u(x, t) = e(xct) , graphed for various times. Note that to keep
the expression x ct constant as t increases you have to increase x also
ux (0, t) = 0
u(l, t) = 60
u(x, 0) = 40
This is a bar of length l, insulated at the left end and held at 60 degrees at the right
end. The initial temperature is 40 degrees all along its length.
14.2 P
74. Describe in whole sentences what the following problem is about.
ut =
ux (0, t) =
ux (l, t) =
u(x, 0) =
uxx
0
0
400
Determine whether there is a steady-state solution to this problem.
75. Describe
and its rate of change is
R
cut dV.
The rate of heat ow through the boundary surface S is
k u n dA.
S
By the divergence theorem this last integral is equal to
(k u) dV.
R
Thus we have
R
cut dV =
(k
u
new
t
u
u
1
x
u
2
3
x
Figure 23: A stencil.
For the second space derivative uxx , remember it is the x derivative of ux . So we
can do this:
u3 u2
u
u2x 1
uxx (x, t) x
x
u3 2u2 + u1
=
.
(x)2
The he
R
Figure 22: A region R inside the main block of material, with some heat energy owing
between R and the rest of the block. Energy conducts into R through its boundary surface
S. There is no heat sour
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Figure 18: Graph of the cubic polynomial a3 + 5a2 + 6a + 1. The roots are approximately
3.2, 1.6, .3.
and this corresponds to the compl
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
12
Figure 20: The graph in Problem 66. Remember 6 is approximately 2.
12
Boundary Value Problems
T ODAY: A change from initial conditions. Boundary values.