Web page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.
The wave equation: utt = [[gamma]]2 uxx, 0 < x < a, 0 < t. (16.1)
Initial conditions: A(u(0,x) = f(x),ut(0,x) = g(x)) 0 < x < a (16.2)
Boundary conditions: A(u(t,0) = 0,u(t,a) = 0) 0 < t. (16.3)
We already know the form for u. We have seen repeatedly that there should be functions [[phi]] and [[theta]] such that
u(t,x) = [[phi]](x - [[gamma]]t) + [[theta]](x + [[gamma]]t).
The initial conditions are
f(x) = u(0,x) = [[phi]](x) + [[theta]](x), 0 < x < a
g(x) = ut(0,x) = - [[gamma]] [[phi]]'(x) + [[gamma]][[theta]]'(x).
Dividing the second equation by [[gamma]] and integrating we get
F(1,[[gamma]]) I(0,x, )g([[xi]])d[[xi]] = - [[phi]](x) + [[theta]](x).
The equations in [[phi]] and [[theta]] can now be solved with appropriate algebraic manipulations to get that
[[theta]](x) = F(1,2) [f(x) + F(1,[[gamma]]) I(0,x, )g([[xi]])d[[xi]]] for 0 < x < a
and
[[phi]](x) = F(1,2) [f(x) - F(1,[[gamma]]) I(0,x, )g([[xi]])d[[xi]]] for 0 < x < a.
You see the problem; these equations give [[phi]] and [[theta]] only for values of x between 0 and a. But, we want to evaluate [[phi]] and [[theta]] at x +/- [[gamma]]t and for any positive t and 0 < x < a. Just as in the previous section, an extension of f and of g must be made. The extensions must be made so that the boundary conditions are preserved. This means that
0 = u(t,0) = [[phi]](-[[gamma]]t) + [[theta]]([[gamma]]t) for all t > 0
and
0 = u(t,a) = [[phi]](a - [[gamma]]x) + [[theta]](a + [[gamma]]x) for all t > 0.
The first of these gives the extension to negative numbers:
F(1,2) [f(-z) - F(1,[[gamma]]) I(0,-z, )g([[xi]])d[[xi]]] + F(1,2) [f(z) + F(1,[[gamma]]) I(0,z, )g([[xi]])d[[xi]]] = 0.
This means that we need the odd extension F of f and the even extension G of
G(z) = I(0,z, )g([[xi]]) d[[xi]].
We now must address how to make the extensions for x > a. The process is similar to that at 0:
F(1,2) [f(a-z) - F(1,[[gamma]]) I(0,a-z, )g([[xi]])d[[xi]]] + F(1,2) [f(a+z) + F(1,[[gamma]]) I(0,a+z, )g([[xi]])d[[xi]]] = 0.
This implies that F(a+z) = - F(a-z). Use that fact that F is already an odd extension to re-write this last requirement as F(a+z) = F(-a+z). With a little consideration, this can be re-written one more time as
F(w) = F(2a + w) (16.4)
for all w. G can be developed in a similar manner. The result is
BBC[(A(F is an odd, 2a periodic extension of f,G is an even, 2a periodic extension of G(z) = I(0,z, )g([[xi]]) d[[xi]])). (16.5)
Finally,
u(t,x) = F(1,2) [F(x+t) + F(x-t)] + F(1,2) [G(x+t) - G(x-t)] (16.6)
Example 16.2:
We take the interval to be [0,5] the initial distribution to be
u(0,x) = 4 exp(-(x-5/2)2)
with ut(0,x) = 0.
This produces a spike as an initial distribution. First, we make the odd, periodic extension with period 5 as in Figure 16.1.
Figure 16.1
We have viewed the solutions as surfaces as in Figure 16.2.
Figure 16.2
But, we also have viewed the solution as a series of snapshots as in Figure 16.3 which shows five views of the solution. It shows the spike after it has split apart and is moving toward the ends and also after it has reflected off the boundary and is moving back across the boundary.
Figure 16.3
An alternate view is to see the graph of the extension of f for the entire real line as a odd, periodic function over the line. In this perspective, we are back to the methods of Section 12. We can think of moving this extended graph in the direction of one set of characteristic curve and right along another. Out view in the "real world" of this solution is only of what happens on the interval [0,5]. The other part of Figures 16.1 is "out-of-this-world."
Example 16.2:
And, there is the problem of having a forced oscillation:
The pumped wave equation: utt - uxx = sin(t), 0 < x < [[pi]].
Initial conditions: A(u(0,x) = 0,ut(0,x) = 0) 0 < x < [[pi]]
Boundary conditions: A(u(t,0) = 0,u(t,[[pi]]) = 0).
To solve this problem, we change the form of the left side with the standard transformation: S(t,x) = x - t and T(t,x) = x + t.Define the function v by u(t,x) = v(S,T) = v(x-[[gamma]]t,x+[[gamma]]t)
and v satisfies the equation
F([[partialdiff]]2v,[[partialdiff]]s[[partialdiff]]t) = sin((T-S)/2).
Integrating, we get
F([[partialdiff]]v,[[partialdiff]]s) = - 2 cos((T-S)/2) + F(S).
Integrate again and get
v(S,T) = - 4 sin((T-S)/2) + f(S) + g(T).
That is u(t,x) = -4 sin(t) + f(x-t) + g(x+t). It remains to find f and g from the initial and boundary conditions.
Exercise
16.1 Make the odd periodic extension of x (1-x) with period 2.
16.2 Show that if n is a positive integer and each of [[gamma]], a, and b is a positive number then each of the functions
sin(F(n [[pi]] x,a)) cos(F(n [[pi]] [[gamma]] t,a))
is a solution for the partial differential equation (16.1), with boundary conditions (16.3).
16.3 Solve the wave equation with boundary conditions (16.3) if g(x) = 0 and
f(x) = F(sin([[pi]]x/a),5) + sin(3[[pi]]x/a).
16.4 Solve the wave equation with boundary conditions (16.3) if g(x) = 0 and
f(x) = BLC{(A( m F(2x,a) 0 < x < F(a,2), m (2 - F(2x,a)) F(a,2) < x < a)).
16.5 Solve the wave equation with boundary conditions (16.3) if f(x) = 0 and g(x) = 1 for 0 < x < a.
16.6 Solve the wave equation with boundary conditions (16.3) if f(x) = 0 and g(x) = sin(5[[pi]]x/a) for 0 < x < a.
16.7 Solve the wave equation with boundary conditions (16.3) if f(x) = 0 and
g(x) = BLC{(A(0 if 0 < x < 2, 1 if 2 < x < 3, 0 if 3 < x < 5)).