Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1995,1996 by Evans M. Harrell II and James V. Herod. All rights reserved.

At Georgia Tech, students took 50 minutes (calculators allowed but no notes). Most students did reasonably well, but few finished early.

1. Find the normal modes (product solutions) for the damped wave equation

u_tt + u_t = u_xx

for 0 < x < 1, with boundary conditions:

u_x(t,0) = 0

u(t,1) = 0

Make sure to give the specific dependence on x and on t.

THE NORMAL MODES ARE:

cos( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi² - 1) t/2

and

sin( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi² - 1) t/2,

n = 1, 2, ...

Here is why. Assuming that u(t,x) = T(t) X(x) and substituting, we find

(T ''(t) + T(t) ) X(x) = T(t) X''(x).

Divide by T(t) X(x). The left side depends only on T, while the right depends only on x, so we can set both equal to a constant, - mu. The eigenvalue equation becomes

- X'' = mu X

X'(0) = X(1) = 0.

The first of these conditions, X'(0) = 0, means that the solutions have to be pure cosines, so, up to a normalization constant,

X(x) = cos( mu^1/2 x).

The other boundary condition means that cos( mu^1/2) = 0, so mu_n^1/2 = (2n-1) \pi /2, n = 1,2,...

The time dependence (according to the ODE solution on the last page of the test) for mu_n is:

T(t) = exp(-t/2) (C₁ cos( ((2n-1) \pi² - 1) t/2 ) + C₂ sin(((2n-1) \pi² - 1) t/2 ) )

The full normal modes would be :

cos( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi² - 1) t/2

and

sin( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi² - 1) t/2

2. Solve the wave equation

u_tt = u_xx

for 0 < x < 1, with boundary conditions

u_x(t,0) = ux_x(t,1) = 0

and initial conditions

u(0,x) = sin²(\pi x)

ut(0,x) = 0

With these Neumann boundary conditions, we know that the general solution is

Because of the second boundary condition, all the b's are zero. The a's are the Fourier cosine coefficients for the function sin²(\pi x), which can be calculated with the aid of the formulae given on the last page of the test. An alternative way to find the Fourier cosine series is to use the trigonometric identity

cos(2 \pi x) = 1 - 2 sin²(\pi x),

which implies than

sin²(\pi x) = 1/2 - (1/2) cos(2 \pi x).

This is the Fourier cosine series! All the other terms in the series have coefficient zero. Hence the future solution is:

u(t,x) = 1/2 - (1/2) cos(2 \pi t) cos(2 \pi x).

It is also possible to solve with d'Alembert's method, and you will get the same equation,

except that you might write it as

1/2 - (1/4) ( cos(2 \pi (x-t)) + cos(2 \pi (x-t)) )

Back to the test.

Back to Chapter VII

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