Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

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) \pi^{2} - 1) t/2

and

sin( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi^{2} - 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) (C_{1} cos( ((2n-1) \pi^{2} - 1) t/2 ) +
C_{2} sin(((2n-1) \pi^{2} - 1) t/2 ) )

The full normal modes would be :

cos( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi^{2} - 1) t/2

and

sin( (2n-1) \pi x /2 ) exp(-t/2) cos( ((2n-1) \pi^{2} - 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) = sin^{2}(\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 sin^{2}(\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 sin^{2}(\pi x),

which implies than

sin^{2}(\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)) )

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