Test 1

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1996, 2010 by Evans M. Harrell II and James V. Herod. All rights reserved.


SAMPLE TEST SOLUTIONS

1.

a) Solve the following problem:

   PDE ut = 4 uxx, for 0 < t, 0 < x < 1

   BC ux(t,0) = 0, ux(t,1) = 0, for 0 < t

   IC u(0,x) = 2 x - x2, for 0 < x < 1

ANSWER:

u(t,x) = 2/3 - 4 Sum exp(-4 n2pi2 t) cos(n pi x)/(n2 pi2).

The way to get this solution is to recall that the separated solutions with these BC are either constants or of the form constant exp(-4 n2 pi2 t) cos(npi x)/(n2. The general solution is a sum of these solutions, with coefficients we could call an. When we put t = 0, we get a Fourier cosine series, and the coefficients are the Fourier cosine coefficients for 2 x - x2. Then use the standard formula for these coefficients.

b) Find the maximum value of u(t,x) for 0 <= t <= 1, 0 <= x <= 1:

We know by the maximum principle that the maximum value occurs either at t=0, t = 1, x = 0, or x = 1, so only these values need to be considered. Physical reasoning can be used, if you wish, to eliminate all possibilities for t > 0 (the rod is insulated and has no sources, so heat cannot appear at the ends while the temperature relaxes to equilibrium). Considering t=0, we find:

ANSWER:

The maximum temperature occurs at x = 1, t = 0

The maximum temperature is umax = 1


2. Some background information:

There is a complete, orthonormal set of functions denoted phin(x), for -infinity < x < infinity, which are eigenfunctions for the ordinary differential equation

- phin'' + x2 phin = (2n+1) phin, n = 0, 1, 2, ....

You may use the notation phin in the answer to this problem.

Consider the following PDE:

    PDE utt = grad2 u - x2 u, for 0 < t, 0 < x < pi 1, 0 < y < pi

BC u(t,x,0) = u(t,x, pi ) = 0, for 0 < t

Find the normal mode with the lowest frequency of vibration (include the time dependence):

ANSWER:

This is a straightforward separation of variables, albeit with a new function. The function Y satisfies the usual eigenvalue equation with 0 Dirichlet BC, so the spatial part of the separated solutions are of the form phin sin(m y), n = 0, 1, ...; m = 1, 2, .... If the full solution is of the form Tnm(t) phin(x) sin(m y), then substituting into the PDE shows that

Tnm''(t) = munm Tnm(t),

where munm = (2 n + 1) + m2. The solutions are sines and cosines of munm1/2. The lowest frequency occurs when m = 0, n = 1:

u(t,x,y) = (A01 cos(21/2 t) + B01 sin(21/2 t) ) phi0(x) sin(y),

or, if you prefer,

u(t,x,y) = (A01 cos(21/2 t) + B01 sin(21/2 t) ) exp(-x2/2) sin(y).

Find the general solution:

ANSWER:

u(t,x,y) = Sum of (Anm cos((2n+1+m2)1/2 t) + Bnm sin((2n+1+m2)1/2 t) ) phin(x) sin(m y), for n = 0, 1, ...; m = 1, 2, ....


3. A slice of pizza is shaped like a sector in cylindrical coordinates,

0 < r < 20 cm,

0 < theta < pi /3 radians

0 < z < 1 cm.

It has come to thermal equilibrium while sitting on a student's computer monitor, so that the temperature on its surface is

u(r, theta , 0) = 30

u(r, theta , 1) = 20

u(r,0,z) = u(r, pi /3,z) = 30 - 10 z

u(20, theta ,z) = 30 - 10 z - 5

This is a low quality pizza consisting of a homogeneous material (independent of position)

a) The partial differential equation for a homogeneous material at thermal equilibrium is Laplace's equation,

    grad2 u = 0

b) Answer the following questions.

Are there useful simplifications involving the boundary conditions? If so, what are they?

Be specific and put the answer here: Most of the boundary conditions can be made homogeneous by redefining the unknown as

    v(t,x) = u - (30 - 10 z).

The only BC which remains nonhomogeneous is the one at r = 20.

Are there useful separated solutions? If so, write the specific ordinary differential equations that the separated solutions satisfy below. Include boundary conditions.

write v(r,theta, z) = R(r) Q(theta) Z(z). Then:

    - Q'' = lambda Q,
    theta(0) = theta(pi/3) = 0

    - Z'' = mu Z,
    Z(0) = Z(1) = 0

    R'' + (1/r) R' = (lambda/r2 + mu) R,
    R(0) bounded

(We don't do anything about the nonhomogeneous boundary before putting the separated solutions together.)

c) Solve the differential equations with the given boundary conditions.

The equation for R is our old friend Bessel's equation, but the eigenvalues lambda are now determined by the conditions on Q, which lead to eigenfunctions of the form sin(3 m theta) and eigenvalues lambdam = 9 m2.


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