Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

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

1.

a) Solve the following problem:

   PDE u_t = 4 u_xx, for 0 < t, 0 < x < 1

   BC u_x(t,0) = 0, u_x(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 n² pi² 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(n pi x)/(n². The general solution is a sum of these solutions, with coefficients we could call a_n. 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 u_max = 1

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

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

You may use the notation \phi n in the answer to this problem.

Consider the following PDE:

    PDE u_tt = 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 \phi_n sin(m y), n = 0, 1, ...; m = 1, 2, .... If the full solution is of the form T_nm(t) \phi_n(x) sin(m y), then substituting into the PDE shows that

T_nm''(t) = \mu_nm T_nm(t),

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

u(t,x,y) = (A₀₁ cos(2^1/2 t) + B₀₁ sin(2^1/2 t) ) \phi₀(x) sin(y),

or, if you prefer,

u(t,x,y) = (A₀₁ cos(2^1/2 t) + B₀₁ sin(2^1/2 t) ) exp(-x²/2) sin(y).

Find the general solution:

ANSWER:

u(t,x,y) = Sum of (A_nm cos((2n+1+m²)^1/2 t) + B_nm sin((2n+1+m²)^1/2 t) ) \phi_n(x) sin(m y), for n = 0, 1, ...; m = 1, 2, ....

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), \Theta(\theta) Z(z). Then:

    - \Theta'' = \lambda \Theta,
    \Theta(0) = \Theta(\pi/3) = 0

    - Z'' = \mu Z,
    Z(0) = Z(1) = 0

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

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

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

Looking at the eigenvalue equations in part b), we see that the equation for Z is familiar, and the solutions are:

    Z_n(z) = sin(n pi z), \mu = (n \pi)².

The equation for \Theta is also of the same form, except that the way the variable \theta enters is somewhat novel. The length of the interval is \pi/3, so:

    \Theta_m(\theta) = sin(3 m \theta), \lambda = (3 m)².

Finally, the equation for R is not Bessel's equation but something simpler, since one term is missing. Its solutions are

    r^a, where a = +/-(\lambda + \mu)^1/2.

Boundedness at r=0 excludes the negative powers, so we are left with a general solution for v:

For brevity let us define a_mn as the exponent of r. In order to determine the coefficients c, we set r=20 and solve for the coefficients in the resulting double Fourier sine series, which has to sum to -5. We use the orthogonality relations for the sets {sin(n \pi z) and {sin(3 m \theta)} to find that

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