NAME:_________________________________

Instructions: Write the answers where indicated (or points will be taken off). You may attach worksheets if they are likely to be helpful: Some credit may be given for a correct method, if it is clearly explained and convincing.

Some possibly helpful integrals done by Maple are given on the last page.

    u(0,y,z) = u(1,y,z) = 0
    u_y(x,0,z) = u_y(x,2,z) = 0
    u(x,y,0) = x, u(x,y,3) = 0

Solution.

The product solutions of Laplace's equation which satisfy all the homogeneous boundary conditions are of the form

for n = 1, 2, ... and m = 0,1,2,.... The general solution will be a double Fourier series of the form

As usual, the hyperbolic sines in the denominator are just I chose to make things work out nicely when we check the nonhomogeneous BC, which fort this problem occuirs at z=0. At that boundary we must have:

Here there is a nice simplification, since there is no dependence on y. This means that only the a₀ term in the cosine series is different from 0, and we are left with the Fourier sine series

With the help of the integrals done by Maple and the usual formula for the Fourier coefficients, we find that

ANSWER:

    u_t = u_xx.

The rod lies on the x-axis from 0 to L, and is pefectly insulated at x=0. At the other end,

it is heated so that:

    u_x(t,L) = exp(- ²t/4).

At t=0, the rod is at temperature 10.

HINTS: The following are solutions of the heat equation:

    exp(-a² t) sin(a x)

    exp(-a² t) cos(a x)

a) Solve for the temperature u(t,x) for t>0, 0 < x < L. Begin with L in the range 1 < L < 2 and then consider the limit as L 2.

Solution.

The first step is to convert this problem into one with homogeneous boundary conditions. We can do this by defining

as suggested in the hint. This function satisfies the heat equation and has homogeneous Neumann boundary conditions

    v_x(t,0) = v_x(t,L) = 0.

Its initial condition is:

The general solution for v uses a Fourier cosine series:

where the coefficients am are the Fourier cosine coefficients for the initial condition.

A calculation with the standard formula shows that

and

The solution for u is thus:

The first term and the n=1 term both become singular as L 2, but their sum has a finite limit, which can be calculated with l'Hôpital's rule. The result is:

b) At what values of t,x for 0 t 100, 0 x L is the temperature maximal? At what values of t,x in these ranges is it minimal?

Solution.

This question requires the maximum principle, and the observation that heat enters at x=L.

    t_max,x_max = 100,L

    t_min,x_min = 0,0

c) What is the equilibrium temperature in the rod when t ?

Solution.

This can either be read off the solution from part a) or else we can recall that the total heat in the rod is

so

Integrating this gives

Since the equilibrium temperature is a constant, it is the coefficient of L in this expression.

> int(x*sin(n*pi*x/L), x=0..L);

                     2
                    L  (-sin(n pi) + n pi cos(n pi))
                  - --------------------------------
                                  2   2
                                 n  pi

> int(x*cos(n*pi*x/L), x=0..L);

                                             2      2
               (cos(n pi) + n pi sin(n pi)) L      L
               ------------------------------- - ------
                            2   2                 2   2
                           n  pi                 n  pi

> int(x^2 *sin(n*pi*x/L), x=0..L);

    3   2   2                                                    3
   L  (n  pi  cos(n pi) - 2 cos(n pi) - 2 n pi sin(n pi))       L
 - ------------------------------------------------------ - 2 ------
                            3   3                              3   3
                           n  pi                              n  pi

> int(x^2 *cos(n*pi*x/L), x=0..L);

    3   2   2                                             
   L  (n  pi  sin(n pi) - 2 sin(n pi) + 2 n pi cos(n pi)) 
 - ------------------------------------------------------ 
                            3   3                         
                           n  pi