March 26, 1996 Key to Exam 2Math 312

1. (50) Consider the problem

   

   1. Derive the equations that must be solved in order to find the ``eigenvalues'' associated with this problem.

      Using separation of variables, we look for solutions of the form u= XT, where X is a function of x only and T is a function of t only. This leads to the pair of equations:

      

      Thus, . Forcing X to satisfy the boundary conditions (2) and (3), leads to the equations:

      

      Solving the first equation for b in terms of a, we are lead to the equation . If either or a is zero, we get that X is also zero. Thus, the only way we can get a nontrivial solution to this problem is for , or .

   2. What are the ``separated'' solutions of this problem? That is, if u(x,t) = X(x)T(t) satisfies (1), (2), and (3), what must u equal?

      The separated solutions are

      

      where the are the solutions to the eigenvalue equation in part (a).

   3. Explain how to solve this problem. Be sure to explain how you would determine the coefficients , if your solution is of the form .

      I would look for a solution of the form . Since the partial differential equation and the boundary conditions are homogeneous, we can expect that u(x,t) will also satisfy the partial differential equation as well as the boundary conditions. Thus, we only have to pick the so that u satisfies the initial condition. Set . Since the differential operator , with the homogeneous boundary conditions is ``self adjoint'', we know that eigenfunctions associated with different eigenvalues are orthogonal, and that the set of these eigenfunctions is complete. Thus, we know that we can pick the so that u satisfies the initial condition and

      

   4. Explain how to solve this problem if instead of (1), (2) and (3) above we had

      

      along with the initial condition (4).

      Notice that the partial differential equation and the boundary conditions are nonhomogeneous. These terms, though, are independent of t. Thus, subtract from the desired soluton u that steady state solution of the partial differential equation which also satisfies the boundary conditions. If v is this function, then

      

      Setting w=u-v, we see that w satisfies equations (1) through (3) and equation (4) is now w(x,0)=1-v(x). Solve for w as we did above, and then add v to w to get u.

2. (30) Solve the problem

   

   where the boundary condition is in polar coordinates.

   We first note that the partial differential equation in nonhomogeneous. Thus, let's look for a simple solution to the nonhomogeous PDE. Seeing that the nonhomogeneity is a function of y alone, leads to setting . Then . Set w = u - v. Then w satisfies Laplace's equation. Separated solutions to Laplace's equation in polar coordinates which remain bounded at the origin are of the form , . Thus, we look for a solution to this problem of the form:

   

   Setting r=2, we see that the terms and are the Fourier series coefficients of the function . Solve for these terms in the usual manner, and then set u =w+v.

3. (20) Work either of the following two problems. You must tell me which one you want graded.
   1. Solve the following problem:

      

      Using separation of variables on the differential equation, we see that separated solutions are of the form:

      

      If the are to satisfy the boundary conditions for all values of t we must have a = 0 and must be an integer multiple of . Thus, we look for a solution of form

      

      The unknowns and are determined from the two initial conditions. Thus, we have:

      

      Both of these equations are Fourier sine series expansions, and are solved for the unknown coefficients in the usual manner.

   2. Use the function , to show that there is only one solution to the following problem:

      

      Suppose that the above problem has two solutions, and . Set . Then w satisfies the wave equation with homogeneous boundary conditions and zero intitial conditions.

      

      The two integral terms cancel each other. Since w is zero for x=0 and x=1 this implies that is also zero for x=0 and x=1. Thus, the term when evaluated at the endpoints 1 and 0 is zero. Thus, the derivative of E is zero, and E is constant, and we have E(t) = E(0). However, if we have zero initial data as we do for our function w, then 0=E(0)=E(t) for all t. Thus, we must have for all values of x and t. This means that w is constant, but w is zero when t= 0, thus, w is identically zero. This of course implies that .

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