May 7, 1996      Key to Final Exam        Math 312

1. (20) Suppose that is periodic with period 2, continuous on the interval [-1,1], and that .
   1. The Fourier series of equals . What are the arguments of the sine and cosine functions, and how do you calculate the coefficients and ? Explain how one derives the formulas for the coefficients.

      Since the function is periodic with period two, the trignometric functions must also be of period two. Thus, the arguments of the sine and cosine functions are . The formulas for the coefficients are:

      

      For an explanation of these formulas, see your text.

   2. What additional conditions, if any, are needed to guarantee that the Fourier series of converges to :
      1. In on the interval [-1,1]?

         Since the function is continuous on the closed interval [-1,1], Its square is bounded and integrable. There is a theorem that says the Fourier series of any square integrable function will converge to the function in the mean square sense.

      2. Pointwise on the interval [-1,1]?

         If we also impose the condition that the derivative of is piecewise continuous, then is piecewise smooth. In this case the Fourier series of will converge at each point to the arithmetic average of the left and right hand limits of at the given point. Since is continuous everywhere, the Fourier series of converges pointwise at every point.

      3. Uniformly on the interval [-1,1]?

         If the derivative of is piecewise continuous, then the Fourier series of converges uniformly to on the interval [-1,1].

   3. What additional conditions if any, are needed to guarantee that the average of the partial sums of the Fourier series of converges pointwise to .

      No addition conditions are needed. Continuity of is sufficient.

2. (35) Use Fourier transform techniques to solve the problem below.

   

   Taking the Fourier transform (in x) of the partial differential equation, we get the following equation for the transform of the solution.

   

   The general solution to this differential equation is . The constant c is U(0), which is the transform of the solution to the partial differential equation at y=0. Thus, , where F is the transform of the initial data. Using the fact that , we have . Thus, the solution to our problem is

   

3. (15) Suppose that and satisfy and , for , where L is the differential operator defined below, and .

   

   Suppose also that both functions satisfy the boundary conditions:

   

   Show that .

   The trick, so to speak, is integration by parts.

   

   Thus, we have the following equation:

   

   To see that the term involving the boundary values of the eigenfunctions equals zero is a routine computation. Since we have assumed that the eigenvalues are different, the factor is not zero. Thus, the inner product of the eigenfunctions must be zero.

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