Science:Math Exam Resources/Courses/MATH257/December 2011/Question 04 (b)
• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •
Question 04 (b) 

Consider the following problem for the heat equation with a timedependent source term, and mixed boundary conditions:
Find the solution to this problem using the method of eigenfunction expansion. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

How can we formulate this problem in terms of eigenfunctions of a similar problem? 
Hint 2 

Consider the related problem first. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. We must solve this eigenfunction expansion. First, following the hints, we consider the problem without a source term and attempt a separation of variables : for some fixed constant . Let's start with the xequation. From the boundary conditions we require to be positive. Therefore To plug in the boundary conditions we first calculate the derivative of : so that . From the boundary condition , we get A = 0. (The possibility that is ruled out by the other boundary condition). Similarly, from , we get Hence, the eigenfunction for is given (after suppressing the arbitrary constant B) by At this point we have where we normally would find for this problem. This would lead us to a solution of the form:
However, recall that the problem we really want to solve is , not just . We use the same eigenfunctions for x and now allow our coefficients to depend on t (which subsumes the terms as well). Thus so that becomes Recall that so we get Further, recall that our eigenfunctions are orthogonal so we can multiply both sides by and then integrate over x to get Also recall that so that Therefore, We can solve this by integrating factor, Recall that and hence so that where we have redefined the constant. To solve for the arbitrary constant recall that We use orthogonality once more to obtain Using the above we get which implies Hence, we finally conclude that for 