Question:
Return once again to the problem of the second-last question.
The correct solution to this problem, that you would find using the so-called method of separation of variables, is:
Now shed some light on the question why this solution is smooth for any arbitrary 0. To do so, first explain why any sum of sines of the form
Next, you are allowed to make use of the fact that the function is still smooth if the coefficients go to zero quickly enough. In particular, if you can show that
Use this to show that above is indeed infinitely smooth for any 0. And show that it is not true for 0, where the solution jumps at the origin.
Answer:
Search through a table book, in the Fourier series section, for the following result:
Using this result, you can show that the boundary condition at 0 is satisfied. The other boundary conditions and the partial differential equation can be verified directly from the form of the solution.
A single sine is an infinitely smooth function. You can differentiate it as many times as you want without getting singularities. Argue that according to calculus, the derivatives of a sum of two functions are just the sum of the derivatives of each separate function. So a sum of two sines is an infinitely smooth functions. And then so is the sum of a sum of two sines and another sine. And then so is the sum of the sum of three sines and another sine.
To show that the coefficients go to zero sufficiently quickly, you can use l’Ho[s]pital. You first need to compare with the generic to see what the coefficients are that you want to be vanishingly small for large .