IBVP: Heat Equation

Question

everyone!

Could someone help me out with a question? There is a PDE, which is the heat equation. There is also a solution. In the solution, it says that an odd extension of the PDE has to be computed. Then it says, that to get a solution we need to multiply the $n$-th term by $e^{-\left(\frac{n\pi}{L}\right)^2\alpha^2t}$.

My questions are:

1. Why an odd extension is used and not an even one?
2. Why do we simply multiply by $e^{-\left(\frac{n\pi}{L}\right)^2\alpha^2t}$ to get a solution?

Thank you in advance for your replies!

Answer 1

1. Because $u(0,t)=0$, which does not hold for even functions $c_n\cos(nx)$ when $c_n\neq0$.

2. That is part of solving the PDE by separation of variables: you multiply every $n$-th $x$-function with its corresponding $t$-function and can sum them all together by the principle of superposition.

Answer 2

1. The wording of the solution is misleading because it implies you can tell whether an odd or even extension can be used just by looking at the initial conditions. In fact, there is no way to tell whether an odd or even extension would be used just by looking at the initial condition (the time condition). Just by changing the boundary conditions, this problem might as well give you an even extension.

   The solution said odd extension because it had already done the calculations for this particular problem in Assignment 4 and found that the solution in space involves only sine functions.

2. The exponential equation is the equation in $t$ only. This pde problem has been solved by the method separation of variables. If you look at the trigonometric term, it is a function of $x$ only. However, the original function is of both $x$ and $t$, which should make sense to you why we multiply the two functions together. To understand more about this method, you'd need to review what you went over in your class.