Solving a wave equation: $a^2\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$

Question

Find product solution $$a^2\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$$ by the method of separation of variables

So first off: $$u(x,t)=X(x)T(t)$$

$$a^2X''(x)T(t)=X(x)T''(t)$$

$$a^2\frac{X''(x)}{X(x)}=\frac{T''(t)}{T(t)}=\lambda$$

First part of the equation:

$$a^2X''(x)-\lambda X(x)=0$$

But I dont know what to do from here on. Lambda can be either less, equal to, or more than $0$, and for these lambdas solutions will be different.

Actually I know what to do. Just got to divide this into 3 cases, but what I dont know is how to merge all these cases into one product solution.

Answer 1

That looks correct. I think there are product solutions for each case. See the way it is done for the heat equation...

...or check out what I just found.

Answer 2

You've got it so far.

For the $T$ equation, you have that $T'' = \alpha^{2} \lambda T$ (you should really put the $\alpha^{2}$ in this equation to make your life easier). You need a function that, when differentiated twice, gives you back the same function times a constant. You seem like you know what that is.

For the $X$ equation, you have $X'' = \lambda X$ - and you can guess a similar solution function because you have the same sort of setup as the T equation.

The issue comes in dealing with the constraints. Depending on those, there are different ways of dealing with the problem.

You can guess a solution of the form $$ \sum_{n} a_{n}x^{n}$$ and go from there, differentiating term by term and combining sums with similar lowest power. Have you learned about Power Series? Frobenius Series?