Is it legitimate to factor the wave equation?

Question

In solving the wave equation $$u_{tt} - c^2 u_{xx} = 0$$ it is commonly 'factored'

$$u_{tt} - c^2 u_{xx} = \bigg( \frac{\partial }{\partial t} - c \frac{\partial }{\partial x} \bigg) \bigg( \frac{\partial }{\partial t} + c \frac{\partial }{\partial x} \bigg) u = 0$$

to get $$u(x,t) = f(x+ct) + g(x-ct).$$

My question is: is this legitimate?

The partial differentiation operators are not variables, but here in 'factoring' they are treated as such.

Also it does not seem that both factors can individually be set to zero to obtain the solution--either one or the other, or both might be zero.

Answer 1

Yes, this logic is totally legitimate. In the language of linear algebra, we have two operators $A$ and $B$ and the wave equation is $$AB \psi = 0, \quad A = \partial_t - c \partial_x, \quad B = \partial_t + c \partial_x.$$ Since $A$ and $B$ commute, they may be simultaneously diagonalized, i.e. there is a basis $\psi_i$ where $$A \psi_i = \lambda_i \psi_i, \quad B \psi_i = \lambda'_i \psi_i$$ so the wave equation becomes $$AB \psi_i = \lambda_i \lambda'_i \psi_i = 0.$$ This is satisfied for $\lambda_i = 0$ or $\lambda'_i = 0$, so the set of solutions to the wave equation is just the union of the nullspaces of $A$ and $B$.

Answer 2

1) differentials are linear operators. if you are working on a space of differntiable functions they satisfy $\frac{\partial}{\partial x}\frac{\partial}{\partial y} = \frac{\partial}{\partial y} \frac{\partial}{\partial x}$. so yes, this is legit.

2) to prove they can be individually set to zero, put them separately into wave equation you started with.

P.S. if your answer feels like a magical guess, try changing the variables $\alpha = x - ct, \beta = x + ct$ then things come up straightforward

Answer 3

Yes, this is appropriate. You can apply the operators in the brackets one after another and get the same result as with the second-order derivatives. This "factoring" would be wrong, however, if, for example, $c$ were a function of $x$ or $t$.

Answer 4

Let us define $$ f(u)=\frac{\partial u}{\partial t}-c\frac{\partial u}{\partial x},\ g(u)=\frac{\partial u}{\partial t}-c\frac{\partial u}{\partial x} $$

Then, wave equation can be expressed by composition of two functions.

$$ f(g(u))=g(f(u))=\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2 u}{\partial x^2}=0 $$

This equation is vaild for all function u, so that

$$ g(u)=f(u)=0 $$

I hope this helps you. '^'