# D'Alembert's Solution to wave equation

1. Dec 30, 2013

### pierce15

Hello,

How does the change of variables $\alpha = x + at , \quad \beta = x - at$ change the differential equation

$$a^2 \frac{ \partial ^2 y}{ \partial x^2 } = \frac{ \partial ^2 y} {\partial t ^2}$$

to

$$\frac{ \partial ^2 y}{\partial \alpha \partial \beta } = 0$$

? I'm having a hard time following the proofs on wolfram alpha, etc

2. Dec 30, 2013

### pierce15

I think I've got it, although I don't think this is a legitimate proof:

$$\frac{ \partial \alpha}{\partial t} = a, \quad \frac{\partial \beta}{\partial t} = -a \implies \frac{1}{-a^2} \frac{ \partial \alpha \partial \beta}{\partial t^2 } = 1$$

By the same logic,

$$\frac{ \partial \alpha \partial \beta}{ \partial x^2} = 1$$

From the wave equation:

$$a^2 \frac{ \partial ^2 y}{\partial x^2} \frac{\partial x^2}{\partial \alpha \partial \beta } = \frac{ \partial ^2 y}{\partial t^2} \frac{\partial t^2}{\partial \alpha \partial \beta} (-a^2)$$

$$\implies \frac{ \partial ^2 y}{\partial \alpha \partial \beta } = 0$$

Is that OK? It's cool to just treat everything like a fraction?