# Equation - Wave Equation Derivation Question

1. Apr 1, 2013

### alejandrito29

equation -- Wave Equation Derivation Question

Hello, my teacher says that if, on a wave equation

$$f(x-ct)=f(e)$$ then
$$\partial_{ee}= \partial_{tt}- c^2 \partial_{xx}$$

but i think that

$$\partial_{t}=\frac{\partial }{\partial e} \frac{\partial e}{\partial t}=-c\frac{\partial }{\partial e}$$

and

$$\partial_{x}=\frac{\partial }{\partial e} \frac{\partial e}{\partial x}=\frac{\partial }{\partial e}$$

then

$$\partial_{tt}- c^2 \partial_{xx}= c^2 \partial_{ee}- c^2 \partial_{ee}=0$$

what is the correct?

2. Apr 2, 2013

### vanhees71

Of course you are right. Indeed the general solution of the homogeneous wave equation in 1+1 dimensions is
$$f(t,x)=f_1(x-c t)+f_2(x+c t).$$
You come to this conclusion by the substitution
$$u_1=x-c t, \quad u_2=x+ c t$$
This gives through the chain rule
$$\partial_{u_1} \partial_{u_2}=\frac{1}{4}(\partial_x^2-\partial_t^2/c^2).$$
This means that you can write the wave equation
$$\left (\frac{1}{c^2} \partial_t^2-\partial_x^2 \right ) f=0.$$
as
$$\partial_{u_1} \partial_{u_2} f=0.$$
This is very easy to integrate successively. Integrating with respect to $u_1$ first gives
$$\partial_{u_2} f = \tilde{f}_2'(u_2)$$
and then
$$f=\tilde{f}_1(u_1)+\tilde{f}_2(u_2) = \tilde{f}_1(x-ct)+\tilde{f}_2(x+c t)$$
with two arbitrary functions that are at least two times differentiable with respect to their arguments.