Equation - Wave Equation Derivation Question

[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?

 

[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.