# equation -- Wave Equation Derivation Question

by alejandrito29
Tags: derivation, equation, wave
 P: 143 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?
 Sci Advisor Thanks P: 2,152 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.

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