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Equation - Wave Equation Derivation Question

by alejandrito29
Tags: derivation, equation, wave
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Apr1-13, 05:44 PM
P: 143
Hello, my teacher says that if, on a wave equation

[tex]f(x-ct)=f(e)[/tex] then
[tex]\partial_{ee}= \partial_{tt}- c^2 \partial_{xx}[/tex]

but i think that

[tex]\partial_{t}=\frac{\partial }{\partial e} \frac{\partial e}{\partial t}=-c\frac{\partial }{\partial e} [/tex]


[tex]\partial_{x}=\frac{\partial }{\partial e} \frac{\partial e}{\partial x}=\frac{\partial }{\partial e} [/tex]


[tex]\partial_{tt}- c^2 \partial_{xx}= c^2 \partial_{ee}- c^2 \partial_{ee}=0[/tex]

what is the correct?
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Apr2-13, 03:10 AM
Sci Advisor
P: 2,478
Of course you are right. Indeed the general solution of the homogeneous wave equation in 1+1 dimensions is
[tex]f(t,x)=f_1(x-c t)+f_2(x+c t).[/tex]
You come to this conclusion by the substitution
[tex]u_1=x-c t, \quad u_2=x+ c t[/tex]
This gives through the chain rule
[tex]\partial_{u_1} \partial_{u_2}=\frac{1}{4}(\partial_x^2-\partial_t^2/c^2).[/tex]
This means that you can write the wave equation
[tex]\left (\frac{1}{c^2} \partial_t^2-\partial_x^2 \right ) f=0.[/tex]
[tex]\partial_{u_1} \partial_{u_2} f=0.[/tex]
This is very easy to integrate successively. Integrating with respect to [itex]u_1[/itex] first gives
[tex]\partial_{u_2} f = \tilde{f}_2'(u_2)[/tex]
and then
[tex]f=\tilde{f}_1(u_1)+\tilde{f}_2(u_2) = \tilde{f}_1(x-ct)+\tilde{f}_2(x+c t)[/tex]
with two arbitrary functions that are at least two times differentiable with respect to their arguments.

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