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Equation  Wave Equation Derivation Question 
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#1
Apr113, 05:44 PM

P: 143

Hello, my teacher says that if, on a wave equation
[tex]f(xct)=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] and [tex]\partial_{x}=\frac{\partial }{\partial e} \frac{\partial e}{\partial x}=\frac{\partial }{\partial e} [/tex] then [tex]\partial_{tt} c^2 \partial_{xx}= c^2 \partial_{ee} c^2 \partial_{ee}=0[/tex] what is the correct? 


#2
Apr213, 03:10 AM

Sci Advisor
Thanks
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(xc t)+f_2(x+c t).[/tex] You come to this conclusion by the substitution [tex]u_1=xc 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] as [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(xct)+\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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