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Consider the following linear wave equation. $$ u_t+cu_x-\gamma u_{xx}+\delta u_{xxx}=0 $$

If we know the following initial data, $$ u(x,0)= 3\cos^2(x)+\sin(x) $$ how to get an explicit solution?

I know that the general solution is: $$ v(x,t)=A\exp( ik[x-(c-\delta k^2)t] )\exp(-\gamma k^2t) $$

Even if I compare $u(x,0)=v(x,0)$, I can not solve it.

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  • $\begingroup$ I think there is a $t$ missing in the second exponential of the general solution. It should be $exp(-\gamma k^2 t)$, right? $\endgroup$ – AugSB Jul 10 '15 at 7:54
  • $\begingroup$ General solution should rather be superposition of such $v_{k}(x,t)$, where the domain of $k$ is dependent of boundary conditions. $\endgroup$ – mikis Jul 10 '15 at 7:58
  • $\begingroup$ Could you answer to my question, not comment? $\endgroup$ – jakeoung Jul 10 '15 at 8:02
  • $\begingroup$ Ok, but I need information, where our function $u$ is defined. Is the domain $\mathbb{R}$ or for example $[0,1]$ ? $\endgroup$ – mikis Jul 10 '15 at 8:11
  • $\begingroup$ Another error: $3\cos^2(x) +\sin(x) = \frac{1}{2} \big(3\cos(2x) + 2\sin(x) + 3\big)$, not what you wrote. $\endgroup$ – AugSB Jul 10 '15 at 8:16
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Assume (for simplicity) that our function is defined on $\mathbb{R}$ with no specific boundary conditions. So there is no restriction on $k$. Hence $k\in \mathbb{R}$ and the general solution is $u(x,t)=\int_{\mathbb{R}}A(k)v_{k}(x,t)dk$ (superposition of modes numbered by $k$). So, $u(x,0)$ is given by $\int_{\mathbb{R}}A(k)e^{ikx}dk$. We can use Fourier transform to calculate $A(k)$ and use it to find solution of our equation with respect to given initial conditions.

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  • $\begingroup$ If there is a boundary condition, is it more complex? $\endgroup$ – jakeoung Jul 10 '15 at 9:05
  • $\begingroup$ For example : if the domain is $[0,1]$ and we have co-called Dirichlet boundary conditions, i.e. $u(0,t)=u(1,t)=0$, then we have "quantization" of $k$, i.e $k$ can be only a natural number and there is no cosine term in $v_k$. It is something like guitar string. We can only have quantized wave modes, and in general solution we have sum instead of integral. $\endgroup$ – mikis Jul 10 '15 at 9:26

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