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The problem is:
\begin{equation} u_{tt} -c^2u_{xx} - a^2 u = 0 \end{equation} with $\hspace{2mm}-\infty < x < \infty $, $ \hspace{2mm} u(x,t) \hspace{2mm}$ bounded as $ x \rightarrow \pm \infty$

and IC $\hspace{2mm} u(x,0) = f(x), \hspace{2mm} u_t(x,0)=0$

I have transformed the equation and worked out the solution \begin{equation} \hat{u}(\xi, t) = \hat{f}(\xi)\cosh\Big(\sqrt{a^2-c^2\xi^2}\Big) \end{equation}but am not sure how to proceed.

user2869037
  • 31
  • Transform it back into a convolution of $f$ with the Fourier transform of $\cosh(\sqrt{a^2-c^2\xi^2})$? – Neal Apr 24 '15 at 20:44
  • Sorry, I'm still not sure how to proceed with this suggestion. I have searched around but don't what the Fourier transform of $ \cosh $ is – user2869037 Apr 24 '15 at 21:07

2 Answers2

0

Hint:

Let $\begin{cases}x_1=\dfrac{ax}{c}\\t_1=at\end{cases}$ ,

Then $u_x=u_{x_1}(x_1)_x+u_{t_1}(t_1)_x=\dfrac{au_{x_1}}{c}$

$u_{xx}=\left(\dfrac{au_{x_1}}{c}\right)_x=\left(\dfrac{au_{x_1}}{c}\right)_{x_1}(x_1)_x+\left(\dfrac{au_{x_1}}{c}\right)_{t_1}(t_1)_x=\dfrac{a^2u_{x_1x_1}}{c^2}$

$u_t=u_{x_1}(x_1)_t+u_{t_1}(t_1)_t=au_{t_1}$

$u_{tt}=(au_{t_1})_t=(au_{t_1})_{x_1}(x_1)_t+(au_{t_1})_{t_1}(t_1)_t=a^2u_{t_1t_1}$

$\therefore a^2u_{t_1t_1}-a^2u_{x_1x_1}-a^2u=0$

$u_{t_1t_1}=u_{x_1x_1}+u$

Similar to PDE - solution with power series:

Consider $u(x_1,0)=F(x_1)$ and $u_{t_1}(x_1,0)=G(x_1)$ ,

Let $u(x_1,t_1)=\sum\limits_{n=0}^\infty\dfrac{u_{t_1}^{(n)}(x_1,0)t_1^n}{n!}$ ,

Then $u(x_1,t_1)=\sum\limits_{n=0}^\infty\dfrac{u_{t_1}^{(2n)}(x_1,0)t_1^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\dfrac{u_{t_1}^{(2n+1)}(x_1,0)t_1^{2n+1}}{(2n+1)!}$

$u_{t_1t_1t_1t_1}=(u+u_{x_1x_1})_{t_1t_1}=u+u_{x_1x_1}+(u+u_{x_1x_1})_{x_1x_1}=u+u_{x_1x_1}+u_{x_1x_1}+u_{x_1x_1x_1x_1}=u+2u_{x_1x_1}+u_{x_1x_1x_1x_1}$

Similarly, $u_{t_1}^{(2n)}=\sum\limits_{k=0}^nC_k^nu_{x_1}^{(2k)}$

$u_{t_1t_1t_1}=(u+u_{x_1x_1})_{t_1}=u_{t_1}+(u_{t_1})_{x_1x_1}$

$u_{t_1t_1t_1t_1t_1}=(u_{t_1}+(u_{t_1})_{x_1x_1})_{t_1t_1}=u_{t_1}+(u_{t_1})_{x_1x_1}+(u_{t_1}+(u_{t_1})_{x_1x_1})_{x_1x_1}=u_{t_1}+(u_{t_1})_{x_1x_1}+(u_{t_1})_{x_1x_1}+(u_{t_1})_{x_1x_1x_1x_1}=u_{t_1}+2(u_{t_1})_{x_1x_1}+(u_{t_1})_{x_1x_1x_1x_1}$

Similarly, $u_{t_1}^{(2n+1)}=\sum\limits_{k=0}^nC_k^n(u_{t_1})_{x_1}^{(2k)}$

$\therefore u(x_1,t_1)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nF^{(2k)}(x_1)t_1^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nG^{(2k)}(x_1)t_1^{2n+1}}{(2n+1)!}$

Community
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doraemonpaul
  • 16,488
-1

The solution is $\hat{u}(\xi, t) $= $\hat{f}(\xi)\cos\Big(\sqrt{-a^2+c^2\xi^2}\Big)$

After that you shoud convert $cos at $ to $(e^{ait }+e^{-ait })/2$

Sara
  • 659