2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Transform it back into a convolution of $f$ with the Fourier transform of $\cosh(\sqrt{a^2-c^2\xi^2})$? $\endgroup$ – Neal Apr 24 '15 at 20:44
  • $\begingroup$ 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 $\endgroup$ – user2869037 Apr 24 '15 at 21:07
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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 $(e^{ait }+e^{-ait })/2$

$\endgroup$
  • $\begingroup$ Sara chera es nadadi??!! $\endgroup$ – k1.M May 17 '15 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.