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How do I solve the equation $\frac{\partial^2}{\partial x\partial t} u(x,t)=\frac{\partial^2}{\partial x^2} u(x,t)$ with the initial condition $u(x,t=0)=\sqrt{\frac{\pi}{2}}\exp(-|x|)$ ? The solution must be using fourier transform. I tried transofrming $x$ and then getting 1st order PDE for $t$, but then I didnt know where to put the initial conditions.

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    $\begingroup$ Do you mean the question specifies you must use the Fourier transform? Or is that how you want to do it? Alternative methods are to integrate with respect to $x$, then use the method of characteristics, or make the substitution $u_{x} = v$. $\endgroup$ Feb 10, 2016 at 13:15
  • $\begingroup$ This should be solved with fourier transform only. $\endgroup$
    – Jonshv
    Feb 10, 2016 at 14:05

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In this case, the Fourier transform is not the simplest method. But, if this is asked , we have to use the Fourier transform.

Notation of Fourier transform of $u(x,t)$ relatively to the variable $x$ : $$\mathscr{F}_x\left(u(x,t) \right)(\omega)=U(\omega,t)$$ $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial x \partial t}$$ $\mathscr{F}_x\left(\frac{\partial^2 u}{\partial x^2} \right)(\omega)=-\omega^2 U(\omega,t)$

$\mathscr{F}_x\left(\frac{\partial^2 u}{\partial x \partial t} \right)(\omega)=-i \omega \frac{\partial U}{\partial t}$

$$-\omega^2 U = -i \omega \frac{\partial U}{\partial t} \quad\to\quad \frac{1}{U}\frac{\partial U}{\partial t} = -i \omega $$ The integration relatively to $t$ leads to : $$U(\omega,t)=C(\omega) e^{-i\omega t}$$ The condition : $u(x,0)=\sqrt{\frac{\pi}{2}}e^{-|x|}\quad$ is Fourier-transformed to$\quad U(\omega,0)=\frac{1}{1+\omega^2}$ $$U(\omega,0)=C(\omega) e^{-i\omega 0} = \frac{1}{1+\omega^2} \quad \to\quad C(\omega)= \frac{1 }{1+\omega^2}$$ $$U(\omega,t)= \frac{e^{-i\omega t} }{1+\omega^2}$$ The inverse Fourier transform leads to $u(x,t)$ : $$u(x,t)=\mathscr{F}_{\omega}^{-1}\left(U(\omega,t) \right)(x)= \mathscr{F}_{\omega}^{-1}\left(\frac{e^{-i\omega t} }{1+\omega^2} \right)(x)= \sqrt{\frac{\pi}{2}}e^{-|x+t|}$$

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Let the eigensolution be $X(x)T(t)$, then $X'(x)T'(t)=X''(x)T(t)$ or

$\displaystyle \frac{X''(x)}{X'(t)}=\frac{T'(t)}{T(t)}=ik$

$X(x)=ae^{ikx}$,

$T(t)=be^{ikt}$

$\displaystyle u(x,0)=\sqrt{\frac{\pi}{2}} e^{-|x|}$

$\displaystyle \phi(k)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u(x,0) e^{-ikx} dx$

$\displaystyle u(x,t)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k) e^{ik(x+t)} dk$

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  • $\begingroup$ is this the fourier transform of $\phi (k)$? $\endgroup$
    – Jonshv
    Feb 11, 2016 at 16:40
  • $\begingroup$ Yes, $\phi(k)$ and $u(x,0)$ are FT pairs. $\endgroup$ Feb 11, 2016 at 19:07

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