# Solving a system of linear PDEs

For $-\infty<x<\infty$, $t>0$, \begin{cases} u_t-v_x=0,\\ v_t-u_x=0,\\ u(x,0)=\frac{1}{1+x^2},\\ v(x,0)=0. \end{cases}

I would like to know the approach to solve this system of equations. Any hints would be appreciated.

The derivation allows to separate into two independant EDP of the Laplace kind :

We take Laplace transforms in time $(t)$, and Fourier transforms in space $(x)$. Taking the Fourier transform of the first two equations we get

\begin{align*} 0&=\frac{\mathrm{d}}{\mathrm{d} t}U(\omega,t) + i \omega V(\omega, t)\\ 0&=\frac{\mathrm{d}}{\mathrm{d} t} V(\omega,t) + i \omega V(\omega,t) \end{align*}

Now we take the Laplace transform and get

\begin{align*} 0&=s\mathcal{U}(\omega, s) - U(\omega,0)+ i\omega \mathcal{V}(\omega, s)\\ 0&=s \mathcal{V}(\omega,s) -V(\omega,0)+ i\omega \mathcal{U}(\omega, s) \end{align*}

Now, by taking the fourier transform (in space) of the initial conditions we learn that $U(\omega,0) = \exp(-|\omega|)\sqrt{\frac{\pi}{2}}$ and $V(\omega,0) = 0$. Plugging in these transformed initial conditions we get

\begin{align*} 0 &= s\mathcal{U}(\omega,s) - \exp(-|\omega|) + i \omega \mathcal{V}(\omega,s)\\ 0 &= s\mathcal{V}(\omega,s) + i\omega \mathcal{U}(\omega,s) \end{align*} Solve the second equation for $\mathcal{V}(\omega,s_)$ and plug it into the first equation, we get

\begin{align*}0= s\mathcal{U}(\omega,s) -\exp(-|\omega|) \sqrt{\frac{\pi}{2} } + \frac{\omega^2}{s} \mathcal{U}(\omega,s) \end{align*}

\begin{align*} \Rightarrow (s^2+\omega^2)\mathcal{U}(\omega,s) = s \exp(-|\omega|) \sqrt{\frac{\pi}{2}} \end{align*}

\begin{align*} \Rightarrow \mathcal{U}(\omega,s) =\frac{ s}{s^2+\omega^2} \exp(-|\omega|) \sqrt{\frac{\pi}{2}} \end{align*}

Now we take the inverse laplace transform and get

$U(\omega, t) = \cos(\omega t) \exp(-|\omega|) \sqrt{\frac{\pi}{2}}$

Now we can take the inverse Fourier transform and get

\begin{align*} u(x,t) = \frac{1}{2} \left(\frac{1}{1-(x-t)^2} + \frac{1}{1-(x+t)^2}\right). \end{align*}

Now, we can recover $v(x,t)$ by using similar techniques and obtain

\begin{align*} v(x,t) = \frac{1}{2}\frac{1}{1+(x+t)^2} - \frac{1}{2} \frac{1}{1+(x-t)^2}. \end{align*}