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The following PDE is given: $u_{tt}(t,x)=u(t,x)+12u_t(t,x)+12u_x(t,x)$, $u(t,0)=Ae^{-t^2}\sin{\omega t}$

May I ask you for hints about how to solve that PDE ? I don't think that I can use the method of characteristics, because there is a second derivative ($u_{tt}$). Separation of variables doesn't seem to be an option either. And also, don't we need a second initial condition if we have a second partial derivative ?

I am truly lost, and would appreciate any hint/help.

Edit: separation of variables seems to work, but the initial condition makes things complicated. If we do so, $A$ would depend on $t$ , which cannot be.

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  • $\begingroup$ But separation of variables does work $$\frac{12X'}{X} = \frac{T''}{T} - 1 - \frac{12 T'}{T}$$ but the right approach to take depends on the boundary conditions, which you haven't supplied. $\endgroup$ – mattos Oct 12 '19 at 4:35
  • $\begingroup$ @mattos well, the boundary condition is $u(t,0)=Ae^{-t^2}\sin{\omega t}$. $\endgroup$ – Poujh Oct 12 '19 at 10:09
  • $\begingroup$ $12X'=kX$ thus $k=12$ thus $T(12+1)=13T=T''-12T'$ thus $T''=-T'=T$. But I don't really see how it matches with the boundary condition $u(t,0)=Ae^{-t^2}\sin{\omega t}$ $\endgroup$ – Poujh Oct 12 '19 at 10:24
  • $\begingroup$ @mattos if i take the derivative with respect to t of $Ae^{-t^2}\sin{\omega t}$, i get $Ae^{-t^2}(\omega \cos{\omega t}-2t\sin{\omega t})$. The second derivative is $-Ae^{-t^2}((b^2-4t^2+2)\sin{\omega t}+4\omega t \cos{\omega t})$. Thus $\omega$ should be $\pi \over 2$ or $3\pi \over 2$, but to hold, $A$ should be $\frac{1}{2t}$ which cannot be. Also $-A$ should be equal to $(b^2-4t^2+2)$ which can also not be. $\endgroup$ – Poujh Oct 12 '19 at 10:34
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TL;DR: The main method to solve PDEs with constant coefficients is Fourier transformation. The trick will be to only Fourier transform wrt. $t$.

  1. The inertial profile is Gaussian $$ u_0(t)~=~Ae^{-t^2}\sin (\omega_0 t)~=~\frac{A}{2i}\sum_{\pm}\pm e^{-t^2\pm i\omega_0 t},\tag{1} $$ so the Fourier transform is also Gaussian $$ \hat{u}_0(\omega) ~=~\int_{\mathbb{R}}\!\mathrm{d}t~e^{-i\omega t} u_0(t) ~\stackrel{(1)}{=}~\frac{\sqrt{\pi}A}{2i} \sum_{\pm}\pm e^{-\frac{1}{4}(\omega \mp\omega_0)^2} .\tag{2} $$

  2. OP's Fourier transformed PDE reads $$ \frac{\partial\hat{u}(\omega,x)}{\partial x} ~=~\left(i\omega -\frac{1}{12}(\omega^2+1)\right)\hat{u}(\omega,x), \tag{3} $$ with unique solution $$ \begin{align}\hat{u}(\omega,x)~\stackrel{(3)}{=}~&\hat{u}_0(\omega)e^{(i\omega -\frac{1}{12}(\omega^2+1))x}\cr ~\stackrel{(2)}{=}~&\frac{\sqrt{\pi}A}{2i} \sum_{\pm}\pm e^{-\frac{1}{4}(\omega \mp\omega_0)^2+(i\omega -\frac{1}{12}(\omega^2+1))x} ,\end{align} \tag{4}$$ which is again Gaussian in $\omega$.

  3. The solution to OP's PDE is given by the inverse Fourier transform $$ \begin{align} u(t,&x)\cr\cr =&\int_{\mathbb{R}}\!\frac{\mathrm{d}\omega}{2\pi}~e^{i\omega t}\hat{u}(\omega,x) \cr \stackrel{(4)}{=}&\frac{A}{4i\sqrt{\frac{x}{12}+\frac{1}{4}}}\cr &\sum_{\pm}\pm \exp\left(-\frac{36 t^2 +3\omega_0^2 x + 72 tx + x(3+37 x) \mp 36 i \omega_0 (t+x)}{12(x+3)}\right)\cr =&\frac{A}{\sqrt{\frac{x}{3}+1}} \exp\left(-\frac{36 t^2 +3\omega_0^2 x + 72 tx + x(3+37 x)}{12(x+3)}\right)\cr &\times \sin \frac {36 \omega_0 (t+x)}{12(x+3)}, \end{align}\tag{5}$$ which is therefore also Gaussian in $t$.

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Hint: it's a parabolic equation with constant coefficients. To put it into the standard form, where $t$ is time variable, one has to swap $x$ and $t$: $u_{t}=u_{xx}/12-u_x-u/12$, $u(x,0)=\varphi(x)=Ae^{-x^2}\sin{\omega x}$. The fundamental solution $G$ of this equation can be written out explicitly. The solution of the Cauchy problem can be represented as a Poisson potential: $$ u(x,t)=\int_{-\infty}^{\infty}G(x-y,t) \varphi(y)\,dy. $$ This integral is an elementary function.

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