$\frac{\partial^2 X_1}{\partial t^2}+\omega_0^2X_1=2\omega_0(1+a_0exp(-\tau))^{-1.5} Sin(3\omega_0t)$ where $a_0$ and $\omega_0$ are constants.

I tried to solve it using methods of characteristics.


For the pde that I want to solve, A=1, B=C=D=E=0, F=$\omega_0^2$, G=$2\omega_0(1+a_0exp(-\tau))^{-1.5} Sin(3\omega_0t)$


so the pde is parabolic everywhere.

$\frac{d\tau}{dt}=\frac{B}{A}$=c, constant which is just $\epsilon$ since $\tau=\epsilon t$

$\implies$ $\xi=\epsilon$

If $\xi$ is constant then the Jacobian of $\xi$ and $\eta$ will always be 0 no matter what function we choose for $\eta$.

So I don't know how to move on from here.

  • $\begingroup$ Since you have derivatives with respect to one variable only, you could work like it was an ordinary differential equation. Just be careful with constants of integration. $\endgroup$ – mickep Oct 18 '15 at 9:09

Assume that $w>0$ is a real positive number.

The characteristic equation of an homogeneous equation $$u''+w^2 u=0$$ is $$r^2+w^2 =0$$ which roots are $r=\pm i w$ therefore the solution is $$u=a\sin x + b \cos x$$

To solve the inhomogeneous equation $$u''+w^2 u=f(x)$$ one uses either method of undetermined coefficients or a variation of parameters. The first assumes that the function $f(x)$ can be written in a specific form, the second involves integration of the $\frac{f}{Wronskian}$

Hope this helps.


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