4
$\begingroup$

I am interested in understanding as much as I can about the following partial differential equation, which is a generalization of the 1D wave equation:

$$\frac{\partial^2 u(x,t)}{\partial t^2} + \zeta(t)\frac{\partial u(x,t)}{\partial t}= \alpha(t)\frac{\partial^2 u(x,t)}{\partial x^2} + \beta(t)\frac{\partial u(x,t)}{\partial x} + \gamma(t)u(x,t) + \alpha(t)f(x),\quad x\in(-\infty,\infty),\quad t\geq 0$$

where $\zeta(t)\geq 0$, $\beta(t)\geq 0$, and $\alpha(t)\geq 0$. This equation has initial data $u(x,0)$ and velocity $\partial_t u(x,0)$. Here are a few examples:

  1. When $\zeta(t) = \beta(t) = \gamma(t) = f(t) = 0$, it is the standard 1D wave equation. Thus, the influence of the initial data travels at a speed $\alpha(t)$.

  2. When $\zeta(t) = \gamma(t) = f(t) = 0$, it is the 1D Klein-Gordon equation. Thus, the influence of the initial data travels at a speed $\leq\alpha(t)$. Should I be thinking about dispersion relations in this case?

  3. When $\beta(t) = \gamma(t) = f(t) = 0$, it is the 1D wave equation with dampening. Should I be thinking about lack of energy conservation in this case?

  4. When $\zeta(t) = \beta(t) = \gamma(t) = 0$ and $f(x) = \delta_0(x)$ ($\delta_0 =$ delta function at origin), it is the 1D wave equation with plucking at $x = 0$?

I would like to understand properties of the solution for any choice of $\alpha, \beta,\gamma,\zeta,f$. What tools are there to help me? When is there a conservation of energy statement? Thank you very much in advance.

$\endgroup$
1
$\begingroup$

When $\beta(t)=k\alpha(t)$ , where $k$ is a constant, the PDE is separable:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T''(t)+\zeta(t)X(x)T'(t)=\alpha(t)X''(x)T(t)+k\alpha(t)X'(x)T(t)+\gamma(t)X(x)T(t)$

$X(x)T''(t)+\zeta(t)X(x)T'(t)-\gamma(t)X(x)T(t)=\alpha(t)X''(x)T(t)+k\alpha(t)X'(x)T(t)$

$X(x)(T''(t)+\zeta(t)T'(t)-\gamma(t)T(t))=\alpha(t)T(t)(X''(x)+kX'(x))$

$\dfrac{T''(t)+\zeta(t)T'(t)-\gamma(t)T(t)}{\alpha(t)T(t)}=\dfrac{X''(x)+kX'(x)}{X(x)}$

$\endgroup$

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.