telegraph equation stability

Question

I have the next equation

$ u_{tt} - k^2 u_{xx} + 2 \alpha u_t = 0 \quad $ with $u = u(x,t)$

this equation is called telegraph equation, and this equation seems like wave equation, so to prove the stability of the solutions I tried to do the things like the wave equation, find a solution type d'alembert and after the stability results easy, I took

$ \eta = t - \frac{1}{k} x $ and $ \nu = t + \frac{1}{k} x $

so

$ \frac{\partial}{\partial t} = \frac{\partial}{\partial \eta} \frac{\partial \eta }{\partial t} + \frac{\partial}{\partial \nu} \frac{\partial nu }{\partial t} = \frac{\partial}{\partial \eta} + \frac{\partial }{\partial \nu} $

$ \frac{\partial}{\partial x} = \frac{\partial}{\partial \eta} \frac{\partial \eta }{\partial x} + \frac{\partial}{\partial \nu} \frac{\partial nu }{\partial x} = \frac{1}{k} ( - \frac{\partial}{\partial \eta} + \frac{\partial }{\partial \nu} ) $

$ \frac{\partial ^2}{\partial t^2} = \frac{\partial^2}{\partial \eta^2} + 2\frac{\partial}{\partial \eta}\frac{\partial }{\partial \nu} + \frac{\partial^2 }{\partial \nu ^2} $

$ \frac{\partial ^2}{\partial X^2} = \frac{1}{k^2}( \frac{\partial^2}{\partial \eta^2} - 2\frac{\partial}{\partial \eta}\frac{\partial }{\partial \nu} + \frac{\partial^2 }{\partial \nu ^2} ) $

so

$ 0 = u_{\eta \eta} +2u_{\eta \nu} + u_{\nu \nu} - k^2 ( \frac{1}{k^2} ( u_{\eta \eta} -2u_{\eta \nu} + u_{\nu \nu} ) ) + 2\alpha( u_{\eta} + u_{\nu} ) \\ \; = 4 u_{\eta \nu} + 2\alpha( u_{\eta} + u_{\nu} ) $

and now I don't what to do ! some hint or idea? exist another method to prove the stability?

Answer 1

Assume the solutions have the form $u(x,t) = X(x)T(t)$, if you replace that into the original PDE you get

$$ X(x)\frac{d^2 T(t)}{dt^2} + 2\alpha X(x)\frac{dT(t)}{dt} = k^2T(t)\frac{d^2X(x)}{dx^2} $$

after rearranging,

$$ \frac{1}{T(t)}\left[\frac{d^2 T(t)}{dt^2} + 2\alpha\frac{dT(t)}{dt}\right] = k^2\frac{1}{X(x)}\frac{d^2X(x)}{dx^2} $$

So you have something that depends exclusively on the variable $t$ on the l.h.s, whereas the r.h.s depends solely on $x$, the only possibility here is that both things are a constant, say $\lambda$

$$ \frac{d^2 X}{dx^2} = \frac{\lambda}{k^2}X ~~~\mbox{and}~~~ \frac{d^2 T}{dt^2} + 2\alpha\frac{dT}{dt} = \lambda T $$

Now you should consider all posible values of $\lambda$, that is $\lambda =0$, $\lambda > 0$ and $\lambda < 0$, so for instance $\lambda > 0$ will result in either exponentially growing or decaying functions with $X$, if you want the oscillatory behavior of the wave function you must consider the case $\lambda = -\mu^2 < 0$, this results in

$$ X(x) = A\cos(\mu x/k) + B\sin(\mu x/k) $$

and more interestingly

$$ \frac{d^2T}{dt^2} + 2\alpha \frac{dT}{dt} = -\mu^2 T $$

can you take it from here?

Answer 2

I'll assume $k, \alpha > 0$, and you're doing this on the domain $-\infty < x < \infty$. Let $$ E(t) = \frac{1}{2} \int_{-\infty}^\infty \left(u_t^2 + k^2 u_x^2\right)\; dx $$ and suppose $u$ and its derivatives go to $0$ sufficiently fast as $|x| \to \infty$ and are sufficiently well behaved that this is finite and we can differentiate under the integral sign. Then $$ E'(t) = \int_{-\infty}^\infty \left(u_t u_{tt} + k^2 u_x u_{xt}\right)\; dx$$ Integrate the second term by parts, and apply the PDE:

$$ \eqalign{E'(t) &= \int_{-\infty}^\infty \left(u_t u_{tt} - k^2 u_{xx} u_t\right)\; dx\cr &= - 2 \alpha \int_{-\infty}^\infty u_t^2 \; dx \le 0}$$

implying stability.