# Telegraph Equation separation of variables

I've been trying to solve the telegraph equation by the method of separation of variables. The equation is given by: \begin{align*} u_{tt}+au_t+bu&=c^2u_{xx}, \quad 00\\ u(x,0) &= f(x), \quad u_t(x,0) = 0\\ u(0,t)=u(l,t)&=0 \end{align*}

I suppose that $$u(x,t)=X(x)T(t)$$, and found the next equations \begin{align*} X^{''}-\lambda X &= 0\\ T^{''}+aT^{'}+(b-\lambda c^2)T&=0 \end{align*}

Also, I found that the non-trivial solution for $$X$$ is when $$\lambda <0$$. Consequently, $$X_n(x)=Asin\left(\frac{n\pi x}{l}\right)$$, with $$\lambda = -\left(\frac{n\pi}{l}\right)^2$$, and $$A$$ constant. Nevertheless, I'm stuck with the solution for $$T$$, I know that the roots of the characteristic polynomial are given by: \begin{align*} r_i=\frac{-a\pm\sqrt{a^2-4(b-\lambda c^2)}}{2} \end{align*}

But I don't know how to treat with that. Thanks :)

## 1 Answer

For each $$\lambda_n=-\left(\frac{n\pi}{l}\right)^2$$, you have two possible values for $$r$$: $$r_n^{\pm}=\frac{-a \pm \sqrt{a^2 - 4(b-\lambda_n c^2)}}{2}.$$ Therefore, $$T_n(t)=A_ne^{r_n^+t}+B_ne^{r_n^-t}.$$ The initial condition $$u_t(x,0)=0$$ implies $$T_n'(0)=0$$, or $$r_n^+A_n+r_n^-B_n=0$$, whose solution is $$A_n=C_nr_n^-, B_n=-C_nr_n^+$$, so that $$T_n(t)=C_n(r_n^-e^{r_n^+t}-r_n^+e^{r_n^-t}).$$ Finally, the constants $$C_n$$ are determined by the condition $$u(x,0)=f(x)$$: $$u(x,0)=\sum_{n=1}^{\infty}T_n(0)\sin\left(\frac{n\pi x}{l}\right) =\sum_{n=1}^{\infty}C_n(r_n^- -r_n^+)\sin\left(\frac{n\pi x}{l}\right)=f(x),$$ hence $$u(x,t)=\sum_{n=1}^{\infty}\frac{f_n}{r_n^- -r_n^+}\left(r_n^-e^{r_n^+t}-r_n^+e^{r_n^-t}\right)\sin\left(\frac{n\pi x}{l}\right),$$ where $$f_n=\frac{2}{l}\int_0^{l}f(\xi)\sin\left(\frac{n\pi \xi}{l}\right)\,d\xi.$$