1
$\begingroup$

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 0<x<l, \quad t>0\\ 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 :)

$\endgroup$
0
1
$\begingroup$

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 $$ f_n=\frac{2}{l}\int_0^{l}f(\xi)\sin\left(\frac{n\pi \xi}{l}\right)\,d\xi. $$

$\endgroup$
0

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.