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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 :)

andres florian
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1 Answers1

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