telegraph equation
TelegraphEquation
2008-05-13 11:59:14
2008-05-13 11:59:14
Topic
wave equation
telegraph equation
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Both the electric voltage and the \PMlinkescapetext{current in a double conductor} satisfy the {\em telegraph equation}
\begin{align}
f_{xx}''-af_{tt}''-bf_t'-cf = 0,
\end{align}
where $x$ is distance, $t$ is time and\, $a,\,b,\,c$\, are non-negative constants.\, The equation is a generalised form of the wave equation.
If the initial conditions are\, $f(x,\,0) = f_t'(x,\,0) = 0$\, and the boundary conditions \,$f(0,\,t) = g(t)$,\, $f(\infty,\,t) = 0$,\, then the Laplace transform of the solution function \,$f(x,\,t)$\, is
\begin{align}
F(x,\,s) = G(s)e^{-x\sqrt{as^2+bs+c}}.
\end{align}
In the special case\, $b^2-4ac = 0$,\, the solution is
\begin{align}
f(x,\,t) = e^{-\frac{bx}{2\sqrt{a}}}g(t-x\sqrt{a})H(t-x\sqrt{a}).
\end{align}
{\em Justification of} (2).\; Transforming the differential equation (1) gives
$$F_{xx}''(x,\,s)-a[s^2F(x,\,s)-sf(x,\,0)-f_t'(x,\,0)]-b[sF(x,\,s)-f(x,\,0)]-cF(x,\,s) = 0,$$
which due to the initial conditions simplifies to
$$F_{xx}''(x,\,s) = (\underbrace{as^2+bs+c}_{K^2})F(x,\,s).$$
The solution of this ordinary differential equation is
$$F(x,\,s) = C_1e^{Kx}+C_2e^{-Kx}.$$
Using the latter boundary condition, we see that
$$F(\infty,\,s) = \int_0^\infty e^{-st}f(\infty,\,t)\,dt \equiv 0,$$
whence\, $C_1 = 0$.\, Thus the former boundary condition implies
$$C_2 = F(0,\,s) = \mathcal{L}\{g(t)\} = G(s).$$
So we obtain the equation (2).