PlanetPhysics/Telegraph Equation

Both the electric voltage and the current in a double conductor satisfy the telegraph equation $$\begin{matrix} f_{xx}-af_{tt}-bf_t'-cf = 0, \end{matrix}$$ 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 $$\begin{matrix} F(x,\,s) = G(s)e^{-x\sqrt{as^2+bs+c}}. \end{matrix}$$ In the special case\, $$b^2-4ac = 0$$,\, the solution is $$\begin{matrix} f(x,\,t) = e^{-\frac{bx}{2\sqrt{a}}}g(t-x\sqrt{a})H(t-x\sqrt{a}). \end{matrix}$$

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).

Justification of (3).\; When the discriminant of the quadratic equation \,$$as^2\!+\!bs\!+\!c = 0$$\, vanishes, the roots coincide to\, $$s = -\frac{b}{2a}$$,\, and\, $$as^2\!+\!bs\!+\!c = a(s+\frac{b}{2a})^2$$.\, Therefore (2) reads $$F(x,\,s) = G(s)a^{-x\sqrt{a}(s+\frac{b}{2a})} = e^{-\frac{bx}{2\sqrt{a}}}e^{-x\sqrt{a}s}G(s).$$ According to the delay theorem, we have $$\mathcal{L}^{-1}\{e^{-ks}G(s)\} = g(t-k)H(t-k),$$ wnere $$H$$ is Heaviside step function.\, Thus we obtain for $$\mathcal{L}^{-1}\{F(x,\,s)\}$$ the expression of (3).