PlanetPhysics/Telegraph Equation

Both the electric voltage and the current in a double conductor satisfy the telegraph equation

$\begin{array}{c}{f}_{x^{″}x}-a{f}_{t^{″}t}-b{f_t}^{\prime }-cf=0,\end{array}$

where $x$ is distance, $t$ is time and\, $a,b,c$\, are non-negative constants.\, The equation is a generalised form of the [[../WaveEquation/|wave equation]].

If the initial conditions are\, $f(x,0)={f_t}^{\prime }(x,0)=0$\, and the [[../PiecewiseLinear/|boundary]] conditions \,$f(0,t)=g(t)$,\, $f(\mathrm{\infty },t)=0$,\, then the [[../2DLT/|Laplace transform]] of the solution [[../Bijective/|function]] \,$f(x,t)$\, is

$\begin{array}{c}F(x,s)=G(s)e^{-x\sqrt{a{s}^2+bs+c}}.\end{array}$

In the special case\, $b^2-4ac=0$,\, the solution is

$\begin{array}{c}f(x,t)=e^{-\frac{bx}{2\sqrt{a}}}g(t-x\sqrt{a})H(t-x\sqrt{a}).\end{array}$

Justification of (2).\; Transforming the [[../DifferentialEquations/|differential equation]] (1) gives $F_{x^{″}x}(x,s)-a[s^{2}F(x,s)-sf(x,0)-{f_t}^{\prime }(x,0)]-b[sF(x,s)-f(x,0)]-cF(x,s)=0,$ which due to the initial conditions simplifies to $F_{x^{″}x}(x,s)=(\underset{K^2}{\underbrace{a{s}^2+bs+c}})F(x,s).$ The solution of this [[../DifferentialEquations/|ordinary differential equation]] is $F(x,s)=C_1{e}^{Kx}+C_2{e}^{-Kx}.$ Using the latter boundary condition, we see that $F(\mathrm{\infty },s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f(\mathrm{\infty },t)dt\equiv 0,$ whence\, $C_1=0$.\, Thus the former boundary condition implies $C_2=F(0,s)=ℒ\{g(t)\}=G(s).$ So we obtain the equation (2).

Justification of (3).\; When the [[../QuadraticFormula/|discriminant]] of the [[../QuadraticFormula/|quadratic equation]] \,$a{s}^2+bs+c=0$\, vanishes, the roots coincide to\, $s=-\frac{b}{2a}$,\, and\, $a{s}^2+bs+c=a(s+\frac{b}{2a}{)}^2$.\, 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 [[../Formula/|theorem]], we have ${ℒ}^{-1}\{e^{-ks}G(s)\}=g(t-k)H(t-k),$ wnere $H$ is Heaviside step function.\, Thus we obtain for ${ℒ}^{-1}\{F(x,s)\}$ the expression of (3).