Show that $v$ and $i$ satisfy the telegraph equation

Question

The voltage $v$ and the current $i$ in an electrical cabe along the $x$-axis satisfy the coupled equations: \begin{equation} i_{x}+Cv_{t}+Gv=0, \tag{1} \end{equation} \begin{equation} v_{x}+Li_{t}+Ri=0, \tag{2} \end{equation}

where $C, G, L$ and $R$ are the capacitance, (leakage) conductance, inductance, and resistance per unit lenght in the cable. Show that $v$ and $i$ both satisfy the telegraph equation:

\begin{equation} u_{xx}=LCu_{tt}+(RC+LG)u_{t}+RGu. \tag{3} \end{equation}

Suppose $i,v\in C^{2}$.

Answer 1

I found the our OP Mateus Rocha's answer a little hard to follow, so I thought I'd add mine to the discussion:

We are given

$i_x + Cv_t + Gv = 0, \tag 1$

$v_x + Li_t + Ri = 0; \tag 2$

then from (1),

$Cv_t + Gv = -i_x, \tag 3$

whence, taking $\partial / \partial t$ of each side,

$Cv_{tt} + Gv_t = -i_{xt}, \tag 4$

multiplying by $L$,

$LCv_{tt} + LGv_t = -Li_{xt}; \tag 5$

likewise from (2),

$Li_t + Ri = -v_x; \tag 6$

we take $\partial / \partial x$:

$Li_{tx} + Ri_x = -v_{xx}, \tag 7$

or

$Li_{tx} = -v_{xx} - Ri_x; \tag 8$

since we assume $i, v \in C^2$, we have

$i_{tx} = i_{xt}, \tag 9$

so (8) becomes

$Li_{xt} = -v_{xx} - Ri_x; \tag{10}$

substituting $i_x$ from (3),

$Li_{xt} = -v_{xx} + R(Cv_t + Gv); \tag{11}$

combining (5) and (11),

$LCv_{tt} + LGv_t = -Li_{xt} = v_{xx} - R(Cv_t + Gv), \tag{12}$

and after a little algebraic rearrangement,

$v_{xx} = LCv_{tt} + (LG + RC)v_t + RGv, \tag{13}$

the sought for result in $v(x, t)$.

The derivation for $i(x, t)$ is very similar, to wit:

we start again from (1), but now take $\partial/\partial x$:

$i_{xx} + Cv_{tx} + Gv_x = 0, \tag{14}$

whence

$i_{xx} = -Cv_{tx} - Gv_x; \tag{15}$

substituting $v_x$ from (2),

$i_{xx} = -Cv_{tx} + G(Li_t + Ri); \tag{16}$

we take $\partial / \partial t$ of (2):

$v_{xt} = - Li_{tt} - Ri_t, \tag{17}$

and insert $v_{xt} = v_{tx}$ from this into (16):

$i_{xx} = C(Li_{tt} + Ri_t) + G(Li_t + Ri); \tag{18}$

a little algebra transforms this to

$i_{xx} = CLi_{tt} + (CR + GL)i_t + Ri, \tag{19}$

and voila! the same equation as (13) with $v$ replaced by $i$!

Answer 2

Note that, $$LCu_{tt}+(RC+LG)u_{t}+RGu=LCu_{tt}+RCu_{t}+LGu_{t}+RGu$$

$$=L(Cu_{tt}+Gu_{t})+R(Cu_{t}+Gu).$$ By $(1)$, $Cv_{t}+Gv=-i_{x}$ and $Cv_{tt}+Gv_{t}=-i_{xt}$. Replacing $u$ by $v$, $$L(Cv_{tt}+Gv_{t})+R(Cv_{t}+Gv)=-Li_{xt}-Ri_{x}. $$ By $(2)$, $Li_{t}+Ri=-v_{x}\Rightarrow Li_{xt}+Ri_{x}=-v_{xx}$, we get the equation $(3)$.

Analogously we prove for $i$.