I was trying to solve the 1D Telegrapher's Equation, with the following initial conditions: $$ V(x, 0) =0 \\ \frac{\partial V}{\partial t}(x, 0) =0 \\ V(x = 0, t) = V_g(t) \\ \frac{\partial V}{\partial t}(x = L, t) = 0 $$ The functions $V(x, t)$ and $I(x, t)$ are the ones to be determined. The equation can be written as a dissipative wave equation or as a coupled differental system of equations: $$ \frac{\partial V}{\partial x} + L\frac{\partial I}{\partial t} = -IR \\ \; \\ \frac{\partial I}{\partial V} + C\frac{\partial V}{\partial t} = -GV \\ \; \\ $$
I was trying to solve the system using the characteristcs method, avoiding solving the full wave equation with Fourier Series because of the boundry conditions. Especially: $V(x = 0, t) = V_g(t)$. I thought that a good first try to simplify the system was to write everyting as: $$ \eta(x, t) = \alpha V(x,t) + i\beta I(x, t) $$ Where $i$ is the imaginary unit. I thought that with this method, the differential equation would become one of this similar form: $$ \gamma \frac{\partial \eta}{\partial x} - i\delta \frac{\partial \eta}{\partial t} = -i\lambda \eta $$ The coefficients: $\alpha, \beta, \gamma, \delta$ and $\lambda$ need to be determined to recover the original system, and would depend on the circuit constants $R, L, C$ and $G$. This equation is easily solvable with the method of characteristics, the problem is, there is no set of coeficients that satisfy the conditions at first glance.
Is there any way of solving the system and applying the initial conditions (avoiding Fourier)? If we cannot solve via method of characteristics the system, can we solve the full wave equation: $$ \frac{\partial^2 V}{\partial x^2} - LC\frac{\partial^2 V}{\partial t^2} = (RC + LG) \frac{\partial V}{\partial t} + R G V \; \\ \frac{\partial^2 I}{\partial x^2} - LC\frac{\partial^2 I}{\partial t^2} = (RC + LG) \frac{\partial I}{\partial t} + R G I $$ with this method?
