I am trying to solve a wave equation (2nd order PDE) in z and t with absorbing boundary conditions, i.e. a boundary condition that relates the partial z and t derivatives. For this particular example, I believe the answer should be zero everywhere, but I'm interested ultimately in other, non-trivial, cases.

Here's my attempt:

  D[V[z, t], {z, 2}] == D[V[z, t], {t, 2}],
  DirichletCondition[V[z, t] == 0, t == 0],
  Derivative[0, 1][V][z, t] == NeumannValue[0, t == 0],
  Derivative[1, 0][V][0, t] == Derivative[0, 1][V][0, t],
  Derivative[1, 0][V][1, t] == -Derivative[0, 1][V][1, t]
 {t, 0, 1}, {z, 0, 1}

which yields the error:

NDSolveValue::overdet: There are fewer dependent variables, {V[z,t]}, than equations, so the system is overdetermined.

I would appreciate suggestions for better ways to resolve this problem, or in general to approach the problem. (I realize the problem is solvable analytically--I'm trying it this way as a step along the way to a harder problem that will not be solvable analytically.)

The application here is modeling an electrical transmission line, and I've tried a few approaches so far, but haven't come across one that works.

I am using Mathematica 11.3.

  • 1
    $\begingroup$ 1. Where are the initial conditions? 2. Are you trying to implement ABC? If so, have you read this post?: mathematica.stackexchange.com/q/128516/1871 $\endgroup$
    – xzczd
    Apr 13, 2019 at 2:54
  • $\begingroup$ The initial conditions disappeared when creating the MWE... whoops. But there is still a problem: when I add an initial condition on the time derivative, it complains that the problem is overspecified. I know I can get that aspect to work when I'm not using ABCs (yes, that's what I'm attempting here), hence the lack of time derivative in the initial conditions currently. I'll update the question to reflect this issue. $\endgroup$ Apr 13, 2019 at 10:12
  • 3
    $\begingroup$ How could V be a function of z and t and then you write D[V[z, t], {x, 2}]? would not this be zero? $\endgroup$
    – Nasser
    Apr 13, 2019 at 10:21
  • 1
    $\begingroup$ The first part of your pde D[V[z, t], {z, 2}] == D[V[z, t], {z, 2}] yields True, Usually I would expect the pde at this place. Usually the NeumannValue is part of the main pde and the DirichletCondition is separated. For me it would be easier to help if you provide the complete pde problem. $\endgroup$ Apr 13, 2019 at 14:21
  • 1
    $\begingroup$ Then I think it's proper to mark this question as a duplicate of that post. $\endgroup$
    – xzczd
    Apr 13, 2019 at 16:01