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I have a wave equation for displacement and velocity, I want to add this boundary condition $v(x=0,\,t>0)=1$

My mathematica code is

sol = 
  NDSolve[
    {1/1000 D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}], 
     D[u[x, t], t] == v[x, t], 
     u[x, 0] == 0, 
     v[x, 0] == 0, v[0, t > 0] == 1, v[L, t] == 0}, 
    {u[x, t], v[x, t]}, {x, 0, L}, {t, 0, 1}]

L is a constant.

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  • $\begingroup$ What is the value of L? Would it be acceptable to just reduce your condition to v[0, t] == 1? $\endgroup$
    – MarcoB
    Jan 20, 2021 at 22:11
  • $\begingroup$ Let try v[0, t] == If[t < 10^-4, 0, 1], but system looks like overdetermined. You don't need v as separate variable if you use second order equation for 'u'. $\endgroup$ Jan 20, 2021 at 22:17
  • $\begingroup$ Can you please tell me how to use NDSolve if v is not a separate variable? And how to plot v if I don't make v as separate variable? $\endgroup$ Jan 20, 2021 at 22:57

1 Answer 1

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Do I understand this correctly: A string at rest between 0 and L with zero displacement and zero velocity at t==0 and where you move the point at x==0 with constant velocity of 1.

Note, we simply define the velocity at t==0 for all x to be zero and the velocity of point x==1 for all times. There is some inconsistency here, MMA will give a warning, but will happily calculate anyway. We are integrating and a point singularity will have no effect. For an example we choose L==1:

L = 1;
sol[x_, t_] = 
  u[x, t] /. NDSolve[{1/1000 D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}],
      u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0, 
      Derivative[0, 1][u][0, t] == 1, u[L, t] == 0}, 
     u[x, t], {x, 0, L}, {t, 0, 1}][[1]];
Plot3D[sol[x, t], {x, 0, L}, {t, 0, 1}, AxesLabel -> {"x", "t", "z"}]

enter image description here

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