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I'm having the following issue that can not find a solution , by using this boundary condition v[0, t] == u[1, t] that Mathematica does not understand, someone please help me! 

NDSolve[{
        D[u[x, t], t] == D[u[x, t], x, x] - u[x, t],
        D[v[x, t], t] == D[v[x, t], x, x] - v[x, t],
        u[x, 0] == 1,
        v[x, 0] == 1,
        u[0, t] == 1,
        Derivative[1, 0][u][1, t] == 0,
        v[0, t] == u[1, t],
        Derivative[1, 0][v][1, t] == 0
        },
      {u[x, t], v[x, t]},
      {x, 0, 1},
      {t, 0, 5}]

I appreciate any help...

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  • $\begingroup$ v[0, t] and u[1, t] could be anything compatible with the boundary conditions (for example 1 - t^n for any n). How could NDSolve[ ] guess what you want? $\endgroup$
    – Dr. belisarius
    Commented Aug 16, 2015 at 6:28
  • $\begingroup$ Can you give us an idea as to what these two equations are trying to solve? Are they meant to be coupled somehow? The significance of the equation itself may allow for clues to help you. $\endgroup$
    – dearN
    Commented Aug 16, 2015 at 11:30
  • $\begingroup$ Yes , they are to be coupled equations for the boundary conditions. Describe the transfer of heat between two different material plates. Thank you, any help. $\endgroup$
    – Jeiveison Gobério Maia
    Commented Aug 16, 2015 at 16:28

1 Answer 1

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NDSolve is complaining, because v[0, t] == u[1, t] appears to link boundary conditions at x == 0 and x == 1. This can be circumvented by redefining x as 1 - x for v, so that the code becomes,

{su, sv} = NDSolveValue[{
        D[u[x, t], t] == D[u[x, t], x, x] - u[x, t],
        D[v[x, t], t] == D[v[x, t], x, x] - v[x, t],
        u[x, 0] == 1, v[x, 0] == 1, u[0, t] == 1,
        Derivative[1, 0][u][1, t] == 0,
        v[1, t] == u[1, t],
        Derivative[1, 0][v][0, t] == 0},
        {u, v}, {x, 0, 1}, {t, 0, 5}];

The solutions for u[x, t] and v[1-x, t] can be spliced together and plotted by

Plot3D[Piecewise[{{su[x, t], x <= 1}, {sv[2 - x, t], x > 1}}], {x, 0, 2}, 
    {t, 0, 5}, PlotRange -> {All, All, {0, 1}}, AxesLabel -> {x, t, u : v}]

enter image description here

Note that, although the surface is continuous at x == 1, its derivative is not. Possibly, the boundary conditions there, as given in the Question, should be revised.

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  • $\begingroup$ Thank you, but this is not an acceptable solution because the boundary condition for v at x = 1 has been modified. What precisely it is actually a way to write properly in Mathematica a jump condition. I am most grateful for the help so far and future! $\endgroup$
    – Jeiveison Gobério Maia
    Commented Aug 17, 2015 at 5:19
  • $\begingroup$ @JeiveisonGobérioMaia Actually, I did not change the physical boundary condition for v at x == 1. Rather, I changed the definition of the first independent variable from x to 1 - x for v, did the computation, and changed it back again in the Plot3D. This redefinition has no impact of the differential equation itself, because it is second order in x, and its only effect on the boundary conditions is to change v[0, t] to v[1, t] and Derivative[1, 0][v][1, t] == 0 to Derivative[1, 0][v][0, t] == 0, which allows NDSolve to handle the calculation. $\endgroup$
    – bbgodfrey
    Commented Aug 17, 2015 at 14:22
  • $\begingroup$ @JeiveisonGobérioMaia Your problem also can be solved with the procedure discussed in Introduction to Method of Lines but a lot more work is involved. See 78493 for another problem involving a non-local boundary condition. $\endgroup$
    – bbgodfrey
    Commented Aug 17, 2015 at 14:35
  • $\begingroup$ Thank you.......................................... $\endgroup$
    – Jeiveison Gobério Maia
    Commented Aug 19, 2015 at 19:06

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