Avoid problem with boundary conditions in NDSolve[ ]?

Questions
, , , ,
1

I’m having the following issue that can not find a solution , by using this boundary condition

 u[0, t] == 1,
 v[0, t] == u[1, t],
 w[0, t] == v[1, t],
 z[0, t] == w[1, t]

that Mathematica does not understand, someone please help me!

s = NDSolve[
  {
   D[u[x, t], t] == -D[u[x, t], x] - 1 v[x, t] w[x, t] z[x, t],
   D[v[x, t], t] == -D[v[x, t], x] - 2 u[x, t] w[x, t] z[x, t],
   D[w[x, t], t] == -D[w[x, t], x] - 3 u[x, t] v[x, t] z[x, t],
   D[z[x, t], t] == -D[z[x, t], x] - 4 u[x, t] v[x, t] w[x, t],
   
   u[x, 0] == 1,
   v[x, 0] == 2,
   w[x, 0] == 3,
   z[x, 0] == 4,
   
   u[0, t] == 1,
   v[0, t] == u[1, t],
   w[0, t] == v[1, t],
   z[0, t] == w[1, t]
   },
  
  {u[x, t], v[x, t], w[x, t], z[x, t]},
  
  {x, 0, 1}, {t, 0, 1}]

I’m getting this error message :

NDSolve::bcedge: Boundary condition v[0,t]==u[1,t] is not specified on a single edge of the boundary of the computational domain. >>

I have already presented a similar problem here, but the reasoning suggested is not a good solution for my case.

What strategies do you know to treat this problem ? I appreciate any help…

2

I wonder:
line 1: From the BC v[0, t] == u[1, t] one concludes that v[0,0]==u[1,0]
line 2: From the BC u[x,0]==1 follows that u[1,0]==1 therefore from line 1 v[0,0]==1
line 3: However from the BC v[x,0]==2 follows that v[0,0]==2 which contradicts with line 2

I didn’t check further for other contradictions (this might be enough to generate problems)
are you sure that your BCs are correct?

yehuda