Error from NDSolve "Cannot find starting value for the variable"

Question

I need help regarding this code. I have used NDSolve to find a solution.

F = Θ + a*Sin[2*π*(x - t)] + 
   b*Sin[2*π*(x - t) + ϕ];
N1 = Sqrt[M^2 + (1/k)];
t = 0.4;
a = 0.2;
b = 0.3;
m = 0.15;
ϕ = π/3;
M = 1;
k = 3;
Nt = 0.8;
Nb = 1;
Pr = 2;
L = 0.1;
Bm = 1;
Bh = 0.5;
Θ = 1.6;
Gr = 1.2;
Br = 0.8;
ζ = 0.02;
Rn = 0.75;
h1 = -1 - m*x - a*Sin[2*π*(x - t) + ϕ]
h2 = 1 + m*x + b*Sin[2*π*(x - t)]
sol = NDSolve[{ψ''''[
     y] - ζ*(6 ψ''[y] *ψ'''[y]*ψ'''[y] + 
       3 ψ''[y]*ψ''[y]* ψ''''[y]) - N1*N1*ψ''[y] +
     Gr*θ'[y] + Br*σ'[y] == 
   0, (1 + Pr*Rn)*θ''[y] + Nb*Pr*σ'[y]*θ'[y] + 
    Nt*Pr*θ'[y]*θ'[y] == 
   0, σ''[y] + Nt/Nb*θ''[y] == 0, ψ[h2] == F/
   2, ψ[h1] == -(F/2), ψ'[h1] == 0, ψ'[h2] == 
   0, σ'[h1] == Bm*σ[h1], σ'[h2] == 
   Bm*(1 - σ[h2]), θ'[h1] == 
   Bh*θ[h1], θ'[h2] == 
   Bh*(1 - θ[h2])}, {ψ, θ, σ}, {y, h1, h2}]

I get the following error

NDSolve::ndsv: Cannot find starting value for the variable θ.

Why do I get this error for θ but not ψ or σ? Further i have used the same code for velocity graphs and they turned out really great without any errors

F = Θ + a*Sin[2*π*(x - t)] + 
   b*Sin[2*π*(x - t) + ϕ];
N1 = Sqrt[M^2 + (1/k)];
x = 0.4;
t = 0.2;
a = 0.3;
b = 0.4;
m = 0.25;
ϕ = 2 π/3;
M = 1;
k = 0.8;
Nt = 0.8;
Nb = 0.4;
Pr = 0.2;
Bm = 4;
Bh = 2;
Θ = 1.5;
Gr = 0.7;
Br = 0;
ζ = 0.002;
Rn = 0.6;
h1 = -1 - m*x - a*Sin[2*π*(x - t) + ϕ]
h2 = 1 + m*x + b*Sin[2*π*(x - t)]
sol = NDSolve[{ψ''''[
      y] - ζ*(6 ψ''[y] *ψ'''[y]*ψ'''[y] + 
        3 ψ''[y]*ψ''[y]* ψ''''[y]) - 
     N1*N1*ψ''[y] + Gr*θ'[y] + Br*σ'[y] == 
    0, (1 + Pr*Rn)*θ''[y] + Nb*Pr*σ'[y]*θ'[y] + 
     Nt*Pr*θ'[y]*θ'[y] == 
    0, σ''[y] + Nt/Nb*θ''[y] == 0, ψ[h2] == F/
    2, ψ[h1] == -(F/2), ψ'[h1] == 0, ψ'[h2] == 
    0, σ'[h1] == Bm*σ[h1], σ'[h2] == 
    Bm*(1 - σ[h2]), θ'[h1] == 
    Bh*θ[h1], θ'[h2] == 
    Bh*(1 - θ[h2])}, {ψ, θ, σ}, {y, h1, h2}];
A1 = Plot[Evaluate[D[ψ[y], y] /. sol], {y, h1, h2}, 
  PlotRange -> All, 
  PlotStyle -> {Darker[Blue, 0.5], Thickness[0.004]}, 
  AxesOrigin -> Automatic, 
  BaseStyle -> {FontFamily -> "Times", FontSize -> 15}, 
  FrameLabel -> {"y", "u"}, Frame -> True, Axes -> False]

My final goal is to make a ContourPlot of the form

ContourPlot[ψ[y], {y, -2, 2}, {x, 0, 2}, Contours -> 60, 
 ColorFunction -> (Hue[#] &), ClippingStyle -> Automatic]

Answer 1

Okay so your NDSolve won't run unless you have x defined, but you want to make a ContourPlot with x as one of the axes. So you need to make a Table, where for every value of x you rerun the NDSolve and then make a Table where you run over the y values.

But there is a hitch, since the domain on which your NDSolve is valid is different for every value of x - so your data won't be rectangular, nor will your contour plot.

sol[x_] := 
 Module[{F, t, N1, a, b, m, ϕ, M, k, Nt, Nb, Pr, L, Bm, 
   Bh, Θ, Gr, Br, Rn, h1, h2, sol, ζ},
  F = Θ + a*Sin[2*π*(x - t)] + 
    b*Sin[2*π*(x - t) + ϕ];
  N1 = Sqrt[M^2 + (1/k)];
  t = 0.4;
  a = 0.2;
  b = 0.3;
  m = 0.15;
  ϕ = π/3;
  M = 1;
  k = 3;
  Nt = 0.8;
  Nb = 1;
  Pr = 2;
  L = 0.1;
  Bm = 1;
  Bh = 0.5;
  Θ = 1.6;
  Gr = 1.2;
  Br = 0.8;
  ζ = 0.02;
  Rn = 0.75;
  h1 = -1 - m*x - a*Sin[2*π*(x - t) + ϕ];
  h2 = 1 + m*x + b*Sin[2*π*(x - t)];
  sol = NDSolve[{ψ''''[
        y] - ζ*(6 ψ''[y]*ψ'''[y]*ψ'''[y] + 
          3 ψ''[y]*ψ''[y]*ψ''''[y]) - 
       N1*N1*ψ''[y] + Gr*θ'[y] + Br*σ'[y] == 0,
     (1 + Pr*Rn)*θ''[y] + Nb*Pr*σ'[y]*θ'[y] + 
       Nt*Pr*θ'[y]*θ'[y] == 
      0, σ''[y] + Nt/Nb*θ''[y] == 0, ψ[h2] == 
      F/2, ψ[h1] == -(F/2), ψ'[h1] == 0,
     ψ'[h2] == 0,
     σ'[h1] == Bm*σ[h1],
     σ'[h2] == Bm*(1 - σ[h2]),
     θ'[h1] == Bh*θ[h1],
     θ'[h2] == Bh*(1 - θ[h2])}
    , {ψ, θ, σ}, {y, h1, h2}];
  {h1, h2, sol}
  ];

Generate the data

Monitor[
 data = Flatten[
    Table[
     {h1, h2, funcs} = sol[x];
     Table[
      {x, y, Last@(ψ[y] /. funcs)}
      , {y, h1, h2, .05}]

     , {x, 0, 2, .05}]
    , 1];
 , {x, y}]

and plot it

ListContourPlot[data, Contours -> 60, ColorFunction -> (Hue[#] &), 
 ClippingStyle -> Automatic]