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Solve the following fourth order ODE?

Posted 6 years ago

I have two fourth order ODE's with eight boundary conditions. I try to solve it but did not get the result. Please anyone can check the code? there is any mistake in it or Mathematica can't solve it. Thanks in advance

Subscript[U, 
  p] = (Sqrt[Da]
      E^(-((Y Sqrt[\[Epsilon]])/Sqrt[
      Da])) (-1 + E^((Y Sqrt[\[Epsilon]])/Sqrt[
       Da])) (E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
        Da]) (1 + E^((Y Sqrt[\[Epsilon]])/Sqrt[Da])) - 
       2 Da (E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) - E^((
          Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) (-1 + E^((
          Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) - 
       2 Sqrt[Da] (E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) - E^((
          2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) \[Epsilon]^(
        3/2) + Subscript[y, 
        p] (-2 E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) + 
          2 Sqrt[Da] E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) \[Epsilon]^(
           3/2) - 2 Sqrt[Da] E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/
           Sqrt[Da]) \[Epsilon]^(3/2) + 
          E^((Sqrt[\[Epsilon]] (Y + Subscript[y, p]))/Sqrt[
           Da]) (-2 + Subscript[y, p]) + 
          E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) Subscript[y,
            p])))/(2 (-Sqrt[Da] + \[Epsilon]^(
       3/2) - \[Epsilon]^(3/2) Subscript[y, p] + 
       E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
        Da]) (Sqrt[Da] + \[Epsilon]^(
          3/2) - \[Epsilon]^(3/2) Subscript[y, p])));

Subscript[U, 
  c] = ((-1 + Y) (Sqrt[Da] + Sqrt[Da] Y - 2 Da \[Epsilon]^(3/2) + 
     4 Da E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) \[Epsilon]^(
      3/2) - Y \[Epsilon]^(3/2) - 
     2 Sqrt[Da] Subscript[y, p] + \[Epsilon]^(3/2) Subscript[y, p] + 
     Y \[Epsilon]^(3/2) Subscript[y, p] - \[Epsilon]^(3/2) 
\!\(\*SubsuperscriptBox[\(y\), \(p\), \(2\)]\) - 
     E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
      Da]) (Sqrt[Da] (1 + Y) + (2 Da + Y) \[Epsilon]^(
         3/2) - (2 Sqrt[Da] + (1 + Y) \[Epsilon]^(3/2)) Subscript[y, 
         p] + \[Epsilon]^(3/2) 
\!\(\*SubsuperscriptBox[\(y\), \(p\), \(2\)]\))))/(
  2 (-Sqrt[Da] + \[Epsilon]^(3/2) - \[Epsilon]^(3/2) Subscript[y, p] +
      E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
      Da]) (Sqrt[Da] + \[Epsilon]^(
        3/2) - \[Epsilon]^(3/2) Subscript[y, p])));

A = FullSimplify[D[Subscript[U, p], {Y, 2}]];

B = FullSimplify[D[Subscript[U, p], {Y, 1}] /. {Y -> 0}];

F = FullSimplify[
   Integrate[Subscript[U, c]/Um, {Y, Subscript[y, p], 1}]];

Fmplify[
 DSolve[{X''''[Y] - (Bi*(1 + k))/k X''[Y] - 
     1/(k*Um) (A - Bi*Subscript[U, p]) == 0, 
   Z''''[Y] - (Bi*(1 + k))/k*Z''[Y] + Bi/k*Subscript[U, p]/Um == 0, 
   X''[Subscript[y, p]] == 1/k*Subscript[U, p]/Um, 
   Z''[Subscript[y, p]] == 0, Z'''[0] - Bi*(Z'[0] - X'[0]) == 0, 
   k*X'''[0] + Bi*(Z'[0] - X'[0]) == B*1/Um, 
   X[Subscript[y, p]] == Z[Subscript[y, p]], 
   F == -k*X'[Subscript[y, p]] - Z'[Subscript[y, p]], X[0] == Z[0], 
   1 == -k*X'[0] - Z'[0] }, {X[Y], Z[Y]}, Y]]
3 Replies

Thanks for your response. I found one mistake in my BC, that Up should be in terms of yp in first BC. So, I corrected it but still don't get the result. Kindly check the corrected code.

Subscript[U, 
  p] = (Sqrt[Da]
      E^(-((Y Sqrt[\[Epsilon]])/Sqrt[
      Da])) (-1 + E^((Y Sqrt[\[Epsilon]])/Sqrt[
       Da])) (E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
        Da]) (1 + E^((Y Sqrt[\[Epsilon]])/Sqrt[Da])) - 
       2 Da (E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) - E^((
          Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) (-1 + E^((
          Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) - 
       2 Sqrt[Da] (E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) - E^((
          2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da])) \[Epsilon]^(
        3/2) + Subscript[y, 
        p] (-2 E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) + 
          2 Sqrt[Da] E^((Y Sqrt[\[Epsilon]])/Sqrt[Da]) \[Epsilon]^(
           3/2) - 2 Sqrt[Da] E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/
           Sqrt[Da]) \[Epsilon]^(3/2) + 
          E^((Sqrt[\[Epsilon]] (Y + Subscript[y, p]))/Sqrt[
           Da]) (-2 + Subscript[y, p]) + 
          E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) Subscript[y,
            p])))/(2 (-Sqrt[Da] + \[Epsilon]^(
       3/2) - \[Epsilon]^(3/2) Subscript[y, p] + 
       E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
        Da]) (Sqrt[Da] + \[Epsilon]^(
          3/2) - \[Epsilon]^(3/2) Subscript[y, p])));

Subscript[U, 
  c] = ((-1 + Y) (Sqrt[Da] + Sqrt[Da] Y - 2 Da \[Epsilon]^(3/2) + 
     4 Da E^((Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[Da]) \[Epsilon]^(
      3/2) - Y \[Epsilon]^(3/2) - 
     2 Sqrt[Da] Subscript[y, p] + \[Epsilon]^(3/2) Subscript[y, p] + 
     Y \[Epsilon]^(3/2) Subscript[y, p] - \[Epsilon]^(3/2) 
\!\(\*SubsuperscriptBox[\(y\), \(p\), \(2\)]\) - 
     E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
      Da]) (Sqrt[Da] (1 + Y) + (2 Da + Y) \[Epsilon]^(
         3/2) - (2 Sqrt[Da] + (1 + Y) \[Epsilon]^(3/2)) Subscript[y, 
         p] + \[Epsilon]^(3/2) 
\!\(\*SubsuperscriptBox[\(y\), \(p\), \(2\)]\))))/(
  2 (-Sqrt[Da] + \[Epsilon]^(3/2) - \[Epsilon]^(3/2) Subscript[y, p] +
      E^((2 Sqrt[\[Epsilon]] Subscript[y, p])/Sqrt[
      Da]) (Sqrt[Da] + \[Epsilon]^(
        3/2) - \[Epsilon]^(3/2) Subscript[y, p])));

A = FullSimplify[D[Subscript[U, p], {Y, 2}]];

B = FullSimplify[D[Subscript[U, p], {Y, 1}] /. {Y -> 0}];

F = FullSimplify[
   Integrate[Subscript[U, c]/Um, {Y, Subscript[y, p], 1}]];

G = FullSimplify[Subscript[U, p] /. {Y -> Subscript[y, p]}];

FullSimplify[
 DSolve[{X''''[Y] - (Bi*(1 + k))/k X''[Y] - 
     1/(k*Um) (A - Bi*Subscript[U, p]) == 0, 
   Z''''[Y] - (Bi*(1 + k))/k*Z''[Y] + Bi/k*Subscript[U, p]/Um == 0, 
   X''[Subscript[y, p]] == 1/k*G/Um, Z''[Subscript[y, p]] == 0, 
   Z'''[0] - Bi*(Z'[0] - X'[0]) == 0, 
   k*X'''[0] + Bi*(Z'[0] - X'[0]) == B*1/Um, 
   X[Subscript[y, p]] == Z[Subscript[y, p]], 
   F == -k*X'[Subscript[y, p]] - Z'[Subscript[y, p]], X[0] == Z[0], 
   1 == -k*X'[0] - Z'[0] }, {X[Y], Z[Y]}, Y]]

The boundary conditions are incorrect. In the case of an exact solution, we have a message

DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution.

In the case of a numerical solution, we have the message `NDSolve::bcnan: Boundary conditions not numerical

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