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Solve the following fourth order ODE?

Posted 5 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}]];

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*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

For what purpose do you want a solution to this problem? Even in the simplest case, the solutions are very cumbersome. In my opinion, a numerical solution is preferable. I will give a working code that performs symbolic calculations for the numerical values of parameters

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]}];

Subscript[y, 
  p] = .001; \[Epsilon] = .001; Da = 1; Bi = 1; k = 1; Um = 1;

 s = DSolve[{D[X[Y], {Y, 4}] - (Bi*(1 + k))/k X''[Y] - 
           1/(k*Um) (A - Bi*Subscript[U, p]) == 0, 
       D[Z[Y], {Y, 4}] - (Bi*(1 + k))/k*Z''[Y] + 
      Bi/k*Subscript[U, p]/Um == 0, 
       Z''[Subscript[y, p]] == 0, 
       X[Subscript[y, p]] == XP, Z[Subscript[y, p]] == XP, 
       F == -k*X'[Subscript[y, p]] - Z'[Subscript[y, p]], X[0] == X0, 
    Z[0] == X0, 
        -k*X'[0] - Z'[0] == 1, 
    Bi*(Z'[0] + X'[0]) == D[Z[Y], {Y, 3}] /. Y -> 0}, {X[Y], Z[Y]}, Y];

x = X[Y] /. s; z = Z[Y] /. s;
x2p = D[x, Y, Y] /. Y -> Subscript[y, p];
x20 = D[x, Y, Y] /. Y -> 0;
x30 = D[x, {Y, 3}] /. Y -> 0;
x10 = D[x, Y] /. Y -> 0;

z10 = D[z, Y] /. Y -> 0;
x0 = First[X0 /. Solve[k*x2p == G/Um, X0]];

xp = First[XP /. Solve[k*x30 + Bi*(z10 - x10) == B/Um, XP]];

xpf = xp /. X0 -> x0;
xp0 = First[XP /. NSolve[XP == xpf, XP]];

x00 = x0 /. XP -> xp0;

This is the result of calculating dependencies XN=X[Y], ZN=Z[Y] with numerical coefficients.

XN = First[x /. {X0 -> x00, XP -> xp0} // FullSimplify]


Out[]= 2.11134*10^8 + (-124770. + 0.25 Y) Y - 
 499.75 Cosh[0.0316228 Y] + 361.302 Cosh[1.41421 Y] + 
 3.94494*10^6 Sinh[0.0316228 Y] + 120.434 Sinh[1.41421 Y]

 ZN = First[z /. {X0 -> x00, XP -> xp0} // FullSimplify]

Out[]= 2.11134*10^8 + (-124812. + 0.25 Y) Y - 
 500.25 Cosh[0.0316228 Y] + 0.000625083 Cosh[1.41421 Y] + 
 3.94889*10^6 Sinh[0.0316228 Y] - 44.5035 Sinh[1.41421 Y]

XN0 = XN /. Y -> 0; ZN0 = ZN /. Y -> 0;
{Plot[XN - XN0, {Y, 0, 1}, AxesLabel -> {"Y", "X-X(0)"}], 
 Plot[ZN - ZN0, {Y, 0, 1}, AxesLabel -> {"Y", "Z-Z(0)"}]}

fig1

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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