Message Boards Message Boards

GROUPS:

Solve the following fourth order ODE?

Posted 1 month ago
237 Views
|
3 Replies
|
1 Total Likes
|

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

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

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]]

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

Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract