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I need help on solving two systems which are related to each other. Basically, first I have to solve a non-linear ODE system. Then, I have to solve another system of equations, I have to ask mathematica to get the values from first system, in order to solve the second system.

The first system is this one:

inf = 50;
s = ParametricNDSolve[{H'[η] == -2*F[η] - ((1 - n)/(1 + n))*η*F'[η], 
    F[η]^2 - (G[η] + 1)^2 + (H[η] + ((1 - n)/(1 + n))*η*F[η])*F'[η] == 
    (F'[η]^2 + G'[η]^2)^((n - 1)/2)*F''[η] + F'[η]*((n - 1)*(F'[η]^2 + G'[η]^2)^((n - 3)/2)
    *(F'[η]*F''[η] + G'[η]*G''[η])), 
    2*F[η]*(G[η] + 1) + (H[η] + ((1 - n)/(1 + n))*η*F[η])*G'[η] == 
    (F'[η]^2 + G'[η]^2)^((n - 1)/2)*G''[η] + G'[η]*((n - 1)*(F'[η]^2 + G'[η]^2)^((n - 3)/2)
    *(F'[η]*F''[η] + G'[η]*G''[η])), 
    F[0] == 0, G[0] == 0, H[0] == 0, 
    F'[inf] == H[inf] F[inf], G'[inf] == H[inf] (G[inf] + 1)}, 
    {F, G, H, F', G'}, {η, 0, inf}, {n, Fp0, Gp0}, 
    Method -> {"Shooting", "StartingInitialConditions" -> 
    {F[0] == 0, G[0] == 0, H[0] == 0, F'[0] == Fp0, G'[0] == Gp0}}];

The value of F[η], G[η] and H[η] that I get from the first system will be used to solve the second system.

I believe there is a lot of mistakes i've done to solve the second question. Hence, I need help. thank you :)

This is the second system:

inf = 20; 
    alphabar=1;
    betabar=1;
    omegabar=0;
    R=1;
    sol =First@ NDSolve[{y1'[\[Eta]] == y2[\[Eta]] , 
y2'[\[Eta]]==(1/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(H[\[Eta]]-((n - 1)*(F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 3)/2)*(F'[\[Eta]]*F''[\[Eta]] + 
                G'[\[Eta]]*G''[\[Eta]])))*y2[\[Eta]]+(1/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(\[ImaginaryI]*R*(alphabar*F[\[Eta]]+betabar*G[\[Eta]]-omegabar)+F[\[Eta]])*y1[\[Eta]]-(2/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(G[\[Eta]]+1)*y5[\[Eta]]+(R/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*F'[\[Eta]]*y3[\[Eta]]+\[ImaginaryI]*(R/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*alphabar*y4[\[Eta]],
y3'[\[Eta]]==-(\[ImaginaryI]*alphabar*(1/R))*y1[\[Eta]]-\[ImaginaryI]*betabar*y5[\[Eta]],
y4'[\[Eta]]==-((\[ImaginaryI]*R*(alphabar*F[\[Eta]]+betabar*G[\[Eta]]-omegabar)+H'[\[Eta]])/R)*y3[\[Eta]]-(H[\[Eta]]/R)*(-\[ImaginaryI]((alphabar*(1/R))*y1[\[Eta]]-\[ImaginaryI]*betabar*y5[\[Eta]]))-(1/R)*(((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2))*y3''[\[Eta]]+y3'[\[Eta]]*(n - 1)((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 3)/2))(F'[\[Eta]]* F''[\[Eta]]+ G'[\[Eta]]*G''[\[Eta]])),
y5'[\[Eta]]==y6[\[Eta]],
y6'[\[Eta]]==(1/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(H[\[Eta]]-((n - 1)*(F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 3)/2)*(F'[\[Eta]]*F''[\[Eta]] + 
                G'[\[Eta]]*G''[\[Eta]])))*y6[\[Eta]]+(1/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(\[ImaginaryI]*R*(alphabar*F[\[Eta]]+betabar*G[\[Eta]]-omegabar)+F[\[Eta]])*y5[\[Eta]]-(2/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*(G[\[Eta]]+1)*y1[\[Eta]]+(R/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*F'[\[Eta]]*y3[\[Eta]]+\[ImaginaryI]*(R/((F'[\[Eta]]^2 + G'[\[Eta]]^2)^((n - 1)/2)))*betabar*y4[\[Eta]],  y1[0] == 0, y2[0] == 0, y3[0] == 0, y4[0] == 0, y5[0] == 0, y6[0] == 0,y1[inf] == 0, y2[inf] == 0, y3[inf] == 0,y4[inf] == 0, y5[inf] == 0, y6[inf] == 0} /. n -> 1.0,
        {y1,y2,y3,y4,y5,y6}, {\[Eta], 0.00001, inf}];

This is the error:

 NDSolve
:Derivative order  1.  in term  y1
(1.)
[η]  should be a non-negative machine-sized integer.

Thank you so much if anyone can help me on this.

Attached is the system:

I replaced η with y, U with F, V with G and W with H.

enter image description here

enter image description here

enter image description here

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  • $\begingroup$ The first part of this question is addressed here. $\endgroup$ – bbgodfrey Feb 3 '17 at 9:27
  • $\begingroup$ Because all but the third ODE in the second block of code are first order, there should be only seven boundary conditions, not twelve. Please specify which you wish to use. Also, the fourth ODE contains y3''[η] as well as y4'[η]. Is this correct? $\endgroup$ – bbgodfrey Feb 3 '17 at 10:23
  • $\begingroup$ Also, since these equations are linear and homogeneous in y the solution is zero. $\endgroup$ – bbgodfrey Feb 3 '17 at 11:23
  • $\begingroup$ Why don't you try to solve the two systems as a one system? $\endgroup$ – zhk Feb 3 '17 at 12:50
  • $\begingroup$ What is i in the 2nd system? $\endgroup$ – zhk Feb 3 '17 at 14:26
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Here I will illustrate with a simple example to solve two couple systems of odes.

Method I

(*The first system*)
Ode1 = x'[t] == 1/y[t];    
Ode2 = y'[t] == x[t];    
ics := {x[0] == 1, y[0] == 2};
sys = Join[{Ode1, Ode2}, ics];    
soln = First@NDSolve[sys, {x, y}, {t, 0, 10}];    
x1[t_] := Evaluate[x[t] /. soln]; (*assign the interpolating function x[t] to x1[t]*)    
y1[t_] := Evaluate[y[t] /. soln]; (*assign the interpolating function y[t] to y1[t]*)    
(*The second system*) 
Ode3 = x2'[t] == x1[t];
Ode4 = y2'[t] == x2[t] + y1[t] + y2[t];
ics2 := {x2[0] == -1, y2[0] == -2};
sys1 = Join[{Ode3, Ode4}, ics2];
soln1 = First@NDSolve[sys1, {x2, y2}, {t, 0, 10}];
Plot[{x1[t], y1[t], {x2[t], y2[t]} /. soln1}, {t, 0, 10}]

enter image description here

Method II

Solve the two systems as one,

Ode3 = x2'[t] == x[t];    
Ode4 = y2'[t] == x2[t] + y[t] + y2[t];    
combsys12 = Join[{Ode1, Ode2, Ode3, Ode4}, ics, ics2];    
combsoln = First@NDSolve[combsys12, {x, y, x2, y2}, {t, 0, 10}];    
Plot[{x[t], y[t], x2[t], y2[t]} /. combsoln, {t, 0, 10}]

enter image description here

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