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gabbbooooo

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1 years, 35 days
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Re-shooting method...

gabbbooooo 5
September 11 2023
0 0

@mmcdara 

good day, please the help for shooting method is still awaiting. Any bother with an idea how to implement it should please guid me.

Thank you for your help


 

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plots; ode1 := (1+1/bet)*(diff(x3(eta), eta))-x2(eta)^2+x1(eta)*x3(eta)+Gr*(y1(eta)+lambda1*y1(eta)^2)+Gc*(z1(eta)+lambda2*z1(eta)^2)-M*x2(eta) = 0

(1+1/bet)*(diff(x3(eta), eta))-x2(eta)^2+x1(eta)*x3(eta)+Gr*(y1(eta)+lambda1*y1(eta)^2)+Gc*(z1(eta)+lambda2*z1(eta)^2)-M*x2(eta) = 0

 (1)
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(diff(y2(eta), eta))*(1+(4*R*(1/3))*(y1(eta)+CT)^3)/Pr+4*R*(y1(eta)+CT)^2*y2(eta)^2+x1(eta)*y2(eta)+Nb*z2(eta)*y2(eta)+Nt*y2(eta)^2-de*(x1(eta)*x2(eta)*y2(eta)+x1(eta)^2*(diff(y2(eta), eta))) = 0

(diff(y2(eta), eta))*(1+(4/3)*R*(y1(eta)+CT)^3)/Pr+4*R*(y1(eta)+CT)^2*y2(eta)^2+x1(eta)*y2(eta)+Nb*z2(eta)*y2(eta)+Nt*y2(eta)^2-de*(x1(eta)*x2(eta)*y2(eta)+x1(eta)^2*(diff(y2(eta), eta))) = 0

 (2)
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ode3 := diff(z2(eta), eta)+Pr*Le*z2(eta)*x1(eta)+Nt*(diff(y2(eta), eta))/Nb-Pr*Le*dc*(x1(eta)*x2(eta)*z2(eta)+x1(eta)^2*(diff(z2(eta), eta))) = 0

diff(z2(eta), eta)+Pr*Le*z2(eta)*x1(eta)+Nt*(diff(y2(eta), eta))/Nb-Pr*Le*dc*(x1(eta)*x2(eta)*z2(eta)+x1(eta)^2*(diff(z2(eta), eta))) = 0

 (3)
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fixedparameter := R = .5, CT = .2, Gr = .1, Gc = .1, lambda1 = .2, lambda2 = .2, Pr = .72, Nb = .5, Le = 1, dc = .2, de = .2, Nt = .5, M = 1.5, bet = 1.5

R = .5, CT = .2, Gr = .1, Gc = .1, lambda1 = .2, lambda2 = .2, Pr = .72, Nb = .5, Le = 1, dc = .2, de = .2, Nt = .5, M = 1.5, bet = 1.5

 (4)
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bcs1 := x1(0) = 0, x2(0) = 1+xi*x3(0), x3(0) = alpha1, y2(0) = alpha2, z1(0) = 1+`ς`*z2(0), y1(0) = 1+Zeta*y2(0), z2(0) = alpha3

x1(0) = 0, x2(0) = 1+xi*x3(0), x3(0) = alpha1, y2(0) = alpha2, z1(0) = 1+varsigma*z2(0), y1(0) = 1+Zeta*y2(0), z2(0) = alpha3

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L := [0., .1, .25]

[0., .1, .25]

 (6)
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Download shhooting.

Re-shooting method...

gabbbooooo 5
September 10 2023
0 0

Good day all, 

base on the advice i recieved from some of the member of this group on the system of equations I pasted few day back , can some one please assist me or show me how to solve the equations using shooting method?. 

your assistant is highly appreciated.

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plots; ode1 := (1+1/bet)*(diff(x3(eta), eta))-x2(eta)^2+x1(eta)*x3(eta)+Gr*(y1(eta)+lambda1*y1(eta)^2)+Gc*(z1(eta)+lambda2*z1(eta)^2)-M*x2(eta) = 0

(1+1/bet)*(diff(x3(eta), eta))-x2(eta)^2+x1(eta)*x3(eta)+Gr*(y1(eta)+lambda1*y1(eta)^2)+Gc*(z1(eta)+lambda2*z1(eta)^2)-M*x2(eta) = 0

 (1)
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(diff(y2(eta), eta))*(1+(4*R*(1/3))*(y1(eta)+CT)^3)/Pr+4*R*(y1(eta)+CT)^2*y2(eta)^2+x1(eta)*y2(eta)+Nb*z2(eta)*y2(eta)+Nt*y2(eta)^2-de*(x1(eta)*x2(eta)*y2(eta)+x1(eta)^2*(diff(y2(eta), eta))) = 0

(diff(y2(eta), eta))*(1+(4/3)*R*(y1(eta)+CT)^3)/Pr+4*R*(y1(eta)+CT)^2*y2(eta)^2+x1(eta)*y2(eta)+Nb*z2(eta)*y2(eta)+Nt*y2(eta)^2-de*(x1(eta)*x2(eta)*y2(eta)+x1(eta)^2*(diff(y2(eta), eta))) = 0

 (2)
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ode3 := diff(z2(eta), eta)+Pr*Le*z2(eta)*x1(eta)+Nt*(diff(y2(eta), eta))/Nb-Pr*Le*dc*(x1(eta)*x2(eta)*z2(eta)+x1(eta)^2*(diff(z2(eta), eta))) = 0

diff(z2(eta), eta)+Pr*Le*z2(eta)*x1(eta)+Nt*(diff(y2(eta), eta))/Nb-Pr*Le*dc*(x1(eta)*x2(eta)*z2(eta)+x1(eta)^2*(diff(z2(eta), eta))) = 0

 (3)
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fixedparameter := R = .5, CT = .2, Gr = .1, Gc = .1, lambda1 = .2, lambda2 = .2, Pr = .72, Nb = .5, Le = 1, dc = .2, de = .2, Nt = .5, M = 1.5, bet = 1.5

R = .5, CT = .2, Gr = .1, Gc = .1, lambda1 = .2, lambda2 = .2, Pr = .72, Nb = .5, Le = 1, dc = .2, de = .2, Nt = .5, M = 1.5, bet = 1.5

 (4)
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bcs1 := x1(0) = 0, x2(0) = 1+xi*x3(0), x3(0) = alpha1, y2(0) = alpha2, z1(0) = 1+`ς`*z2(0), y1(0) = 1+Zeta*y2(0), z2(0) = alpha3

x1(0) = 0, x2(0) = 1+xi*x3(0), x3(0) = alpha1, y2(0) = alpha2, z1(0) = 1+varsigma*z2(0), y1(0) = 1+Zeta*y2(0), z2(0) = alpha3

 (5)
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L := [0., .1, .25]

[0., .1, .25]

 (6)
 

 

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Download shhooting.

Re-shooting method...

gabbbooooo 5
September 06 2023
0 0

@Carl Love 

Thank you for your detail explanations, let me try to solve the problem using shooting method. You and other member will help me check and correct it incase i did not succeed. Thank you in advance.

Re-inorrect boundary condition...

gabbbooooo 5
September 05 2023
0 0

@dharr

Thank you so much you are absolutely correct. I have corrected it though the solution is still not feasible  at some points.

Re-Solution up to xi=0.25...

gabbbooooo 5
September 05 2023
0 0

@mmcdara 

thank you so much once again, please dont be offeended, i say the "1" is replaced by "Lambda" in the new code and i cant see the value of lamda through out. what is the value of lambda please?

thank you in advance.

Re-critical value...

gabbbooooo 5
September 05 2023
0 0

@mmcdara 

Thank you so much for your prompt response, I will read through to understand what you did, since i am a new user. The author I am trying to validate my work with used shooting method to reduce the ODE to IVP. May be i should be thinking on how to implement the shooting method in maple

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