snhaseela2

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9 years, 22 days

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These are replies submitted by snhaseela2

@tomleslie Yes, finally got the solution. i've made mistakes. Oh my, I'm so careless. Btw, Thanks to you and preben!

@Preben Alsholm 

It worked perfectly! I truly appreciate your replies, guidance and clear explanation. It was very helpful and understandable. At this moment, i'm implementing these steps for my current case study :) Thanks a lot, again!

 

@Preben Alsholm 

Thanks alot sir! I'll try my best to find the solution and i think i did it :)

But sir, based on ur statement ----> " But I think that you should try the method used by dsolve/numeric/bvp first as you also did and since that was successful (after help from Tom Leslie!) I would recommend staying with that. "


here is my curiosity : Do you mean that I don't need to use 'shooting method' to solve this type of problem? If the answer is yes, can i say that 'shooting method' is one of the optional method maybe used to figure out the solution? (hope you understand what i mean)

@Preben Alsholm 

Thanks so much for your details explanation sir! i'll read and study more on the odeplot.

Related to this problem (since i'm right now eager to learn about shooting method):

then, I try to solve this problem by using shooting method which is constructed from the following paper:

http://www.sciencedirect.com/science/article/pii/S1876107015003405

But I get stuckBefore that here is my worksheet:  


restart

Shootlib := "C:\\shooting/":

libname := Shootlib, libname:

with(Shoot):

with(plots):

lambda := 1.0

1.0

(1)

Pr := .70

.70

(2)

Sc := .7

.7

(3)

R := .1

.1

(4)

m := 2.0

2.0

(5)

g := .1

.1

(6)

Kr := .2

.2

(7)

M := 1

1

(8)

FNS := {A(eta), B(eta), C(eta), D(eta), E(eta), F(eta), G(eta), H(eta)}:

ODE := {diff(A(eta), eta) = B(eta), diff(B(eta), eta) = C(eta), diff(C(eta), eta) = D(eta), diff(D(eta), eta) = [C^2-(1+lambda)*(MB+B^2-AC)/beta]/A, diff(E(eta), eta) = F(eta), diff(F(eta), eta) = (3*Pr/(3+4*R))(M*B*E-g*E-A*F), diff(G(eta), eta) = H(eta), diff(H(eta), eta) = Sc*(m*B*G+Kr*G-A*H)};

{diff(A(eta), eta) = B(eta), diff(B(eta), eta) = C(eta), diff(C(eta), eta) = D(eta), diff(D(eta), eta) = [C^2-2.0*(MB+B^2-AC)/beta]/A, diff(E(eta), eta) = F(eta), diff(F(eta), eta) = .6176470587, diff(G(eta), eta) = H(eta), diff(H(eta), eta) = 1.40*B*G+.14*G-.7*A*H}

(9)

IC := {A(0) = 0, B(0) = 1, C(0) = 0, D(0) = S[2], E(0) = 1, F(0) = S[3], G(0) = 1, H(0) = S[4]};

{0 = S[2], A(0) = 0, B(0) = 1, C(0) = 0, E(0) = 1, F(0) = S[3], G(0) = 1, H(0) = S[4]}

(10)

L := 5;

5

(11)

BC := {B(L) = L, F(L) = 0, G(L) = 0};

{B(5) = 5, F(5) = 0, G(5) = 0}

(12)

infolevel[shoot] := 1:

S := shoot(ODE, IC, BC, FNS, [S[1] = 1, S[2] = 1, S[3] = 1, S[4] = 1])

Error, (in Shoot:-shoot) invalid initial conditions, must be given at one point

 

NULL



Download SatrayanaIVP_try.mwSatrayanaIVP_try.mw

 

The question is :

1) At my first attempt after some study about shoot - i thought i need to choose the 'right' starting values of the control parameter (S1,S2,S3 and S4). But the problem is, i don't understand how to pick up  that 'value' randomly. 

2) Really hope that someone will answer/teach me. Thanks a bunch!

 

 

@tomleslie 

Thank you tom for the reply and some fixes :)

May i ask you one thing - why do we need to use that 'maxmesh value'? And once the default value for maxmesh has been selected (as you did before), how do we know when the 'default maxmesh value' is right?

 

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