Why does my code for shooting method using 'ODE45' or 'ODE23s' does not converge to the boundary value.?
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    naygarp
 el 1 de Mayo de 2018
  
    
    
    
    
    Comentada: Torsten
      
      
 el 4 de Mayo de 2018
            I have used shooting method with ' ode45' or ' ode23s'.
But , the solution doesn't converge and it takes a lot of time.
The equations are
f"=g(g^2+gamma^2)/(g^2+lambda*gamma^2) ------ (1)
g'= (1/3)*f'^2-(2/3)*(f*f")+ Mn*f' ------------------------(2)
t"+Rd*t"+ 2*Pr*f*t'/3+ Nb*t'*p'+Nt*(t')^2= 0------(3)
p"+(2*Lew*f*p')/3+ Nt*t"/Nb= 0 ------------------------(4)
With the initial and boundary conditions
f(0)=0, f'(0)=1, t(0)=1, p(0)=1
f'(infinity)=0, t(infinity)=0, p(infinity)=0
The code for shooting method using ode45 is
function shooting_method
clc;clf;clear;
global lambda gama Pr Rd Lew Nb Nt Mn
gama=1;  
Mn=1;
Rd=0.1;
Pr=10;
Nb=0.2;
Lew=10;
Nt=0.2;
lambda=0.5;
x=[1 1 1];
options= optimset('Display','iter');
x1= fsolve(@solver,x);
end
function F=solver(x)
options= odeset('RelTol',1e-8,'AbsTol',[1e-8 1e-8 1e-8 1e-8 1e-8 1e-8 1e-8]);
[t,u]= ode45(@equation,[0 10],[0 1 x(1) 1 x(2) 1 x(3)],options)
s=length(t);
F= [u(s,2),u(s,4),u(s,6)];
%deval(0,u)
plot(t,u(:,2),t,u(:,4),t,u(:,6));
end
function dy=equation(t,y)
global lambda gama Pr Rd Lew Nb Nt Mn
dy= zeros(7,1);
dy(1)= y(2);
dy(2)= y(3)*(y(3)^2+gama^2)/(y(3)^2+lambda*gama^2);
dy(3)= y(2)^2/3-(2*y(1)*y(3)*(y(3)^2+gama^2))/(3*(y(3)^2+lambda*gama^2))+Mn*y(2);
dy(4)= y(5);
dy(5)= -(2*Pr*y(1)*y(5))/(3*(1+Rd)) - (Nb*y(5)*y(7))/(1+Rd) - (Nt*y(5)^2)/(1+Rd);
dy(6)= y(7);
dy(7)= -((2*Lew*y(1)*y(7))/3)+ (Nt/Nb)*((2*Pr*y(1)*y(5))/(3*(1+Rd)) + (Nb*y(5)*y(7))/(1+Rd) + (Nt*y(5)^2)/(1+Rd));
end
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Respuesta aceptada
  Torsten
      
      
 el 2 de Mayo de 2018
        Try to start with the solution you get from "bvp4c" for the vector x.
Best wishes
Torsten.
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Más respuestas (1)
  Jan
      
      
 el 2 de Mayo de 2018
        
      Editada: Jan
      
      
 el 2 de Mayo de 2018
  
      You want to get:
 f'(infinity)=0, t(infinity)=0, p(infinity)=0
but you integrate on the interval [0, 10]. 10 is not infinity. It is possible, that there is no possible start value, which reaches the wanted final point at the time 10.
Another problem can be the standard limitation of the single shooting methods: if a certain parameter causes a trajectory with Inf or NaN values, convergence is impossible. Then a multiple-shooting approach can help. Ask Wikipedia for details.
You use ode45 or ode23s? One is for non-stiff, the other for stiff systems. Using them by trial and error seems to be a very relaxed method of applied mathematics.
8 comentarios
  Torsten
      
      
 el 4 de Mayo de 2018
				function F=solver(x)
options= odeset('RelTol',1e-8,'AbsTol',[1e-8 1e-8 1e-8 1e-8 1e-8 1e-8 1e-8]);
[t,u] = ode15s(@equation,[0 4],[0 1 x(1) 1 x(2) 1 x(3)],options)
F= [u(end,2),u(end,4),u(end,6)];
y1 = u(1,:);  % should be equal to [0 1 x(1) 1 x(2) 1 x(3)]
plot(t,u(:,5),t,u(:,7));
end
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