Bvp4c Error; cannot solve collocation equations- singular Jacobian
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The eqns to be solved: f'''+ff''+1-f'^2-A(f'-1+(n/2)f'')+M(1-f')=0 (1/Pr)Y''+fY'-2f'Y-A(2Y+(n/2)Y')+Ec(M(1-f')^2+f''^2)=0 w/BC's: f(0)=0,f'(0)=lambda,Y(0)=1,f'(inf)=>1,Y(inf)=>0 it is known that there are two solutions where one can be found at infinity=10 and will add infinity=40 when I can fix error. Later will also be running for different values of lambda. I have checked and rechecked equation, I have also played with the initial guesses. I would appreciate any guidance.
global Pr A M Ec lambda
Pr=1; A=0.1; M=0.05; Ec=1; lambda=-1.33;
%
rlow=0;
rhigh=10;
N=50;
options=bvpset('stats','on','RelTol',1e-9);
solinit=bvpinit(linspace(rlow,rhigh,N),[0, -1, 0, 1 ,0]);
%
sol=bvp4c(@projode,@mybcs,solinit,options);
function dy=projode(n,y)
global Pr A M Ec
dy=[y(2); y(3); A*y(2)+A*n*y(3)/2+M*y(2)+y(2)^2-y(1)*y(3)-1-A-M; y(4);...
Pr*(2*y(2)*y(4)+A*2*y(4)+A*n*y(5)/2-y(1)*y(5)-Ec*M*(1-y(2))^2-Ec*y(3)^2)];
function dy=projode(n,y)
global Pr A M Ec
dy=[y(2); y(3); A*y(2)+A*n*y(3)/2+M*y(2)+y(2)^2-y(1)*y(3)-1-A-M; y(4);...
Pr*(2*y(2)*y(4)+A*2*y(4)+A*n*y(5)/2-y(1)*y(5)-Ec*M*(1-y(2))^2-Ec*y(3)^2)];
function res=mybcs(ya,yb)
global lambda
res=[ya(1) ya(2)-lambda ya(4)-1 yb(2)-1 yb(4)];
采纳的回答
Torsten
2017-11-13
Replace
function dy=projode(n,y)
global Pr A M Ec
dy=[y(2); y(3); A*y(2)+A*n*y(3)/2+M*y(2)+y(2)^2-y(1)*y(3)-1-A-M; y(4);...
Pr*(2*y(2)*y(4)+A*2*y(4)+A*n*y(5)/2-y(1)*y(5)-Ec*M*(1-y(2))^2-Ec*y(3)^2)];
by
function dy=projode(n,y)
global Pr A M Ec
dy=[y(2); y(3); A*y(2)+A*n*y(3)/2+M*y(2)+y(2)^2-y(1)*y(3)-1-A-M; y(5);...
Pr*(2*y(2)*y(4)+A*2*y(4)+A*n*y(5)/2-y(1)*y(5)-Ec*M*(1-y(2))^2-Ec*y(3)^2)];
Best wishes
Torsten.
2 个评论
Shelly
2017-11-13
naygarp
2017-11-17
Hey Shelly I have a similar problem as yours
How do I modify your code in mine, please help
Here's my equations(coupled) need to use shooting method with RK-4
(1+2M*eta)f''' + 2M*f"+ f*f" -f'^2 - k1*f' + lambda*theta=0 ---------- (1)
(1+2M*eta)theta" + 2M*theta' + Pr(f*theta'-f'*theta)=0 -------(2)
'f' and 'theta' are functions of 'eta', eta is an independent variable
3 initial conditions are given: eta=0, f(0)=0, f'(0)=1,theta(0)=1
Say I reduce these equations (1) and (2) to five ode (shooting method)
f'=z ; f(0)=0 -----(3)
z'=p ; z(0)=1 ------(4)
p'= (-2M*f"-f*f"+f'^2+k1*f'-lamda*theta)/(1+2*M*eta) ;p(0)= (guess value) -----(5)
theta'= q ; theta(0)=1 -----(6)
q' = (-2M*theta'-Pr(f*theta'-f'*theta))/(1+2*M*eta) ; q(0)= (guess value) ------(7)
The boundary conditions that needs to be satisfied are: f'(eta=10)=> 0 and theta(eta=10)=>0 as eta=>10
Given:
M= 1
k1= 0.1
lamda= 0.1
Pr= 0.7
taking step length: h= 0.01
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