boundary conditions at infinity

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mallela ankamma rao
评论： mallela ankamma rao 2024-3-29
Good afternoon one and all. Good afternoon torsten sir
Sir my code is working by taking the boundary conditions as x = linspace(0,1,100) but doesnot working at x = linspace(0,Inf,100);
I request you sir please help me how to write the code for 3 sets of boundary conditions at η ≥0 , η =0 and η→∞.
I attached my equations image and my code error below.
My code :
x = linspace(0,Inf,100);
yinit = [0 1 0 1 0 1 ];
init = bvpinit(x,yinit);
options = bvpset('RelTol',1e-8,'AbsTol',1e-12);
sol = bvp4c(@odefun, @bcfun, init,options);
plot(sol.x, sol.y)
hold on
function dydx = odefun(x,y) % equations being solved
x0=1;x1=1;x2=1;x3=1;x4=1;x5=1;x6=1;s=1;Da=1;Gr=1;Gc=1;prs=1;Ra=1;Qr=1;Sc=1; Kr=1; E=1;n=1;s=1;
us = y(1);
usy = y(2);
theta = y(3);
thetay = y(4);
phi = y(5);
phiy = y(6);
dydx = zeros(6,1);
dydx(1) = usy;
dydx(2) = ((1/x0)*(x1*s*us)+(x0/Da)*us-x2*Gr*theta-x3*Gc*phi-(E/s)*(exp(-s*n)));
dydx(3) = thetay;
dydx(4) = ((1/x5)*(x4*prs*theta)+Ra*theta-Qr*theta);
dydx(5) = phiy;
dydx(6) = (Sc/x6)*(s*phi+Kr*phi);
end
function res = bcfun(yL,yR)
s=1;k=1;
res = zeros(6,1);
res(1) = yL(1)-(1/(2*1i))*(1/(s-1i*k)-1/(s+1i*k));
res(2) = yL(3)-(1/s^2)*(1-exp(-s));
res(3) = yL(5)-(1/s^2)*(1-exp(-s));
res(4)= yR(1);
res(5)= yR(3);
res(6)= yR(5);
end
ERROR:
Error using bvp4c (line 248)
Unable to solve the collocation equations -- a singular Jacobian encountered.
Error in bvpmodel (line 12)
sol = bvp4c(@odefun, @bcfun, init,options);

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回答（1 个）

Torsten
Torsten 2024-3-26
编辑：Torsten 2024-3-26
Ran in:
You must use a sufficiently large number instead of "Inf" here:
x = linspace(0,Inf,100)
x = 1x100
0 Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf
  7 个评论
显示 5更早的评论
Torsten
Torsten 2024-3-28
I suggest solving equations (8), (9) and (10) directly using "pdepe".
mallela ankamma rao
mallela ankamma rao 2024-3-29
Thank you sir. i will try in pdepe method

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