function res = mat4bc(ya,yb)
res = [ ya(1)-1
yb(2)
ya(3)
yb(4) ];
The above is for 2 boundary conditions. I need the coding for 3 conditions.
function sol=ex4
ex4init=bvpinit(linspace(0,100),[0 1 0 -1]);
sol=bvp4c(@ex4ode, @ex4bc, ex4init)
end
function f=ex4ode(x,y)
f=[y(2)
-x*x*y(2)
y(4)
-x*x*y(4)
];
end
function res=ex4bc(ya,yb,yc)
res=[ya(1)-3.462*10^(-22)
ya(3)-2.017
yb(1)-1
yb(3)
yc(2)+yc(4)
];
end
This is my program. when I run this program, I'm getting an error. Can someone tell me what is the error here?

6 Kommentare

Torsten
Torsten am 27 Jun. 2019
You can define boundary conditions for the solution variables at x=a (using ya) and at x=b (using yb). So I don't understand what yc is meant to represent in your code.
Maybe you could include the mathematical description of your problem.
Joy Salomi
Joy Salomi am 27 Jun. 2019
These are my boundary conditions.
At x=infinity, u=1, v=0
At x=0, u=3.462*10^-22 and v=2.017
At x=lambda, du/dx = - dv/dx
what is the matlab code for this?
Torsten
Torsten am 27 Jun. 2019
Your solutions for u and v are already determined by the four conditions at x=0 and x=infinity. You can't impose a fifth boundary condition.
Joy Salomi
Joy Salomi am 27 Jun. 2019
Okay sir. Thank you so much for your response. I'll try.
Torsten
Torsten am 27 Jun. 2019
Bearbeitet: Torsten am 27 Jun. 2019
Maybe of interest for you:
This code gives you the solution for your system:
function main
fun = @(t,c1,c2)c1*gamma(1/3)*gammainc(t.^3/3,1/3,'upper')/3^(2/3)+c2;
c2u = 1;
c1u = (3.462e-22-1)*3^(2/3)/gamma(1/3);
c2v = 0;
c1v = 2.017*3^(2/3)/gamma(1/3);
t = linspace(0,20,200);
u = fun(t,c1u,c2u);
v = fun(t,c1v,c2v)
plot(t,u,t,v)
end
Maybe your question was to determine lambda where du/dx = -dv/dx ?
Joy Salomi
Joy Salomi am 4 Mär. 2020
Thank you for your reply Sir.

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