sol2 = bvp4c(@bvpexam2, @bcexam2, sol);
7 views (last 30 days)
Show older comments
how to get second solution from this code.I ploted first solution for different values of alpha.
function Ibrardual
clc
clear all
Nt=0.5; Nb=0.5; Le=2; Pr=1; alpha=-0.2138; s=1; A=1;
%% solution in structure form
%First solution
sol = bvpinit(linspace(0,2,10), [1 0 0 0 0 0 0]);
sol1 = bvp4c(@bvpexam2, @bcexam2, sol);
x1 = sol1.x;
y1=sol.y;
plot(x1,y1(3,:),'b');
hold on
result=(A)^-1/2*y1(3,1)
%% Here I define residual of boundary conditions
function res = bcexam2(y0, yinf)
res= [y0(1)-s; y0(2)-alpha; y0(4)-1; y0(6)-1;
yinf(2); yinf(4);yinf(6)];
end
%% First order ODEs are define here
function ysol = bvpexam2(x,y)
yy1 = -A*(y(1)*y(3)-(y(2))^2)-y(2);
yy2 = -Pr*(A*y(1)*y(5)+Nb*y(5)*y(7)+Nt*(y(5))^2);
yy3 = (-Le*A*(y(1)*y(7))-(Nt/Nb)*( -Pr*(A*y(1)*y(5)+Nb*y(5)*y(7)+Nt*(y(5))^2)));
ysol = [y(2); y(3); yy1;y(5);yy2;y(7);yy3];
end
end
0 Comments
Accepted Answer
Torsten
on 1 Jul 2024
Ibrardual()
function Ibrardual
Nt=0.5; Nb=0.5; Le=2; Pr=1; Alpha=[-0.2138;-0.4]; s=1; A=1;
%% solution in structure form
%First solution
hold on
for i = 1:numel(Alpha)
alpha = Alpha(i);
sol = bvpinit(linspace(0,2,10), [1 0 0 0 0 0 0]);
sol1 = bvp4c(@bvpexam2, @bcexam2, sol);
x1 = sol1.x;
y1 = sol1.y;
plot(x1,y1(3,:),'b');
end
hold off
%% Here I define residual of boundary conditions
function res = bcexam2(y0, yinf)
res= [y0(1)-s; y0(2)-alpha; y0(4)-1; y0(6)-1;
yinf(2); yinf(4);yinf(6)];
end
%% First order ODEs are define here
function ysol = bvpexam2(x,y)
yy1 = -A*(y(1)*y(3)-(y(2))^2)-y(2);
yy2 = -Pr*(A*y(1)*y(5)+Nb*y(5)*y(7)+Nt*(y(5))^2);
yy3 = (-Le*A*(y(1)*y(7))-(Nt/Nb)*( -Pr*(A*y(1)*y(5)+Nb*y(5)*y(7)+Nt*(y(5))^2)));
ysol = [y(2); y(3); yy1;y(5);yy2;y(7);yy3];
end
end
0 Comments
More Answers (0)
See Also
Categories
Find more on Calculus in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!