# sol2 = bvp4c(@bvpexam2, @bcexam2, sol);

5 views (last 30 days)
Yasir on 1 Jul 2024
Edited: Walter Roberson on 1 Jul 2024
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

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

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