# How can draw this kind of graph?

4 views (last 30 days)
gkb on 13 Nov 2023
Edited: DGM on 20 Nov 2023
how can we draw a velocity profile of the 3rd-order non-linear boundary value problem that converges for different values? give me code for u'''+u*u''+s*[Gr*G+Gm*H-M*u']=0 where s=0.02, Gm=10, Gr=4 and M=0.1 with boundary condition u=1,u'=0.001; G=0.001; H=0.001 when t=0 and u'=0; G-0; H=0 when t= infinity, where H=2 and G=4 within one graph for Different values of Gm
Torsten on 13 Nov 2023
Use bvp4c or bvp5c to set up the problem.
Sam Chak on 13 Nov 2023
@gkb, I'm confused.
You only introduced ONE differential equation in the question.
where the parameters other than the state variables u, , , are constants.
Why do F, G, and H magically appear in the context? Can we settle the first and then post another question for the system of differential equations F, G, H?

Syed Sohaib Zafar on 19 Nov 2023
Edited: DGM on 20 Nov 2023
code_gkb_Matlab
%%%%%%%%%%%%%%%
function code_gkb_Matlab
global eps Gr Gm M Pr Jh Sc S0
%eq1
eps=0.001; Gr=0.5; M=0.5;
% eq2
Pr=1.4; Jh=0.3; %eps M
% eq3
Sc=0.7; S0=0.2;
val=[0.1 0.2 0.3 0.4];
for i = 1:1:4;
Gm = val(i);
solinit = bvpinit(linspace(0,6),[0 1 0 1 0 1 0]);
sol= bvp4c(@shootode,@shootbc,solinit);
eta = sol.x;
f = sol.y;
figure (1)
plot(eta,f(2,:));
xlabel( '\eta');
ylabel('\bf f'' (\eta)');
hold on
end
lgd=legend('Gm = 0.1','Gm = 0.2','Gm = 0.3','Gm = 0.4');
end
function dydx = shootode(eta,f);
global eps Gr Gm M Pr Jh Sc S0
dydx = [f(2)
f(3)
-f(1)*f(3)-eps*(Gr*f(4)+Gm*f(6)-M*f(2))
f(5)
-Pr*f(1)*f(5)+Jh*Pr*((1/eps)*f(3)^2+M*f(2)^2)
f(7)
2*Sc*f(2)*f(6)-Sc*f(1)*f(7)-S0*Sc*(-Pr*f(1)*f(5)+Jh*Pr*((1/eps)*f(3)^2+M*f(2)^2))
];
end
function res = shootbc(fa,fb)
global eps
res = [fa(1)-1; fa(2)-eps; fa(4)-eps; fa(6)-eps; fb(2)-0; fb(4)-0; fb(6)-0];
end
DGM on 20 Nov 2023
Edited to add formatting and to make it run.

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