# How can draw this kind of graph?

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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
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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?

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### Answers (1)

Syed Sohaib Zafar on 19 Nov 2023
Edited: DGM on 20 Nov 2023
Here's your required code
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
##### 1 CommentShow -1 older commentsHide -1 older comments
DGM on 20 Nov 2023
Edited to add formatting and to make it run.

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