How can draw this kind of graph?
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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
6 件のコメント
Sam Chak
2023 年 11 月 13 日
@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?

回答 (1 件)
Syed Sohaib Zafar
2023 年 11 月 19 日
編集済み: DGM
2023 年 11 月 20 日
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
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