yy3 is generating a straight line while it should generate a curve, i don't understand what is wrong!!!

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Fareeha
Fareeha 2023-8-11
评论： Torsten 2023-8-11
function ABC
% Initialization of paramters
Pr=6.2;
M=4;
Bi=0.3;
beta=0.2;
ks1=0.5;
ks2=0.1;
rhos1=0.2;
rhos2=0.3;
n=3;
kbf=0.598;
phi1=0.1;
phi2=0.04;
rhocps1=0.3;
rhocps2=0.1;
rhof=997.1;
kf=0.613;
rhocpf=4179;
A1=((1-phi1)^(2.5)).*((1-phi2)^(2.5));
B1=(1-phi2).*((1-phi1).*rhof + rhos1.*phi1) + rhos2.*phi2;
A2=B1*(1/rhof); % rho_Hnf/rho_f
B2=ks2+(n-1)*kbf-(n-1)*(kbf-ks2).*phi2;
B3=ks2+(n-1)*kbf+(kbf-ks2).*phi2;
C5=B2/B3; % k_Hnf/k_bf Nanofluid Constant
B4=ks1+(n-1)*kf-(n-1)*(kf-ks1).*phi1;
B5=ks1+(n-1)*kf+(kf-ks1).*phi1;
C6=B4/B5; % k_bf/k_f
A4=C5*C6; % k_Hnf/k_f
B6=(1-phi2)*((1-phi1)*rhocpf+phi1*rhocps1)+phi2*rhocps2; % rhocp_Hnf
B7=1/rhocpf;
A3=B6*B7; % rhocp_Hnf/rhocp_f
% Initial Condition Input
sol = bvpinit(linspace(0,8,100), [1 0 0 0 0 0 0 0]);
% solution in structure form
sol1 = bvp4c(@bvpexam2, @bcexam2, sol);
x1 = sol1.x;
y1 = sol1.y;
%%% Plotting of the temperature
plot(x1, y1(7, :))
hold on
function res=bcexam2(y0, yinf)
res=[y0(2)-1;y0(5)-beta;y0(1)+y0(4);y0(8)+(Bi/A4)*(1-y0(7)); yinf(2);yinf(5);yinf(7); yinf(4)];
end
function dydx = bvpexam2(~,y)
yy1=A1*(A2*(y(2)^2-(y(1)+y(4))*y(3))+M*y(2));
yy2 = A1*(A2*(y(5)^2-(y(1)+y(4))*y(6))+M*y(5));
yy3 =-(A3/A4)*Pr*(y(1)+y(4))*y(8);
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(7);yy3];
end
end
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Torsten
Torsten 2023-8-11
编辑：Torsten 2023-8-11
As initial condition for y(7), you set y(7) = 0, and you define the differential equation for y(7) as dy(7)/dx = y(7). The solution is y(7) = 0 for all x - and that's what is plotted.
The line
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(7);yy3];
must read
dydx= [y(2);y(3);yy1;y(5);y(6);yy2;y(8);yy3];
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Fareeha
Fareeha 2023-8-11
@Torsten i have another question
in function
function res=bcexam2(y0, yinf)
res=[y0(2)-1;y0(5)-beta;y0(1)+y0(4);y0(8)+(Bi/A4)*(1-y0(7)); yinf(2);yinf(5);yinf(7)];
end
I removed one BC, now i have 8 vectors in dydx and 7 BC, in such condition how could code work?
Torsten
Torsten 2023-8-11
You need as many boundary conditions as there are first-order differential equations. Otherwise, you get a one-dimensional solution manifold.
Consider
y '' = c with y(0) = 1
The solution is
y(x) = 1 + a1*x + c/2*x^2
with arbitrary parameter a1.

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