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how to ues the exponentials and derivatives in bvp4c

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Syed Mohiuddin
Syed Mohiuddin 2024-3-14
移动：Torsten 2024-3-16
I have a coupled non-linear differential equations
u''-b1*(t')*(u')+(1+b1*t)*[G1*F1*t+G2*F1*p-F3*P]=0;
t''-b2*(t')^2+B*F6*(u')^2+(b2-b1)*t*B*F6*(u')^2-b2*b1*B*F6*t^2*(u')^2=0;
p''- A*p=0
and the boundary conditions are
u=0,t=1+m,p=1+n at y=-1 and u=0,t=1,p=1 at y=1.
The code is:
clc;
p=0.01;
Betaf= 207;
Betas = 17;
Beta = 0.5;
kof = 0.613;
kos = 400;
m = -1;
b2 = 0.5;
b1 = 0.5;
G1 = 5;
G2 = 5;
A = 0.5;
Rhof = 997.1;
Rhos = 8933;
P1 = 0.5;
n=0.5;
B=0.01;
A1 = (1-p).^2.5;
A2 = 1/(1 + 1/Beta);
A3 = (1-p)+p.*((Rhos.*Betas)./(Rhof.*Betaf));
F1 = A2.*A3;
F3 = A1.*A2;
F4 = (kos + 2*kof - 2*p.*(kof - kos))/(kos + 2*kof + p.*(kof - kos));
F5 = (1 + 1/Beta)./A1;
F6 = F5./F4;
dydx=@(x,y)[y(4);
y(5);
y(6);
(b1.*y(4).*y(5)-(1+b1.*y(2)).*(G1.*F1.*y(2)+G2.*F1.*y(3)-F3.*P1));
b2.*y(5).^2-B.*F6.*y(4).^2-(b2-b1).*y(2).*B.*F6.*y(4).^2+b2.*b1.*B.*F6.*y(2).^2*y(4).^2;
A.*y(3)];
BC=@(ya,yb)[ya(1);yb(1);ya(2)-(1+m);yb(2)-1.0;ya(3)-(1+n);yb(3)-1.0];
yinit=[0.01;0.01;0.01;0.01;0.01;0.01];
solinit=bvpinit(linspace(-1,1,50),yinit);
S=bvp4c(dydx,BC,solinit)
I need to find T1= e^(-b1*(1+m))*(du/dy), here (du/dy) should be found at y = -1
T2= e^(-b1)*(du/dy), here (du/dy) should be found at y = 1
N1= -(1+b2*(1+m))*(dt/dy), here (dt/dy) should be found at y = -1
N1= -(1+b2)*(dt/dy), here (dt/dy) should be found at y = 1
how to write T1,T2,N1,N2 in the program
Please help me to find these values
Thank you

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采纳的回答

Torsten
Torsten 2024-3-16
编辑：Torsten 2024-3-16
yp1 = deval(S,1)
ym1 = deval(S,-1);
T1= exp(-b1*(1+m))*ym1(4);
T2= exp(-b1)*yp1(4);
N1= -(1+b2*(1+m))*ym1(5);
N2= -(1+b2)*yp1(5)

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