how to ues the exponentials and derivatives in bvp4c

2 次查看（过去 30 天）
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

请先登录，再回答此问题。

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)

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Biotech and Pharmaceutical 的更多信息

标签

产品


版本

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by