# How to get the derivate using bvp4c

4 views (last 30 days)
Syed Mohiuddin on 26 Nov 2022
Commented: Torsten on 27 Nov 2022
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;
P = 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.*P));
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;
Alpha.*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 like to know how to get the derivative value du/dy at y= -1 and y=1 from the above program

Torsten on 26 Nov 2022
I assumed Alpha = A in your code.
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;
P = 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.*P));
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)
S = struct with fields:
solver: 'bvp4c' x: [-1 -0.9388 -0.8776 -0.8163 -0.7551 -0.6939 -0.6327 -0.5714 -0.5102 -0.4490 -0.3878 -0.3469 -0.3265 -0.3061 -0.2449 -0.1837 -0.1224 -0.0612 6.9389e-18 0.0612 0.1020 0.1429 0.1837 0.2449 0.3061 0.3673 0.4286 0.4898 0.5510 0.6122 0.6735 … ] y: [6×37 double] yp: [6×37 double] stats: [1×1 struct]
plot(S.x,S.y(1,:))
S.y(4,1) % u'(-1)
ans = 9.1263
S.y(4,end) % u'(1)
ans = -5.8619
Torsten on 27 Nov 2022
So you should come to the result that
S.y(5,1) is dt/dy at y = -1
S.y(5,end) is dt/dy at y = 1
S.y(6,1) is dp/dy at y = -1
S.y(6,end) is dp/dy at y = 1

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