Integration using BVP4C

I have a coupled non-linear differential equations

(d^2 f)/(dy^2 )+m2*g2*dB/dy-2*i*R2*g1*f - g3*G1*y - R4*g1 = 0

(d^2 B)/(dy^2 )+t4/(1-i*H1)*df/dy=0

Boundary conditions are

f=0 at y=0

f=C1 at y=1

And

dB/dy-(t4/(P1* (1-i*H1 ) ))* B=0 at y=0

dB/dy+(t4/(P2 (1-i*H1 ) ))* B=0 at y=1

While I run the program I get the value of U1 using the boundary conditions (y=0 and y=1), but now i need to get the integration of U1, between the limits 0 to 1.

In BVP4c, the solution is obtained using the boundary conditions (y=0 and y=1), but now how to get the solution? Please help me

Matlab programs is enclosed for your reference

close all

clc

p=1;

P1=2;

P2=2;

b1=0.00021;

b2=0.000058;

S1=0.005;

S2=580000;

G1=2;

m2=20;

R1=997.1;

R2=3;

C1=1;

R3=4420;

B=0.5;

H1=0.25;

K1=3;

R4=1;

t1=(1./((1-p).^2.5));

t2=(1-p)+(p.*(R3./R1));

t3=(1-p)+p.*((R3.*b2)./(R1.*b1));

S=(S2./S1);

t4=1-((3*(1-S).*p)./((2+S)+(1-S).*p));

g1=t2./t1;

g2=1/t1;

g3=t3./t1;

m1=(t4./(P1.*(1-1i.*H1)));

m2=(t4./(P2.*(1-1i.*H1)));

dydx=@(x,y)[y(3);

y(4);

-m2.*g2.*y(4)+2.*1i.*R2.*g1.*y(1)+g3.*G1.*x+R4.*g1;

(-t4./(1-1i.*H1)).*y(3)];

BC = @(ya,yb)[ya(1);yb(1)-C1;ya(4)-m1.*ya(2);yb(4)+m2.*yb(2)];

yinit = [0.01;0.01;0.01;0.01];

solinit = bvpinit(linspace(0,1,50),yinit);

U1 = bvp4c(dydx,BC,solinit);

hold on

[Torsten]

What is U1 ? Is it f, df/fy, d^2f/dy^2, B, dB/dy, d^2B/dy^2 ? Or all these functions ?

[Syed Mohiuddin]

U1=f

[Torsten]

Then add the differential equation

dU1/dy = f, U1(0) = 0

for a new function U1 to your system.

U1(1) then yields integral_{y=0}^{y=1} f(t) dt.

[Syed Mohiuddin]

Thank you, but in bvp4c :

U1 = bvp4c(dydx,BC,solinit);

The above step yields numerical value, but i want U1 in terms of y, so that i can integrate

sir, you have mentioned U1(0)=0, how do you consider it?

Please help me in this regard,

[Syed Mohiuddin]

Thanks a lot

[Syed Mohiuddin]

where is the option to accept the answer?

[Torsten]

Thanks a lot

You are welcome. I moved my comment to an answer (which now can be accepted).

Answer 1

[Torsten]

close all
clc
p=1;
P1=2;
P2=2;
b1=0.00021;
b2=0.000058;
S1=0.005;
S2=580000;
G1=2;
m2=20;
R1=997.1;
R2=3;
C1=1;
R3=4420;
B=0.5;
H1=0.25;
K1=3;
R4=1;
t1=(1./((1-p).^2.5));
t2=(1-p)+(p.*(R3./R1));
t3=(1-p)+p.*((R3.*b2)./(R1.*b1));
S=(S2./S1);
t4=1-((3*(1-S).*p)./((2+S)+(1-S).*p));
g1=t2./t1;
g2=1/t1;
g3=t3./t1;
m1=(t4./(P1.*(1-1i.*H1)));
m2=(t4./(P2.*(1-1i.*H1)));
dydx=@(x,y)[y(3);
    y(4);
    -m2.*g2.*y(4)+2.*1i.*R2.*g1.*y(1)+g3.*G1.*x+R4.*g1;
    (-t4./(1-1i.*H1)).*y(3);
    y(1)];
BC = @(ya,yb)[ya(1);yb(1)-C1;ya(4)-m1.*ya(2);yb(4)+m2.*yb(2);ya(5)];
yinit = [0.01;0.01;0.01;0.01;0];
solinit = bvpinit(linspace(0,1,50),yinit);
U1 = bvp4c(dydx,BC,solinit);  
hold on
% plot y1(x) = f(x)
plot(U1.x,real(U1.y(1,:)),'r')
% plot y5(x) = integral_{t=0}^(t=x} f(t) dt
plot(U1.x,real(U1.y(5,:)),'b')
hold off
grid on

% print y5(1) = integral_{t=0}^(t=1} f(t) dt
U1.y(5,end)

ans = 0.5000 - 0.0000i