how to write a code for two input -one output function (f=@(v,r)) using bisection method?

%declaration of constants and variables
pi=3.14;
G=6.67*10^(-11);M=2*10^30;c=10^4;rho=10^(-21);n=2.5;
gamma=1+(1/n);

dotm=pi*G^2*M^2*(rho/c^3)*(2/(5-3*gamma))^((5-3*gamma)/(2*gamma-2));
u=(dotm/(4*pi*rho))*c^(2*n);

%span of v
v1=10;
v2=10^5;
tol=10^-10;
v=v1:c:v2;

%span of r
r1=.1*7*10^8;
r2=3.8*7*10^8;
r=r1:7*10^8:r2;

%function declaration
f=@(v,r)((v.^2)./2)+(n.*((u./(v.*r.^2)).^(1./n)))-((G*M)./r)-(n.*c.^2);

[Jan]

A bi-section method is meaningful in one dimension only.

What is the prupose of defining v and r as vectors, but as inputs of f also?

Just a note: While 6.67*10^(-11) is an expensive power operation, 6.67e-11 is a cheap constant and easier to read.

[Nabagata Chaudhury]

The input vectors are the value of r and v.for each value of r we have to find the corresponding two values of v by bisection method.then have to run the loop for each value of r.how to solve?

[Torsten]

And why do you give values to v if the vector is unknown ?

[Nabagata Chaudhury]

Yes, that's a good point.undersuch situation how to solve?

[Nabagata Chaudhury]

If f(v1,r) and f(v2,r) should be of opposite sign according to initial condition of bisection method

[Jan]

Before we can suggest some code, you have to mention explicitly, what you want to calculate. I guess, you wanted to use the bisection method to find a root of the function? Then an optimization tool should be applicable for the 2-dim case.

Answer 1

[Jan]

pi = 3.14;  % Brrrr
G  = 6.67e-11; M = 2e30; c = 1e4; rho = 1e-21; n = 2.5;
gamma = 1 + 1/n;
dotm = pi*G^2*M^2*(rho/c^3) * (2/(5-3*gamma))^((5-3*gamma)/(2*gamma-2));
u = dotm / (4*pi*rho) * c^(2*n);
v1 = 10;
v2 = 1e5;
r1 = 0.7e8;  % .1*7*10^8;
r2 = 3.8*7e8;
f = @(y) y(1)^2 / 2 + (n * ((u / (y(1)*y(2)^2)).^(1/n))) - (G*M)/y(2) - n*c^2;
x = fmincon(f, [v1+v2, r1+r2]/2, [], [], [], [], [v1, r1], [v2, r2])

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

x = 1×2

1.0e+07 * 0.0100 7.0000