BVP PROBLEM how can we take 'c' in x-axis(continuous range.[-1.3:0.02:-1.1]) and to find f"(0)

function main
    S = 1; c  = -1.25; Pr = 0.7; n  = -0.8;
    x  = [3 -1];
    x1 = fsolve(@solver, x);
    function F = solver(x)
        [t, u] = ode45(@equation,[0,5], [S, c, x(1), 1, x(2)]);
             F = [u(end, 2)-1  u(end, 4)];
     figure(1)
    plot(t, u(:,2));
    hold on
    end 
    function dy = equation(t, y)
        dy    = zeros(5,1);
        dy(1) = y(2);
        dy(2) = y(3);
        dy(3) = y(2)^2 - y(1) * y(3) - 1;
        dy(4) = y(5);
        dy(5) = Pr * (n * y(2) * y(4) - y(1) * y(5));
    end
end

[Torsten]

Is f = y(1) ?

[MINATI]

Dear Torsten

yes f = y(1)

Sorry for seeing late

Answer 1

[Torsten]

function main

    S = 1; c  = [-1.3:0.02:-1.1]; Pr = 0.7; n  = -0.8;
    sol = zeros(numel(c),1);
    x  = [3 -1];
    
    for i = 1:numel(c)
      c_actual = c(i);  
      x1 = fsolve(@solver, x);
      sol(i) = x1(1);
      x = x1;
    end
    plot(c,sol)

    function F = solver(x)
        [t, u] = ode45(@equation,[0,5], [S, c_actual, x(1), 1, x(2)]);
        F = [u(end, 2)-1 , u(end, 4)];
    end 

    function dy = equation(t, y)
        dy    = zeros(5,1);
        dy(1) = y(2);
        dy(2) = y(3);
        dy(3) = y(2)^2 - y(1) * y(3) - 1;
        dy(4) = y(5);
        dy(5) = Pr * (n * y(2) * y(4) - y(1) * y(5));
    end

end

[MINATI]

Dear Torsen

Thanks for your time and effort. This modification looks loke OK!

But

1. when I gave different values of 'n' with "hold on" option, single curve came instead of multiple.
2. Is plot(c,sol) plots here for (f'), If so boundary condition is not matching
3. Another is to find f"(0)

Need help

[Torsten]

plot (c,sol) plots f''(0) against f'(0).

I don't know what you mean with

when I gave different values of 'n' with "hold on" option, single curve came instead of multiple.

[MINATI]

plot (c,sol) plots f''(0) against f'(0).

But I need f '(0) vs 'c' with above given range.

when I gave different values of 'n' with "hold on" option, single curve came instead of multiple.

For n=input('n=') I set -0.3, -0.5, -0.8 but instead of three curve only one curve came in one figure.

[Torsten]

c = f'(0), so a plot of f'(0) vs. c gives you the main diagonal.

sol(i) in the code from above contains f''(0) for given c(i) - so plot what you like with the arrays sol and c.

To see why you only get one plot for three values of n you will have to show the code you are using.

[MINATI]

Actually first code was OK for different values of 'n' but need f ' vs c graph.

Here is the modified (your) code for reference:

function main

S = 1; c = [-1.3:0.02:-1.1]; Pr = 0.7; %n = -0.8;

n=input('n=')

sol = zeros(numel(c),1);

x = [3 -1];

for i = 1:numel(c)

c_actual = c(i);

x1 = fsolve(@solver, x);

sol(i) = x1(1);

x = x1;

end

plot(c,sol)

hold on

function F = solver(x)

[t, u] = ode45(@equation,[0,5], [S, c_actual, x(1), 1, x(2)]);

F = [u(end, 2)-1 , u(end, 4)];

end

function dy = equation(t, y)

dy = zeros(5,1);

dy(1) = y(2);

dy(2) = y(3);

dy(3) = y(2)^2 - y(1) * y(3) - 1;

dy(4) = y(5);

dy(5) = Pr * (n * y(2) * y(4) - y(1) * y(5));

end

end

And also I didnt get you here:

sol(i) in the code from above contains f''(0) for given c(i) - so plot what you like with the arrays sol and c.

AS

I am not so strong as you in programming, perhaps

So please modify if possible

[Torsten]

A plot of f' vs. c does not make sense since c equals f'.

If you want to plot for different values of n, you will have to make a second loop over n as I did one loop over c.

[MINATI]

But in first program 'c' is a constant and then I need to vary it in X-axis with the range

c = [-1.3:0.02:-1.1]; where (f') varies with (eta) in X-axis previously.

[Torsten]

S=f(0),

c=f'(0),

x(1) = f''(0)

1 = g(0)

x(2) = g'(0)

Or do I misunderstand something ?

[MINATI]

yes you are right but if I want to draw f' w.r.t Pr = [-1.3:0.02:-1.1];(which is not in boundary condition part) then what to do?

[Torsten]

But you prescribe f' (c array). Thus you can just plot c against Pr without any calculation.

[MINATI]

Ok but I am still in confusion

f' does not satisfy boundary condition here also.

[MINATI]

Dear Torsten

Give a last try

[Torsten]

I did not answer since I don't understand your question.

Can you elaborate ?

[MINATI]

I am working on this attached paper. For reference you can follow:

http://dx.doi.org/10.1016/j.jppr.2017.07.007

Want to solve Eq.(11)(12) with B.Cs (13)(14)

In Fig. (2-5), in X-axis c/a is taken.

In other figs., dual solution concept is used i.e if we change x = [3 -1]; as anything like x = [5 -2]; or any other, we will get SECOND solution.

[MINATI]

Please have a look