Need help while defining ode function for bvp4c/bvp5c/ode45 solve

I wanted to define system of ode functions for my higher order problems. I have differential equations are :

x=cosy

z=siny

y'=sqrt((lambda*f*p*siny/x)+(siny^2/x^2)-(2/lambda^2)+(2*cosy/lambda)-(3*a*p^2*lambda^2/2))

y"=(-y'*x'/lambda*x)-(cosy/x)-(f*p'/4)-(f*x'*p/4*x)+(y'*cosy/x)+(siny/lambda)+(lambda*f*p*cosy/4*x)+(siny*cosy/x^2)+(mu_1*(sin(y) - mu_2*cos(y)) / (2 * x^2)

where, y,x,p are functions of s. and f,lambda and a are constants.

while defining ode function for my bvp solve, I write the code as

function dydx = odefun2(t, y, params)

lambda = params.lambda;

a = params.a;

f = params.f;

mu_1=params.mu_1;

mu_2=params.mu_2;

y1 = y(1); % y

y2 = y(2); % x

y3 = y(3); % z

y4 = y(4); % p

y5 = y(5); % p_dot

y6 = cos(y1); %x_dot

y7 = sin(y1); %z_dot

ydot = sqrt((lambda * f * y4 * sin(y1) / y2) + (sin(y1)^2 / y2^2) - (2 / lambda^2) + (2 * cos(y1) / lambda) - (3 * a * (y4^2) * lambda^2)/2);

yddot = -((ydot * cos(y1) / (lambda * y2))) - (cos(y1) / y2) - (0.5 * y5 / 4) - (0.5 *cos(y1) * y4 / (4 * y2)) + (ydot * cos(y1) / y2) + (sin(y1) / lambda) + (0.5 * lambda * y4 * cos(y1) / (4 * y2)) + (sin(y1) * cos(y1) / y2^2) + (mu_1*(sin(y1) - mu_2*cos(y1)) / (2 * y2));

dydx = [y5; y6; y7; ydot; yddot];

end

But I have no equation for p'. And I am not sure that how can I relate x and z with x' and z' respectively. Also I don't have any differential equation for p'. how I should relate p and p'?

need some suggestions.

[Torsten]

You cannot work with two different differential equations for one and the same solution variable (in your case y).

Or is it an implicit equation for p that

d/dt (y'(t)) = y''(t)

or written out

d/dt (sqrt((lambda*f*p*siny/x)+(siny^2/x^2)-(2/lambda^2)+(2*cosy/lambda)-(3*a*p^2*lambda^2/2))) = 
(-y'*x'/lambda*x)-(cosy/x)-(f*p'/4)-(f*x'*p/4*x)+(y'*cosy/x)+(siny/lambda)+(lambda*f*p*cosy/4*x)+(siny*cosy/x^2)+(mu_1*(sin(y) - mu_2*cos(y)) / (2 * x^2)

?

[Md Sojib]

So, You are suggesting that bvp4c/bvp5c is applicable for only one differential equation with higher order?

[Torsten]

No. I say that you can't solve a system of differential equations that contains contradictory equations.

Say you have a system

y'' = 1
y' = 3*t

Then differentiating the last equation gives

y'' = d/dt (3*t) = 3

contradicting the first equation

y'' = 1

Or did you arrive at the equation for y'' by just differentiating y' with respect to t ? Then the equation for y'' is superfluous.

[Md Sojib]

Thank you. In that case, is there any alternative way to solve it numerically? I want to give a try. Because I got these equations from derivation.

Actually, I am developing a shape model of a surface by appling calculus of variation method. From that derivation, I got these differential equations which were the common factor of different variational functions.

[Torsten]

You didn't answer my questions.

[Md Sojib]

I have already mentioned it my last comment, the equation of y" was not directly came from dy'/dt. lets say, y=y0+eta (y1) ; x=x0+eta (x1) ; p=p0+eta (p1) so on for other functions. I applied these to a specific equation to get minimized shape of that surface. The above equations are the: (differential equation 1) y1 +(differential equation 2)x1 ... so on. I hope I clarify your questions.Please, Let me know If you want to know more on that.

[Torsten]

The equations for x and z make no problem because they don't need to be solved. Once you have y, you also have x and z. And expressions for x or z in the equation for y can be substituted by cos(y) and sin(y).

So you need to solve for y and p. What are the equations and what are the boundary/initial conditions ?

[Md Sojib]

boundary conditions:

x(0)-R*sin(alpha);

z(0)+R*cos(alpha)/2;

y(0)-alpha;

y'(inf)=0;

p(0)=0;

p'(inf)=0

The equations are mentioned in the code.

[Torsten]

I tried to explain that you cannot use both equations for y because they contradict each other.

And you didn't supply an equation for p. Or - a question you did not yet answer - should p be deduced from the equation

d/dt (y'(t)) = y''(t)

or - written out -

d/dt ( sqrt((lambda*f*p*siny/x)+(siny^2/x^2)-(2/lambda^2)+(2*cosy/lambda)-(3*a*p^2*lambda^2/2)) ) =
(-y'*x'/lambda*x)-(cosy/x)-(f*p'/4)-(f*x'*p/4*x)+(y'*cosy/x)+(siny/lambda)+(lambda*f*p*cosy/4*x)+(siny*cosy/x^2)+(mu_1*(sin(y) - mu_2*cos(y)) / (2 * x^2)

?

[Md Sojib]

I don't have any separate equation for p or p'. But after derivation, differential equations contains these two terms.

[Torsten]

Then you will have to dive into the derivation of the equations again. If you don't know the equation for a function within your problem formulation, how do you want to solve it ?