Md Sojib about 8 hours ago
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Commented: Torsten about 2 hours ago
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.
11 Comments Show 9 older commentsHide 9 older comments
Show 9 older commentsHide 9 older comments
Torsten 3 minutes ago
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Edited: Torsten less than a minute ago
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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 less than a minute ago
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So, You are suggesting that bvp4c/bvp5c is applicable for only one differential equation with higher order?
Torsten 1 minute ago
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Edited: Torsten less than a minute ago
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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 8 minutes ago
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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 4 minutes ago
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You didn't answer my questions.
Md Sojib 7 minutes ago
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Edited: Md Sojib 2 minutes ago
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 2 minutes ago
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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 5 minutes ago
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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 5 minutes ago
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Edited: Torsten 5 minutes ago
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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 3 minutes ago
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Edited: Md Sojib 1 minute ago
I don't have any separate equation for p or p'. But after derivation, differential equations contains these two terms.
Torsten 6 minutes ago
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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 ?
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