solve multiple algebric integral equations
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Hi I have three equations and three unknowns I need to solve these.
Two of the equations are integral equations
Could Matlab do that ?
unknowns are a,b,c. equations:
0 = integral (c^2/(a-(x-1/2)*b-1/2*sin(pi*x))^2) dx , from x=0 to x=1.
0 = integral (x-1/2)*(1-c)^2/(1-a +(x-1/2)*b-1/2*sin(pi*x))^2 dx , from x=0 to x=1.
0 = c^2-2*c^2*a-a^2-a*b+b^2/4-2*a*b*c ..
I need to solve these three equations to find three unknowns a,b,c.
x is double variable from 0 to 1 with the increnment of 0.01. So we have 101 x values.
Could I find the answer numerically?
I have R2014b version.
4 Comments
John D'Errico
on 31 Jan 2015
Edited: John D'Errico
on 31 Jan 2015
It COULD. But then again, maybe it could not. You never know. And how can we possibly know if you do not show us what they are?
Are you looking for a symbolic solution, or a numerical one?
What version of MATLAB do you have? The symbolic TB? The optimization TB?
The point is you need to be more forthcoming with details, else we cannot help.
The crystal ball is so cloudy today.
Matt J
on 31 Jan 2015
I wonder if the equations are correct. The first equation only has a solution if c=0, since the integrand is non-negative everywhere. If c=0, then the 3rd equation implies one of 2 possible relationships between a and b
a = b*(sqrt(2)-1)/2
or
a = b*(-sqrt(2)-1)/2
Meva
on 1 Feb 2015
Accepted Answer
You can try FSOLVE, if you have the Optimization Toolbox and if the dependence of the equations on the unknown parameters is smooth.
Otherwise, if you have just a few unknowns, you can try FMINSEARCH, using it to minimize the violation of your equations. For example, to solve the equations
a+b=1,
a-b=-1,
you could do
violation=@(x) norm([sum(x)-1;-diff(x)+1]);
ab=fminsearch( violation, [1,1])
a=ab(1);
b=ab(2);
except you would use your actual equations, of course.
2 Comments
Meva
on 1 Feb 2015
Well, the documentation for fsolve should probably be all you need. It gives simple examples and everything. Basically, you need to write a function that creates a vector of all right hand side values of your equations.
function F=myobjective(parameters)
a=parameters(1);
b=parameters(2);
c=parameters(3);
eq1=@(x) @(x)c.^2./(a-(x-1./2).*b-1./2.*sin(pi.*x)).^2
eq2=@(x) (x-1./2).*(1-c).^2./(1-a+(x-1./2).*b-1./2.*sin(pi.*x)).^2
F(1) = integral(eq1, 0,1);
F(2) = integral(eq2,0,1);
F(3) = c^2-2*c^2*a-a^2-a*b+b^2/4-2*a*b*c
end
solution = fsolve(@myobjective, pInitial,...)
More Answers (1)
Matt J
on 31 Jan 2015
I wonder if the equations are correct. The first equation only has a solution if c=0, since the integrand is non-negative everywhere. If c=0, then the 3rd equation implies one of 2 possible relationships between a and b
a = b*(sqrt(2)-1)/2
or
a = b*(-sqrt(2)-1)/2
This reduces the 2nd equation to an equation in 1 variable, and you can solve with fzero.
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