solve multiple algebric integral equations

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Meva
Meva 2015-1-31
编辑： Matt J 2015-2-1
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.
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Matt J
Matt J 2015-1-31
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
Meva 2015-2-1
Hi Mett,
The equations are not like that actually so complicated.
I just have simplified them in order to represent.

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Matt J
Matt J 2015-2-1
编辑：Matt J 2015-2-1
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 个评论
Meva
Meva 2015-2-1
Thanks Matt,
I have optimisation toolbox.
Next step will be how to use this toolbox to solve my problem.
Matt J
Matt J 2015-2-1
编辑：Matt J 2015-2-1
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,...)

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更多回答（1 个）

Matt J
Matt J 2015-1-31
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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