solve multiple algebric integral equations

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Meva 2015-1-31

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编辑： Matt J 2015-2-1

采纳的回答： Matt J

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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:
= integral (c^2/(a-(x-1/2)*b-1/2*sin(pi*x))^2) dx , from x=0 to x=1.
= 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.
= 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 2015-1-31

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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 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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Answer 1

Matt J 2015-2-1

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编辑：Matt J 2015-2-1

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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.

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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 2015-2-1

编辑：Matt J 2015-2-1

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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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Answer 2

Matt J 2015-1-31

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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.