Roots of a two variables equation involving a numerical integration

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Daniel Tamayo 2020-4-12

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评论： J. Alex Lee 2020-4-17

I'm looking for a solution of a of two variables equation problem,

The issue is that '

' and '

' are integrals of a certain kind of functions 'f' and 'g' involving the variables x, y and p, in a very complicate way (since the integration is only on p, it eventually disappears), that include logarithms of the three variables, like the example function

below

I have tried to use a double for loop for the variables x and y, so that I can integrate numerically over p only. However, it has not been efficient and one of the problems I have is that for certain configurations of values of x,y and p, negative numbers are obtained that generate an indeterminacy in the logarithm.

In summary, I must find the pairs of numbers (x, y) that satisfy the following type of equation, and plot them

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Daniel Tamayo 2020-4-13

Thanks for your answer! sorry about the length...

Where,

and

Ameer Hamza 2020-4-14

Yes, these are quite long. If you can write them in MATLAB, maybe we can suggest a solution with fsolve, fmincon or some other optimization algorithm.

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Answer 1

J. Alex Lee 2020-4-14

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Assuming x and y are scalars and there is a solution, use fsolve, not an optimization algorithm. You'll need initial guesses. The beauty of numerical solution is that you can literally just define the functions you have written out, and call them. The numerical integration can be done with matlab's in-built integral() function, which according to docs can even take Inf as an integration limit (sounds cool, i have never tried this). The structure would look something like

X0 = [x0,y0]
X = fsolve(@(X)myresidual(X),X0)
function res = myresidual(X)
x = X(1);
y = X(2);
res = (y-x) .* (1+integral(@(p)f(x,y,p),0,Inf)) + integral(@(p)g(x,y,p),0,Inf);
end
function out = f(x,y,p)
% and you call other functions to do the other complicated sub-parts like eps1, L1p, L1m, etc.
end
function out = g(x,y,p)
end
function out = L1(x,y,M,e1)
end
% etc

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John D'Errico 2020-4-17

I think the problem you are tripping over is that fsolve returns only one solution. But here, you have apparently two variables and one equation. (Your commented response to Ameer is a bit confusing though.)

That means there will generally be infinitely many solutions, think of it as a curve in the (x,y) plane. This is often called an implicit function, much like a circle in the plane.

But finding all solutions if an analytical solution fails to exist is impossible, since there are infinitely many of them. fsolve can never do that, since it can solve for only one solution at a time, based on the starting values to the problem.

If you just want a plot, then fimplicit is a great way to solve these problems.

IF you just want a list of a FINITE set of solutions for whatever reason, then one thing is to just write a loop, fixing x at a sequence of values. Then solve for y, as essentially a function of x. Save all solutions found for y as a function of x.

A problem may arise with this approach, even for a simple case - the circle. Here you would find for any given x, there are 0, 1 or 2 solutions for y. The real world sucks, and sometimes there is no simple answer to a problem.

Another solution might be to use a contour plot, here the proper tool will be contourc. The idea of contourc is it will have you evaluate the overall function for a large grid of points. Then you will tell it to search for the piecewise linear curve through that grid where the function is exactly zero (determined by a local linear interpolation.) The result will be a segmented linear approximation to the solutions of your problem, in the (x,y) plane. The quality of this approimation will be a function of how fine was your grid as provided to contourc.

John D'Errico 2020-4-17

One thing that I love is the ability of contour and contourc to solve problems of this sort, as well as return the solution locus as an approximation.

Path following is another nice idea, one that is also often overlooked. I even considered writing a toolbox for the purpose once.

J. Alex Lee 2020-4-17

Indeed, I am using contourX more and more in lieu of actually solving for conditions when the problem is challenging. ironically i am hell bent on analytical solutions when my problem reduces to a cubic polynomial.

I never bothered to google it until now, but there's a nice wiki page with links to toolboxes in different languages

https://en.wikipedia.org/wiki/Numerical_continuation

including 1 for matlab, though i haven't at all looked at it:

https://sourceforge.net/projects/matcont/

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