# Finding Numerically the solution line for an equation

by LayMuon
Tags: equation, line, numerically, solution
 P: 148 I have a function which depends say upon two variables: $f(x,y)$. I need to find numerically the solution line $y(x)$ of the equation $f(x,y) = 0$. The function itself for arbitrary set of {x,y} may evaluate to some complex number (it involves lots of square roots) but for the solution I am seeking the imaginary part on the solution line should be zero. I understand that one in principal can scan through the whole region of {x,y} to get the curve but I need numerically efficient way to write the code. Thanks.
 Homework Sci Advisor HW Helper Thanks P: 9,789 Is f(x,y) differentiable analytically? I'll assume not, but I will assume it's continuous. If you can feed in a feasible range for y(x0) then you could program a binary chop on the values. (Do I need to explain that?) Having found one point on the line, (x0, y0), you could then search for solutions of f(x,y) = 0 where the |(x,y)-(x0, y0)| = some small r. Again, you could do a binary chop on the direction from (x0, y0) to (x,y).
 P: 148 What is "binary chop"?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 Finding Numerically the solution line for an equation If $f(x_0, y_0)> 0$ and $f(x_1, y_1)< 0$ then, because f is continuous, there must be a point between $(x_0, y_0)$ and $(x_1, y_1)$ where f is 0. We have no idea where so checking one point is as good as another- use the midpoint, $(x_2, y_2)= ((x_0+ x_1)/2, (y_0+ y_1)/2)$. Now look at $f(x_2, y_2)$. If it is 0, we are done! If not, it is either positive or negative. If it is positive, then, since $f(x_1, y_1)< 0$, there must be a solution between them. If it is negative, then, since $f(x_0, y_0)> 0$, there be a solution between them. In either case, define $(x_3, y_3)$ to be the mid point of that interval. That is, I think, what haruspex is calling a "binary chop". I would call it a "binary search".
 P: 148 My concern is that f(x,y) does not evaluate necessarily to real value, it might as well be complex, so it being positive or negative is not well defined.
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P: 9,789
 Quote by LayMuon My concern is that f(x,y) does not evaluate necessarily to real value, it might as well be complex, so it being positive or negative is not well defined.
That's not a problem. 0 is 0, and the modulus of f is zero if and only if f is 0.
P: 148
 Quote by haruspex That's not a problem. 0 is 0, and the modulus of f is zero if and only if f is 0.
But then if you are taking |f(x,y)| how do you implement the comparisons of where it is positive and where negative?
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P: 9,789
 Quote by LayMuon But then if you are taking |f(x,y)| how do you implement the comparisons of where it is positive and where negative?
Ooops! Yes, you're right, sorry. Let me think some more.
 Homework Sci Advisor HW Helper Thanks P: 9,789 OK, instead of doing the binary chop based on the sign of f, you can look for local minima of |f|. This is a bit trickier because you need to track three 'current' points. E.g. if |f(x,y0)| > |f(x,y1)| < |f(x,y2)|, y0 < y1 < y2, then you know there is a min between y0 and y2. So next pick y3 = (y0 + y2). If |f(x,y3)| > |f(x,y1)| then you know there's one between y3 and y2. But it might turn out to be > 0, and there could be a min between y0 and y3.
P: 148
 Quote by haruspex OK, instead of doing the binary chop based on the sign of f, you can look for local minima of |f|. This is a bit trickier because you need to track three 'current' points. E.g. if |f(x,y0)| > |f(x,y1)| < |f(x,y2)|, y0 < y1 < y2, then you know there is a min between y0 and y2. So next pick y3 = (y0 + y2). If |f(x,y3)| > |f(x,y1)| then you know there's one between y3 and y2. But it might turn out to be > 0, and there could be a min between y0 and y3.
but in that case you would just find local minima of the function, not zeros.
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P: 9,789
 Quote by LayMuon but in that case you would just find local minima of the function, not zeros.
Since |f| ≥ 0, every zero will be a local minimum. So if you find all the minima you will find all the zeroes.
 P: 148 But not all local minima are the zeros
 P: 148 And another problematic thing with what you suggest is that you have to literally scan all x values, the problem is like for single variable case. I was more looking to be able to dispense with scanning x fully.
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