Evaluating implicit functions numerically

Question

A simple question.

Are there any methods on how to evaluate an implicit function f(x,y) = 0 numerically?

Does one fix one of the variables and then use newtons method or similar on f(x0,y) and then iterate?

Thanks

Answer 1

I suppose that evaluate stands for solve.

What you can do is to consider that you want to minimize $$\Phi=\frac 12\left[f(x,y)\right]^2$$ (hoping, for sure, that you will get zero as the final result !).

Then, since you look for the minimum, you have to solve $$\frac{\partial \Phi}{\partial x}=f(x,y)\,\frac{\partial f(x,y)}{\partial x}=0$$ $$\frac{\partial \Phi}{\partial y}=f(x,y)\,\frac{\partial f(x,y)}{\partial y}=0$$ and, just as in Newton method, use Taylor expansions truncated to first orders. So, you have two linear equations for two unknown variables.

Just search for function minimization (this is wide domain).

Answer 2

You can also convert the problem to a differential equation where the initial condition follows from one particular solution for $y$ for one particular $x$. Given some arbitrary pair $(x,y)$ for which you have $f(x,y) = 0$, you can find a new infinitesimally shifted solution pair $(x+\delta x,y + \delta y)$, by demanding that $f(x+\delta x,y+\delta y) = 0$. To first order in $\delta x$ and $\delta y$, we then have:

$$f(x,y) + \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y \approx 0 $$

Using $f(x,y) = 0$, we get:

$$\dfrac{\delta y}{\delta x} \approx -\frac{ \dfrac{\partial f}{\partial x} }{ \dfrac{\partial f}{\partial y} }$$

In the limit of $\delta x \to 0$ the l.h.s. becomes the derivative of y w.r.t. x and the relation becomes exact:

$$\dfrac{d y}{d x} =-\frac{ \dfrac{\partial f}{\partial x} }{ \dfrac{\partial f}{\partial y} }$$

You thus have an ordinary differential equation that you can solve numerically using standard methods.