Suppose I have some function $F(x,y) = (x-x_0)^2 + (y-y_0)^2$

The variables $x_0 ,y_0$ are my 'targets' for $x,y$, i.e. I want to determine $x,y$ such that $F(x,y) = 0$

Now, $x$ and $y$ are obtained from numerically solving a system of ordinary differential equations. For different initial conditions $\alpha, \beta$, we get different values for $x,y$.

The aim is then to find $\alpha, \beta$ such that $F = 0$. I had thought some sort of gradient descent algorithm would work, but I can't seem to frame this problem as a gradient descent one.

How would I go about determining $\alpha, \beta$?

-----Clarifications in response to comments----

$(\alpha, \beta)$ are some initial conditions which are related through a transformation to initial conditions on ODEs $\frac{dx}{d\lambda}, \frac{d y}{d \lambda}$ for som parameter $\lambda$. These ODEs can be solved numerially and then evaluated at a particular point $\lambda$, to produce $x,y$. Ultimately I want to determine $\alpha, \beta$ such that $x = x_0$ and $y=y_0$

I hope this is more clear

  • $\begingroup$ So presumably you can't "see" $x_0$ and $y_0$. Is there anything else you can say about $F$? Because in some sense, obviously, the solution is $(x_0,y_0)$ but it would appear you're aware of this too $\endgroup$ – Squirtle Feb 20 '18 at 17:38
  • $\begingroup$ You also need to specify what system of ODEs you're talking about. Are $\alpha$ and $\beta$ just different names for instantiations of $x$ and $y$? $\endgroup$ – Squirtle Feb 20 '18 at 17:42
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    $\begingroup$ It is not clear what you are asking. What do you mean by $x,y$ are obtained from an ODE? Where do $\alpha,\beta$ come from? Do you mean you have some function $g(\alpha,\beta)$ that produces $x,y$ and you are trying to solve $F \circ G (\alpha,\beta) = 0$? $\endgroup$ – copper.hat Feb 20 '18 at 17:43

This is only to get you started.... not actually an answer, since it's not clear what you're asking:

If you want to apply the gradient method, you need to take the partial derivatives.

$$\frac{\partial F}{\partial x} = 2(x-x_0)\partial_x(x-x_0)=2(x-x_0)$$ $$\frac{\partial F}{\partial y} = 2(y-y_0)\partial_y(y-y_0)=2(y-y_0)$$


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