0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.