# State change of input-output system

Edited

Given a computer model $$F:\mathbb{R}^3 \to \mathbb{R}$$, with inputs $$x, w$$ and $$z$$, and output $$y=F(x,w,z)$$, where for any input, we are able to evaluate the output, my goal is to tune the inputs to achieve a certain state change (change in output).

More precisely:

Goal: Given an input $$x_0, w_0, z_0$$ and an output $$y_0$$, I want to increase my output to a state $$y_1$$.

Question: How to change my inputs to achieve the new state $$y_1$$.

My approach: \begin{align} dy &= \frac{\partial F}{\partial x}(x_0, w_0, z_0)dx + \frac{\partial F}{\partial w}(x_0, w_0, z_0) dw + \frac{\partial F}{\partial z}(x_0, w_0, z_0)dz \\ y_1-y_0 &= \frac{\partial F}{\partial x}(x_0, w_0, z_0) (x_1 - x_0) + \frac{\partial F}{\partial w}(x_0, w_0, z_0) (w_1 - w_0)+ \frac{\partial F}{\partial z}(x_0, w_0, z_0) (z_1 - z_0) \end{align} Since $$x_0, w_0, z_0, y_0, y_1$$ are given, and assuming we are able to compute from the model $$F$$ the numerical approximation of the derivative of $$F$$ with respect to each input at the point $$(x_0, w_0, z_0)$$, the new values $$x_1, w_1, z_1$$ then should lie on the plane given by the second equation above.

I am also aware that I made a linear approximation that is only exact when $$F$$ is linear with respect to the inputs.

I am going in the right direction in formulation the problem ? If not, any suggestions?

• This is a (possibly) nonlinear equation. There are many methods available to solve them: Newton iteration, Levenberg-Marquardt method, fixed-point iteration, pseudo-transient continuation, etc. If you have a good initial guess and you can evaluate the Jacobians, I suggest Newton's method. Otherwise (no good initial guess), Levenberg-Marquardt might work. Oct 15 at 11:59
• Is $y$ and input or an output? The first line is confusing in this regard, as is the statement "pair of inputs" when your function seems to be taking three arguments. Oct 16 at 1:14
• @WolfgangBangerth Sorry for the confusion ... I edited the question Oct 16 at 8:19
• @cos_theta The derivatives are evaluated numerically and hence the equation is always linear in $x_1, w_1, z_1$ .. Oct 16 at 14:48
• the problem, as you have formulated it now, does not have a unique solution. What is this for exactly, that might help us suggest some extra conditions. Oct 16 at 16:44