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Consider the following general set of 2 ODE's

$$\dot{x}=\Theta(\dot{x} )f_1(x,y)+(1-\Theta(\dot{x}))f_2(x,y)$$ $$\dot{y}=(1-\Theta(\dot{y}))g_1(x,y)+\Theta(\dot{y})g_2(x,y)$$

where $\Theta(x)$ is the unit step function.

what it means is when $\dot{x}>0$ then the dynamics are govern by $f_1$ and when $\dot{x}<0$ then the dynamics are govern by $f_2$. With $\dot{y}$ it is the other way around.

Now let's consider that I can make in $f_1,f_2,g_1,g_2$ the transformations $x-y\rightarrow d$ and $y-y\rightarrow 0$, which leaves me with

$$\dot{x}=\Theta(\dot{x} )f_1(d)+(1-\Theta(\dot{x}))f_2(d)$$ $$\dot{y}=(1-\Theta(\dot{y}))g_1(d)+\Theta(\dot{y})g_2(d)$$

I would like to apply the transformation over the derivatives, such that I get an equation for $\dot{d}$ as well

$$\dot{d}=\dot{x}-\dot{y}=\Theta(\dot{x} )f_1(d)+(1-\Theta(\dot{x}))f_2(d)-(1-\Theta(\dot{y}))g_1(d)-\Theta(\dot{y})g_2(d)$$

If I apply the transformation on the unit step function as well, it seems that I lose information about the system

$$\dot{d}=\Theta(\dot{d} )f_1(d)+(1-\Theta(\dot{d}))f_2(d)-(1-\Theta(0))g_1(d)-\Theta(0)g_2(d)$$

The unit step can be defined when $\Theta(0)=\frac{1}{2}$ or $\Theta(0)=1$. In both cases I lose the information about $\dot{y}$.

Is there any way to make a transformation of such a system to $\dot{d}$?

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    $\begingroup$ You lost information when you redefined $f_1(x,y) \to f_1(x-y)$, which isn't always true. $\endgroup$ – Dylan May 20 at 8:43
  • $\begingroup$ @Dylan, I assume here that the functional form of $f_1$ can do so, for example, $f_1(x,y)=4(x-y)+e^{x-y}$. $\endgroup$ – jarhead May 20 at 9:20
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    $\begingroup$ Then I'm not sure what you're asking? Solve the reduced system for $d$, then plug it back into the original equations of $\dot x$ and $\dot y$ $\endgroup$ – Dylan May 20 at 9:43
  • $\begingroup$ @Dylan, yes but when you substitute $x\rightarrow x-y=d$ and $y\rightarrow y-y=0$ you end up with $\Theta(\dot{y}) \rightarrow \Theta(0)$ and lose the sign of the derivative which is also dynamic. $\endgroup$ – jarhead May 20 at 9:50
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    $\begingroup$ That should tell you that the substitution isn't valid. You can't reduce one variable to $0$ without justifying it. $\endgroup$ – Dylan May 20 at 15:45

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