# Coordinate transformation in ODE's with a unit step function

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 $$\dot{d}$$?

• You lost information when you redefined $f_1(x,y) \to f_1(x-y)$, which isn't always true. – Dylan May 20 at 8:43
• @Dylan, I assume here that the functional form of $f_1$ can do so, for example, $f_1(x,y)=4(x-y)+e^{x-y}$. – jarhead May 20 at 9:20
• 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 $\dot y$ – Dylan May 20 at 9:43
• @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. – jarhead May 20 at 9:50
• That should tell you that the substitution isn't valid. You can't reduce one variable to $0$ without justifying it. – Dylan May 20 at 15:45