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Let $y:[0,1] \rightarrow \mathbb R$ be a differentiable function. Let $\alpha, \beta,\gamma \in \mathbb{R}$ be some fixed parameters.

The equation is the following:

$$y'(x) = y(x)\left( \alpha - \beta \textbf{1}_{\{y(x) > \gamma \}} \right)$$ with initial condition $y(0)=y_0\in \mathbb{R}$.

Of course one can write: $$y(x)=y_0 \exp \left\{ \alpha x - \beta \int_0^x \textbf{1}_{\{y(z) > \gamma\}}dz\right\}$$ but I wonder if it is possible to find a better expression.

Does anyone see any close form or manner to get a form as close as possible?

Thank you!

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It seems like you need to break this down into many cases. For example, if $0<y(0)=y_0<\gamma$, then you have $y'(x)=\alpha y(x)$ for some short interval $[0,x]$ and if say $\alpha<0$, then the solution is just $y(x)=y_0e^{\alpha x}$ for $x>0$. So it would help to specify what $\alpha,\beta,\gamma,y_0$ are and how they relate to each other in size.

If $\alpha-\beta>0$, $0<y_0<\gamma$ then the solution is $y(x)=y_0e^{\alpha x}$ until $y(x_c)=\gamma$, which implies $x_c=\frac{1}{\alpha}\ln(\gamma/y_0)$. Then the solution is $y(x)=\gamma\exp((\alpha-\beta)(x-x_c)) $. The solution is parametric in $x$, I don't know how you would want this to be more explicit. Also, there is a discontinuity of the derivative at $x_c$ meaning that the solution should formally be interpreted as an integral one. This means that technically speaking, it's ill defined at $x_c$.

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  • $\begingroup$ Yes, $y(0)=y_0>0$ and let's say $y_0 < \gamma$ and $\alpha,\beta>0$. $\endgroup$ – Martingalo Feb 5 '15 at 20:56
  • $\begingroup$ So in this case, your solution will be $y(x)=y_0 e^{\alpha x}$ up till $y(x)=\gamma$ (which you can solve for). Then $\beta$ is turned on, in which case you need to know if $\alpha-\beta>0$ or $\alpha-\beta<0$. Again, you need more info. $\endgroup$ – Alex R. Feb 5 '15 at 20:58
  • $\begingroup$ Yes let us say $\alpha - \beta >0$. But how can you solve it completely? Without saying $y(x)=\gamma$? $\endgroup$ – Martingalo Feb 5 '15 at 21:09
  • $\begingroup$ OK, see edit... $\endgroup$ – Alex R. Feb 5 '15 at 21:20
  • $\begingroup$ This is very explicit. I will need to take a look at it carefully. Thanks a lot! $\endgroup$ – Martingalo Feb 5 '15 at 21:23

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