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Let $A,B$ be $n\times n$ matrices. I am interested in the following ODE in $\mathbb{R}^n$

$$ \frac{dx_t}{dt}=Ax_t+Bx^+_t $$

where $x_t^+=(x^+_{1,t},...,x^+_{n,t})$ and $(\cdot)^+$ is the rectifier: $r^+=max\{0,r\}.$

Does this type of ODE have a name? And are there any known criterias for stability? Has it been studied by anyone in general?

The closest i have found is the "Threshold-Linear networks" studied here for example. I appreciate any reference similar to this system.

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  • $\begingroup$ What is $x_tdt$? $\endgroup$ Commented Aug 19, 2019 at 13:14
  • $\begingroup$ I just meant the derivative of $x$. I have changed it. $\endgroup$
    – Conformal
    Commented Aug 19, 2019 at 13:36

1 Answer 1

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Let me change slightly your notations: you want to study the system of ODE $$ \dot x=Ax+Bx_+,\quad \text{where $x_+=(x_{j,+})_{1\le j\le n}$}\tag{$\ast$}. $$ A first remark is that the flux $F(x)=Ax+Bx_+$ is globally Lipschitz-continuous, i.e. has a globally bounded (distribution) derivative. Since $\Vert B x_+ \Vert\le C\vert x\vert$, the general theory of ODE is providing unique solutions for a given initial data and in fact a flow $\phi(t,y)$ so that the unique solution of $(\ast)$ with value $y$ at $t=0$ is $\phi(t,y)$, $$ \dot\phi(t,y)=A\phi(t,y)+B\phi(t,y)_+, \quad \phi(0,y)=y. $$ By the way, that flow is also Lipschitz-continuous with respect to the $y$ variables. A second remark is that you can get rid of the linear part by conjugation: setting $ x=e^{tA} X, $ you get $$ e^{tA} \dot X+Ae^{tA} X=Ae^{tA} X+B(e^{tA} X)_+, \quad\text{i.e}\quad \dot X=e^{-tA}B(e^{tA} X)_+. $$ A third remark is that for $n=1$, you can separate the variables and get formally $$ \frac{dx}{ax+bx_+}=dt\quad\text{i.e.}\quad \frac{dX}{X_+}=bdt\quad\text{i.e.}\quad \ln (X_+(t)/X_+(0))=bt\quad\text{i.e.}\quad X_+(t)=e^{bt}X_+(0). $$ In fact we see that in one dimension $ \phi(t,y)=e^{ta}e^{bt} y\mathbf 1_{\mathbb R_+}(y)+ \mathbf 1_{\mathbb R_-}(y)e^{ta} y. $

Now for studying stability, you may for instance start with the case where $A$ is a self-adjoint real-valued matrix: you can diagonalize $A$ (and thus $e^{tA}$) in an orthonormal basis of $\mathbb R^n$. Since the matrix $e^{tA}$ is diagonal and positive, you get that $$ (e^{tA} X)_+=e^{tA} X_+, $$ and thus the system is $ \dot X=e^{-tA}Be^{tA} X_+. $ You need to study the spectrum of the matrix $e^{-tA}Be^{tA}$ and say if the matrix $B$ commutes with $A$, you are reduced to study the stability of $ \dot X=BX_+, $ and we may note that $$ \frac{d}{dt}\Vert X\Vert^2=2\langle B X_+, X\rangle\le C\Vert X\Vert^2, $$ and Gronwall's inequality gives you $\Vert X(t)\Vert^2\le e^{Ct}\Vert X(0)\Vert^2$.

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