According to Springer's Encyclopedia of Math entry on [differential inequalities](https://www.encyclopediaofmath.org/index.php/Differential_inequality), T. Wazewski proved in 1950 the following theorem:



> 
> Consider the system of differential inequalities given by
>  $$ \dot{x}\_i (t) > f\_i(t,x\_1, \ldots, x\_n), \ i = 1, \ldots, n. $$
>  If each $f\_i$ is non-decreasing with respect to the arguments $x\_j$
>  (for all $j \ne i$), the following estimate is valid:
>  $$ x\_i(t) > z\_i(t)\ \text{for}\ t\_0 < t < t\_M,\ i = 1, \ldots, n, \tag{A} \label{A} $$ where $z$ is the solution of the differential equation 
> 
> 
> $$ \dot{z}(t) = f(t,z),\ z(t\_0) = x(t\_0). $$
> 
> 
> 


(1) Can we drop the assumption that each $f\_i$ is non-decreasing? If not, it is possible to construct an explicit counter-example?


(2) If the assumption that each $f\_i$ is non-decreasing is not necessary, what are some current results that guarantee that \eqref{A} holds with weaker assumptions on $f$?



On Wazewski's theorem on system of differential inequalities
In dimension 1 it is true without the ordered condition, assuming that $x$ and $z$ are suitably differentiable. 


*Proof*: 


At $t = t\_0$, we have $x(t\_0) = z(t\_0)$, and $x'(t\_0) > z'(t\_0)$ by assumption. This implies that $x > z$ on some interval $(t\_0, t\_m)$ for $t\_m$ sufficiently close to $t\_0$. Let $T$ be the infimum of all times where $x(t) \leq z(t)$. Then by continuity we have $x(T) = z(T)$ and $x(t) > z(t)$ for all $t\in (t\_0,T)$. Since $x$ and $z$ are differentiable, we have that this means $x'(T) - z'(T) \leq 0$, a contradiction. 




---


In dimension 2 or higher there are *many* counterexamples. 


Define $f(x,y)$ to be the following function. 


1. In the first quadrant, $f(x,y) = (0,0)$ if $x < 1$ and $(0, 1-x)$ if $x \geq 1$.
2. In the second quadrant, $f(x,y) = (0,0)$.
3. In the third quadrant, $f(x,y) = (0,0)$ if $y > -1$ and $(0,1+y)$ if $y \leq -1$.
4. In the fourth quadrant, take $f(x,y) = (0,0)$ within the unit circle, and outside the function $\frac{r-1}{r} (y,-x)$ where $r = \sqrt{x^2 + y^2}$.


One can check easily that $f$ is continuous. 


Consider the solution to $\dot{z} = f(z)$ with $z(0) = (0,0)$. The solution is trivially $z(t) = z(0)$. 


Consider now the curve $\zeta(t)\in \mathbb{R}^2$ described as follows. 


1. $\zeta(t) = (t,t)$ for $t \in [0,9]$. Smoothly join, observing the condition that $\zeta'\_1 > 0$ and $\zeta'\_2 \geq -1$ to
2. $\zeta(t) = (t,20-t)$ for $t \in [11,39]$. Smoothly join, observing the condition that $\zeta'\_1 > -1/50$ and $\zeta'\_2 > -1$ to
3. $\zeta(t) = ( (40-t)/100 + 40, (t-40)/1000000 - 20)$ for $t\in [40,4041]$.


You can check that $\zeta'(t) > f(\zeta)$ throughout. But $\zeta\_2(4041) < z\_2(4040) = 0$ and $\zeta\_1(4041) < 0 = z(4041)$. 


(If you just want one component violated, $t = 21$ will do in the example above; the example can be also made much simpler in this case: for instance


Let $f(x,y) = (-y,x)$. Let $z(0) = \zeta(0) = (0,0)$. Let $z$ solve $z' = f(z)$ and let $\zeta'$ solve $\zeta' = f(z) + (1,1)$. It is pretty simple to show that $\zeta$ enters the second quadrant in finite time.) 


