# On Wazewski's theorem on system of differential inequalities

According to Springer's Encyclopedia of Math entry on differential inequalities, 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 $f$?

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.)