Comparison theorem for systems of ODE

Question

Let vector-function $x(t)$ satisfy a differential equation $$ \dot x = f(x), $$ and a vector-function $y($t) satisfy a differential inequality $$ \dot y \leq f(y) $$ with starting positions $y(0) < x(0)$. If a function $f(x)$ satisfies the property: $$ f_{i}(x_1+\alpha_1,\ldots,x_{i-1}+\alpha_{i-1},x_i,x_{i+1}+\alpha_{i+1},\ldots,x_{n}+\alpha_{n}) \geq f_{i}(x_1,\ldots,x_n) $$ for any $\alpha_{1} \geq 0, \ldots, \alpha_{n} \geq 0$ (i.e. it is quasimonotone), then $y(t) \leq x(t)$ for any $t>0$. Function $f(x)$ is smooth.

Is there a name for such theorem? Please help me to proof it or give me a reference.

Answer 1

This theorem is known in Russia as the Chaplygin lemma. It can be proved as follows. Suppose that it isn't true. Then let $$ t^{*} = \inf \{ t \geqslant 0 \mid \exists i\colon y_{i}(t) > x_{i}(t) \} < \infty $$ By definition of $t^{*}$ we have that $y_{i}(t^{*}) = x_{i}(t^{*})$ and for any $j \neq i$ we have $y_{j}(t^{*}) \leqslant x_{j}(t^{*})$. Then by the quasimonotony propertie we have $$ f_{i}(y(t^{*})) \leqslant f_{i}(x(t^{*})) \tag{0} $$ On the other hand by the definition of $t^{*}$ there exists some small $\delta > 0$ such that $$ y_{i}(t^{*}+\Delta t) > x_{i}(t^{*} + \Delta t) \tag{1} $$ for any $0 < \Delta t < \delta$. Then $$ \dot y_{i}(t^{*}) \geqslant \dot x_{i}(t^{*}) = f_{i}(x(t^{*})) \tag{2} $$ because the opposite inequality implies contradiction with $(1)$. There may occur two different situations.

1. $\dot y_{i}(t^{*}) < f_{i}(y(t^{*}))$. From $(2)$ it follows that $f_{i}(y(t^{*})) > f_{i}(x(t^{*}))$. This is a contradiction with $(0)$.
2. $\dot y_{i}(t^{*}) = f_{i}(y(t^{*}))$. Then consider a solution $y_{\varepsilon}(t)$ of differential inequality $$ \dot y_{\varepsilon} \leqslant f(y_{\varepsilon})-\varepsilon $$ From the first case we have that $y_{\varepsilon}(t) \leqslant x(t)$ for any $t$. Then let $\varepsilon \to 0^{+}$ and use that the solution depends continuously of parameter $\varepsilon$.

Theorem is proved.