Cauchy problem and unique solution Let $f: \Omega \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ be a continuous function on the open subset $\Omega$ which is non-increasing on the second variable, i.e., $f(t,x) \geq f(t,y)$, whenever $(t,x),(t,y) \in \Omega$, and $x \leq y$. Prove that any solution $\varphi : [t_0, \alpha] \rightarrow \mathbb{R}$ to te cauchy problem
$$
x'=f(t,x),\, x(t_0)=x_0
$$
is unique in the sense that if $\psi :[t_0, \beta] \rightarrow \mathbb{R}$ is another solution to this cauchy problem, then $\varphi = \psi$ on $[t_0, \gamma)$, where $\gamma=\min\{\alpha, \beta \}$.
 A: Since $x(t) = \varphi(t)$ and $x(t) = \psi(t)$ are solutions to the Cauchy problem, we then have
$$\begin{aligned}\varphi'(t) &= f(t,\varphi(t)), \; \varphi(t_0) = x_0 \\ \psi'(t) &= f(t,\psi(t)), \; \psi(t_0) = x_0.\end{aligned} $$
Let $t_1 = \inf\{t \in [t_0,\gamma):  \varphi(t) \neq \psi(t)\}$. Note that $t_1 > t_0$ as $\varphi(t_0) = x_0 = \psi(t_0)$ and by continuity of $\varphi - \psi$. At $t = t_1$, it is necessarily true that $\varphi'(t_1) \neq \psi'(t_1)$ and $\varphi(t_1) = \psi(t_1)$. Without loss of generality, suppose that $\varphi'(t_1) > \psi'(t_1)$.
$$ $$
If $t_1 \neq \gamma$, by continuity of $(\varphi - \psi)'$ (since $f$ is continuous), there exists an open neighbourhood of $t_1$, say $(t_1 - \varepsilon,t_1 + \varepsilon)$ (for sufficiently small $\varepsilon > 0$) such that $(\varphi - \psi)' > 0$, ie $$\varphi'(t) > \psi'(t). \tag{1}$$ Integrating from $t_1$ to $t$ and using $\varphi(t_1) = \psi(t_1)$, this implies that $$\varphi(t) > \psi(t).$$
Now, define $y(t) = \varphi(t) - \psi(t)$. By the given ODEs, on $t \in [t_1, t_1 + \varepsilon)$, we have
$$y'(t) = f(t,\varphi(t)) - f(t, \psi(t)) \leq 0$$
by the non-increasing property in the second variable for $f$. However, note that $$y'(t) = \varphi'(t) - \psi'(t) \leq 0 \tag{2}$$ by definition. $(1)$ thus contradicts $(2)$.
This implies that $t_1 = \gamma$, and thus $\varphi(t) = \psi(t)$ for all $t \in [t_0,\gamma)$.
