Solution of homogeneous differential equation stays above zero solution

Question

Let $I \subseteq R$ be an interval such that $0 \in I$ and $f \in C^1(\mathbb{R},\mathbb{R})$ with $f(0) = 0$. For some $t_0 \in I, y_0 \in \mathbb{R}$, consider the initial value problem $$ \begin{cases} y'(t) = f(y(t)), \\ y(t_0) = y_0. \end{cases} $$ Show that for every solution $y \colon I \rightarrow \mathbb{R}$ with $y(0) = y_0, y_0 > 0$, it holds: $$ \forall t \in I \colon y(t) > 0. $$

Answer 1

Note that, since $f(0) = 0$, the zero function is a global solution for $y_0 = 0$ for any starting point $t_0 \in I$.

Let $y \colon I \rightarrow \mathbb{R}$ be a solution with $y(0) = y_0 > 0$. If the proposition is false, there exists a $T \in I$, such that $y(T) \le 0$. Since $y$ is a solution, it is continuous and by the intermediate value theorem, we can find a $t_0$ between $0$ and $T$ with $y(t_0) = 0$. As $f$ is continuously differentiable, by Picard-Lindelöf the solution to every $(t_0, y_0) \in I \times \mathbb{R}$ is unique. But since $0$ is a solution to $y_0 = 0$ and any $t_0 \in I$, $y$ must be the zero function. Then, we have the contradiction $$ 0 = y(0) = y_0 > 0. $$ Therefore, the assumption is false, and $y(t) > 0$ for all $t \in I$.