[proofplan] We show that every forward trajectory in a bounded forward-invariant interval converges to an equilibrium. Global forward existence follows from boundedness. The key one-dimensional constraint is that $f$ has constant sign between consecutive equilibria, forcing monotonicity of $y(t)$. A bounded monotone [function](/page/Function) converges, and the [limit](/page/Limit) must be an equilibrium since a nonzero [derivative](/page/Derivative) would drive the trajectory out of $\overline{I}$. [/proofplan] [step:Establish global forward existence from boundedness] Since $I = (a,b)$ is forward-invariant and bounded, the solution $y(t)$ remains in the compact set $\overline{I} = [a, b]$ for all forward time. By the extension theorem for ODEs, a solution that stays in a compact set cannot blow up in finite time, so $y(t)$ is defined for all $t \ge 0$. [/step] [step:Show $y(t)$ is eventually monotone using the intermediate value theorem] The function $t \mapsto y(t)$ is monotone on any time interval where $f(y(t), \mu)$ does not change sign. Partition $I$ into sub-intervals by the equilibria $\{y^* \in I : f(y^*, \mu) = 0\}$. On each open sub-interval between consecutive equilibria (or between an equilibrium and an endpoint of $I$), the sign of $f$ is constant by the [intermediate value theorem](/theorems/629): since $f(\cdot, \mu)$ is continuous and does not vanish on the sub-interval, it cannot change sign. Therefore $\dot{y}(t) = f(y(t), \mu)$ has constant sign on any time interval during which $y(t)$ remains in such a sub-interval, making $y(t)$ strictly monotone there. Once $y(t)$ enters a sub-interval between consecutive equilibria, it remains there: if $y(t)$ reached an equilibrium $y^*$, uniqueness of solutions would force $y(t) \equiv y^*$ for all subsequent time, which is consistent with convergence. Hence $y(t)$ is eventually monotone. [/step] [step:Prove convergence to an equilibrium] Since $y(t)$ is bounded (in $\overline{I}$) and eventually monotone, the limit $\ell := \lim_{t \to +\infty} y(t)$ exists. By [continuity](/page/Continuity) of $f$, the derivative satisfies $\dot{y}(t) = f(y(t), \mu) \to f(\ell, \mu)$ as $t \to +\infty$. If $f(\ell, \mu) \neq 0$, then $|\dot{y}(t)| \ge c > 0$ for some constant $c$ and all sufficiently large $t$, which would force $|y(t) - y(T)| \ge c(t - T) \to \infty$ as $t \to \infty$, contradicting the fact that $y(t)$ remains in $\overline{I}$. Hence $f(\ell, \mu) = 0$, and $y^* := \ell$ is an equilibrium in $\overline{I}$. [/step]