Let \(f:\mathbb R\to\mathbb R\) be bounded and continuously differentiable. Show that every solution of \[ y'(x)=f(y(x)) \] is monotone.
Solution
Let \(y\) be a solution on an interval \(I\).
1. If \(y'(x_0)=0\) at some \(x_0\in I\), then \(y\) is constant on\(I\).
From the differential equation, \[ 0=y'(x_0)=f(y(x_0)). \] Set \(c\coloneqq y(x_0)\). Then the constant function \[ z(x)\equiv c \] also satisfies \(z'=f(z)\).
Because \(f\) is \(C^1\), it is locally Lipschitz. Hence the initial-value problem \[ u'=f(u), \qquad u(x_0)=c \] has locally unique solutions. Thus \(y=z\) on a neighborhood of \(x_0\).
More generally, let \[ E\coloneqq\{x\in I:y(x)=c\}. \] The set \(E\) is closed by continuity. If \(x\in E\), then \(f(y(x))=f(c)=0\), so the same local-uniqueness argument shows that \(y\equiv c\) near \(x\); hence \(E\) is open in \(I\). Since \(I\) is connected and \(E\) is nonempty, \(E=I\).
2. Every nonconstant solution has derivative of one strict sign on\(I\).
If \(y\) is nonconstant, step 1 shows that \[ y'(x)\ne0 \qquad(x\in I). \] The derivative \[ y'=f\circ y \] is continuous. A continuous nonvanishing real-valued function on the connected interval \(I\) has constant sign. Thus either \(y'(x)>0\) for every \(x\in I\) or \(y'(x)<0\) for every \(x\in I\).
3. Every solution is monotone.
A constant solution is monotone. By step 2, every nonconstant solution is either strictly increasing or strictly decreasing.
4. Q.E.D.
Step 3 proves the assertion.