Let \(n>1\) be an integer. Is there a differentiable function \[ f:[0,\infty)\to\mathbb R \] such that \[ f'=f^n \] and \(f(0)>0\)?
Solution
No such function exists.
1. If such a function existed and\[ a\coloneqq f(0)>0, \]then\[ f(t)>0 \]for every \(t\geq0\).
Suppose instead that \(f\) vanishes somewhere. Since \(f(0)>0\) and \(f\) is continuous, let \(\tau>0\) be the first zero. Then \[ f(t)>0 \qquad 0\leq t<\tau. \] On this interval, \[ f'(t)=f(t)^n>0, \] so \(f\) is strictly increasing on \([0,\tau)\). Hence \[ f(t)\geq f(0)=a>0 \] there, and continuity gives \(f(\tau)\geq a>0\), contradicting \(f(\tau)=0\).
2. For every \(t\geq0\),\[ f(t)^{1-n} = a^{1-n}-(n-1)t. \]
By step 1, \(f(t)>0\), so \(f^{1-n}\) is differentiable. The chain rule and the differential equation give \[ \frac{d}{dt}f(t)^{1-n} =(1-n)f(t)^{-n}f'(t) = 1-n. \] Integrating from \(0\) to \(t\) yields the displayed identity.
3. The identity in step 2 is impossible on all of \([0,\infty)\).
Take \[ T\coloneqq\frac{a^{1-n}}{n-1}>0. \] Step 2 gives \[ f(T)^{1-n}=0. \] But step 1 gives \(f(T)>0\), so \(f(T)^{1-n}>0\). This is a contradiction.
4. Q.E.D.
Step 3 contradicts the existence of the required differentiable function. Therefore the answer is no.