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The following question requires the use of an easy theorem of calculus, but I am failing to see which one.

Let $ g $ be non-constant and $ C^{1} $ on some interval $ I $. Show that for some subinterval $ J \subseteq I $, there exists a continuous function $ f $ such that the differential equation $ y' = f(y) $ has the solution $ y = g(x) $ on $ J $.

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Don't you mean $y = g(x)$? –  Javier Badia Jan 15 '13 at 7:28
    
Sorry! Mistake in notation! Corrected. –  user44069 Jan 15 '13 at 7:30
    
Isn't this just the fundamental theorem of calculus, in that differentiation and integration are loosely "inverse" operations? Since $g$ is $C^1$ we can always try $f(x)=g'(x)$ and $f$ will be continuous. –  kigen Jan 15 '13 at 7:33
    
@Stefan: Did you actually mean ‘$ y'(x) = f(x) $’ instead of ‘$ y'(x) = f(y(x)) $’? –  Haskell Curry Jan 15 '13 at 8:25
    
I am not really sure. I typed the problem as I have it in front of me. –  user44069 Jan 15 '13 at 8:30
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1 Answer

  • Suppose that $ g: I \to \mathbb{R} $ is non-constant and $ g \in {C^{1}}(I) $.

  • Observe that $ g': I \to \mathbb{R} $ cannot be identically $ 0 $ on $ I $, otherwise by the Mean-Value Theorem, $ g $ would be a constant function.

  • Hence, there exists an $ x_{0} \in I $ such that $ g'(x_{0}) \neq 0 $. Without loss of generality, let us assume that $ g'(x_{0}) > 0 $.

  • As $ g' $ is continuous on $ I $, there exists an open subinterval $ J $ of $ I $ such that $ x_{0} \in J $ and $ g' > 0 $ on $ J $. Intuitively speaking, as $ g'(x_{0}) > 0 $, the continuity of $ g' $ ensures that $ g'(x) > 0 $ for all points $ x \in I $ that are near $ x_{0} $.

  • By the Mean-Value Theorem, $ g|_{J} $ must be strictly increasing.

  • Hence, $ (g|_{J})^{-1}: g[J] \to J $ exists and is continuous on $ g[J] $, which is an open interval.

  • We now need to find a continuous $ f: g[J] \to \mathbb{R} $ such that $$ \forall x \in J: \quad g'(x) = f(g(x)). $$

  • If such an $ f $ exists, then it is necessary that $$ \forall x \in g[J]: \quad g'({(g|_{J})^{-1}}(x)) = f(g({(g|_{J})^{-1}}(x))) = f(x). $$

  • This implies that $ f = g' \circ (g|_{J})^{-1} $, which is continuous on $ g[J] $ because it is the composition of two continuous functions.

  • A quick check shows that $ f $ as defined does satisfy the requirements.

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