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E5

[1QR] Prerequisites:[19S].Let su fix \(x_ 0,t_ 0∈ℝ\), and a bounded and continuous function \(f:ℝ→ℝ\), with \(f(x_ 0)=0\) but \(f(x){\gt}0\) for \(x≠ x_ 0\). We want to study the autonomous problem

\[ \begin{cases} {x}’ (t) = f(x(t))~ ~ , \\ x (t_ 0 ) = x_ 0 ~ ~ .\end{cases} \]

Note that \(x≡ x_ 0\) is a possible solution. Show that if, for \(\varepsilon {\gt}0\) small,  1

\begin{eqnarray} ∫_{x_ 0}^{x_ 0 + 𝜀} \! \frac{1}{f (y)}\, {\mathbb {d}}y = ∞ \label{eq:Osgood_ dei_ poveri_ dx} \\ ∫_{x_ 0-\varepsilon }^{x_ 0 } \! \frac{1}{f (y)}\, {\mathbb {d}}y=∞ \label{eq:Osgood_ dei_ poveri_ sx} \end{eqnarray}

then \(x≡ x_ 0\) is the only solution; while otherwise there are many class \(C^ 1\) solutions: describe them all.

Solution 1

[1QS]

Conditions ?? and ?? are a special case of Osgood uniqueness condition, see Problem 2.25 in [ .

  1. If the condition holds for a \(\varepsilon {\gt}0\) then it holds for every \(\varepsilon {\gt}0\), since \(f{\gt}0\) far from \(x_ 0\).
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  • Osgood uniqueness condition
  • ODE
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