23.1 Autonomous problems
- E418
[1QR] Prerequisites:3.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.
Hidden solution: [UNACCESSIBLE UUID ’1QS’]
Conditions ?? and ?? are a special case of Osgood uniqueness condition, see Problem 2.25 in [ 25 ] .
- E418
[1QV] Set \(𝛼{\gt}1\) and consider
\[ \begin{cases} x’ (t) = |x(t)|^𝛼~ ~ , \\ x (t_ 0 ) = x_ 0 ~ ~ \end{cases} \]with \(x_ 0,t_ 0∈ℝ\) fixed. Show that there is existence and uniqueness of the solution; calculate the maximal definition interval; Use the variable separation method to explicitly calculate solutions. (Since the equation is autonomous, one could assume that \(t_ 0=0\), but the example is perhaps clearer with a generic \(t_ 0\)).
Hidden solution: [UNACCESSIBLE UUID ’1QW’]
- E418
[1QX]What happens in the previous exercise in the case \(𝛼∈ (0,1)\)?
Hidden solution: [UNACCESSIBLE UUID ’1QY’]
- E418
[1QZ] Prerequisites:2.Let us fix \(𝛼{\gt}1\), and consider again
\[ \begin{cases} x’ (t) = |x(t)|^𝛼~ ~ , \\ x (0 ) = 1 ~ ~ \end{cases} \]We have seen in 2 that this ODE admits a maximal solution \(x : I_𝛼 →ℝ\). Fixed \(t∈ℝ\), show that \(t∈ I_𝛼\) for \(𝛼{\gt}1\) close to \(1\), and that \(\lim _{𝛼→ 1+} x(t)=e^ t\).
Note that \(e^ t\) is the only solution of \(x' (t) = |x(t)|\) with \(x (0 ) = 1\).
Hidden solution: [UNACCESSIBLE UUID ’1R0’]