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+--- Thread: Recurrence relations and differential equations (/showthread.php?tid=1819)
Recurrence relations and differential equations - Natsugou - 05/06/2026
I'd like to hear your thoughts on why recurrence relations are considered the discrete version of differential equations.
I'm trying to generalize recurrence relations and autonomous differential equations.
I think that \(\mathbb{R}_+\)-actions are the straightforward continuous version of \(\mathbb{N}\)-actions, so I checked whether an orbit of a \(\mathbb{R}_+\)-action is always a solution (an integral curve) of an autonomous differential equation, and the opposite.
\(\gamma: \mathbb{R} \to \mathbb{R}^n\) is an orbit of a \(\mathbb{R}\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u \in \mathbb{R}\).
\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action if and only if \(\gamma(s) = \gamma(t) \implies \gamma(s+u) = \gamma(t+u)\) for all \(u > 0\).
\(\gamma\) is a solution of an autonomous differential equation \(\gamma'(t) = F(\gamma(t))\) if and only if \(\gamma(s) = \gamma(t) \implies \gamma'(s) = \gamma'(t)\). (Define \(F(x) := \gamma'(t)\) when \(x = \gamma(t)\). The \(F\) is well-defined on \(\gamma(\mathbb{R})\) because of the hypothesis.)
From the Picard–Lindelöf theorem, if \(F: \mathbb{R}^n \to \mathbb{R}^n\) is Lipschitz continuous, a solution \(\gamma: \mathbb{R} \to \mathbb{R}^n\) of a differential equation \(\gamma'(t) = F(\gamma(t))\) is always an orbit of a \(\mathbb{R}\)-action.
Trivially an orbit of a \(\mathbb{R}\)-action is an orbit of a \(\mathbb{R}_+\)-action.
If an orbit of a \(\mathbb{R}_+\)-action is differentiable, it's always a solution of an autonomous differential equation. (The proof is described below.)
Therefore let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, then
\[
\begin{array}{ll}
& \text{there exists a Lipschitz continuous \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)} \\
\implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}\)-action} \\
\implies & \text{\(\gamma\) is an orbit of a \(\mathbb{R}_+\)-action} \\
\implies & \text{there exists \(F: \mathbb{R}^n \to \mathbb{R}^n\) such that \(\gamma'(t) = F(\gamma(t))\)}.
\end{array}
\]
The opposites are not true.
An orbit of a \(\mathbb{R}\)-action that is not a solution of an autonomous differential equation of a Lipschitz continuous function:
Since
\[
\gamma(t) =
\begin{cases}
-t^2 & (t \leq 0) \\
t^2 & (t \geq 0)
\end{cases}
\]
is injective, the \(\gamma: \mathbb{R} \to \mathbb{R}^1\) is an orbit of \(\mathbb{R}\)-action.
Assume there exists a Lipschitz continuous \(F: \mathbb{R}^1 \to \mathbb{R}^1\) such that \(\gamma'(t) = F(\gamma(t))\).
Then there exists \(L \in \mathbb{R}\) such that \(|\gamma'(t_1) - \gamma'(t_2)| = |F(\gamma(t_1)) - F(\gamma(t_2))| \leq L|\gamma(t_1) - \gamma(t_2)|\) for all \(t_1, t_2 \in \mathbb{R}\).
Let \(t_1 \neq 0\) and \(t_2 = 0\).
We see \(\gamma'(t) = 2|t|\), so \(2|t_1| \leq L|t_1^2|\). Therefore
\[
\left|\frac{1}{t_1}\right| = \left|\frac{t_1}{t_1^2}\right| \leq \frac{L}{2}.
\]
\(\left|\frac{1}{t_1}\right| \longrightarrow \infty\) when \(t_1 \longrightarrow 0\), that is a contradiction.
A differentiable orbit of a \(\mathbb{R}_+\)-action that is not an orbit of a \(\mathbb{R}\)-action:
\[
\gamma: \mathbb{R} \to \mathbb{R}^1, \quad
\gamma(t) =
\begin{cases}
t^2 & (t \leq 0) \\
0 & (t \geq 0)
\end{cases}
\]
is an example of it.
A solution of an autonomous differential equation that is not an orbit of a \(\mathbb{R}_+\)-action:
\[
\gamma: \mathbb{R} \to \mathbb{R}^1, \quad
\gamma(t) =
\begin{cases}
0 & (t \leq 0) \\
t^2 & (t \geq 0)
\end{cases}
\]
is a solution of \(\gamma'(t) = F(\gamma(t))\) where
\[
F: \mathbb{R}^1 \to \mathbb{R}^1, \quad
F(x) =
\begin{cases}
0 & (x \leq 0) \\
2\sqrt{x} & (x \geq 0),
\end{cases}
\]
while \(\gamma(-1) = \gamma(0)\) and \(\gamma(-1+1) = 0 \neq 1 = \gamma(0+1)\).
Proof of that a differentiable orbit of a \(\mathbb{R}_+\)-action is a solution of an autonomous differential equation:
We'll prove the contraposition.
Let \(\gamma: \mathbb{R} \to \mathbb{R}^n\) is differentiable, \(\gamma(t) = (\gamma_1(t), \dots, \gamma_n(t))\). (\(\gamma'(t)\) denotes \((\gamma_1'(t), \dots, \gamma_n'(t))\).)
Assume \(\gamma\) is not a a solution of an autonomous differential equation.
Then there are \(t_1, t_2 \in \mathbb{R}\) such that \(\gamma(t_1) = \gamma(t_2)\) and \(\gamma'(t_1) \neq \gamma'(t_2)\).
Therefore \(\gamma'_i(t_1) \neq \gamma'_i(t_2)\) for some \(i\).
Define \(x = \gamma_i(t_1) = \gamma_i(t_2)\), \(\alpha = \gamma_i'(t_1)\), and \(\beta = \gamma_i'(t_2)\). We can assume \(\alpha < \beta\) w.l.o.g.
We know
\[
\forall \varepsilon_1 > 0, \exists \delta_1 > 0, \forall h \in (-\delta_1, \delta_1), \left|\frac{\gamma_i(t_1+h)-x}{h} - \alpha\right| < \varepsilon_1
\]
and
\[
\forall \varepsilon_2 > 0, \exists \delta_2 > 0, \forall h \in (-\delta_2, \delta_2), \left|\frac{\gamma_i(t_2+h)-x}{h} - \beta\right| < \varepsilon_2.
\]
Thus
\[
x + \alpha h - \varepsilon_1|h| < \gamma_i(t_1+h) < x + \alpha h + \varepsilon_1|h|,
\]
\[
x + \beta h - \varepsilon_2|h| < \gamma_i(t_2+h) < x + \beta h + \varepsilon_2|h|.
\]
Since \(\beta - \alpha > 0\) there are \(\varepsilon, \varepsilon' \in \mathbb{R}\) such that \(0 < \varepsilon < \varepsilon' < \beta - \alpha\).
Substitute \(\varepsilon_1 = \varepsilon\) and \(\varepsilon_2 = \beta - \alpha - \varepsilon'\), then for any \(0 < h < \min\{\delta_1, \delta_2\}\),
\begin{align*}
\gamma_i(t_1+h) &< x + \alpha h + \varepsilon_1|h| \\
&= x + \alpha h + \varepsilon h \\
&< x + \alpha h + \varepsilon'h \\
&= x + \beta h - (\beta - \alpha - \varepsilon')h \\
&= x + \beta h - \varepsilon_2|h| \\
&< \gamma_i(t_2+h).
\end{align*}
That is \(\gamma(t_1+h) \neq \gamma(t_2+h)\).
Therefore \(\gamma\) is not an orbit of a \(\mathbb{R}_+\)-action.
From these results, I lost sight of what is the relation between an autonomous differential equation and a \(\mathbb{R}_+\)-action (or a \(\mathbb{N}\)-action).
I was hoping you could help me find that relation.