Global Existence Theorem

• These notes discuss the theorem of global existence and provides an example

Theorem (Global Existence)

• If $f \in C^1 (\mathbb{R} \times \mathbb{R}^n, \mathbb{R}^n)$ and there exists a continuous function $M: \mathbb{R} \rightarrow \mathbb{R}$ such that $|f(t,y)| \leq M(t)(1 + |y|)$ for all $(t,y) \in \mathbb{R} \times \mathbb{R}^n$, then for any $(t_0, y_0) \in \mathbb{R} \times \mathbb{R}^n$, there exists a unique $y \in C^1(\mathbb{R}, \mathbb{R}^n)$ such that $y'(t) = f(t, y(t))$ for all $t \in \mathbb{R}$ and $y(t_0) = y_0$.

Proof

• Let $y$ be the solution to the initial value problem defined on the maximal interval of existence $(t_, t^)$.
• The goal is to show that $t_* = -\infty$ and $t^* = \infty$.
• Assume $t^* < \infty$, and show that this leads to a contradiction.
• From the integral equation: $y(t) = y_0 + \int_{t_0}^t f(s, y(s)) ds$
• Then $|y(t)| \leq |y_0| + \int_{t_0}^t |f(s, y(s))| ds \leq |y_0| + \int_{t_0}^t M(s)(1 + |y(s)|) ds = |y_0| + \int_{t_0}^t M(s) ds + \int_{t_0}^t M(s) |y(s)| ds$
• Let $\mu(t) = |y_0| + \int_{t_0}^t M(s) ds$, then $\mu'(t) = M(t)$. Thus, $|y(t)| \leq \mu(t) + \int_{t_0}^t \mu'(s) |y(s)| ds$
• By Gronwall's inequality: $|y(t)| \leq \mu(t) e^{\int_{t_0}^t \mu'(s) ds} = \mu(t) e^{\mu(t) - \mu(t_0)}$
• Since $\mu(t) = |y_0| + \int_{t_0}^t M(s) ds$, $\mu(t)$ is well-defined up to $t^$, and $\mu(t)$ is bounded on $[t_0, t^]$.
• Thus, $|y(t)|$ is bounded on $[t_0, t^*)$.
• So $\lim_{t \to t^{*-}} y(t)$ exists.
• Let $y^* = \lim_{t \to t^{*-}} y(t)$.
• Extend the solution beyond $t^$ by setting $y(t^) = y^*$.
• This contradicts the assumption that $(t_, t^)$ is the maximal interval of existence.
• Therefore, $t^* = \infty$. Similarly, it can be shown that $t_* = -\infty$.

Example

• Consider the initial value problem: $y' = y^2$, $y(0) = 1$
• $f(y) = y^2$ is not globally Lipschitz.
• Check if $|f(y)| = |y^2| \leq M(1 + |y|)$ for some $M$.
• This implies $y^2 \leq M + M|y|$, or $y^2 - My - M \leq 0$.
• The roots of $y^2 - My - M = 0$ are $\frac{M \pm \sqrt{M^2 + 4M}}{2}$.
• So if $y \in [\frac{M - \sqrt{M^2 + 4M}}{2}, \frac{M + \sqrt{M^2 + 4M}}{2}]$, then $y^2 - My - M \leq 0$.
• But this does not hold for all $y \in \mathbb{R}$, so we cannot find such an $M$.
• Solving the ODE: $\int \frac{dy}{y^2} = \int dt \implies -\frac{1}{y} = t + c$
• Using the initial condition $y(0) = 1$, we get $c = -1$.
• $-\frac{1}{y} = t - 1 \implies y = \frac{1}{1 - t}$
• The maximal interval of existence is $(-\infty, 1)$.

Notes discussing the global existence theorem proof. Includes the theorem statement, the proof, and an example. The theorem gives conditions for solutions of differential equations to exist for all time.