Fundamental ODE Theorems (Existence + Uniqueness)

Lee Smooth Manifolds

Lemma) Let $J\subset \mathbb{R}$ be an open interval, and suppose the differentiable function $u:J\to {\mathbb{R}}^n$ satisfies the following inequality for all $t\in J$:

$$|u^{\prime }(t)|\le f(|u(t)|)$$

where $f:[0,\infty )\to [0,\infty )$ is Lipschitz continuous. If for some $t_0\in J$, $v:[0,\infty )\to [0,\infty )$ is a differentiable real-valued function satisfying the initial-value problem

$$v^{\prime }(t)=f(v(t)),v(0)=|u(t_0)|$$

then for all $t\in J$,

$$|u(t)|\le v(|t-t_0|)$$

Proof) WLOG assume $t_0=0$. We prove the theorem for $t\in J^{+}=\{t\in J|t>0\}$; the rest of the theorem follows from substituting $t$ with $-t$ throughout.

On the open subset of $J^{+}$ in which $|u(t)|>0$, $|u(t)|$ is differentiable and

$$\frac{d}{dt}|u(t)|=\frac{d}{dt}(u(t)\cdot u(t){)}^{1/2}=\frac{1}{2}(u(t)\cdot u(t){)}^{-1/2}(2u(t)\cdot u^{\prime }(t))\le |u^{\prime }(t)|\le f(|u(t)|)$$

Let $A$ be a Lipschitz constant for $f$ and consider the function $w(t)=e^{-At}(|u(t)|-v(t))$. It suffices to show that $w(t)\le 0$ for all $t\in J^{+}$. If some $t\in J^{+}$ satisfies $w(t)>0$, which implies $|u(t)|>v(t)\ge 0$. Thus, $w(t)$ is differentiable and

$$\frac{d}{dt}w(t)\le -A{e}^{-At}(|u(t)|-v(t))+e^{-At}(f(|u(t)|)-f(v(t)))\le 0$$

Assume some $t^{\prime }\in J^{+}$ satisfies $w(t^{\prime })>0$, and let $\tau =\sup \{t\in [0,t^{\prime })|w(t)\le 0\}$ (note $w(0)=0$). By continuity, $w(\tau )=0$ and for all $\gamma \in (\tau ,t^{\prime }]$, $w(\gamma )>0$. $w$ is differentiable in the interval $(\tau ,t^{\prime })$, and there exists some $\gamma \in (\tau ,t^{\prime })$ such that $w^{\prime }(\gamma )(t^{\prime }-\tau )=w(t^{\prime })>0\Rightarrow w^{\prime }(\gamma )>0$. This is an immediate contradiction, implying $w(t)\le 0$ for all $t\in J^{+}$.

(Existence) Let $U\subset {\mathbb{R}}^n$ be an open subset and $V:U\to {\mathbb{R}}^n$ be locally Lipschitz continuous. Let $(t_0,x_0)\in \mathbb{R}\times U$ be given. There exists an open interval $J_0\subset \mathbb{R}$ containing $t_0$ and an open subset $U_0\subset U$ containing $x_0$ such that for each $c\in U_0$, there is a $C^1$ map $y:J_0\to U$ satisfying the following initial value problem:

$$\dot{y}(t)=V(y^1(t),y^2(t),\dots ,y^n(t)),y(t_0)=c$$

Proof) Given $(t_0,x_0)$, we can shrink $U$ as needed to make $V$ Lipschitz continuous with Lipschitz constant $C$. Note that a $C^1$ map $y$ being a solution to the IVT with some $c\in U$ is equivalent to $y$ being a $C^1$ solution to the integral equation

$$y(t)=c+{\int }_{t_0}^{t}V(y(s))ds$$

If we define the map $I_c$ on the function space of continuous functions $J_0\to U$ such that

$$I_{c}y(t)=c+{\int }_{t_0}^{t}V(y(s))ds$$

the problem boils down to finding a fixed point of $I_c$ for some $c\in U$. If a fixed point exists, it is forced to be $C^1$ by the fundamental theorem of calculus, and we are done.

Given $(t_0,x_0)$, choose $r>0$ such that $\overline{B(x_0,r)}\subset U$, and let $M$ be the supremum of $|V|$ on the compact set $\overline{B(x_0,r)}$. Choose $\delta ,ϵ>0$ such that $\delta <r/2$ and $ϵ<\min (r/(2M),1/C)$, and set $J_0=(t_0-ϵ,t_0+ϵ)$. For any $c\in B(x_0,\delta )$, let ${\mathcal{M}}_c$ be the set of continuous maps $y:J_0\to \overline{B(x_0,r)}$ such that $y(t_0)=c$ equipped with the supremum metric topology (which is well-defined as the codomain is bounded). As ${\mathcal{M}}_c$ is a closed subset of a complete function space, it is complete. It is on ${\mathcal{M}}_c$ that we will apply the Banach fixed point theorem; it suffices to show ${\mathcal{M}}_c$ is closed under $I_c$ and that $I_c$ is a contraction mapping.

First, if $y\in {\mathcal{M}}_c$, $I_{c}y\in {\mathcal{M}}_c$. $I_{c}y$ clearly satisfies $I_{c}y(t_0)=c$, and $I_{c}y$ is continuous. For any $t\in J_0$,

$$|I_{c}y(t)-x_0|\le |c-x_0|+{\int }_{t_0}^t|V(y(s))|ds\le \delta +Mϵ<r$$

Thus, $I_{c}y\in {\mathcal{M}}_c$. Furthermore, if $y,\tilde{y}\in {\mathcal{M}}_c$,

$$d(Iy,I\tilde{y})=\underset{t\in J_0}{\sup }|{\int }_{t_0}^{t}V(y(s))ds-{\int }_{t_0}^{t}V(\tilde{y}(s))ds|\le \underset{t\in J_0}{\sup }{\int }_{t_0}^t|V(y(s))-V(\tilde{y}(s))|ds$$ $$\le \underset{t\in J_0}{\sup }{\int }_{t_0}^{t}C|y(s)-\tilde{y}(s)|ds\le Cϵd(y,\tilde{y})$$

as $Cϵ<1$, this is a contraction mapping, and we are done by the Banach fixed point theorem.

(Uniqueness) Let $U\subset {\mathbb{R}}^n$ be an open subset and $V:U\to {\mathbb{R}}^n$ be locally Lipschitz continuous. For any $t_0\in \mathbb{R}$ and $c\in U$, any two differentiable solutions $y,\tilde{y}$ (with open interval domains) to the IVP are equal on their common domain.

Proof) Suppose $y,\tilde{y}:J_0\to U$ are both differentiable functions that solve the IVT, where $J_0$ is an open interval of $\mathbb{R}$. Let $J_1$ be an open interval containing $t_0$ such that $\overline{J_1}\subset J_0$. Note that $y(\overline{J_1})\cup \tilde{y}(\overline{J_1})$ is compact in $U$, and thus there is a Lipschitz constant $C$ for $V$ on that subspace. This implies, for any $t\in \overline{J_1}$,

$$|\frac{d}{dt}(\tilde{y}(t)-y(t))|=|V(\tilde{y}(t))-V(y(t))|\le C|\tilde{y}(t)-y(t)|$$

By Lemma, for any $t\in J_1$,

$$|\tilde{y}(t)-y(t)|\le e^{C|t-t_0|}|\tilde{y}(t_0)-y(t_0)|=0$$

By continuity, $y(t)=\tilde{y}(t)$ for all $t\in \overline{J_1}$. As for any $t_1\in J_0$ there is a closed interval that covers $t_1$ and $t_0$, this implies $y(t)=\tilde{y}(t)$ for all $t\in J_0$. Note that uniqueness has a slightly broader scope than existence does. Furthermore, by continuity we can make our argument work for any differentiable solution on any interval domain that isn’t necessarily open.

(Remark) When discussing ODE solutions we will really only consider solutions with open interval domains in the time dimension because then every point is an interior point in time and so the calculus is neat without weird one-sided derivatives.