Math 63CM Notes
===
Because we're starting to lose track of theorems *again*.

There will be no linear algebra. (If you want to write it, go ahead!)
## ODEs
*Based off Kaya’s notes. Thanks Kaya!*
*Aw ofc any time!*
### Lipschitz and friends
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**Setup.** Let $U\subset \mathbb{R}^n$ be open, and $F:U\rightarrow \mathbb{R}^n$.

Also for the last one we will need a $G:\mathbb{R}^n\times [0,T]\rightarrow \mathbb{R}^n$
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**Def.** $F$ is *Lipschitz* if there exists $C$ such that
$$|F(x)-F(y)|< C|x-y|$$

**Def.** $F$ is *locally Lipschitz* if for each point, there exists a ball around $p$ in which $F$ is Lipschitz.

**Def.** $\{f_n\}$, a sequence of functions, is *uniformly bounded* if there exists a single bound $M$ for all $f_n$:
$$|f_n(x)| < M\qquad \forall x\in U, n\in\mathbb{N}$$

**Def.** $\{f_n\}$, a sequence of functions, is *uniformly Lipschitz* if there exists a single Lipschitz constant $C$ for all $f_n$:
$$|f_n(x)-f_n(y)|<C|x-y|\qquad \forall x,y\in U, n\in\mathbb{N}$$

**Def.** $\{f_n\}$, a sequence of functions, if *uniformly convergent* (across $x$) if
$$\lim_{N\rightarrow\infty}\sup_{n\ge N, x} |f_n(x) - f(x)| = 0$$

**Def.** $G$ is *uniformly Lipschitz across $t$* if there exists a single Lipschitz constant $C$ for all $G(\cdot,t)$:
$$|G(x,t)-G(y,t)|<C|x-y|\qquad x,y\in\mathbb{R}^n, t\in [0,T]$$

### Existence / Uniqueness
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**Setup.** Let $U\subset \mathbb{R}^n$ be open and $F:U\rightarrow \mathbb{R}^n$ be continuous. $x_0\in U$. Consider
$$x'(t)=F(x(t)),\qquad x'(0)=x_0$$
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**Thm.** (Peano Existence Theorem) For some $\delta>0$, exists solution on $[0,\delta]$.

To prove this, we also used:

**Thm.** (Arzela-Ascoli) If $K\subset \mathbb{R}^d$ is compact, and $f_n:K\rightarrow \mathbb{R}$ is uniformly Lipschitz and uniformly bounded, then $f_n$ has a subsequence that converges uniformly.

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*Note.* The general version uses *equicontinuous* instead of uniformly Lipschitz.
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**Thm.** Suppose $F$ is locally Lipschitz, and $x,y$ are solutions on interval $I$. If $x,y$ match at any time $t_0\in I$ then they match along all of $I$.

**Def.** Let $\mathcal{A}$ be the family of intervals around $x_0$ for which there exists a solution. By the above, we can "stitch" the solutions for two intervals $I,J\in\mathcal{A}$ together for a solution on $I\cap J$. Let the *maximal interval* $J_{x_0}$ (w.r.t $x_0$) be
$$J_{x_0} = \bigcup_{I\in \mathcal{A}} I$$

**Thm.** If the maximal interval is $(\alpha,\beta)$ with $\beta<\infty$, then as $t\nearrow \beta$, either $x$ approaches the boundary of $U$ or $\frac{1}{|x|}\rightarrow 0$.

**Thm.** (Picard Iteration) Exists $\delta>0$ such that there exists a unique solution $x$ on the interval $[0,\delta]$.
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Might need the very first fact from "Flows".
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### Solving ODEs
I'm lazier than integrating factors are hard.

### Non-autonomous systems
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**Setup.** Let $F:\mathbb{R}^n\times [0,T]\rightarrow\mathbb{R}^n$, $x_0\in\mathbb{R}^n$. Consider:
$$x'(t)=F(x(t),t)\qquad x'(0)=x_0$$
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**Thm.** If $F$ is Lipschitz then there exists a unique solution on $[0,T]$. (Follows immediately from autonomous case!)
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Where did we jump from existence over $[0,\delta]$ for some $\delta>0$ to existence over $[0,T]$?
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**Thm.** $F$ is uniformly Lipschitz across $t$.Then there exists a unique solution on $[0,T]$.

*Corollary*. $A:[0,T]\rightarrow \mathbb{R}^{n\times n}$ is continuous. Then there exists a unique solution to
$$x'(t) = A(t)x(t)\qquad x(0)=x_0$$

*Corollary*. Under same hypothesis (i.e. F uniformly Lipschitz across t?), there is a unique solution $M:[0,T]\rightarrow \mathbb{R}^{n\times n}$ where
$$M'(t) = A(t)M(t)\qquad M(0)=I$$

**Thm.** (Taylor) $F$ is a $C^1$ map. If $y$ is in a small enough closed ball around $p$ such that
$$|DF(y)-DF(p)|_{\mathrm{op}}\le \varepsilon$$ then
$$|F(y) - F(p)-DF(p)(y-p)|\le \varepsilon|y-p|$$

### Flows
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**Setup.** Let $U\subset \mathbb{R}^n$ be open and $F:U\rightarrow \mathbb{R}^n$ be locally Lipschitz. $x_0\in U$. Consider
$$x'(t)=F(x(t)),\qquad x'(0)=x_0$$
:::

**Thm.** Let $x$ be a solution on $[0,T)$. For any $\varepsilon>0$, there exists $\delta$ such that any maximal solution $y$ with initial condition $y_0\in (x-\delta,x+\delta)$, $[0,T]\subset J_{y_0}$ and $|y(t)-x(t)|<\varepsilon$ (For all $t\in [0,J]$).

**Def.** The *flow* is basically all solutions across all initial conditions: $\Phi(x_0,t)$ where $\Phi(x_0,\cdot)$ is a **maximal** solution for the initial condition $x_0$.

We know $[0,T]\in J_{y_0}$ for all $y_0\in \overline{B(x_0, \varepsilon)}$ (?) and $\Phi(y_0,t)$ depends continuously on $y_0$ for all $y_0\in \overline{B(x_0, \varepsilon)}$ (?)
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help
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**Thm.** If $F$ is $C^1$ then $\Phi:\overline{B(x_0, \varepsilon)}\times [0,T]$ is $C^1$.

**Thm.** Ok wtf is this theorem (Isn't it just saying that the flow associated with a $C^1$ vector field is also $C^1$? -Kaya)

### Liouville's Theorem
**Thm.** Suppose $M(t),A(t)\in\mathbb{R}^{n\times n}$:
$$M'(t) = A(t)M(t)$$
Then $$\frac{d}{dt}(\det M(t)) = (\operatorname{tr} A(t))(\det M(t))$$

**Corollary.** Suppose $F:U\subset\mathbb{R}^n \rightarrow\mathbb{R}^n$ is $C^1$, and $\varphi$ is the flow. Then $\operatorname{tr} DF(x)=0$ for all $x\in U$ implies that $\varphi_t$ is "volume-preserving": for compact $K$, $\operatorname{vol} \varphi_t(K)$ is constant.

### Behaviour
[to be continued]

## Lyapunov stuff
Informal, theorem-only version of [notes](http://web.stanford.edu/class/math63cm/lyapunov.pdf).

**Def.** For $x'(t)=F(x(t))$ and eq. pt $F(p)=0$,
- $p$ is *stable* if for any ball containing $p$, if you start close enough to $p$ you'll never leave the ball
- $p$ is *asymptotically stable* if starting close enough to $p$ means you'll fly towards $p$.

**Setup.** $U\subseteq\mathbb{R}$ open, $F:U\rightarrow \mathbb{R}^n$ is $C^1$, $p$ is an eq. pt. Everything will be "in some ball"

**Thm.** (Lyapunov 1) If there exists $V$ ($C^1$) defined in some ball around $p$ such that
(1) $V$ is strictly minimal at $p$
(2) $\nabla V\cdot F\le 0$
Then $p$ is a stable equilibrium.

**Thm.** (La Salle/Kravsovski) Furthermore, define $K$ is the set of all points where $\nabla V\cdot F=0$ (*restricted to some ball*). If the only point that doesn't leave $K$ is $p$, then $p$ is asymptotically stable.

**Thm.** (Lyapunov 2) Similar to Lyapunov 1, but if we know
(2') $\nabla V\cdot F<0$ except at $p$
then $p$ is asymptotically stable.

## Gradient Flows
Well gradient flow is trivial, for Hamiltonian flows check out [2017 notes](http://web.stanford.edu/~jluk/math63CMspring17/HamiltonianSystems.170412.pdf).

## Calculus of Variations
[Brian White's Notes](http://web.stanford.edu/class/math63cm/calculus-of-variations.pdf)

I guess just know the derivation and time-independent formulation.

## Limit sets
*Thanks Jensen!*

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Fix a solution $x$ to $x'=F(x)$.
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**Def.** $\Omega$ is the set of all $\omega$-limit points, i.e., points expressible as $\lim_{k\rightarrow \infty}x(t_k)$ where $t_k\rightarrow \infty$.

**Thm.** If $\Omega$ is bounded, $\lim_{t\rightarrow \infty} \mathrm{dist}(x(t),\Omega) = 0$.

*Proof.* (Sketch) Let $\Omega_{\varepsilon}$ denote all points within distance $\varepsilon$ of $\Omega$. If the statement was not true, there exists arbitrarily late points inside and outside $\Omega_{\varepsilon}$ for some $\varepsilon > 0$, so using IVT there are infinitely many points on $\partial \Omega_\varepsilon$. However, since this is bounded, by (B-W) there exists a convergent subsequence on $\partial \Omega_\varepsilon$ (disjoint from $\Omega$), contradiction.

*Cor.* If $\Omega=\{p\}$, $x(t)\rightarrow p$.

**Thm.** If $\Omega$ is bounded, it is "connected".

*Proof.* If we can split $\Omega$ into two sets $\Omega_1,\Omega_2$ where $\mathrm{dist}(\Omega_1,\Omega_2) > 0$, then we can apply the previous argument to $(\Omega_1)_{\varepsilon}$ and $(\Omega_2)_{\varepsilon}$ for some small enough $\varepsilon$.

*Cor.* If $\Omega$ are made of isolated points, then it's a single point.

**Thm.** $\Omega\cap U$ is invariant under flow $\varphi_t$.

*Proof.* Consider a $\omega$-limit point $y$ (with $x(t_k)\rightarrow y$). Since solutions are continuous w.r.t the initial conditions,
$$\varphi_t(x(t_k)) = \varphi_{t+t_k}(x_0) \rightarrow \varphi_t(y)$$
so $\varphi_t(y) \in \Omega$.

*Cor.* If $\Omega$ is a single point, then it is an equilibrium point (since the flow is stuck there).

**Thm.** If there exists $L:U\rightarrow \mathbb{R}$ where $\nabla L\cdot F\le 0$, then $L$ is constant on $\Omega\cap U$.

*Proof.* $\frac{d}{dt}L(x(t)) < 0$ so $L\circ x$ is monotone decr. and the lower limit exists (possibly $-\infty$). Now if $x(t_k)\rightarrow y\in \Omega$, then $:(x(s_k))\rightarrow L(y)$ so $L(y)$ is the lower limit.

**Cor.** Since $\Omega\cap U$ is invariant under the flow, $\frac{d}{dt}L(\phi_t(y)) = \nabla L(y)\cdot F(y) = 0$.

**Cor.** If $\Omega\cap U$ is nonempty, then $L$ is bounded below (because $L(y)$ is some finite value).

**Thm.** (Global Lyapunov) If $F$ is locally Lipschitz and:
- Levels sets are compact. ($\mu$-level set is $\{x\in U: L(x)\le \mu\}$)
- $\nabla L \cdot F < 0$ except at isolated points $E$.

Then the flow $\varphi_t(x_0)$ exists for all $t\ge 0$ and $\varphi_t(x_0)\rightarrow e\in E$.

*Proof.* $\nabla L\cdot F < 0$, so $L$ is nonincreasing and remains insice the compact level sets. Existence theorem gives us solution exists for $t\ge 0$.

$\Omega$ is non-empty by (B-W), and from previous corollary any $\omega$-limit point $y$ satisfies $\frac{d}{dt}(L(\varphi_t(y))) = 0$ so $y\in E$. Since $\Omega$ is connected, it is therefore a single point $\{e\}$, and hence an equilibrium point, so the solution $\varphi_t(x)$ tends to it.

### Planar Flows

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Work in $\mathbb{R}^2$, assume $x=F'(x)$.
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**Def.** A *transversal* (through a point $p$) is a line segment $S$ for which $F(s):s\in S$ is never parallel to $S$.

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Let $S$ be a transversal.
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**Thm.** If $a\in S$, then there exists $r>0$ and smooth $h:B_r(a)\rightarrow\mathbb{R}$ such that $h(a)=0$ and $\varphi_{h(x)}(x) \in S$ for all $x\in B_r(a)$.

*i.e. there's a sufficiently small ball where all points end up along the transversal after a backward or forward adjustment in time, and the adjustment is smooth as a function of location.*

*Proof.* Choose $r>0$ such that $F$ points in the same direction across $S$ (going from one side to the other). Going forwards or backwards in time, $x$ leaves this disc on the opposite side of where it started, so finish by IVT, then use implicit function theorem to get $h(x) = t$.

**Thm.** If $a\in S$ is an $\omega$-limit, then there exists a $\omega$-limit sequence on $S$ converging to $a$ (call this the *crossing*).

*Proof.* Exists a sequence $x(t_k)\rightarrow a$, so just use the previous theorem to adjust the times by $h(x(t_k))$, and since $h(x(t_k))\rightarrow h(x) =0$, this new sequence tends to $h(x)$.

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This is not technically correct, but eh.
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**Thm.** This crossing is discrete (i.e. no continuous chunks).

**Thm.** This crossing is monotonic w.r.t time.

**Thm.** $S$ has at most one $\omega$-limit point.

*Proof.* Use monotonicity.

**Thm.** If $\Omega\subset U$ is compact and $\Gamma\subset \Omega$ is a periodic orbit, then $\Omega=.\Gamma$.

*Proof.* For each $p\in \Gamma$, define $S_p$ to be a small transversal through $p$ (length $2\varepsilon$ centered on $p$) normal to $\Gamma$. We claim that there's a choice of $\varepsilon$ where this is indeed a transversal for all $p$ (below). To finish, $\Omega\cap S_p = \{p\}$, so $\Omega\cap \bigcup_{p \in Gamma}S_p = \Gamma$, so $\Omega$ contains nothing else within $\varepsilon$ of $\Gamma$.

> Showing that $\varepsilon$ exists: take $p\in \Gamma$, then $F(p)\neq 0$ (because it's part of a periodic orbit). Take a small disc $\overline{B_{\varepsilon(p)}(p)}$ such that for any $q_1,q_2$ inside, $F(q_1)\cdot F(q_2) > 0$. By compactness, find a finite subcover of $\Gamma$ using half-sized discs ($\overline{B_{\varepsilon(p_i)/2}(p_i)}$) and let $\varepsilon = \min\{\varepsilon(p_i)/2\}$. So for any $q$ on $S_p$, there's a $p_i$ at most $\varepsilon$ away from $p$ and so both $p,q$ are at most $2\varepsilon \le \varepsilon(p_i)$ from $p_i$, so $F(p)\cdot F(q) > 0$, so $S_p$ is indeed a transversal.

**Thm.** (Poincare-Bendixson) If $\Omega$ is a non-empty, compact subset of $U$ with no equilibrium points, then $\Omega$ is a periodic orbit.

*Proof.* Let $y\in \Omega$. Consider $\overline{\Omega}$ - the $\omega$-limit set for the flow starting at $y$. $\overline{\Omega}$ is nonempty by B-W (since the flow starting at $y$ remains in compact $\Omega$). Since $\Omega$ also contains its own limit points, so $\overline{\Omega} \subseteq \Omega$.

Consider $q\in \overline{\Omega}$ (not an equilibrium point by assumption), so we can find a transversal $S_q$. The fact that $q$ is a $\omega$-limit of $\varphi_t(y)$ means that exists a crossing between $\varphi_t(y)$ and $S_q$, but the crossing points all lie on $\Omega \cap S_q$ (which must be a single point, $q$), so $\varphi_{t_k}(y)= q$ for all $k$ and thus this solution is periodic.

So $\Omega$ contains a periodic orbit $\Gamma$, thus $\Omega = \Gamma$.

**Thm.** (Generalized P-B) Suppose that $\Omega \subset U$ is compact. Let $E\subset \Omega$ be equilibrium points contained in $\Omega$, and suppose $E$ is finite.
- Case 1: $\Omega = E$ (and so $\Omega$ is a single point by connectedness).
- Case 2: $E = \varnothing$ (so $\Omega$ is a periodic orbit by P-B).
- Case 3: Both $E$ and $\Omega \setminus E$ are nonempty: $\Omega\setminus E$ is a set of trajectories between points in $E$.

*Proof.* (Case 3 only) Firstly, $\Omega$ does not contain a periodic orbit (otherwise $E = \varnothing$). Let $y\in \Omega \setminus E$. Consider $\overline{\Omega}$ - the $\omega$-limit set of $\varphi_t(y)$ (which is non-empty by B-W), and $\overline{\Omega}\subseteq \Omega$ as shown previously. Now, if $\overline{\Omega}$ does not have equilibrium points, it is a periodic orbit by P-B so $\Omega$ has a periodic orbit (contradiction), so $\overline{\Omega}$ is a single point which $\varphi_t(y)$ converges to. By similar arguments with the $\alpha$-limit of $\varphi_t(y)$, it must converge backwards in time as well.

## Miscellaneous "Obvious Facts"
**Fact.** (Derivative commutes with limits) Let $f_n\rightarrow f$ uniformly, and $f_n'\rightarrow g$ uniformly. Then $f'=g$.

**Fact.** Given a sequence $\{x_n\}\subset \mathbb{R}^n$, and every subsequence contains a sub-subsequence that converges to $x$, then the original sequence converges to $x$.

**Fact.** Locally Lipschitz over a compact set implies Lipschitz over the whole set.