Ordinary Differential Equations/Blow-ups and moving to boundary

Proposition (concatenation of solutions to ODEs):

Assume we have a continuous function $f:[0,1]\times \mathbb {R} ^{n}\to \mathbb {R} ^{n}$ and two functions $x_{1}:[t_{0}-\gamma ,t_{0}]\to \mathbb {R} ^{n}$, $x_{2}:[t_{0},t_{0}+\delta ]\to \mathbb {R} ^{n}$ satisfying

${\begin{cases}x_{1}'(t)=f(t,x_{1}(t))&t\in [t_{0}-\gamma ,t_{0}]\\x_{1}(t_{0})=x_{0}\end{cases}}$

and

${\begin{cases}x_{2}'(t)=f(t,x_{2}(t))&t\in [t_{0},t_{0}+\delta ]\\x_{2}(t_{0})=x_{0}\end{cases}}$

respectively. Then the function

$x:[t_{0}-\gamma ,t_{0}+\delta ]\to \mathbb {R} ^{n},x(t):={\begin{cases}x_{1}(t)&t\in [t_{0}-\gamma ,t_{0}]\\x_{2}(t)&t\in [t_{0},t_{0}+\delta ]\end{cases}}$

solves

${\begin{cases}x'(t)=f(t,x(t))&t\in [t_{0}-\gamma ,t_{0}+\delta ]\\x(t_{0})=x_{0}.\end{cases}}$

Proof: We prove differentiability at $x_{0}$ as follows: We claim that the derivative of $x$ is given by $f(t_{0},x_{0})$. To prove our claim, we note that

${\frac {x_{2}(t_{+})-x_{1}(t_{-})}{t_{+}-t_{-}}}={\frac {x_{2}(t_{+})-x_{2}(t_{0})}{t_{+}-t_{0}}}{\frac {t_{+}-t_{0}}{t_{+}-t_{-}}}+{\frac {x_{1}(t_{0})-x_{1}(t_{-})}{t_{0}-t_{-}}}{\frac {t_{0}-t_{-}}{t_{+}-t_{-}}}\to f(t_{0},x_{0}),t_{+},t_{-}\to 0$

where $t_{+}\in [t_{0},t_{0}+\delta ]$ and $t_{-}\in [t_{0}-\gamma ,t_{0}]$; this is because

${\frac {t_{+}-t_{0}}{t_{+}-t_{-}}}+{\frac {t_{0}-t_{-}}{t_{+}-t_{-}}}=1$.

In the case where both $t_{+},t_{-}$ are both contained in the same of the two intervals $[t_{0},t_{0}+\delta ]$, $[t_{0}-\gamma ,t_{0}]$, the convergence is clear anyhow. $\Box$

Definition (maximal interval of existence):

Let an ordinary differential equation

$(*){\begin{cases}x'(t)=f(t,x(t))&\\x(t_{0})=x_{0}\end{cases}}$

be given. The maximal interval of existence is defined to be the interval $(a,b)$, where

$a:=\inf\{t\in \mathbb {R} |\exists x:(a,t_{0}]\to \mathbb {R} ^{n}:x{\text{ satisfies }}(*)\}$ and $b:=\sup\{\}$.

Proposition (existence of maximal interval of existence in the continuous case):

Definition:

Let an ordinary differential equation

${\begin{cases}x'(t)=f(t,x(t))&\\x(t_{0})=x_{0}\end{cases}}$

be given, where $f$ is continuous. The maximal interval of existence around $(t_{0},x_{0})$ is the maximal (w.r.t. set inclusion) interval $I$ such that $t_{0}\in I$ and there exists a solution $x$ defined on $I$ to the equation above.

Note that only the preceding theorem on concatenation of solutions ensures that the definition of a maximal interval of existence makes sense, since otherwise it might happen that there are two intervals $I_{1}=(a_{1},b_{1})$ and $I_{2}=(a_{2},b_{2})$ ($a_{1}<a_{2}<t_{0}<b_{1}<b_{2}$) such that $t_{0}$ is contained within both intervals and a solution is defined on both intervals, but the solutions are incompatible in the sense that none can be extended to the "large" interval $(a_{1},b_{2})$. The theorem on concatenation makes sure that this can never occur.

We now aim to prove that if we walk along the solution graph $(t,x(t))$ as $t$ approaches the endpoints of the maximal interval of existence $I$, then in a sense we move towards the boundary of $D$, where $D$ is required to be open and is the domain of definition of $f$. This shall mean that for any compact set $K\subset D$, if we pick $t$ large or small enough, $(t,x(t))$ is outside $K$. The proof is longer and needs preparation.

Proposition (Containing a compact set in a canonical compact set):

Let $f:D\to \mathbb {R} ^{n}$ be the right hand side of a differential equation, where $D$ is open (in fact, only the set $D$ matters for this lemma). Set

$D_{k}:=\left\{x\in D|\operatorname {dist} (x,\partial D)>{\frac {1}{k}}\wedge \|x\|<k\right\}$

for $k\in \mathbb {N}$. If $K\subseteq D$ is any compact set, then for sufficiently large $k$, $K\subset D_{k}$.

Proof: Since $K$ is compact, it is bounded. Therefore, $K\subseteq B_{k_{1}}(0)$ for a sufficiently large $k_{1}\in \mathbb {N}$. Furthermore, due to the above, $K$ has a certain minimum distance $\epsilon$ to the boundary, and we may choose $k_{2}\in \mathbb {N}$ such that ${\frac {1}{k_{2}}}<\epsilon$. Choose $k:=\max\{k_{1},k_{2}\}$. Then

$K\subseteq B_{k_{1}}(0)\subseteq B_{k}(0)$ and $\forall x\in K:\operatorname {dist} (x,\partial D)>\epsilon >{\frac {1}{k_{2}}}>{\frac {1}{k}}$.

Hence, $K\subset D_{k}$. $\Box$

Proposition (solutions of ODEs extend up to the boundary):

Let $f:D\to \mathbb {R} ^{n}$ be the right hand side of a differential equation (where $D$ is open), and let $x:I\to \mathbb {R} ^{n}$ be a solution to that equation, where $I=(a,b)$ is the interior of the maximal interval of solution. If $t$ is sufficiently close to either $a$ or $b$, then for each compact $K\subset D$ the point $(t,x(t))$ lies outside $K$.

Proof: Suppose otherwise. Then without loss of generality, we have a sequence $t_{1}>t_{2}>t_{3}>\cdots >t_{k}>\cdots$ such that $t_{k}\to a,k\to \infty$ and $(t_{k},x(t_{k}))\in K$ (an analogous supposition for the other end of the interval $b$ is led to a contradiction analogously). Since $K$ is compact, the sequence $(t_{k},x(t_{k}))$ has an accumulation point $(a,x^{*})$. We claim that in fact

$\lim _{t\to a}(t,x(t))=(a,x^{*})$.

Pick $m\in \mathbb {N}$ such that $K\subset D_{m}$. Let $\epsilon >0$ be arbitrary. We may restrict ourselves to $\epsilon >0$ sufficiently small such that $B_{\epsilon }((a,x^{*}))\subseteq D_{m}$. Since $f$ is continuous, it is bounded on the compact ${\overline {B_{\epsilon }((a,x^{*}))}}$, say by $M>0$. Now pick $\delta <\epsilon /(2M)$ such that $(a+\delta ,x(a+\delta ))\in B_{\epsilon /2}((a,x^{*}))$. If we assume that $x$ leaves $B_{\epsilon }((a,x^{*}))$ for $|t-a|<\delta$, the intermediate value theorem applied to the function

$(a,a+\delta )\to \mathbb {R} ,t\mapsto \|(a,x^{*})-(t,x(t))\|$

yields the existence of an $s\in (a,a+\delta )$ such that $\|(a,x^{*})-(t,x(t))\|=\epsilon$. But

${\begin{aligned}\|(a,x^{*})-(t,x(t))\|&\leq \|(a,x^{*})-(a+\delta ,x(a+\delta ))\|+\|(a+\delta ,x(a+\delta ))-(t,x(t))\|\\&<\epsilon /2+\|(a+\delta ,x(a+\delta ))-(t,x(t))\|\\&=\epsilon /2+\left\|\int _{t}^{a+\delta }f(s,x(s))ds\right\|\leq \epsilon /2+\delta M<\epsilon /2+\epsilon /2<\epsilon \end{aligned}}$,

contradiction.

Hence, $\lim _{t\to a}(t,x(t))\in K$. But on the other hand, by Peano's existence theorem and concatenation of solutions we may extend the solution at $a+\mu$ for every $\mu >0$ to the left by a fixed amount (namely for $\gamma =\max\{{\frac {1}{2m}},{\frac {1}{2m{\mathcal {M}}}}\}$, where ${\mathcal {M}}:=\min _{(t,x)\in {\overline {D_{2m}}}}|f(t,x)|$ which exists due to continuity of $f$ and compactness of ${\overline {D_{2m}}}$), and doing so for sufficiently small $\mu$ yields the contradiction that $(a,b)$ is not the maximal interval of existence. $\Box$

Corollary:

Let $f:D\to \mathbb {R} ^{n}$ be the right hand side of a differential equation for the special case $D=(c,d)\times \mathbb {R} ^{n}$ for an interval $J=(c,d)$. Let $I=(a,b)$ be the maximal interval of existence of a solution around $(t_{0},x_{0})\in D$. Then either $a=c$ or $\|x(t)\|\to \infty$ as $t\to a$. Similarly, either $b=d$ or $\|x(t)\|\to \infty$ as $t\to b$.

Proof:

From the preceding theorem, the solution eventually leaves every compact $K\subset D$ as $t\to a$ or $t\to b$. In particular, this holds for the compact sets ${\overline {D_{k}}}$. But to leave this implies either $\|(t,x(t))\|\geq k$ or $|c-t|\leq 1/k$ or $|d-t|\leq 1/k$, since the distance of $(t,x(t))$ to $\partial D$ is exactly the distance of $t$ to the nearest of the interval endpoints $c$, $d$. Hence, if not $a=c$, then $\|x(t)\|\to \infty$ as $t\to a$, and the analogous statement for $b$ and $d$.$\Box$