关于本讲资料的说明

这门课在录像标题里把最后一讲标为「期末复习」，但剪藏所附的转录并不是一份覆盖全课程的回顾。它的实际内容是：

1. 复述 Picard–Lindelöf 定理的框架，把上一讲没算完的压缩常数补完；
2. 讨论区间连通性，并给出一个「局部相同推出整体相同」的引理；
3. 用这两件事完成解的整体唯一性；
4. 用 $y'=y^2$ 的显式解说明解会在有限点爆破。

因此本笔记按「知识地图 + 核心论证 + 典型题型 + 易错点」来组织，但内容严格对应转录，不冒充一份完整的期末范围梳理。真正的期末范围应以课程大纲与讲义为准。

时间轴

知识地图：证明由三层组成

Picard–Lindelöf 的整个证明可以拆成三层，每层解决一个独立的问题。

第一层：完备性。 空间 $C([-\delta,\delta])$ 配 sup 度量是 Cauchy 完备的；闭球是完备空间的闭子集，所以闭球也完备。这一层回答「极限存在吗」。

第二层：压缩性。 积分算子 $T$ 在闭球上把距离乘以一个严格小于 $1$ 的常数。这一层回答「迭代收敛吗，极限是不是不动点」。

第三层：连通性。 区间不能分成两个非空的开闭子集。这一层回答「局部唯一能否拼成整体唯一」。

前两层在上一讲完成，第三层是本讲的主线。

把压缩常数算完

先补齐上一讲最后一步。

设定。 $f$ 可导、$f'$ 连续，$g$ 连续，初值 $y(0)=a$。取

$$R=|a|+2, \qquad L=\max_{|z|\le R}|f'(z)|,$$

并把区间收窄为

$$\delta=\min\Bigl(\delta_0,\ \frac{1}{2L}\Bigr),$$

其中 $\delta_0$ 是上一讲使 $T$ 映入闭球的那个常数。因为 $\delta\le\delta_0$，$T$ 仍把半径 $R$ 的闭球映入自身。

压缩估计。 取球内两个函数 $y_1,y_2$。常数 $a$ 与 $g$ 在相减时抵消：

$$\bigl(Ty_1\bigr)(x)-\bigl(Ty_2\bigr)(x) =\int_0^x\Bigl[f\bigl(y_1(s)\bigr)-f\bigl(y_2(s)\bigr)\Bigr]\,ds .$$

取绝对值并把绝对值放进积分：

$$\bigl|(Ty_1)(x)-(Ty_2)(x)\bigr| \le\left|\int_0^x\bigl|f(y_1(s))-f(y_2(s))\bigr|\,ds\right| .$$

固定 $s$。$y_1(s)$、$y_2(s)$ 都在 $[-R,R]$ 内，由中值定理，存在介于两者之间的 $\sigma$，使

$$f(y_1(s))-f(y_2(s))=f'(\sigma)\bigl(y_1(s)-y_2(s)\bigr).$$

所以

$$\bigl|f(y_1(s))-f(y_2(s))\bigr| \le L\,\bigl|y_1(s)-y_2(s)\bigr| \le L\,d(y_1,y_2).$$

代回：

$$\bigl|(Ty_1)(x)-(Ty_2)(x)\bigr| \le L\,d(y_1,y_2)\,|x| \le \delta L\,d(y_1,y_2).$$

由 $\delta\le1/(2L)$，

$$\delta L\le\frac12 .$$

对一切 $x$ 成立，取上确界：

$$d(Ty_1,Ty_2)\le\frac12\,d(y_1,y_2).$$

压缩常数是 $1/2$，与 $y_1,y_2$ 无关。上一讲开头定义的 $\delta=\min(\delta_0,1/(2M+1))$ 也能得到同样的结论，只是这里把常数写得干净些：

$$\delta=\frac{1}{2L} \quad\Longrightarrow\quad c=\frac12 .$$

到这里得到的结论是局部的：在 $(-\delta,\delta)$ 内，$T$ 有唯一不动点，因此 ODE 有唯一解。区间之外还没有讨论。

连通性

要把局部唯一性变成整体唯一性，需要区间的一个拓扑性质。

定义。 度量空间 $X$ 称为连通的，如果 $X$ 中同时既开又闭的子集只有 $\varnothing$ 与 $X$ 本身。

任何空间里，空集与全集都既开又闭，所以这个定义的意思是：没有别的「开闭两栖」子集。

定理。 区间 $I\subseteq\mathbb R$ 配通常度量是连通的。

证明

设 $A\subseteq I$ 非空、既开又闭。要证 $A=I$。

为书写简单，先设 $I$ 是闭区间；开区间与半开区间的论证只需把端点讨论稍作改动。

取 $x_0\in A$。定义

$$B=\Bigl\{\,y\in A : y\ge x_0,\ [x_0,y]\subseteq A\,\Bigr\}.$$

$B$ 非空，因为 $x_0\in B$（区间 $[x_0,x_0]$ 就是一个点）。令

$$y_0=\sup B .$$

分三步。

第一步：$y_0\in A$。 由上确界的定义，存在序列 $y_n\in B$ 使 $y_n\to y_0$。每个 $y_n\in A$，而 $A$ 闭，所以极限 $y_0\in A$。

第二步：$y_0\in B$，即 $[x_0,y_0]\subseteq A$。 任取 $z\in[x_0,y_0]$。因为 $y_n\to y_0$，当 $n$ 足够大时 $y_n\ge z$。而 $y_n\in B$ 意味着 $[x_0,y_n]\subseteq A$，于是 $z\in A$。所以整个 $[x_0,y_0]$ 都在 $A$ 中，$y_0\in B$。

第三步：$y_0$ 必须是 $I$ 的右端点。 反设不是。$A$ 是开集，而 $y_0\in A$，所以存在 $\varepsilon>0$，使

$$(y_0-\varepsilon,\ y_0+\varepsilon)\subseteq A .$$

取 $y_1\in(y_0,y_0+\varepsilon)$。由第二步 $[x_0,y_0]\subseteq A$，再并上 $(y_0,y_1]$，得到 $[x_0,y_1]\subseteq A$，于是 $y_1\in B$，但 $y_1>y_0=\sup B$，矛盾。

所以 $y_0$ 是 $I$ 的右端点，$[x_0,\text{右端点}]\subseteq A$。

对左半边做同样的论证（把序反过来，或对 $-I$ 使用上面的结果），得到 $[\text{左端点},x_0]\subseteq A$。两部分合起来，$A=I$。$\square$

三步各用了一条性质：第一步用闭，第二步用 $B$ 的定义，第三步用开。缺任何一条，论证都断掉——这也解释了为什么「既开又闭」是关键条件。

引理：局部相同推出处处相同

连通性的用处体现在下面这个引理上。

引理。 设 $I$ 是区间，$u_1,u_2:I\to\mathbb R$ 连续。假设：

1. 存在 $x_0\in I$，使 $u_1(x_0)=u_2(x_0)$；
2. 若在某点 $x\in I$ 有 $u_1(x)=u_2(x)$，则存在 $\varepsilon>0$（可以依赖 $x$），使 $u_1=u_2$ 在整个 $(x-\varepsilon,x+\varepsilon)$ 上成立。

则 $u_1=u_2$ 在 $I$ 上处处成立。

证明

定义

$$h(x)=u_1(x)-u_2(x).$$

$u_1,u_2$ 连续，所以 $h$ 连续。令

$$A=\{x\in I : h(x)=0\}$$

为零点集。

$A$ 非空。 条件 1 给出 $h(x_0)=0$，所以 $x_0\in A$。

$A$ 闭。 设 $x_n\in A$、$x_n\to x$。由 $h$ 连续，

$$h(x)=\lim_{n\to\infty}h(x_n)=0,$$

所以 $x\in A$。

$A$ 开（相对于 $I$）。 设 $x\in A$。条件 2 给出 $\varepsilon>0$，使 $h=0$ 在 $(x-\varepsilon,x+\varepsilon)$ 上成立，即这个邻域整个落在 $A$ 中。所以 $A$ 中每点都是内点。

$A$ 非空、既开又闭，而 $I$ 连通，所以 $A=I$。于是 $h\equiv0$，即 $u_1=u_2$。$\square$

引理的内容可以说是「解析延拓」的离散版本：只要两个连续函数在一点相碰，并且每次相碰都能向外扩一圈，它们就必然在整条区间上重合。唯一性的全部工作，就是验证条件 2。

唯一性：把局部拼成整体

定理（Picard–Lindelöf 的唯一性部分）。 设 $y_1,y_2$ 都是初值问题

$$y'=f(y)+g(x), \qquad y(0)=a$$

在某个含 $0$ 的区间 $I$ 上的解。则 $y_1=y_2$。

证明。 两个解在 $0$ 处相等：

$$y_1(0)=a=y_2(0).$$

这满足引理的条件 1。剩下要验证条件 2。

设 $y_1(x_0)=y_2(x_0)$ 对某个 $x_0\in I$ 成立。把问题重新看作以 $x_0$ 为起点的初值问题：此时两个解的初始值相同，且都满足同一个方程。取上一讲给出的那个 $\delta>0$（只依赖 $f,g$ 与初值，与起点无关），在 $(x_0-\delta,x_0+\delta)$ 上，压缩映射定理给出唯一解。而 $y_1$、$y_2$ 都是这个小区间上的解，所以它们必须相同：

$$y_1=y_2 \quad\text{在 }(x_0-\delta,x_0+\delta)\text{ 上}.$$

这正是引理的条件 2。由引理，$y_1=y_2$ 在 $I$ 上处处成立。$\square$

两件事在这里合流：压缩映射定理提供「一点附近的唯一性」，连通性把它传播到整条区间。

爆破：为什么定理只能局部

最后一个问题是：既然证明里费了很大力气把区间收窄到 $\delta$，这是技术上的保守，还是真实现象？

答案是后者。看一个显式例子。

解 $y'=y^2$

取

$$f(y)=y^2, \qquad g=0,$$

并考虑初值 $y(x_0)=a$，其中 $a>0$。方程是

$$y'=y^2 .$$

在解为正的区间上，两边除以 $y^2$：

$$\frac{y'}{y^2}=1 .$$

左边是一个导数：由链式法则，

$$\left(-\frac1y\right)'=\frac{y'}{y^2}.$$

从 $x_0$ 到 $x$ 积分：

$$-\frac{1}{y(x)}+\frac{1}{y(x_0)}=x-x_0 .$$

代入 $y(x_0)=a$：

$$\frac{1}{y(x)}=\frac1a-(x-x_0) =\frac{1-a(x-x_0)}{a}.$$

取倒数：

$$y(x)=\frac{a}{1-a(x-x_0)} =\frac{a}{1+ax_0-ax}.$$

补充：转录把分母记成了「$1+ax_0-x$」，漏掉了 $x$ 前的因子 $a$。按验算，分母是 $1+a x_0-a x$；在 $x=x_0$ 处它等于 $1$，给出 $y(x_0)=a$，可以核对。

解在哪里失效

看分母 $1-a(x-x_0)$。它在

$$x=x_0+\frac1a$$

处为零。当 $x$ 从这个点左侧趋近它时，$y(x)\to+\infty$。所以这个解的最大存在区间是

$$\Bigl(-\infty,\ x_0+\frac1a\Bigr),$$

到右端点就爆破了。

为什么不能延拓过去

设 $t^*=x_0+1/a$。反设存在一个解能延伸到 $t^*$ 的右侧。由唯一性，它在 $(x_0,t^*)$ 上必须与上面的公式一致；但公式在 $x\to t^{*-}$ 时趋于 $+\infty$，而解在 $t^*$ 附近应当取有限值并连续。矛盾。所以任何解都无法越过 $t^*$。

直观上，$y'=y^2$ 让「$y$ 越大，增长越快」：一旦 $y$ 变大，导数变得更大，曲线在有限时间内冲到无穷，切线趋于竖直。这不是证明的保守，而是方程本身的行为。

结论：局部性不可去除。对某些 $f,g$，解的最大存在区间可以任意小，甚至只比一个点长一点。

典型题型

按本讲与合作内容，考试里可能出现三类问题。

题型一：验证压缩性并给出常数

给定 $f$ 与初值，要求证明积分算子在某个闭球上是压缩的。

步骤固定：

1. 取 $R=|a|+2$，算 $L_1=\max_{|z|\le R}|f(z)|$、$L_2=\max_{|z|\le1}|g(z)|$；
2. 取 $\delta_0=\min(1,1/(L_1+1),1/(L_2+1))$，验证 $|Ty(x)|\le|a|+\delta_0L_1+\delta_0L_2<R$；
3. 取 $L=\max_{|z|\le R}|f'(z)|$，把区间收窄到 $\delta=1/(2L)$；
4. 用中值定理得 $d(Ty_1,Ty_2)\le\delta L\,d(y_1,y_2)\le\frac12d(y_1,y_2)$。

题型二：用连通性延拓唯一性

给定两个解，要求证明它们在共同定义域上相等。

步骤：在相碰点附近用局部唯一性；把「相碰」的集合写成 $h^{-1}(0)$；证明它非空、闭（$h$ 连续）、开（局部唯一性）；用区间连通性收尾。

题型三：解简单 ODE 并求最大存在区间

例如 $y'=y^2$、$y(x_0)=a>0$。分离变量，积分，解出 $y$，找分母为零的点，说明爆破。要点是最后一步要论证「不能延拓」，而不是只说「公式在这里没定义」。

易错点

• 把 $c\le1$ 当成压缩：必须是 $c<1$。$c=1$ 时迭代不保证收敛，恒等映射就是反例。
• 在全体连续函数上做压缩：不行。$f'$ 可能无界，必须先限制到闭球。
• 用开球：开球不完备，Cauchy 列的极限可能跑出去。
• 混淆「局部唯一」与「整体唯一」：压缩映射定理只给球内唯一，整体唯一必须靠连通性论证。
• 把爆破理解成「公式不好看」：爆破是解本身趋于无穷，任何延拓都不可能，不是换一种写法就能绕过。
• 连通性证明里漏掉「闭」或「开」：第一步靠闭，第三步靠开，缺一不可。

理解检验

1. 在连通性定理的证明中，为什么 $y_0=\sup B$ 一定属于 $A$？为什么它又属于 $B$？

参考答案：$y_0$ 是 $B$ 的上确界，故存在 $y_n\in B$ 使 $y_n\to y_0$。每个 $y_n\in A$，而 $A$ 闭，所以极限 $y_0\in A$。要证 $y_0\in B$，还需 $[x_0,y_0]\subseteq A$：任取 $z\in[x_0,y_0]$，因为 $y_n\to y_0$，当 $n$ 大时 $y_n\ge z$；而 $[x_0,y_n]\subseteq A$，故 $z\in A$。于是整个 $[x_0,y_0]$ 在 $A$ 中，按 $B$ 的定义 $y_0\in B$。

2. 证明「局部相同推出处处相同」的引理，并指出连通性用在哪里。

参考答案：令 $h=u_1-u_2$，零点集 $A=h^{-1}(0)$。$A$ 非空：条件 1 给出一点。$A$ 闭：若 $x_n\in A$、$x_n\to x$，由 $h$ 连续，$h(x)=\lim h(x_n)=0$。$A$ 开：若 $x\in A$，条件 2 给出一个邻域整个落在 $A$ 中。于是 $A$ 非空且既开又闭，区间连通意味着 $A$ 只能是整个区间，故 $h\equiv0$。连通性用在最后一步：它排除了「零点集只覆盖一部分区间」的可能。

3. 用 Picard–Lindelöf 与上面的引理，证明 ODE 的解在共同定义域上唯一。

参考答案：设 $y_1,y_2$ 定义在含 $0$ 的区间 $I$ 上，且都满足 $y(0)=a$。于是它们在 $0$ 处相等，引理的条件 1 成立。若它们在 $x_0$ 处相等，则以 $x_0$ 为起点、初值 $y_1(x_0)=y_2(x_0)$ 重新应用 Picard–Lindelöf，得到在 $(x_0-\delta,x_0+\delta)$ 上解唯一，即 $y_1=y_2$ 在该邻域内成立，这就是条件 2。由引理，$y_1=y_2$ 在整个 $I$ 上成立。

4. 解 $y'=y^2$、$y(0)=1$，写出最大存在区间，并严格说明为什么不能延拓。

参考答案：分离变量得 $y'/y^2=1$，而 $(-1/y)'=y'/y^2$，从 $0$ 到 $x$ 积分得 $-1/y(x)+1/y(0)=x$。代入 $y(0)=1$ 得 $1/y(x)=1-x$，即 $y(x)=1/(1-x)$。最大存在区间是 $(-\infty,1)$。反设解能延伸到 $1$ 的右侧：由唯一性它必须在 $(0,1)$ 上与 $1/(1-x)$ 一致，但该式在 $x\to1^-$ 时趋于 $+\infty$，与解在 $1$ 附近取有限值矛盾。故无法延拓。

5. 如果去掉 Picard–Lindelöf 中对区间长度的限制，定理会变成什么？它为什么是假的？

参考答案：去掉限制后的断言会是「解在整个 $\mathbb R$ 上存在且唯一」。这是假的，反例就是 $y'=y^2$、$y(0)=1$：它的解 $y=1/(1-x)$ 在 $x\to1^-$ 时爆破，根本不存在定义在整个 $\mathbb R$ 上的解。因此定理只能断言在某个 $\delta>0$ 的小区间上存在唯一解；$\delta$ 的大小依赖 $f$、$g$ 与初值，不能统一取定。

与前后课的关系

本讲把 Picard–Lindelöf 证明的最后两块补上：压缩常数算完，整体唯一性由连通性得到，爆破例子说明局部性不可避免。至此，实分析课程的主线闭合：

实数公理 → 极限与级数 → 连续性与紧性 → 导数与积分 → 一致收敛 → 度量空间与完备性 → 压缩映射 → 微分方程。

每一个箭头都对应一个具体定理，而整条链的终点，是一个纯分析的结论：只要 $f$ 足够光滑，局部解存在且唯一。

A note on this lecture's source

The recording is titled "Review for the Final Exam", but the transcript attached to the clipping is not a course-wide review. Its actual content is:

1. a restatement of the Picard–Lindelöf framework and completion of the contraction estimate left unfinished last time;
2. a discussion of connectedness of intervals and a lemma saying that two continuous functions agreeing locally agree globally;
3. the use of both to prove global uniqueness;
4. the explicit solution of $y'=y^2$, showing blow-up at a finite point.

This note is therefore organized as a knowledge map, the core arguments, typical problem types and common pitfalls, but it follows the transcript strictly and does not pretend to be a complete exam syllabus. For the actual exam scope, the course syllabus and lecture notes are authoritative.

Timeline

Knowledge map: three layers

The proof of Picard–Lindelöf decomposes into three layers, each answering a separate question.

Layer one: completeness. The space $C([-\delta,\delta])$ with the sup metric is Cauchy complete; a closed ball is a closed subset of a complete space, hence complete. This layer answers "does the limit exist".

Layer two: contraction. The integral operator $T$ multiplies distances by a constant strictly below $1$ on the ball. This layer answers "does the iteration converge, and is the limit a fixed point".

Layer three: connectedness. An interval cannot be split into two nonempty open-and-closed pieces. This layer answers "can local uniqueness be glued into global uniqueness".

Layers one and two were completed last time; layer three is the spine of this lecture.

Completing the contraction estimate

First, the step left open last time.

Setup. $f$ is differentiable with $f'$ continuous, $g$ is continuous, and $y(0)=a$. Take

$$R=|a|+2, \qquad L=\max_{|z|\le R}|f'(z)|,$$

and shrink the interval to

$$\delta=\min\Bigl(\delta_0,\ \frac{1}{2L}\Bigr),$$

where $\delta_0$ is the constant from last time that made $T$ map the ball into itself. Since $\delta\le\delta_0$, $T$ still maps the ball of radius $R$ into itself.

The contraction estimate. Take $y_1,y_2$ in the ball. The constants $a$ and the terms $g$ cancel:

$$\bigl(Ty_1\bigr)(x)-\bigl(Ty_2\bigr)(x) =\int_0^x\Bigl[f\bigl(y_1(s)\bigr)-f\bigl(y_2(s)\bigr)\Bigr]\,ds .$$

Take absolute values and move the absolute value inside:

$$\bigl|(Ty_1)(x)-(Ty_2)(x)\bigr| \le\left|\int_0^x\bigl|f(y_1(s))-f(y_2(s))\bigr|\,ds\right| .$$

Fix $s$. Both $y_1(s)$ and $y_2(s)$ lie in $[-R,R]$, so by the mean value theorem there is $\sigma$ between them with

$$f(y_1(s))-f(y_2(s))=f'(\sigma)\bigl(y_1(s)-y_2(s)\bigr).$$

Hence

$$\bigl|f(y_1(s))-f(y_2(s))\bigr| \le L\,\bigl|y_1(s)-y_2(s)\bigr| \le L\,d(y_1,y_2).$$

Substituting back,

$$\bigl|(Ty_1)(x)-(Ty_2)(x)\bigr| \le L\,d(y_1,y_2)\,|x| \le \delta L\,d(y_1,y_2).$$

Since $\delta\le1/(2L)$,

$$\delta L\le\frac12 .$$

This holds for every $x$; taking the supremum,

$$d(Ty_1,Ty_2)\le\frac12\,d(y_1,y_2).$$

The contraction constant is $1/2$, independent of $y_1,y_2$. The previous lecture defined $\delta=\min(\delta_0,1/(2M+1))$ and reaches the same conclusion, only with a messier constant:

$$\delta=\frac{1}{2L} \quad\Longrightarrow\quad c=\frac12 .$$

What has been proved is local: inside $(-\delta,\delta)$ the map $T$ has a unique fixed point, hence the ODE has a unique solution. Outside that interval nothing has been said.

Connectedness

Turning local uniqueness into global uniqueness needs a topological property of intervals.

Definition. A metric space $X$ is connected if the only subsets that are both open and closed are $\varnothing$ and $X$.

In any space the empty set and the whole space are both open and closed, so the definition says there are no other "open-and-closed" subsets.

Theorem. An interval $I\subseteq\mathbb R$ with the usual metric is connected.

Proof

Let $A\subseteq I$ be nonempty and both open and closed. We show $A=I$.

For simplicity assume first that $I$ is closed; for open and half-open intervals only the endpoint discussion changes.

Pick $x_0\in A$ and define

$$B=\Bigl\{\,y\in A : y\ge x_0,\ [x_0,y]\subseteq A\,\Bigr\}.$$

$B$ is nonempty since $x_0\in B$ (the interval $[x_0,x_0]$ is a single point). Let

$$y_0=\sup B .$$

Three steps.

Step 1: $y_0\in A$. By definition of the supremum there is a sequence $y_n\in B$ with $y_n\to y_0$. Each $y_n\in A$ and $A$ is closed, so the limit $y_0\in A$.

Step 2: $y_0\in B$, i.e. $[x_0,y_0]\subseteq A$. Take $z\in[x_0,y_0]$. Since $y_n\to y_0$, for large $n$ we have $y_n\ge z$. And $y_n\in B$ means $[x_0,y_n]\subseteq A$, so $z\in A$. Hence the whole of $[x_0,y_0]$ lies in $A$, so $y_0\in B$.

Step 3: $y_0$ must be the right endpoint of $I$. Suppose not. Since $A$ is open and $y_0\in A$, there is $\varepsilon>0$ with

$$(y_0-\varepsilon,\ y_0+\varepsilon)\subseteq A .$$

Take $y_1\in(y_0,y_0+\varepsilon)$. By Step 2, $[x_0,y_0]\subseteq A$; adjoining $(y_0,y_1]$ gives $[x_0,y_1]\subseteq A$, so $y_1\in B$. But $y_1>y_0=\sup B$, a contradiction.

So $y_0$ is the right endpoint of $I$ and $[x_0,\text{right endpoint}]\subseteq A$.

Running the same argument on the left (reverse the order, or apply the result to $-I$) gives $[\text{left endpoint},x_0]\subseteq A$. Together, $A=I$. $\square$

Each step uses one property: Step 1 uses closedness, Step 2 uses the definition of $B$, Step 3 uses openness. Remove any of them and the argument collapses — which is exactly why "both open and closed" is the right hypothesis.

Lemma: agreeing locally forces agreeing everywhere

Connectedness is used through the following lemma.

Lemma. Let $I$ be an interval and $u_1,u_2:I\to\mathbb R$ continuous. Assume:

1. $u_1(x_0)=u_2(x_0)$ for some $x_0\in I$;
2. whenever $u_1(x)=u_2(x)$ at some $x\in I$, there is $\varepsilon>0$ (allowed to depend on $x$) with $u_1=u_2$ on the whole interval $(x-\varepsilon,x+\varepsilon)$.

Then $u_1=u_2$ everywhere on $I$.

Proof

Define

$$h(x)=u_1(x)-u_2(x).$$

Since $u_1,u_2$ are continuous, so is $h$. Let

$$A=\{x\in I : h(x)=0\}$$

be the zero set.

$A$ is nonempty. Condition 1 gives $h(x_0)=0$, so $x_0\in A$.

$A$ is closed. If $x_n\in A$ and $x_n\to x$, continuity gives

$$h(x)=\lim_{n\to\infty}h(x_n)=0,$$

so $x\in A$.

$A$ is open in $I$. If $x\in A$, condition 2 gives $\varepsilon>0$ with $h=0$ on $(x-\varepsilon,x+\varepsilon)$, i.e. that neighbourhood lies entirely in $A$. So every point of $A$ is interior.

Thus $A$ is nonempty, open and closed, and $I$ is connected, so $A=I$. Hence $h\equiv0$ and $u_1=u_2$. $\square$

The lemma is a discrete version of analytic continuation: two continuous functions that meet at one point, and whose coincidence always spreads to a neighbourhood, must coincide on the entire interval. All the work in proving uniqueness goes into verifying condition 2.

Uniqueness: gluing the local into the global

Theorem (uniqueness part of Picard–Lindelöf). Let $y_1,y_2$ both solve

$$y'=f(y)+g(x), \qquad y(0)=a$$

on some interval $I$ containing $0$. Then $y_1=y_2$.

Proof. The two solutions coincide at $0$:

$$y_1(0)=a=y_2(0).$$

This is condition 1 of the lemma. It remains to check condition 2.

Suppose $y_1(x_0)=y_2(x_0)$ for some $x_0\in I$. Regard the problem as an initial value problem starting at $x_0$: both solutions have the same initial value and satisfy the same equation. Take the $\delta>0$ from the previous lecture (depending only on $f,g$ and the initial value, not on the starting point). On $(x_0-\delta,x_0+\delta)$ the contraction mapping theorem gives a unique solution. Since $y_1$ and $y_2$ both solve on that small interval, they must coincide:

$$y_1=y_2 \quad\text{on }(x_0-\delta,x_0+\delta).$$

This is condition 2. By the lemma, $y_1=y_2$ on all of $I$. $\square$

Two results meet here: the contraction mapping theorem supplies uniqueness near a point, and connectedness propagates it to the whole interval.

Blow-up: why the theorem must be local

One question remains: since the proof worked hard to shrink the interval to $\delta$, is that technical caution or a real phenomenon?

It is real. Here is an explicit example.

Solving $y'=y^2$

Take

$$f(y)=y^2, \qquad g=0,$$

and the initial condition $y(x_0)=a$ with $a>0$. The equation is

$$y'=y^2 .$$

On an interval where the solution is positive, divide by $y^2$:

$$\frac{y'}{y^2}=1 .$$

The left side is a derivative: by the chain rule,

$$\left(-\frac1y\right)'=\frac{y'}{y^2}.$$

Integrate from $x_0$ to $x$:

$$-\frac{1}{y(x)}+\frac{1}{y(x_0)}=x-x_0 .$$

Substituting $y(x_0)=a$,

$$\frac{1}{y(x)}=\frac1a-(x-x_0) =\frac{1-a(x-x_0)}{a}.$$

Taking reciprocals,

$$y(x)=\frac{a}{1-a(x-x_0)} =\frac{a}{1+ax_0-ax}.$$

Note: the transcript writes the denominator as "$1+ax_0-x$", dropping the factor $a$ in front of $x$. Checking the algebra, the denominator is $1+a x_0-a x$; at $x=x_0$ it equals $1$ and gives $y(x_0)=a$.

Where the solution fails

The denominator $1-a(x-x_0)$ vanishes at

$$x=x_0+\frac1a .$$

As $x$ approaches that point from the left, $y(x)\to+\infty$. So the maximal interval of existence is

$$\Bigl(-\infty,\ x_0+\frac1a\Bigr),$$

and the solution blows up at the right endpoint.

Why it cannot be extended

Let $t^*=x_0+1/a$. Suppose a solution extended past $t^*$. By uniqueness it would have to agree with the formula above on $(x_0,t^*)$; but that formula tends to $+\infty$ as $x\to t^{*-}$, whereas a solution near $t^*$ should take finite values and be continuous. Contradiction. So no solution can cross $t^*$.

Intuitively, $y'=y^2$ makes "the larger $y$ is, the faster it grows": once $y$ becomes large, the derivative becomes larger still, and the curve runs to infinity in finite time with a vertical tangent. This is not the proof being conservative; it is the equation's own behaviour.

Conclusion: locality cannot be removed. For some $f,g$ the maximal interval of existence can be arbitrarily small.

Typical problem types

Three kinds of questions follow from this material.

Type one: verify the contraction and produce a constant

Given $f$ and an initial value, prove that the integral operator is a contraction on some closed ball.

The recipe is fixed:

1. take $R=|a|+2$, compute $L_1=\max_{|z|\le R}|f(z)|$ and $L_2=\max_{|z|\le1}|g(z)|$;
2. take $\delta_0=\min(1,1/(L_1+1),1/(L_2+1))$ and check $|Ty(x)|\le|a|+\delta_0L_1+\delta_0L_2<R$;
3. take $L=\max_{|z|\le R}|f'(z)|$ and shrink the interval to $\delta=1/(2L)$;
4. use the mean value theorem to get $d(Ty_1,Ty_2)\le\delta L\,d(y_1,y_2)\le\frac12 d(y_1,y_2)$.

Type two: extend uniqueness by connectedness

Given two solutions, prove they agree on their common domain.

The recipe: use local uniqueness near a point where they meet; write the set of meeting points as $h^{-1}(0)$; prove it nonempty, closed ($h$ continuous) and open (local uniqueness); finish with connectedness of the interval.

Type three: solve a simple ODE and find the maximal interval

For instance $y'=y^2$, $y(x_0)=a>0$. Separate variables, integrate, solve for $y$, locate the zero of the denominator, and explain the blow-up. The essential last step is an argument that the solution cannot be extended, not merely the observation that the formula is undefined.

Common pitfalls

• Treating $c\le1$ as a contraction: it must be $c<1$. For $c=1$ the iteration need not converge, and the identity map is a counterexample.
• Claiming a contraction on all continuous functions: not true. $f'$ may be unbounded; the closed ball is necessary.
• Using an open ball: open balls are not complete, and a Cauchy sequence may converge outside.
• Confusing local and global uniqueness: the contraction theorem gives uniqueness inside the ball only; global uniqueness needs the connectedness argument.
• Reading blow-up as "an ugly formula": the solution itself tends to infinity, and no extension exists; no rewriting avoids it.
• Dropping "closed" or "open" in the connectedness proof: Step 1 needs closedness, Step 3 needs openness; neither can be omitted.

Comprehension check

1. In the connectedness proof, why does $y_0=\sup B$ belong to $A$? Why does it belong to $B$?

Reference answer: Since $y_0$ is the supremum of $B$, there are $y_n\in B$ with $y_n\to y_0$. Each $y_n\in A$ and $A$ is closed, so the limit $y_0\in A$. To get $y_0\in B$ we also need $[x_0,y_0]\subseteq A$: take $z\in[x_0,y_0]$; since $y_n\to y_0$, for large $n$ we have $y_n\ge z$, and $[x_0,y_n]\subseteq A$, so $z\in A$. Hence all of $[x_0,y_0]$ lies in $A$, and by the definition of $B$, $y_0\in B$.

2. Prove the lemma that agreeing locally forces agreeing everywhere, and identify where connectedness is used.

Reference answer: Let $h=u_1-u_2$ and let $A=h^{-1}(0)$ be the zero set. $A$ is nonempty by condition 1. It is closed: if $x_n\in A$ and $x_n\to x$, continuity gives $h(x)=\lim h(x_n)=0$. It is open: if $x\in A$, condition 2 supplies a neighbourhood contained in $A$. So $A$ is nonempty, open and closed; connectedness of the interval forces $A$ to be the whole interval, hence $h\equiv0$. Connectedness is used in that last step: it rules out the possibility that the zero set covers only part of the interval.

3. Use Picard–Lindelöf and the lemma to prove uniqueness on the common domain.

Reference answer: Let $y_1,y_2$ be defined on an interval $I$ containing $0$, both satisfying $y(0)=a$. They agree at $0$, so condition 1 of the lemma holds. If they agree at $x_0$, apply Picard–Lindelöf again with starting point $x_0$ and initial value $y_1(x_0)=y_2(x_0)$; the theorem gives a unique solution on $(x_0-\delta,x_0+\delta)$, so $y_1=y_2$ on that neighbourhood, which is condition 2. The lemma then gives $y_1=y_2$ on all of $I$.

4. Solve $y'=y^2$, $y(0)=1$, give the maximal interval, and prove rigorously that it cannot be extended.

Reference answer: Separating variables gives $y'/y^2=1$, and since $(-1/y)'=y'/y^2$, integrating from $0$ to $x$ gives $-1/y(x)+1/y(0)=x$. With $y(0)=1$, $1/y(x)=1-x$, so $y(x)=1/(1-x)$. The maximal interval is $(-\infty,1)$. Suppose a solution extended to the right of $1$: by uniqueness it would agree with $1/(1-x)$ on $(0,1)$, but that expression tends to $+\infty$ as $x\to1^-$, contradicting the finite continuous value a solution must take near $1$. Hence no extension exists.

5. What would the theorem say without the restriction on the length of the interval, and why would it be false?

Reference answer: Without the restriction the claim would be "a solution exists and is unique on all of $\mathbb R$". This is false; the counterexample is $y'=y^2$, $y(0)=1$, whose solution $y=1/(1-x)$ blows up as $x\to1^-$, so no solution exists on all of $\mathbb R$. Hence the theorem can only assert existence and uniqueness on some small interval $(-\delta,\delta)$, with $\delta$ depending on $f$, $g$ and the initial value and not uniformly choosable.

Relation to neighbouring lectures

This lecture supplies the last two pieces of the Picard–Lindelöf proof: the contraction constant is finished, global uniqueness follows from connectedness, and the blow-up example shows locality is unavoidable. The main line of the analysis course now closes:

real number axioms → limits and series → continuity and compactness → derivatives and integrals → uniform convergence → metric spaces and completeness → contraction mappings → differential equations.

Each arrow corresponds to a concrete theorem, and the whole chain ends in a purely analytic statement: as soon as $f$ is smooth enough, a local solution exists and is unique.