问题：微分方程一定有解吗

考虑初值问题

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

其中 $f$ 可导且 $f'$ 连续，$g$ 连续，$a$ 是给定实数。

这不是一个可以套公式的问题。$f$ 可以任意非线性，显式解通常写不出来。真正要问的是两个定性问题：

• 存在性：给定 $a$，是否至少存在一个解？
• 唯一性：如果有两个解，它们是否必须相同？

这两个问题在整个科学与工程里反复出现：方程描述某个自然过程，首先得确认它至少有一个解，其次得确认这个解是唯一的，否则预测就没有意义。

Picard–Lindelöf 定理回答的正是这两个问题，而且答案是「局部」的：存在某个 $\delta>0$，在 $(-\delta,\delta)$ 上有解，且解唯一。为什么只能局部，最后一讲会用 $y'=y^2$ 的爆破说明。

证明的路线出人意料：用度量空间。具体说，把解写成某个积分算子的不动点，再对这个算子使用压缩映射定理。

时间轴

两个工具

工具一：sup 度量与完备性

在区间 $I$ 上的连续函数空间 $C(I)$ 上取

$$d(h_1,h_2)=\max_{x\in I}|h_1(x)-h_2(x)| .$$

前面已经证明：$\bigl(C([a,b]),d\bigr)$ 是 Cauchy 完备的：每个 Cauchy 列都收敛到某个连续函数。

还要用到一个简单事实：完备空间的闭子集也完备。若 $B$ 是完备空间 $X$ 的闭子集，$(x_n)\subset B$ 是 Cauchy 列，则在 $X$ 中它收敛到某个 $x$；因为 $B$ 闭，$x\in B$。所以 $B$ 中的 Cauchy 列在 $B$ 中收敛。

工具二：压缩映射

定义。 设 $(X,d)$ 是度量空间，$T:X\to X$。若存在常数 $c<1$，使得

$$d\bigl(T(x),T(y)\bigr)\le c\,d(x,y) \quad\text{对一切 }x,y\in X,$$

则称 $T$ 是压缩映射。

这里的 $c$ 必须严格小于 $1$。等于 $1$ 不够：恒等映射取 $c=1$，它每个点都是不动点，结论完全不成立。

定理一：压缩映射至多一个不动点

定理。 若 $T$ 是压缩映射，则 $T$ 至多有一个不动点。

证明。 设 $x$、$y$ 都是不动点，即

$$T(x)=x, \qquad T(y)=y .$$

由压缩性，

$$d(x,y)=d\bigl(T(x),T(y)\bigr)\le c\,d(x,y).$$

把右边移到左边：

$$(1-c)\,d(x,y)\le0 .$$

因为 $1-c>0$ 且 $d(x,y)\ge0$，只能有 $d(x,y)=0$，即 $x=y$。$\square$

这个证明只用了两条：压缩性和度量的正定性。它甚至不需要空间完备。完备性是为下一个定理的存在性准备的。

定理二：压缩映射定理

定理。 设 $(X,d)$ 是 Cauchy 完备的度量空间，$T:X\to X$ 是压缩映射。则 $T$ 有唯一不动点。

唯一性已经由定理一给出，剩下要证存在性：必须造出一个不动点。

第一步：从任意一点开始迭代

任取 $x\in X$，定义

$$x_0=x, \qquad x_1=T(x), \qquad x_2=T^2(x), \qquad\dots$$

即 $x_n=T^n(x)$。这里不需要像 Newton 法那样先选一个好的初始点——任何点都可以，只是收敛速度可能慢一些。

目标：证明 $(x_n)$ 收敛，且极限 $x_\infty$ 满足 $T(x_\infty)=x_\infty$。

第二步：相邻两项的误差按几何级数缩小

对任意 $l\ge1$，

$$\begin{aligned} d(x_l,x_{l+1}) &=d\bigl(T^l(x),T^{l+1}(x)\bigr)\\ &=d\bigl(T(T^{l-1}x),T(T^{l}x)\bigr)\\ &\le c\,d\bigl(T^{l-1}x,T^{l}x\bigr). \end{aligned}$$

也就是说，每往后退一层，误差就乘一个 $c$。反复使用：

$$d(x_l,x_{l+1})\le c^{\,l}\,d(x_0,x_1).$$

第三步：任意两项的误差

设 $n\ge0$，$k\ge1$。先在映射层面迭代 $n$ 次：

$$\begin{aligned} d(x_n,x_{n+k}) &=d\bigl(T^n x,T^n(T^k x)\bigr)\\ &\le c^n\,d(x,T^k x). \end{aligned}$$

现在估计 $d(x,T^k x)$，用三角不等式把它拆成 $k$ 段相邻差：

$$\begin{aligned} d(x,T^k x) &\le\sum_{j=0}^{k-1}d(x_j,x_{j+1})\\ &\le\sum_{j=0}^{k-1}c^j\,d(x,T x) \le\frac{d(x,Tx)}{1-c}. \end{aligned}$$

最后一步用了几何级数求和：$c<1$，所以 $\sum_{j=0}^{\infty}c^j=1/(1-c)$，部分和不超过它。

合并两式：

$$d(x_n,x_{n+k}) \le c^n\,\frac{d(x,Tx)}{1-c}.$$

右端与 $k$ 无关，且当 $n\to\infty$ 时 $c^n\to0$。所以对任意 $\varepsilon>0$，存在 $N$，使 $n\ge N$ 时对一切 $k\ge1$ 都有 $d(x_n,x_{n+k})<\varepsilon$。这正是 Cauchy 列的定义。

这里 $c<1$ 是决定性的：若 $c=1$，$c^n$ 不衰减，什么也证不出来。

第四步：取极限

$X$ 完备，所以 $(x_n)$ 收敛。记极限为

$$x_\infty=\lim_{n\to\infty}x_n .$$

第五步：极限是不动点

要证 $T(x_\infty)=x_\infty$。考虑 $T$ 作用在序列上：

$$T(x_0),T(x_1),T(x_2),\dots$$

因为 $T(x_n)=x_{n+1}$，这串就是原序列去掉首项，所以它也收敛到 $x_\infty$。

另一方面，用压缩性比较 $T(x_\infty)$ 与 $T(x_n)$：

$$d\bigl(T(x_\infty),T(x_n)\bigr)\le c\,d(x_\infty,x_n).$$

因为 $x_n\to x_\infty$，右端趋于 $0$，所以 $T(x_n)\to T(x_\infty)$。但已知 $T(x_n)=x_{n+1}\to x_\infty$。度量空间中极限唯一，因此

$$T(x_\infty)=x_\infty .$$

于是 $x_\infty$ 是不动点。$\square$

这个证明的结构值得记住：先用压缩性造出 Cauchy 列，再用完备性取到极限，最后用压缩性把「$T$ 与极限可交换」建立起来。最后一步实际上是在证明 $T$ 连续：压缩映射自动是连续映射。

把 ODE 变成不动点问题

接下来是定理与微分方程的连接。

映射 $T$ 的定义

设 $I$ 是含 $0$ 的区间。对连续函数 $y:I\to\mathbb R$，定义

$$\bigl(T(y)\bigr)(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

逐条检查性质。

$T(y)$ 连续。 $y$ 连续，$f$ 连续（可导），$g$ 连续，所以被积函数连续；可变上限积分连续。

$T(y)$ 可导，且导数为 $f(y)+g$。 由微积分基本定理，

$$\bigl(T(y)\bigr)'(x)=f\bigl(y(x)\bigr)+g(x).$$

所以 $T$ 的像不只在连续函数空间里，还在连续可导函数空间里。

初值自动满足。

$$\bigl(T(y)\bigr)(0)=a+\int_0^0\cdots=a .$$

不动点等价于解

若 $y$ 是 $T$ 的不动点，即 $y=T(y)$，则

$$y(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

右边可导，且由微积分基本定理 $y'(x)=f(y(x))+g(x)$；又 $y(0)=a$。所以 $y$ 是 ODE 的解。

反之，若 $y$ 是解，即 $y'=f(y)+g$ 且 $y(0)=a$，把 $y'$ 从 $0$ 到 $x$ 积分：

$$y(x)-y(0)=\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

代入 $y(0)=a$ 得 $y=T(y)$。所以 $y$ 是不动点。

两边都是微积分基本定理的直接应用。于是：

$$\text{ODE 的解}\quad\longleftrightarrow\quad T\text{ 的不动点}.$$

限制空间：为什么不能直接在全体连续函数上做

$T$ 在全体 $C(I)$ 上不是压缩映射。问题出在这一项：

$$\bigl|f(y_1(s))-f(y_2(s))\bigr|\le M\,|y_1(s)-y_2(s)| ,$$

这里的 $M$ 是 $|f'|$ 在某个范围上的最大值。但如果 $y_1,y_2$ 的取值可以任意大，$f'$ 就可能无界，$M$ 就不存在，压缩性失效。

这与 Newton 法里必须选一个好初始点是同一类现象：映射只在局部是压缩的。解决办法是把空间限制到一个闭球，在球上 $|f'|$ 有最大值，于是压缩常数可以算出来。

闭球

固定半径 $R>0$，定义

$$B_R=\Bigl\{\,y\in C([-\delta_0,\delta_0]) : |y(x)|\le R\ \text{对一切 }x\,\Bigr\}.$$

这是连续函数空间中、以零函数为中心、半径为 $R$ 的闭球。

为什么必须闭？因为闭子集才继承完备性。开球不闭，Cauchy 列的极限可能落在球外，压缩映射定理的完备性前提就断了。

四个常数的定义

先取

$$R=|a|+2 .$$

多出来的 $2$ 是留给积分项的余量：$a$ 本身加上一个严格小于 $2$ 的量，仍在球内。

因为 $f$ 连续、$g$ 连续，在紧集上有最大值。定义

$$L_1=\max_{|z|\le R}|f(z)|, \qquad L_2=\max_{|z|\le1}|g(z)| .$$

再定义

$$\delta_0=\min\Bigl(1,\ \frac{1}{L_1+1},\ \frac{1}{L_2+1}\Bigr).$$

三个选择各有用途：

• $1$：保证区间 $[-\delta_0,\delta_0]$ 落在 $[-1,1]$ 中，这样 $g$ 的最大值 $L_2$ 才适用；
• $1/(L_1+1)$：保证 $\delta_0L_1<1$；
• $1/(L_2+1)$：保证 $\delta_0L_2<1$。

分母加 $1$ 只是为了防止 $L_1$ 或 $L_2$ 为零时除以零，同时让不等式变成严格的。

第一步：$T$ 把闭球映入自身

要证：若 $y\in B_R$，则 $T(y)\in B_R$。

$T(y)$ 连续已经知道。只需估计 $|x|\le\delta_0$ 时的绝对值。设 $|x|\le\delta_0$。

由三角不等式和积分绝对值不等式，

$$\begin{aligned} \bigl|T(y)(x)\bigr| &\le |a|+\left|\int_0^x f(y(s))\,ds\right| +\left|\int_0^x g(s)\,ds\right|\\ &\le |a|+\left|\int_0^x\bigl|f(y(s))\bigr|\,ds\right| +\left|\int_0^x|g(s)|\,ds\right| . \end{aligned}$$

现在逐项放缩。

第一项。 $y\in B_R$ 意味着 $|y(s)|\le R$，所以 $f(y(s))$ 的自变量落在 $|z|\le R$ 内，由 $L_1$ 的定义，

$$\bigl|f(y(s))\bigr|\le L_1 .$$

第二项。 $s$ 位于 $0$ 与 $x$ 之间，而 $|x|\le\delta_0\le1$，所以 $|s|\le1$，由 $L_2$ 的定义，

$$|g(s)|\le L_2 .$$

长度。 积分区间的长度是 $|x|\le\delta_0$。

于是

$$\bigl|T(y)(x)\bigr|\le |a|+\delta_0L_1+\delta_0L_2 .$$

代入 $\delta_0$ 的两个上界：

$$\delta_0L_1\le\frac{L_1}{L_1+1}<1, \qquad \delta_0L_2\le\frac{L_2}{L_2+1}<1 .$$

所以

$$\bigl|T(y)(x)\bigr|<|a|+2=R .$$

对一切 $|x|\le\delta_0$ 成立。因此 $T(y)\in B_R$，即

$$T:B_R\to B_R .$$

这一步的意义：迭代不会跑出球。若映射把点送到球外，压缩映射定理就没法迭代下去——这正是课程强调空间必须「稳定」的原因。

第二步：压缩性

还需要一个常数。定义

$$M=\max_{|z|\le R}|f'(z)| .$$

$f'$ 连续，在紧集上有最大值，所以 $M$ 是有限的非负实数。再把区间收窄为

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

在 $[-\delta,\delta]$ 上重新取半径仍为 $R$ 的闭球，仍记作 $B_R$。因为 $\delta\le\delta_0$，第一步的「映入自身」依然成立。

取 $y_1,y_2\in B_R$，估计 $T(y_1)-T(y_2)$。两式的常数 $a$ 相减抵消，$g$ 也抵消：

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

取绝对值并把绝对值放进积分：

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

用中值定理处理被积函数。 固定 $s$，把 $y_1(s)$、$y_2(s)$ 看成两个数，记为 $z_1,z_2$。因为 $y_1,y_2\in B_R$，两者都落在 $[-R,R]$。中值定理给出：存在介于 $z_1$、$z_2$ 之间的 $z_3$，使

$$f(z_1)-f(z_2)=f'(z_3)\,(z_1-z_2).$$

于是

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

代回积分：

$$\bigl|T(y_1)(x)-T(y_2)(x)\bigr| \le M\,d(y_1,y_2)\,|x| \le \delta M\,d(y_1,y_2).$$

由 $\delta\le1/(2M+1)$，

$$\delta M\le\frac{M}{2M+1}<\frac12 .$$

所以对一切 $x$ 都有

$$\bigl|T(y_1)(x)-T(y_2)(x)\bigr|\le\frac12\,d(y_1,y_2).$$

取上确界（此时 $\sup=\max$，因为 $T(y_1)-T(y_2)$ 连续）：

$$d\bigl(T(y_1),T(y_2)\bigr)\le\frac12\,d(y_1,y_2).$$

压缩常数可以取 $c=1/2$，严格小于 $1$。$\square$

补充：课程在讲到这里时刚好下课，最后「$\le\delta M\le\tfrac12$」的收尾是在下一讲开头用同样的估计完成的；这里把两步合在一起写完整。下一讲把 $M$ 记作 $L$，并把 $\delta$ 取成 $1/(2L)$，效果相同。

结论

把所有拼起来：

1. $B_R$ 是完备空间 $C([-\delta,\delta])$ 的闭子集，故完备。
2. $T:B_R\to B_R$。
3. $T$ 在 $B_R$ 上是压缩映射，压缩常数 $1/2$。
4. 由压缩映射定理，$T$ 有唯一不动点 $y$。
5. 不动点等价于 ODE 的解。

于是得到了 Picard–Lindelöf 定理的存在性与局部唯一性：在 $(-\delta,\delta)$ 上，给定初值 $a$ 的 ODE 有且只有一个解。

这里证出的是「在一个小球内唯一」。若有两个解都定义在更大的区间上，在球内它们必须相同；球外怎么办，需要下一讲的连通性论证。

本节结论

• 压缩映射上至多一个不动点；完备空间中的压缩映射恰有一个。
• 构造不动点的迭代方法：任取一点开始，误差按 $c^n$ 缩小。
• 完备性的作用是取极限；压缩性的作用是保证 Cauchy 且 $T$ 连续。
• ODE 的解等价于积分算子 $T$ 的不动点。
• $T$ 在全体连续函数上不压缩，必须限制到闭球。
• 常数 $R=|a|+2$、$\delta_0=\min(1,1/(L_1+1),1/(L_2+1))$ 保证 $T$ 映入自身；$\delta=\min(\delta_0,1/(2M+1))$ 保证压缩，常数为 $1/2$。

理解检验

1. 证明压缩映射至多有一个不动点，并指出证明中哪一步用到 $c<1$。

参考答案：设 $x,y$ 都是不动点，则 $d(x,y)=d(Tx,Ty)\le c\,d(x,y)$，移项得 $(1-c)d(x,y)\le0$。因为 $d\ge0$ 且 $1-c>0$，只能 $d(x,y)=0$，故 $x=y$。$c<1$ 用在「$1-c>0$」这一步：若 $c=1$，不等式变成 $0\le0$，什么都推不出来（恒等映射就是反例）。

2. 为什么迭代序列 $(T^n x)$ 是 Cauchy 列？估计式中的 $c^n$ 从哪来？

参考答案：先在映射层面迭代 $n$ 次，压缩性给出 $d(T^n x,T^n(T^k x))\le c^n d(x,T^k x)$；这里每用一次压缩性就提出一个 $c$，用 $n$ 次得到 $c^n$。再对 $d(x,T^k x)$ 用三角不等式拆成 $k$ 段相邻差，每段被 $c^j d(x,Tx)$ 控制，求和后用几何级数放大为 $d(x,Tx)/(1-c)$。合起来 $d(T^n x,T^{n+k}x)\le c^n d(x,Tx)/(1-c)$，与 $k$ 无关且随 $n$ 趋于 $0$，故为 Cauchy 列。

3. 为什么 $T$ 在全体连续函数空间上不是压缩映射？限制到闭球解决了什么？

参考答案：压缩性的估计需要 $|f(y_1(s))-f(y_2(s))|\le M|y_1(s)-y_2(s)|$，而 $M=\max|f'|$ 只在有界区域上才有有限值。若 $y$ 的取值可以任意大，$f'$ 可能无界，$M$ 不存在，估计失效。限制到闭球 $B_R$ 后，球内 $|y(x)|\le R$，$f'$ 在 $[-R,R]$ 上连续、取到最大值 $M$，压缩性才能证明。

4. 常数 $R=|a|+2$ 中的 $2$，与 $\delta_0$ 中分母的「$+1$」，各起什么作用？

参考答案：$2$ 是留给两个积分项的余量。估计给出 $|Ty(x)|\le|a|+\delta_0L_1+\delta_0L_2$，而 $\delta_0L_1<1$、$\delta_0L_2<1$，两项合计严格小于 $2$，所以 $|Ty(x)|<R$，映射留在球内。分母的 $+1$ 有两个作用：避免 $L_1$ 或 $L_2$ 为零时除以零；把 $\delta_0L_i\le L_i/(L_i+1)$ 变成严格小于 $1$，从而得到严格不等式。

5. 为什么必须用闭球而不是开球？

参考答案：压缩映射定理要求空间 Cauchy 完备。完备空间的闭子集完备，而开子集一般不完备：Cauchy 列可能收敛到边界点上，而边界点不在开球里。用开球时，迭代序列的极限可能落到球外，定理的前提失效。闭球则保证极限仍在球内，「映入自身」与「完备」两条同时成立。

与前后课的关系

上一讲把求导与极限的交换讲完，并把常微分方程改写成了积分方程。本讲完成技术工作：压缩映射定理的完整证明、积分算子的构造、闭球上的自映射与压缩性，最终得到局部存在唯一性。

下一讲有两件事：第一，用区间连通性把「球内唯一」提升为「共同定义域上唯一」；第二，用 $y'=y^2$ 的显式解说明解确实可能在有限点爆破，所以定理只能局部。那两件事合起来，才是完整的 Picard–Lindelöf 定理。

Does a differential equation always have a solution?

Consider the initial value problem

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

where $f$ is differentiable with $f'$ continuous, $g$ is continuous, and $a$ is a given real number.

This is not a problem with a plug-in formula. The nonlinearity $f$ can be arbitrary, and explicit solutions usually do not exist. The right questions are qualitative:

• Existence: given $a$, is there at least one solution?
• Uniqueness: if there are two solutions, must they coincide?

These questions recur throughout science and engineering: an equation models some natural process, and one first needs to know that a solution exists at all, then that it is unique — otherwise prediction is meaningless.

The Picard–Lindelöf theorem answers exactly these questions, and the answer is local: there is some $\delta>0$ such that a solution exists on $(-\delta,\delta)$ and is unique there. Why it can only be local is shown in the final lecture by the blow-up of $y'=y^2$.

The route of the proof is unexpected: it uses metric spaces. Specifically, the solution is written as a fixed point of an integral operator, and the contraction mapping theorem is applied to that operator.

Timeline

Two tools

Tool one: the sup metric and completeness

On the space $C(I)$ of continuous functions on an interval $I$, take

$$d(h_1,h_2)=\max_{x\in I}|h_1(x)-h_2(x)| .$$

It was proved earlier that $\bigl(C([a,b]),d\bigr)$ is Cauchy complete: every Cauchy sequence converges to a continuous function.

One more simple fact is needed: a closed subset of a complete space is complete. If $B$ is closed in a complete space $X$ and $(x_n)\subset B$ is Cauchy, then it converges in $X$ to some $x$; since $B$ is closed, $x\in B$. So Cauchy sequences in $B$ converge inside $B$.

Tool two: contractions

Definition. Let $(X,d)$ be a metric space and $T:X\to X$. If there is a constant $c<1$ with

$$d\bigl(T(x),T(y)\bigr)\le c\,d(x,y) \quad\text{for all }x,y\in X,$$

then $T$ is a contraction.

The constant must be strictly less than $1$. Equality is not enough: the identity map has $c=1$ and fixes every point, so the conclusion would collapse.

Theorem one: a contraction has at most one fixed point

Theorem. If $T$ is a contraction, it has at most one fixed point.

Proof. Let $x$ and $y$ be fixed points:

$$T(x)=x, \qquad T(y)=y .$$

By the contraction property,

$$d(x,y)=d\bigl(T(x),T(y)\bigr)\le c\,d(x,y).$$

Move the right side over:

$$(1-c)\,d(x,y)\le0 .$$

Since $1-c>0$ and $d(x,y)\ge0$, necessarily $d(x,y)=0$, hence $x=y$. $\square$

Only two facts are used: the contraction property and positive definiteness. Completeness is not needed here; it is there for existence in the next theorem.

Theorem two: the contraction mapping theorem

Theorem. Let $(X,d)$ be a Cauchy complete metric space and $T:X\to X$ a contraction. Then $T$ has a unique fixed point.

Uniqueness is already covered by theorem one; existence must be constructed.

Step 1: iterate from an arbitrary point

Take any $x\in X$ and define

$$x_0=x, \qquad x_1=T(x), \qquad x_2=T^2(x), \qquad\dots$$

that is, $x_n=T^n(x)$. Unlike Newton's method, no good starting point is needed — any point works, though convergence may be slower.

Goal: show that $(x_n)$ converges and that its limit $x_\infty$ satisfies $T(x_\infty)=x_\infty$.

Step 2: consecutive errors shrink geometrically

For any $l\ge1$,

$$\begin{aligned} d(x_l,x_{l+1}) &=d\bigl(T^l(x),T^{l+1}(x)\bigr)\\ &=d\bigl(T(T^{l-1}x),T(T^{l}x)\bigr)\\ &\le c\,d\bigl(T^{l-1}x,T^{l}x\bigr). \end{aligned}$$

Each step back multiplies the error by $c$. Iterating,

$$d(x_l,x_{l+1})\le c^{\,l}\,d(x_0,x_1).$$

Step 3: the distance between any two terms

Let $n\ge0$ and $k\ge1$. First iterate the contraction $n$ times at the level of maps:

$$\begin{aligned} d(x_n,x_{n+k}) &=d\bigl(T^n x,T^n(T^k x)\bigr)\\ &\le c^n\,d(x,T^k x). \end{aligned}$$

Now estimate $d(x,T^k x)$ by the triangle inequality, splitting it into $k$ consecutive differences:

$$\begin{aligned} d(x,T^k x) &\le\sum_{j=0}^{k-1}d(x_j,x_{j+1})\\ &\le\sum_{j=0}^{k-1}c^j\,d(x,T x) \le\frac{d(x,Tx)}{1-c}. \end{aligned}$$

The last step is the geometric series: since $c<1$, $\sum_{j=0}^{\infty}c^j=1/(1-c)$, and partial sums do not exceed it.

Combining,

$$d(x_n,x_{n+k}) \le c^n\,\frac{d(x,Tx)}{1-c}.$$

The right side does not depend on $k$, and $c^n\to0$ as $n\to\infty$. Hence for every $\varepsilon>0$ there is $N$ such that for $n\ge N$ and every $k\ge1$, $d(x_n,x_{n+k})<\varepsilon$. That is exactly the Cauchy condition.

The hypothesis $c<1$ is decisive: if $c=1$, $c^n$ does not decay and nothing follows.

Step 4: take the limit

$X$ is complete, so $(x_n)$ converges. Write

$$x_\infty=\lim_{n\to\infty}x_n .$$

Step 5: the limit is a fixed point

We show $T(x_\infty)=x_\infty$. Apply $T$ to the sequence:

$$T(x_0),T(x_1),T(x_2),\dots$$

Since $T(x_n)=x_{n+1}$, this is the original sequence with the first term removed, so it also converges to $x_\infty$.

On the other hand, compare $T(x_\infty)$ with $T(x_n)$ using the contraction property:

$$d\bigl(T(x_\infty),T(x_n)\bigr)\le c\,d(x_\infty,x_n).$$

As $x_n\to x_\infty$, the right side tends to $0$, so $T(x_n)\to T(x_\infty)$. But $T(x_n)=x_{n+1}\to x_\infty$. Limits in a metric space are unique, hence

$$T(x_\infty)=x_\infty .$$

So $x_\infty$ is a fixed point. $\square$

The structure is worth remembering: the contraction property produces a Cauchy sequence, completeness supplies the limit, and the contraction property again shows that $T$ commutes with the limit. The last step is in effect a proof that a contraction is continuous.

Turning the ODE into a fixed point problem

Now the connection to differential equations.

The map $T$

Let $I$ be an interval containing $0$. For a continuous $y:I\to\mathbb R$, define

$$\bigl(T(y)\bigr)(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

Check the properties one by one.

$T(y)$ is continuous. $y$ is continuous, $f$ is continuous (differentiable), $g$ is continuous, so the integrand is continuous; an integral with variable upper limit is continuous.

$T(y)$ is differentiable with derivative $f(y)+g$. By the fundamental theorem of calculus,

$$\bigl(T(y)\bigr)'(x)=f\bigl(y(x)\bigr)+g(x).$$

So the image of $T$ lies not merely in continuous functions but in continuously differentiable ones.

The initial value is automatic.

$$\bigl(T(y)\bigr)(0)=a+\int_0^0\cdots=a .$$

Fixed points are exactly solutions

If $y$ is a fixed point, $y=T(y)$, then

$$y(x)=a+\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

The right side is differentiable, and by the fundamental theorem $y'(x)=f(y(x))+g(x)$; also $y(0)=a$. So $y$ solves the ODE.

Conversely, if $y$ is a solution, $y'=f(y)+g$ and $y(0)=a$, integrate $y'$ from $0$ to $x$:

$$y(x)-y(0)=\int_0^x\bigl[f(y(s))+g(s)\bigr]\,ds .$$

Substituting $y(0)=a$ gives $y=T(y)$. So $y$ is a fixed point.

Both directions are direct applications of the fundamental theorem of calculus. Hence

$$\text{solutions of the ODE}\quad\longleftrightarrow\quad\text{fixed points of }T .$$

Restricting the space: why not work on all continuous functions

$T$ is not a contraction on all of $C(I)$. The obstruction is the term

$$\bigl|f(y_1(s))-f(y_2(s))\bigr|\le M\,|y_1(s)-y_2(s)| ,$$

where $M$ is the maximum of $|f'|$ on some range. But if $y_1,y_2$ take arbitrarily large values, $f'$ may be unbounded, $M$ may not exist, and the contraction property fails.

This is the same phenomenon as the need for a good starting point in Newton's method: the map is only locally contracting. The remedy is to restrict the space to a closed ball, on which $|f'|$ has a maximum and the contraction constant can be computed.

The closed ball

Fix a radius $R>0$ and define

$$B_R=\Bigl\{\,y\in C([-\delta_0,\delta_0]) : |y(x)|\le R\ \text{for all }x\,\Bigr\}.$$

This is the closed ball of radius $R$ about the zero function in the space of continuous functions.

Why closed? Because only a closed subset inherits completeness. An open ball is not closed, a Cauchy sequence may converge to a boundary point outside it, and the completeness hypothesis of the contraction theorem would fail.

Four constants

First take

$$R=|a|+2 .$$

The extra $2$ is slack for the integral terms: $a$ plus a quantity strictly below $2$ still lies inside the ball.

Since $f$ and $g$ are continuous, they attain maxima on compact sets. Define

$$L_1=\max_{|z|\le R}|f(z)|, \qquad L_2=\max_{|z|\le1}|g(z)| .$$

Then set

$$\delta_0=\min\Bigl(1,\ \frac{1}{L_1+1},\ \frac{1}{L_2+1}\Bigr).$$

Each choice has a job:

• $1$: ensures $[-\delta_0,\delta_0]\subset[-1,1]$, so that the bound $L_2$ for $g$ applies;
• $1/(L_1+1)$: ensures $\delta_0L_1<1$;
• $1/(L_2+1)$: ensures $\delta_0L_2<1$.

The $+1$ in the denominators only guards against dividing by zero when $L_1$ or $L_2$ vanishes, and it makes the inequalities strict.

Step one: $T$ maps the ball into itself

We show: if $y\in B_R$ then $T(y)\in B_R$.

Continuity of $T(y)$ is known. It remains to bound the absolute value for $|x|\le\delta_0$. Let $|x|\le\delta_0$. By the triangle inequality and the integral inequality,

$$\begin{aligned} \bigl|T(y)(x)\bigr| &\le |a|+\left|\int_0^x f(y(s))\,ds\right| +\left|\int_0^x g(s)\,ds\right|\\ &\le |a|+\left|\int_0^x\bigl|f(y(s))\bigr|\,ds\right| +\left|\int_0^x|g(s)|\,ds\right| . \end{aligned}$$

Now bound each piece.

First piece. $y\in B_R$ means $|y(s)|\le R$, so the argument of $f$ lies in $|z|\le R$; by the definition of $L_1$,

$$\bigl|f(y(s))\bigr|\le L_1 .$$

Second piece. $s$ lies between $0$ and $x$, and $|x|\le\delta_0\le1$, so $|s|\le1$; by the definition of $L_2$,

$$|g(s)|\le L_2 .$$

Length. The interval of integration has length $|x|\le\delta_0$.

Hence

$$\bigl|T(y)(x)\bigr|\le |a|+\delta_0L_1+\delta_0L_2 .$$

Substituting the two upper bounds on $\delta_0$,

$$\delta_0L_1\le\frac{L_1}{L_1+1}<1, \qquad \delta_0L_2\le\frac{L_2}{L_2+1}<1 .$$

Therefore

$$\bigl|T(y)(x)\bigr|<|a|+2=R$$

for all $|x|\le\delta_0$. So $T(y)\in B_R$, i.e.

$$T:B_R\to B_R .$$

Why this matters: the iteration must not leave the ball. If the map sent points outside, the contraction iteration could not be continued — exactly why the course insists that the space be stable under the map.

Step two: the contraction property

One more constant. Define

$$M=\max_{|z|\le R}|f'(z)| .$$

$f'$ is continuous and attains a maximum on a compact set, so $M$ is a finite nonnegative number. Now shrink the interval to

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

On $[-\delta,\delta]$, take again the ball of radius $R$, still denoted $B_R$. Since $\delta\le\delta_0$, the "maps into itself" property still holds.

Take $y_1,y_2\in B_R$ and estimate $T(y_1)-T(y_2)$. The constants $a$ cancel, and so do the $g$ terms:

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

Take absolute values and move the absolute value inside the integral:

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

Apply the mean value theorem to the integrand. Fix $s$ and view $y_1(s)$, $y_2(s)$ as two numbers $z_1,z_2$. Since $y_1,y_2\in B_R$, both lie in $[-R,R]$. The mean value theorem gives a $z_3$ between $z_1$ and $z_2$ with

$$f(z_1)-f(z_2)=f'(z_3)\,(z_1-z_2).$$

Hence

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

Substituting back into the integral,

$$\bigl|T(y_1)(x)-T(y_2)(x)\bigr| \le M\,d(y_1,y_2)\,|x| \le \delta M\,d(y_1,y_2).$$

Since $\delta\le1/(2M+1)$,

$$\delta M\le\frac{M}{2M+1}<\frac12 .$$

Therefore, for every $x$,

$$\bigl|T(y_1)(x)-T(y_2)(x)\bigr|\le\frac12\,d(y_1,y_2).$$

Taking the supremum (here $\sup=\max$, since $T(y_1)-T(y_2)$ is continuous),

$$d\bigl(T(y_1),T(y_2)\bigr)\le\frac12\,d(y_1,y_2).$$

So $c=1/2$ works, strictly less than $1$. $\square$

Note: the lecture reached this point just as time ran out; the final step $\delta M\le\tfrac12$ is completed at the start of the next lecture using the same estimate. Here the two halves are written together. The next lecture writes $M$ as $L$ and takes $\delta=1/(2L)$, with the same effect.

Conclusion

Assembling the pieces:

1. $B_R$ is a closed subset of the complete space $C([-\delta,\delta])$, hence complete.
2. $T:B_R\to B_R$.
3. $T$ is a contraction on $B_R$ with constant $1/2$.
4. By the contraction mapping theorem, $T$ has a unique fixed point $y$.
5. Fixed points are exactly solutions of the ODE.

This proves existence and local uniqueness in the Picard–Lindelöf theorem: on $(-\delta,\delta)$ the ODE with initial value $a$ has exactly one solution.

What is proved is uniqueness "inside one small ball". If two solutions are defined on a larger interval, they must agree inside the ball; what happens outside requires the connectedness argument of the next lecture.

What this lecture establishes

• A contraction has at most one fixed point; in a complete space it has exactly one.
• The iteration starts from an arbitrary point and errors shrink like $c^n$.
• Completeness supplies the limit; contraction supplies the Cauchy property and continuity of $T$.
• Solutions of the ODE are exactly fixed points of the integral operator $T$.
• $T$ is not a contraction on all continuous functions and must be restricted to a closed ball.
• The constants $R=|a|+2$ and $\delta_0=\min(1,1/(L_1+1),1/(L_2+1))$ make $T$ map the ball into itself; $\delta=\min(\delta_0,1/(2M+1))$ makes it a contraction with constant $1/2$.

Comprehension check

1. Prove that a contraction has at most one fixed point, and identify where $c<1$ is used.

Reference answer: If $x,y$ are fixed points, then $d(x,y)=d(Tx,Ty)\le c\,d(x,y)$, so $(1-c)d(x,y)\le0$. Since $d\ge0$ and $1-c>0$, necessarily $d(x,y)=0$ and $x=y$. The hypothesis $c<1$ enters through "$1-c>0$": if $c=1$ the inequality becomes $0\le0$, which yields nothing (the identity map is a counterexample).

2. Why is the iteration $(T^n x)$ Cauchy? Where does $c^n$ come from?

Reference answer: Iterating the contraction $n$ times at the level of maps gives $d(T^n x,T^n(T^k x))\le c^n d(x,T^k x)$; each use of the contraction pulls out one factor $c$, and $n$ uses give $c^n$. Then $d(x,T^k x)$ is split by the triangle inequality into $k$ consecutive differences, each bounded by $c^j d(x,Tx)$, and the sum is dominated by the geometric series $d(x,Tx)/(1-c)$. Hence $d(T^n x,T^{n+k}x)\le c^n d(x,Tx)/(1-c)$, independent of $k$ and tending to $0$ with $n$, which is the Cauchy condition.

3. Why is $T$ not a contraction on the whole space of continuous functions? What does restricting to a closed ball fix?

Reference answer: The contraction estimate needs $|f(y_1(s))-f(y_2(s))|\le M|y_1(s)-y_2(s)|$, and $M=\max|f'|$ is finite only on a bounded region. If $y$ can take arbitrarily large values, $f'$ may be unbounded, $M$ may not exist, and the estimate fails. On the closed ball $B_R$ we have $|y(x)|\le R$, so $f'$ is continuous on the compact set $[-R,R]$, attains a maximum $M$, and the contraction property can be proved.

4. What jobs do the $2$ in $R=|a|+2$ and the $+1$ in the denominators of $\delta_0$ do?

Reference answer: The $2$ is slack for the two integral terms: the estimate gives $|Ty(x)|\le|a|+\delta_0L_1+\delta_0L_2$ with $\delta_0L_1<1$ and $\delta_0L_2<1$, so the two terms total strictly less than $2$ and $|Ty(x)|<R$, keeping the map inside the ball. The $+1$ in the denominators does two things: it avoids division by zero when $L_1$ or $L_2$ vanishes, and it turns $\delta_0L_i\le L_i/(L_i+1)$ into a strict inequality below $1$.

5. Why must the ball be closed rather than open?

Reference answer: The contraction mapping theorem requires a Cauchy complete space. A closed subset of a complete space is complete, but an open subset generally is not: a Cauchy sequence may converge to a boundary point lying outside the open ball. With an open ball the limit of the iteration could fall outside, and the theorem's hypothesis would fail. With a closed ball the limit stays inside, so both "maps into itself" and "complete" hold.

Relation to neighbouring lectures

The previous lecture finished the interchange of differentiation and limits and rewrote the ODE as an integral equation. This lecture completes the technical work: a full proof of the contraction mapping theorem, the construction of the integral operator, the self-map and contraction properties on a closed ball, and finally local existence and uniqueness.

The next lecture does two things: first, use connectedness of intervals to upgrade "unique inside the ball" to "unique on the common domain"; second, use the explicit solution of $y'=y^2$ to show that solutions really can blow up at a finite point, so the theorem can only be local. Together they form the complete Picard–Lindelöf theorem.