Two Proofs of the Cauchy-Peano Theorem

Theorem 1.1 (Cauchy-Peano Existence Theorem)  
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ b continuous in a neighborhood of the point $(x_0, y_0) \in \mathbb{R}^2$. Then there exists a $\alpha > 0$ such that the IVP

$$y' = f(x,y), \, y(x_0) = y_0$$ (1)

has a solution $\phi$ on the interval $I := [x_0 - \alpha, x_0 + \alpha]$. That is, there exists a $\phi = \phi(x)$ defined on $I$ such that

$$\phi'(x) = f(x,\phi(x)), \, x \in I$$

and $\phi(x_0) = y_0$.

Remark: In general, $\phi$ is not unique. If $f$ is Lipschitz continuous with respect to $y$, then uniqueness follows from the Picard theorem / Picard iterates.
Remark 2: We give two proofs to show the differences in the two approaches. The first one is the approximation procedure, and the second is the topological / fixed point method.

Proof 1:
Since $f$ is continuous in a neighborhood of $(x_0,y_0)$, there exists $a > 0$ such that $f$ is continuous in the closed square

$$Q := \{ (x,y) \in \mathbb{R}^2 : \vert x - x_0 \vert \leq a, \vert y - y_0\vert \leq a \}.$$

Let $M = \max_Q \vert f(x,y)\vert$ (which exists as $f$ is a continuous function on a compact set). Set $\alpha := \frac{a}{M}$ (we have assumed, WLOG that $M \geq 1$).
We know, from previous courses, that $\phi = \phi(x)$ is a solution of the IVP (1) if and only if $\phi$ satisfies the integral equation

$$\phi(x) = y_0 + \int_{x_0}^x f(t,\phi(t)) dt$$ (2)

(Consequence of FTC)
Since $f$ is uniformly continuous on $Q$ (a compact set), given $\epsilon > 0$, there exists a $\delta = \delta(\epsilon )$ such that $\vert x - \hat x\vert < \delta$ and $\vert y - \hat y\vert < \delta$ implies $\vert f(x,y) - f(\hat x, \hat y)\vert < \epsilon$ for all $(x,y), (\hat x, \hat y) \in Q$. Therefore, let $\epsilon = \epsilon _n = \frac{1}{n}$ and $\delta = \delta_n$.
We choose points $x_j^{(n)}$ with $x_0^{(n)} = x_0$ and $x_{k(n)} = x_0 + \alpha$ (where $0 \leq j \leq k(n)$), with $\vert x_{j+1}^{(n)} - x_j^{(n)}\vert \leq \frac{\delta_n}{M}$.
We define the polygonal approximation $\phi_n$ on $[x_0,x_0 + \alpha]$ by $\phi_n(x_0) = y_0$ and $\phi_n'(x) = f(x_0,y_0), x_0 \leq x \leq x_1^{(n)}$. Then, $y_1^{(n)} = \phi_n(x_1^{(n)}$, and we continue this iterative definition to get that $\phi_n'(x) = f(x_j^{(n)},y_j^{(n)})$ for $x_j^{(n)} \leq x \leq x_{j+1}^{(n)}$.
Note that $\phi_n$ is piecewise $C^1$ (continuous and has, perhaps, a jump discontinuity in the derivative at the partition points). We then define

$$\Delta_n(t) = \begin{cases}\phi_n'(t) - f(t,\phi_n(t)), & x_j^{(n)} < t < x_{j+1}^{(n)} \\ 0, & t = x_j^{(n)} \end{cases}$$

Then notice that

$$\phi_n(x) = y_0 + \int_{x_0}^x \phi_n'(t) dt = y_0 + \int_{x_0}^x \left[ f(t, \phi_n(t)) + \Delta_n(t) \right] dt$$

We claim that $\vert\Delta_n(t)\vert < \frac1n$. Note,

$$\phi_n'(t) = f(x_j^{(n)}, y_j^{(n)}), \, t \in [x_j^{(n)}, x_{j+1}^{(n)}]$$

Therefore,

$$\Delta_n(t)\vert = \vert f(x_j^{(n)}, y_j^{(n)}) - f(t,\phi_n(t)) \vert, \, t \in [x_j^{(n)}, x_{j+1}^{(n)}]$$

Further note

$$\vert t - x_j^{(n)}\vert \leq \vert x_{j+1}^{(n)} - x_j^{(n)} \leq \delta_n$$

$$\vert y_j^{(n)} - \phi_n(t) \vert \leq \vert y_j^{(n)} - y_{j+1}^{(n)} \vert \leq M \vert x_j^{(n)} - x_{j+1}^{(n)} \vert$$

$$\leq \delta_n$$

Thus, from our $\epsilon - \delta$ condition, $\vert\Delta_n(t)\vert \leq \epsilon _n = \frac1n$.
By the Ascoli-Arzela Theorem, there exists a uniformly convergent subsequence of $\{ \phi_n \}$ which converges to $\phi(t) = \lim_{k \rightarrow \infty} \phi_{n_k}(t)$ for $t \in [x_0 - \alpha, x_0 + \alpha]$. That is,

$$\phi_{n_k}(x) = y_0 + \int_{x_0}^x \left[ f(t,\phi_{n_k}(t)) + \Delta_{n_k}(t) \right] dt$$

$$\phi(t) = y_0 + \int_{x_0}^x f(t,\phi(t)) dt,$$

which is exactly what we needed to show a solution to the DE $\square$.

Proof 1, Variant 2:
Recall, from the first proof,

$$Q := \{ (x,y): \vert x - x_0\vert \leq a, \vert y - y_a\vert \leq a \} \subset U$$

where $U$ is a neighborhood of $(x_0,y_0)$, throughout which $f$ is continuous.

$$M := \max_Q \vert f (x,y) \vert$$

$$\alpha := \frac{a}{M}$$

We define the sequence $\Psi_n$ as follows:

$$\Psi_n(x) := \begin{cases}y_0, & x \leq x_0 \\ y_0 + \int_{x_0}^x f(t,\Psi_n(t - \frac\alpha{n})), & x_0 \leq x \leq x_0 + \alpha \end{cases}$$

(where $n \geq 1$). This is well defined - on the interval $[x_0, x_0 + \frac{\alpha}{n}]$, $x - \frac{\alpha}{n} \leq x_0$, so $\Psi_n$ is defined
So again, we have a sequence $\{ \Psi_n(x) \}$ defined on $[x_0,x_0 + \alpha]$. By the Ascoli-Arzela theorem, there exists a uniformly convergent subsequence - say it converges to $\Psi(x)$.

$$\Psi_{n_k}(x) = y_0 + \int_{x_0}^x f(t, \Psi_{n_k}(t - \frac{\alpha}{n_k})) dt$$

$$\Psi(x) = y_0 + \int_{x_0}^x f(t, \Psi(t)) dt$$

It follows that $\Psi$ is also a solution to the DE.

Proof 2: (Topological Proof)
We look for a fixed point of the equation

$$(Tu)(x) := y_0 + \int_{x_0}^x f(t, u(t)) dt$$

Note that $T:\mathcal{C}\rightarrow \mathcal{C}$, where

$$\mathcal{C}:= \{ u$$   continuous on $$I = [x_0 - \alpha, x_0 + \alpha] \}$$

where $\alpha = \frac{a}{M}$, $M = \max_Q \vert f(x,y)\vert$.
$\mathcal{C}$ is a vector space over $\mathbb{R}$ and $\Vert u \Vert = \displaystyle \max_{x \in I} \vert u(x)\vert$ is a norm on $\mathcal{C}$. Convergence in this norm is uniform convergence on $I$. This means that if $u_n \rightarrow u$ in $\mathcal{C}\implies \Vert u_n - u \Vert \rightarrow 0$ as $n \rightarrow \infty$, which implies $\vert u_n(x) - u(x) \vert \rightarrow 0$ uniformly on $I$.
We look for a fixed point of $T$ - want to find $u \in \mathcal{C}$ with $Tu = u \equiv u$ solves the IVP (1).
We define the set $A \subset \mathcal{C}$ by

$$A := \{ u \in sC: \, \vert u(x) - y_0\vert \leq a, \, \vert u(x_1) - u(x_2)\vert \leq M \vert x_1 - x_2\vert \}$$

Then, $A$ is a closed subset of $\mathcal{C}$, as $u_n \in A \implies \displaystyle \lim_{n \rightarrow \infty} u_n = \hat u \in A$ (follows from the uniform convergence in $\mathcal{C}$). Also, $A$ is convex -

$$u, v \in A \implies w = tu + (1 - t)v \in A, \forall 0 \leq t \leq 1$$

(this follows straight from the definition and triangle inequality). Thus, by the Ascoli-Arzela theorem, $A$ is compact.

The operator maps $T : A \rightarrow A$. Suppose $\vert u(x) - y_0\vert \leq a$. Then,

$$\vert Tu(x) - y_0\vert = \vert \int_{x_0}^x f(t, u(t) dt \vert \leq M \vert x - x_0\vert\leq M \alpha = a$$

Similarly, suppose $\vert u(x_1) - u(x_2)\vert \leq M \vert x_1 - x_2\vert$. Then,

$$\vert Tu(x_1) - Tu(x_2)\vert = \vert \int_{x_0}^{x_1} \vert f(t,u(t)) dt - \int_{x_0}^{x_2} f(t,u(t)) dt \vert$$

$$\leq \vert \int_{x_1}^{x_2} f(t, u(t)) dt \leq M \vert x_1 - x_2 \vert$$

A fixed point of $T$ solves the IVP (1), and $T$ has a fixed point as a consequence of the following theorem.

Theorem 1.2   Schader-Tychonoff Theorem
If $T: X \rightarrow X$ is continuous and if $A \subset X$ is a convex compact subset of the normed linear space $X$ and $T(A) \subset A$, then $T$ has a fixed point in $A$.

Continuity of $T$ is a consequence of the (uniform) continuity of $f$ on $Q$, i.e., $u_n \rightarrow u$ implies $Tu_n \rightarrow T u$.

Friday, 1-14-2004