This problem is taken from Royden and Fitzpatrick's Real Analysis, Fourth Edition (Chapter 10, Problem 44). The text proves the Picard Local Existence Theorem for a single differential equation. The exercise is to state and prove an analogous theorem when $O$ is an open subset of $\mathbf{R} \times \mathbf{R}^n$ containing the point $(x_0, \mathbf{y}_0)$, $\mathbf{g}: O \to \mathbf{R}^n$ is continuous, and the system of differential equations is \begin{align*} \mathbf{f}'(x) &= \mathbf{g}(x, \mathbf{f}(x))\\ \mathbf{f}(x_0) &= \mathbf{y}_0 \end{align*}

I thought I could just use a fixed point argument analogous to the proof in the single equation case (that also seems to be the approach taken in the Wikipedia article). But the question includes the following hint: Approximate $\mathbf{g}$ by a Lipschitz mapping and then use the Arzelà–Ascoli Theorem. I'm lost on how to approach the problem with this hint.


1 Answer 1


This is a bit technical. I will try to outline a proof. Let $\mathbb{R}^{n}$ and $\mathbb{R}^{n+1}$ be both endowed with the maximum norm. First consider an open ball $B((x_0,y_0),r)$ in $\mathbb{R}^{n+1}$ such that $C:=\overline{B((x_0,y_0),r)}\subseteq O$. Now $g$ is bounded on the compact set $C$ by $b> 0$, say: $$ \|g(x,y)\|_\infty \le b \quad ((x,y) \in C). $$ By the multivariate Weierstraß Approximation Theorem there is a sequence of polynomials $(p_m)_{m=1}^\infty$ such that $$ \|p_m(x,y)-g(x,y)\|_\infty \le \frac{1}{m} \quad ((x,y) \in C,~ m \in \mathbb{N}). $$ In particular $$ \|p_m(x,y)\|_\infty \le 1+b \quad ((x,y) \in C, ~ m \in \mathbb{N}). $$ By the Picard-Lindelöf Theorem each IVP $$ u'(x)=p_m(x,u(x)), \quad u(x_0)=y_0 $$ has a solution $u_m:I:=[x_0-\delta,x_0+\delta] \to \mathbb{R}^{n}$ with $\delta:=r/(1+b)$. Note that $(x,u_m(x)) \in C$ $(x \in I)$. Each $u_m$ is Lipschitz continuous on $I$ with constant $1+b$ and $u_m(x_0)=y_0$. Hence $\{u_m:m \in \mathbb{N}\}$ is a bounded and equicontinuous subset of $C(I, \mathbb{R}^{n})$, hence relatively compact by Arzela Ascoli. Let $(u_{m_k})$ be a uniformly convergent subsequence of $(u_m)$ with limit $v\in C(I, \mathbb{R}^{n})$, say. Note that then also $g(x,u_{m_k}(x)) \to g(x,v(x))$ uniformly on $I$ as $k \to \infty$. Now for $x \in I$: $$ \|v(x)-y_0-\int_{x_0}^x g(t,v(t))dt\|_\infty $$ $$ \le \|v(x)-u_{m_k}(x)\|_\infty + \|\int_{x_0}^x p_{m_k}(t,u_{m_k}(t))-g(t,u_{m_k}(t))dt\|_\infty + \|\int_{x_0}^x g(t,u_{m_k}(t))-g(t,v(t))dt\|_\infty $$ $$ \le \|v(x)-u_{m_k}(x)\|_\infty + \frac{1}{m_k}|x-x_0| + \|\int_{x_0}^x g(t,u_{m_k}(t))-g(t,v(t))dt\|_\infty \to 0 \quad (k \to \infty). $$ Thus $$ v(x)=y_0 + \int_{x_0}^x g(t,v(t))dt \quad (x \in I). $$

Edit: I meanwhile looked in the book of Royden and Fitzpatrick, and in fact the exercise is a bit missleading. The function $g$ there is assumed to be continuous (nothing else). In this exercise the authors talk about "a form of Picard's existence theorem" which is incorrect then; it is Peano's existence theorem. If in addition $g(x,y)$ is assumed to be Lipschitz in $y$ an approximation by Lipschitz continuous functions makes no sense to me. Moreover the proofs of the Picard Lindelöf theorem for $n=1$ and $n>1$ are almost the same, just by replacing the absolut value by a norm. So my answer shows a way to prove Peano's theorem via the Picard Lindelöf theorem, but after all I am not sure what this exercise is asking for. I think that this exercise is just poorly designed.

  • 1
    $\begingroup$ @LutzLehmann Right. I have changed that. $\endgroup$
    – Gerd
    Jan 8, 2023 at 11:17
  • $\begingroup$ Maybe I’m misunderstanding, but it seems like you are using a multiequation Picard-Lindelöf Theorem when asserting the existence of a solution $u_m$. Isn’t that the result we’re trying to prove? $\endgroup$
    – David
    Jan 8, 2023 at 14:53

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