Picard Iterates Converge Uniformly

Question

I have a homework question that asks to show that the Picard iterates $$ \phi_{n+1}(t) = \int_0^t 1 + \phi_n^2(s) \, ds, \quad \quad \phi_0(t) = 0 $$ converge uniformly on any compact interval $[-r, r] \subseteq (- \pi/2, \pi/2)$.

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I was thinking about stating the fact that those Picard iterates represent the initial value problem $$ x' = 1 + x^2 = f(x), \quad \quad x(0) = 0, $$ whose solution is of course $x(t) = \tan(t)$. That is, $$ \phi_{n+1}(t) = \int_0^t f(\phi_n(s)) \, ds, \quad \quad \phi_0(t) = 0. $$ Then use the fact that $f$ is Lipschitz on any compact interval to get a bound $$ \| \phi_{n+1} - \phi_n \| \leq K \| \phi_n - \phi_{n-1} \| $$ for some constant $K$ (where $\| \cdot \|$ is the sup norm). Then I could bound $ \| \phi_n - \phi_m \|$ and show that $\{ \phi_i \}$ is a Cauchy sequence and I'd be done. However, I think $K$ has to be less than $1$ for that argument to work and I don't think I can show that. Also, that argument doesn't take into account the restriction $[-r, r] \subseteq (- \pi/2, \pi/2)$, only the fact that $f$ is Lipschitz on any compact interval.

I'd really appreciate a push in the right direction.

Answer 1

This is a quite elegant problem. (It is solved, if I remember well, in Fritz John's ODEs, Courant Institute Lecture Notes.)

Hints. It is easy to show inductively that $$ 0\le \varphi_n(t)\le \varphi_{n+1}(t)\le\tan t,\quad \text{for}\,\, t\in [0,\pi/2), $$ and $$ 0\ge \varphi_n(t)\ge \varphi_{n+1}(t)\ge\tan t,\quad \text{for}\,\, t\in (-\pi/2,0]. $$ Next use the fact that: If $K$ compact and $\psi_n:K\to\mathbb R$, $n\in\mathbb N$, a monotone and bounded sequence, then $\{\psi_n\}_{n\in\mathbb N}$ uniformly convergent. (This is Theorem 7.13, page 150, in W. Rudin's Principles of Mathematical Analysis.)

Hence $\{\varphi_n\}_{n\in\mathbb N}$ converges uniformly in every compact subset of $(-\pi/2,\pi/2)$.

Note. Lipschitz continuity has not been used!

Answer 2

We want the map

$$h(g)(t) = \int_0^t f(g(s)) ds$$

to be a contraction on an appropriate subset of $(C[-T,T],\| \cdot \|_\infty)$, where $T>0$. To get this, note that $f$ is Lipschitz on any compact interval with Lipschitz constant $L$. In particular, it is Lipschitz on $[-M,M]$ with $L=2M$. (Why?) Therefore, given a value of $T$ we have

$$\| h(g_1) - h(g_2) \|_\infty \leq L T \| g_1 - g_2 \|_\infty$$

So $h$ is a contraction if $T<1/L$. Thus if $|g_1|,|g_2|$ are both at most $M$ on $[-T,T]$, then this is a contraction if $T<1/(2M)$.

Now we don't know a priori how fast the solution $x$ grows, so we can't necessarily calculate $L(T)$ in order to find a suitable value of $T$. (We could if $f$ were globally Lipschitz, but it isn't.) But we don't need to. Instead, we get an interval $[-T_1,T_1]$ with $T_1>0$ where everything is OK. Then we consider Picard iteration on the same ODE with the initial condition $(T_1,x(T_1))$. This gives us a $T_2$, which lets us extend to a new IVP with initial condition $(T_2,x(T_2))$, and so the process repeats.

To make this formal, prove by induction that given any interval $[-T_n,T_n]$ where we have a solution, we can always get $[-T_{n+1},T_{n+1}] \supset [-T_n,T_n]$ where we also have a solution. Then we get a solution on $\bigcup_{n=1}^\infty [-T_n,T_n]$. Now try to prove that this union must necessarily be all of $(-\pi/2,\pi/2)$. (Hint: if it weren't, then $x$ wouldn't blow up at the end, and we could continue to an even larger domain. Why is this impossible?)

If you have some concern about uniformity with these multiple intervals, note that there are only ever finitely many contained in any particular $[-r,r]$ where we have a solution, so we can take $N$ as a maximum of the finitely many $N_k$ that were given on each piece.