Picard's existence theorem states that if $U$ is an open subset of $\mathbb{R}^2$ and $f$ is a continuous function on $U$ that is Lipschitz continuous with respect to the second variable then there is some $h>0$ such that the initial value problem

$$y'=f(x,y), \\y(a)=b$$

has a unique solution on an interval $(a-h,a+h)$ in $U$.

Now assume that I wanted to solve $$y'=y, \\ y(0)=1$$ using the fixed point theorem. Solving the above problem is equivalent to solving $$\phi'(x)=\int_0^x\phi(t)dt+1$$ for $\phi$.

Now consider the map $F:C([-r,r])\rightarrow C([-r,r])$ given by

$$(F(\psi))(x)=\int_0^x\psi(t)dt+1$$ where $r>0$.

Since $C([-r,r])$ is complete, if we only proved that $F$ is a contraction then we could use the fixed point theorem to find the solution. But $F$ is a contraction only if $2r<1$. So choose $r<\frac{1}{2}$.

Using successive approximations we find out that $y=e^x$ is the solution on an interval $(-\frac{1}{2},\frac{1}{2})$. But obviously this is not the maximal open interval on which $y=e^x$ is our solution. I wonder how we can proceed from here to conclude that $y=e^x$ is the solution on whole $\mathbb{R}$.

  • $\begingroup$ Here this is a linear differential equation with constant coefficients, so the solution are defined on whole $\mathbb{R}$ "automatically". $\endgroup$ – Balloon Dec 24 '15 at 15:05
  • $\begingroup$ @Baloown I know. But I asked it it is possible to find the solution on whole $\mathbb{R}$ repeating/modyfing the reasoning using successive approximations. $\endgroup$ – luka5z Dec 24 '15 at 15:15
  • $\begingroup$ Oh I see. Perhaps you can use that : if one of the iterated of $F$ (i.e. one of the $F\circ\dots\circ F$) is a contraction, then $F$ has a unique fixed point. So you could choose a $r$ bigger an show that $e$ is indeed solution on all compact sets as you wanted. $\endgroup$ – Balloon Dec 24 '15 at 15:25
  • $\begingroup$ I rather thought of glueing solutions. I mean using fixed point theorem on overlapping intervals of length less than $1$ and then using uniqueness property of the solution concluding that the whole solution is unique. $\endgroup$ – luka5z Dec 24 '15 at 15:33
  • $\begingroup$ But in all cases your interval should contain $0$ to answer the Cauchy's problem, so you can't work on $[1/2,1]$ for example ? $\endgroup$ – Balloon Dec 24 '15 at 15:37

The global solution of an initial value problem (IVP) $$x'=f(t,x),\quad x(t_0)=x_0\tag{1}$$ is not obtained via Picard's theorem, but through analytic continuation. The grand picture is as follows:

We are given an open domain $\Omega\subset{\mathbb R}\times{\mathbb R}^n$ and a continuous function $$f:\quad \Omega\to{\mathbb R}^n,\qquad(t,x)\mapsto f(t,x)$$ ($n=1$ in the sequel) which is locally Lipschitz continuous with respect to the second variable. This means that for every point $(t_0,x_0)\in\Omega$ there is a rectangular window $W$ with center $(t_0,x_0)$ and a constant $C$ such that $$|f(t,x)-f(t,x')|\leq C|x-x'|\qquad\bigl((t,x),(t,x')\in W\bigr)\ .$$ An $f\in C^1(\Omega)$ automatically fulfills these conditions. Picard's theorem then guarantees that for every $(t_0,x_0)$ there is a window $$[t_0-h,t_0+h]\times [x_0-q,x_0+q]\subset W$$ and a micro-solution $$\phi_0:\quad[t_0-h,t_0+h]\to [x_0-q,x_0+q] \tag{2}$$ of $(1)$.

The micro-solution $(2)$ is continuous up to $t_1:=t_0+h$, and even satisfies $\phi_0'(t_1-)=f\bigl(t_1,\phi_0(t_1)\bigr)$. The uniqueness part of Picard's theorem then allows to conclude that immediately to the left of $t_1$ this $\phi_0$ coincides with the micro-solution $\phi_1$ of the IVP $$x'=f(t,x),\quad x(t_1)=\phi_0(t_1)\ .$$ Note that $\phi_1$ is defined up to the point $t_2:=t_1+h'$ for some $h'>0$, hence extends $\phi_0$ further to the right. Concatenating $\phi_0$ and $\phi_1$, and proceeding in this way forever (and similarly to the left of $t_0$) we arrive at the global solution $$\phi_*:\quad I_* \to{\mathbb R}$$ of the IVP $(1)$. The interval $I_*$ on which $\phi_*$ is defined may be infinite, and in general depends on the initial point $(t_0,x_0)$. About this $\phi_*$ one can say the following:

(i) Global uniqueness: Any solution $\phi$ of $(1)$, defined in some open interval $I\subset{\mathbb R}$, is part of $\phi_*$. This means that $I\subset I_*$, and that $\phi(t)=\phi_*(t)$ for all $t\in I$.

(ii) The graph of $\phi_*$ extends to infinity, or to the boundary of $\Omega$. More precisely: Given any compact set $K\subset\Omega$ there are points $\bigl(t,\phi_*(t)\bigr)\notin K$.

These things are proven in detail in books titled "Theory of differential equations".

  • $\begingroup$ Thank you very much! I would upvote this answer 100 times if I could. So clear. $\endgroup$ – luka5z Dec 27 '15 at 11:42
  • $\begingroup$ So I was right saying that if $f$ is globally Lipschitz with respect to second variable, then there is global unique solution? $\endgroup$ – luka5z Dec 27 '15 at 12:31
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    $\begingroup$ @luka5z: Yes; but the assumption "globally Lipschitz" is too strong. It is seldom fulfilled. An example are linear ODEs with bounded coefficients. $\endgroup$ – Christian Blatter Dec 27 '15 at 12:48

First I prove a lemma :

If $F$ has one iteration which is a contraction, then $F$ has a unique fixed point.

Proof : If $F^n$ is a contraction then it has a unique fixed point $\psi$. The equalities $$F^n\circ F(\psi)=F\circ F^n(\psi)=F(\psi)$$ show that, by unicity of fixed point, $F(\psi)=\psi,$ and so existence of fixed point for $F.$ For unicity, we see that the existence a new fixed point $\varphi$ implies that $$F^n(\varphi)=F\circ\dots\circ F\circ F(\varphi)=F\circ\dots\circ F(\varphi)=\dots=\varphi$$ and so $\varphi=\psi.$ $\square$

Now let $r\in\mathbb{R}^+,$ and let's show that one of the iterated of $F$ is a contraction on $C([0,r]).$ We have that : $$|(F(\psi)-F(\varphi))(t)|=|\int_0^t\psi(s)-\varphi(s)|\mathrm{d}s\leq t\,||\psi-\varphi||_\infty$$ and so $$||F(\psi)-F(\varphi)||_{\infty}\leq r\cdot||\psi-\varphi||_{\infty}$$ so $F$ is $r-$lipschitz. Now see that $$|(F^2(\psi)-F^2(\varphi))(t)|\leq\int_0^t|F(\psi(s))-F(\varphi(s))|\mathrm{d}s\leq\int_0^t s||\psi-\varphi||_\infty\mathrm{d}s=\frac{t^2}{2}||\psi-\varphi||_{\infty}.$$

One can see that $$||F^n(\psi)-F^n(\varphi)||_\infty\leq\frac{r^n}{n!}||\psi-\varphi||_\infty$$ and as it exists $n\in\mathbb{N}$ such that $n!>r^n$ we establish our result.

  • $\begingroup$ The statement of your lemma doesn't make a lot of sense. Do you mean "if $F$ has one iteration which is a contraction, then $F$ has a unique fixed point."? $\endgroup$ – johnny09 Mar 29 at 2:09
  • $\begingroup$ @johnny09 Exactly! $\endgroup$ – Balloon Mar 29 at 7:44
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    $\begingroup$ @johnny09 I have corrected it, thanks a lot for the remark. $\endgroup$ – Balloon Mar 29 at 7:49

For ODE defined on a stripe $(a,b)×\Bbb R^n$ with a global Lipschitz constant $L$ there, one can use the modified supremum norm $$ \|y\|_L=\sup e^{-2L·|t-t_0|}\|y(t)\| $$ where the initial condition is at $t_0\in (a,b)$. One can then show that in this norm the Picard iteration is contractive for continuous functions over the whole of $(a,b)$, giving the maximally possible interval at once.


Existence is usually related to compactness in some or the other way. In your case you'd try to estimate the derivate of the solution on the interval for which you have a solution, $(s,t)$, say, and try to show that the solution is bounded and has bounded derivates on $[s,t]$. You then continue to solve in a neighbourhood of these points and extend your solution. In case you can generalize this procedure independently of $s, t$ then you can conclude that the set on which the solution exists is closed. Picard's theorem tells you it's open, so it's connected. Picard also gives you uniqueness, as a nice addon.

Now the only subset of $\mathbb{R}$ which is closed and open is $\mathbb{R}$ itself, so you are done (in this case. Of course proving compactness prove is often not possible).

  • $\begingroup$ In other words, if $f(x,y)$ is Lipschitz continuous with respect to $y$ on whole $\mathbb{R}$ then are we guaranteed that unique solution exists on whole $\mathbb{R}$? So we can interate through whole $\mathbb{R}$ glueing solutions on small intervals? $\endgroup$ – luka5z Dec 24 '15 at 15:52
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    $\begingroup$ @luka5z Not quite. The Lipschitz condition allows you only to apply Picard's theorem. Deriving bounds for the solution and the derivative is something which you have to read off from the equation itself. Lipshitz wrt to $y$ does not, in general, imply bounded with respect to $x$, and it also does not imply bounds for the solution itself. In case $y^\prime=y$ you trivially have a bound for $y^\prime$ if you have one for $y$ and vice versa you can conclude from this that $y$ is uniformly continuous, hence $y^\prime $ is hence it admits a continuous extensiion at the boundary of the interval. $\endgroup$ – Thomas Dec 24 '15 at 16:16
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    $\begingroup$ @luka5z (continued). So in this case you can conclude: $y$ bounded $\Rightarrow y^\prime$ admits continuous extension to closure of interval and then you can apply the conclusion of my response. (It remains to show $y$ is bounded on bounded intervals). In general this is more involved. $\endgroup$ – Thomas Dec 24 '15 at 16:18
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    $\begingroup$ @luka5z Honestly I don't know and would need to check. I tried to outline the general procedure how you would extend a local solution to a global one. This is by thinking of the differential equation as a feedback machine. If $f$ is nice and $y^\prime= f(x,y)$ and $y$ is nice, then $y^\prime $ is nice. This implies that $y$ is even nicer than you have known. So, because $f$ is nice, also $y^\prime $ is even nicer. It may be so nice that you can conclude it is coninuous on $[s,t]$ instead of $(s,t)$. Now you can start over with initial point(s) $s, t$. To ensure this catches $\mathbb{R}$ $\endgroup$ – Thomas Dec 24 '15 at 18:26
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    $\begingroup$ @luka5z (continued) and does not stop somewhere you use a topological argument and argue with connectedness. This works fine with $y^\prime = y$, but for that case it may actually be too heavy machinery. $\endgroup$ – Thomas Dec 24 '15 at 18:28

Your equation is $$ \phi'(x) = \phi(x),\;\;\; \phi(0)=1. $$ You can see that $\phi'(x)=\phi(x),\; \phi(a)=C$ has a unique solution on any interval $[a,a+h]$ for sufficiently small $h$. Because of this, if you solve $\phi_1'=\phi_1,\;\;\phi(a_1)=C$ on $[a_1,b_1]$, then you can next solve $\phi_2'=\phi_2,\;\;\; \phi_2(a_2)=\phi_1(b_1)$, glue the two functions together on $[a_1,b_2]$ and conclude that $$ \phi(x) = \int_{a_1}^{x}\phi'(t)dt+C,\;\;\; a_1 \le x \le b_2. $$ If you let $\mathscr{C}=\{ y \ge a_1 : \mbox{integral equation solution extends to } [a,y] \}$, then you can show that $\mathscr{C}$ is both open and closed in $[a,\infty)$, which leads to the conclusion that $\mathscr{C}=[a_1,\infty)$. Showing the set is open uses the above extension argument; you can use a continuity argument to extend solutions on $[a,y)$ to $[a,y]$ in order show that $\mathscr{C}$ is closed. Arguing in this way leads to global existence and uniqueness, knowing only that you have a local Lipschitz condition.

Picard iteration has an added feature that can be characterized by showing that an operator $T^n$ is contractive for some $n \ge 1$. For example, if you iterate $$ \phi(x)=\int_{a}^{x}\phi(t_1)dt_1+A \\ = \int_{a}^{x}\int_{a}^{t_1}\phi(t_2)dt_2dt_1+A(x-a)+A, $$ then you can stop after $n$ iterations so long as the following integral operator is contractive: $$ L_n\phi = \int_{a}^{x}\int_{a}^{t_1}\int_{a}^{t_{2}}\cdots\int_{a}^{t_{n-1}}\phi(t_n)dt_n dt_{n-1}\cdots dt_1. $$ For $a \le x \le b$, $$ \|L_n\phi\| \le \frac{(b-a)^{n}}{n!}\|\phi\| $$ For fixed $b$ and large enough $n$, the norm coefficient $\frac{(b-a)^{n}}{n!}$ can be made arbitrarily small because it is the general term of a convergent exponential series.

  • $\begingroup$ Thank you very much, that is a nice answer. It is much clearer for me now. $\endgroup$ – luka5z Dec 27 '15 at 11:43

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