The Picard Iteration

Now we can start actually closing in on a solution to our initial value problem. Recall the setup:

$\displaystyle\begin{aligned}v'(t)&=F(v(t))\\v(0)&=a\end{aligned}$

The first thing we’ll do is translate this into an integral equation. Integrating both sides of the first equation and using the second equation we find

$\displaystyle v(t)=a+\int\limits_0^tF(v(s))\,ds$

Conversely, if $v$ satisfies this equation then clearly it satisfies the two conditions in our initial value problem.

Now the nice thing about this formulation is that it expresses $v$ as the fixed point of a certain operation. To find it, we will use an iterative method. We start with $v_0(t)=a$ and define the “Picard iteration”

$\displaystyle v_{i+1}(t)=a+\int\limits_0^tF(v_i(s))\,ds$

This is sort of like Newton’s method, where we express the point we’re looking for as the fixed point of a function, and then find the fixed point by iterating that very function.

The one catch is, how are we sure that this is well-defined? What could go wrong? Well, how do we know that $v_i(s)$ is in the domain of $F$ ? We have to make some choices to make sure this works out.

First, let $B_\rho$ be the closed ball of radius $\rho$ centered on $a$ . We pick $\rho$ so that $F$ satisfies a Lipschitz condition on $B_\rho$ , which we know we can do because $F$ is locally Lipschitz. Since this is a closed ball and $F$ is continuous, we can find an upper bound $M\geq\lVert F(x)\rVert$ for $v\in B_\rho$ . Finally, we can find a $c<\min\left(\frac{\rho}{M},\frac{1}{K}\right)$ , and the interval $J=[-c,c]$ . I assert that $v_k:J\to B_\rho$ is well-defined.

First of all, $v_0(t)=a\in B_\rho$ for all $t\in J$ , so that’s good. We now assume that $v_i$ is well-defined and prove that $v_{i+1}$ is as well. It’s clearly well-defined as a function, since $v_i(t)\in B_\rho$ by assumption, and $B_\rho$ is contained within the domain of $F$ . The integral makes sense since the integrand is continuous, and then we can add $a$ . But is $v_{i+1}(t)\in B_\rho$ ?

So we calculate

$\displaystyle\begin{aligned}\left\lVert\int\limits_0^tF(v_i(s))\,ds\right\rVert&\leq\int\limits_0^t\lVert F(v_i(s))\rVert\,ds\\&\leq\int\limits_0^tM\,ds\\&\leq Mc\\&\leq\rho\end{aligned}$

which shows that the difference between $v_{i+1}(t)$ and $a$ has length smaller than $\rho$ for any $t\in J$ . Thus $v_{i+1}:J\to B_\rho$ , as asserted, and the Picard iteration is well-defined.

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May 5, 2011 - Posted by John Armstrong | Analysis, Differential Equations

4 Comments »

It’s really an interesting coincidence that I’ve just done that as one of my practice exam questions… Good timing

Comment by anonymous | May 5, 2011 | Reply
[...] that we’ve defined the Picard iteration, we have a sequence of functions from a closed neighborhood of to a closed neighborhood of . [...]

Pingback by The Picard Iteration Converges « The Unapologetic Mathematician | May 6, 2011 | Reply
[...] convergence of the Picard iteration shows the existence part of our existence and uniqueness theorem. Now we prove the uniqueness [...]

Pingback by Uniqueness of Solutions to Differential Equations « The Unapologetic Mathematician | May 9, 2011 | Reply
[...] like to go back and give a different proof that the Picard iteration converges — one which is closer to the spirit of Newton’s method. In that case, we [...]

Pingback by Another Existence Proof « The Unapologetic Mathematician | May 10, 2011 | Reply

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