The Unapologetic Mathematician

Mathematics for the interested outsider

The Picard Iteration

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


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.

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


  1. 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

  2. […] 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

  3. […] 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

  4. […] 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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