The Unapologetic Mathematician

Mathematics for the interested outsider

The Picard Iteration Converges

Now that we’ve defined the Picard iteration, we have a sequence of functions v_i:J\to B_\rho from a closed neighborhood of 0\in\mathbb{R} to a closed neighborhood of a\in\mathbb{R}^n. Recall that we defined M to be an upper bound of \lVert F\rVert on B_\rho, K to be a Lipschitz constant for F on B_\rho, c less than both \frac{\rho}{M} and \frac{1}{K}, and J=[-c,c].

Specifically, we’ll show that the sequence converges in the supremum norm on J. That is, we’ll show that there is some v:J\to B_\rho so that the maximum of the difference \lVert v_k(t)-v(t)\rVert for t\in J decreases to zero as i increases. And we’ll do this by showing that the individual functions v_i and v_j get closer and closer in the supremum norm. Then they’ll form a Cauchy sequence, which we know must converge because the metric space defined by the supremum norm is complete, as are all the L^p spaces.

Anyway, let L=\lVert v_1-v_0\rVert_\infty be exactly the supremum norm of the difference between the first two functions in the sequence. I say that \lVert v_{i+1}-v_i\rVert_\infty\leq(cK)^iL. Indeed, we calculate inductively

\displaystyle\begin{aligned}\lVert v_{i+1}(t)-v_i(t)\rVert&\leq\int\limits_0^t\lVert F(v_i(s))-F(v_{i-1}(s))\rVert\,ds\\&\leq K\int\limits_0^t\lVert v_i(s)-v_{i-1}(s)\rVert\,ds\\&\leq K\int\limits_0^t(cK)^{i-1}L\,ds\\&\leq(cK)(cK)^{i-1}L\\&=(cK)^iL\end{aligned}

Now we can bound the distance between any two functions in the sequence. If i<j are two indices we calculate:

\displaystyle\begin{aligned}\lVert v_j-v_i\rVert_\infty&=\left\lVert\sum\limits_{k=i}^{j-1}v_{k+1}-v_k\right\rVert_\infty\\&\leq\sum\limits_{k=i}^{j-1}\lVert v_{k+1}-v_k\rVert_\infty\\&\leq\sum\limits_{k=i}^{j-1}(cK)^kL\end{aligned}

But this is a chunk of a geometric series; since cK<1, the series must converge, and so we can make this sum as small as we please by choosing i and j large enough.

This then tells us that our sequence of functions is L^\infty-Cauchy, and thus L^\infty-convergent, which implies uniform pointwise convergence. The uniformity is important because it means that we can exchange integration with the limiting process. That is,

\displaystyle\lim\limits_{k\to\infty}\int_0^tv_k(s)\,ds=\int\limits_0^t\lim\limits_{k\to\infty}v_k(s)\,ds=\int\limits_0^tv(s)\,ds

And so we can start with our definition:

\displaystyle v_{k+1}(t)=a+\int\limits_0^tF(v_k(s))\,ds

and take the limit of both sides

\displaystyle\begin{aligned}v(t)&=\lim\limits_{k\to\infty}v_{k+1}(t)\\&=\lim\limits_{k\to\infty}\left(a+\int\limits_0^tF(v_k(s))\,ds\right)\\&=a+\int\limits_0^t\lim\limits_{k\to\infty}F(v_k(s))\,ds\\&=a+\int\limits_0^tF(v(s))\,ds\end{aligned}

where we have used the continuity of F. This shows that the limiting function v does indeed satisfy the integral equation, and thus the original initial value problem.

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May 6, 2011 - Posted by | Analysis, Differential Equations

1 Comment »

  1. [...] convergence of the Picard iteration shows the existence part of our existence and uniqueness theorem. Now we [...]

    Pingback by Uniqueness of Solutions to Differential Equations « The Unapologetic Mathematician | May 9, 2011 | Reply


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