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,


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


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.

May 6, 2011 Posted by | Analysis, Differential Equations | 1 Comment