# '2nd order' Picard Iteration

I'm self-studying differential equations using MIT's publicly available materials. One of the problem set exercises deals with what I'm calling a second order Picard Iteration. To be explicit, we are given the initial value problem \begin{align} x'' = f(t,y), \qquad x(t_0) = x_0, \qquad x'(t_0) = x_1 \end{align} where $f$ is continuous and Lipschitz on a rectangle $|t -t_0| \leq T$ and $|x - x_0| \leq K$ and a sequence of functions given by \begin{align} x_n(t) & = \begin{cases} x_0 & n = 0\\ x_0 + x_1(t - t_0) + \int_{t_0}^t(t-s)f(x_{n-1}(s),x)ds & n \geq 1 \end{cases} \end{align} We are to show that $\lbrace x_n \rbrace$ converges to a solution to the IVP on the interval $|t - t_0| \leq \min \left(T, \frac{K}{B}\right)$, where $M$ is the maximum value on the rectangle in question and $B = |x_1| + \frac{MT}{2}$. It seems to me that we are to proceed in a manner analogous to the proof of the local existence theorem; I am, however, getting stuck on the inductive step.

Here's what I have. In essence, what we want to do is show that $\lbrace x_n \rbrace$ converges uniformly and then use the fact that uniform convergence implies that we can move the limit from the outside to the inside of an integral sign. So \begin{align} |x_1(t) - x_0(t)| & = \left|(t - t_0)x_1 + \int_{t_0}^t(t-s)f(x_0,t)ds\right|\\ & \leq |(t - t_0)x_1| + \int_{t_0}^t|t-s||f(x_0,x)ds\\ & \leq T|x_1| + M\int_{t_0}^t|t-s|ds\\ & \leq T|x_1| + \frac{MT^2}{2}\\ & \leq TB \end{align} and therefore \begin{align} |x_2(t) - x_1(t)| & \leq \int_{t_0}^t|t-s||f(x_1(s),t) - f(x_2(s),t)|ds\\ & \leq L\int_{t_0}^t|t-s||x_1(s) - x_0(s)|ds\\ & \leq TBL\int_{t_0}^t|t-s|ds\\ & \leq TBL\left(\frac{T^2}{2}\right)\\ \end{align} And here we reach the inductive step, where my troubles begin. Assume we have \begin{align} |x_n(t) - x_{n-1}(t)| \leq TBL^{n-1}\left(\frac{T}{2}\right)^{2(n-1)} \end{align} Then we obtain \begin{align} |x_{n+1}(t) - x_{n-1}(t)| \leq TBL^n\left(\frac{T}{2}\right)^{2n} \end{align} We want the right side of this inequality to go to $0$, but of course it doesn't, unless $T < 2$. Now, in the proof of the local existence theorem, the analogous denominator is $n!$, which suggests that instead of $2^{2n}$ we should have $(2n)!$, but I can't see how we'd obtain that, given that each iteration of $\int_{t_0}^t(t-s)$ is going to yield $\frac{(t - t_0)^2}{2}$, so that we're doubling the denominator each time.

Any thoughts would be appreciated.

• Is what you are doing here any different from applying "first order" Picard iteration to the system of equations $x'=y,y'=f(t,x,y)$? If not, then you're thinking too hard, because all you actually need to check is the Lipschitz condition for the function $(x,y) \mapsto (y,f)$. – Ian Jul 24 '16 at 18:08
• @Ian I'm not sure. What would $y$ and $f(t,x,y)$ be in this case? (My suspicion is that it's more complicated than what you suggest, but kind of for non-math reasons: the exercise specifically mentions the constant $B$, which naturally pops out of the calculation for $|x_1(t) - x_0}|$. – solitaireartist Jul 24 '16 at 18:16
• $x,y$ are just the names of the variables; $x$ encodes the "position" while $y$ encodes the "velocity". Consequently $f(t,x,y)$ is whatever $x''$ is in your equation. (This is a very standard trick.) – Ian Jul 24 '16 at 18:21
• @Ian I see (vaguely) what you're doing. However, at this point in the course we don't know about applying Picard iteration to systems of equations. Moreover, the value $B = |x_1| + \frac{TB}{2}$, referenced in the statement of the problem, falls out of the procedure I'm using to bound $|x_1(t) - x_0(t)|$ so naturally that I can't help but believe it's not a coincidence. – solitaireartist Jul 24 '16 at 20:03