= Picard iteration
{c}
{title2=$y_{n+1}=\mathcal P y_n$}
<Picard iteration> replaces a tentative trajectory by the accumulated <integral> of its proposed rate. For an <initial value problem> $y'=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting <function> by
$$
(\mathcal P u)(t)=y_0+\int_{t_0}^t f(s,u(s))\,ds,\qquad y_0(t)\equiv y_0,\qquad y_{n+1}=\mathcal P y_n.
$$
Here $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A <fixed point> of $\mathcal P$ solves this <integral equation>, and the <fundamental theorem of calculus> gives the original <ordinary differential equation>.
For $y'=y$, $y(0)=1$, successive <integrals> give
$$
\begin{aligned}
y_0(t)&=1,\\
y_1(t)&=1+t,\\
y_2(t)&=1+t+\frac{t^2}{2},\\
y_3(t)&=1+t+\frac{t^2}{2}+\frac{t^3}{6},\\
y_4(t)&=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24},\\
y_n(t)&=\sum_{k=0}^n\frac{t^k}{k!}.
\end{aligned}
$$
At $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx2.71828183$. The fourth <polynomial> misses by about $0.00994850$. On $0\leq t\leq1$, every omitted term of the <exponential series> is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving <uniform convergence> to the <exponential function> on this <interval>.
\Image[/picard-iteration-exponential.png]
{title=Picard iteration builds the exponential solution by repeated integration}
{description=The first five <polynomials> $y_0,\ldots,y_4$ approach the <exponential function> on $0\leq t\leq1$. Endpoint values are listed in the legend; the last error is approximately $0.00994850$.}
The beauty of <Picard iteration> is that each proposed trajectory supplies a better rate history for the next one. For this example, one <integration> adds exactly the next term of the <exponential series>. In general, the <integrals> may be difficult to compute, but the construction still explains local existence and uniqueness.
The short <contraction mapping> argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times\overline B_r(y_0)\subset U$ on which
$$
\|f(t,z)\|\leq M,\qquad \|f(t,u)-f(t,v)\|\leq L\|u-v\|.
$$
The second bound is the uniform local <Lipschitz condition>. Set $J=[t_0-h,t_0+h]$ with $0<h\leq a$, $Mh\leq r$ and $Lh<1$. In the <function space> of <continuous functions> $J\to\mathbb R^d$, use the <supremum norm> $\|u\|_\infty=\sup_{t\in J}\|u(t)\|$. On its <closed ball> $\|u-y_0\|_\infty\leq r$, direct <integral> estimates give
$$
\|\mathcal P u-y_0\|_\infty\leq Mh\leq r,\qquad \|\mathcal P u-\mathcal P v\|_\infty\leq Lh\|u-v\|_\infty.
$$
Thus the map preserves the <closed ball> and is a <contraction mapping> there. This <closed ball> is a <complete metric space>, so the <contraction mapping theorem> gives convergence to exactly one <fixed point>. Any other local <solution of a differential equation> stays in the same ball after shortening its <interval>, and hence agrees there.
For $y'=y$, take $r=1$, so $|y|\leq2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx0.64872127<1$. The plotted <interval> $[0,1]$ is longer than this simple guarantee: the explicitly computed <exponential series> converges there anyway. The condition $Lh<1$ is sufficient for this one-step <contraction mapping> argument, rather than a necessary lifespan bound.
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