Past exam of the differential equations 1 course of the University of Oxford / 2021 / 2 / a

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.