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

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.

With the standard constant starting Picard iterate, $y_0(x)=1$, the next two Picard iterates areand, for $x>0$,ThusThe restriction $x>0$ for $y_2$ is essential for the integral from $1$ to $x$: the preceding iterate $y_1$ vanishes at $0$, where the new integrand $s+1/s$ is undefined.

The rectangle avoids the singular line $y=0$ exactly when $0<k<1$. There $y>0$, and the largest $|x|$ on $[1-h,1+h]$ is $1+h$. Since $1+1/y$ decreases with $y$, the smallest feasible uniform bound isIt is attained at $x=1+h$, $y=1-k$. For the Lipschitz constant, the mean value theorem givesThe quotients approach this upper bound as $x=1+h$ and $y,z\downarrow 1-k$, so the smallest feasible constant isSubstitution into part (a) givesThe contraction mapping restriction is automatic for every M-test-admissible $h$ precisely when $A(k)<B(k)$. Because $0<k<1$,Thereforeis exactly the range where the two proofs guarantee all the same closed symmetric intervals, including the largest M-test interval.

For completeness, set $h_M=(\sqrt{1+4A(k)}-1)/2$ and $h_C=(\sqrt{1+4B(k)}-1)/2$. The M-test allows $0<h\le h_M$; the contraction mapping theorem allows $0<h\le h_M$ with $h<h_C$. At $k=2-\sqrt{2}$, these have the same supremum $h_M=h_C$, but contraction excludes the endpoint $h=h_M$. For $2-\sqrt{2}<k<1$, it gives the strictly shorter range $0<h<h_C<h_M$. For a particular $h$, both proofs work exactly when both displayed inequalities hold, even for these larger $k$.

Near $(1,1)$ the solution of a differential equation has $y>0$, so the separable differential equation givesEquivalently its trajectory lies on the level setOn $y>0$, the function $G(y)=y-\log (1+y)$ is strictly increasing because $G^{\prime }(y)=y/(1+y)>0$, and its range is $(0,\infty )$. Thus the positive branch exists exactly whereThe connected component containing $x=1$ isThe implicit function theorem gives a smooth solution of a differential equation at every such $x$. It remains finite at every finite $x>x_{\ast }$, because $G$ increases without bound. As $x\downarrow x_{\ast }$, we have $y\downarrow 0$, and the original right side $x+x/y$ becomes singular. Hence this is the maximal interval $(x_{\ast },\infty )$ through the prescribed initial condition.

In particular, the solution is not defined for all $x\in (0,2)$. It cannot pass through $x=x_{\ast }$ as a classical solution, since its required value there is $y=0$.

Treat both sides of $a$ at once: for $\sigma \in \{-1,1\}$ set $w(t)=v(a+\sigma t)$, $0\le t\le h$. Nonnegativity givesDefine $W(t)=ct+A{\int }_0^{t}w(r)dr$. By the fundamental theorem of calculus, $W^{\prime }(t)=c+Aw(t)\le c+AW(t)$ and $W(0)=0$. The integrating factor $e^{-At}$ givesSince $w\le W$, this proves the affine Gronwall estimate

Continuous dependence of an ODE solution on parameters here means continuity of the map $\beta \mapsto y_{\beta }$ into $C([a-h,a+h])$ with its supremum norm: for every ${\beta }_0\in I$ and $\epsilon >0$, all sufficiently close $\beta \in I$ satisfySubtract the two integral equations and use the uniform Lipschitz constant $L$ on the rectangle. With $v(x)=|y_{\beta }(x)-y_{\gamma }(x)|$ we obtainPart (i), with $c=|\beta -\gamma |$ and $A=L$, therefore proves, for $L>0$,The case $\beta =\gamma$ follows from uniqueness or the same estimate with zero forcing. If $L=0$, the direct integral estimate gives $\Vert y_{\beta }-y_{\gamma }{\Vert }_{\infty }\le h|\beta -\gamma |$. In either case the map is even Lipschitz continuous, and hence continuous, throughout $I$ in the stated supremum norm.