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

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 $y=0$.