Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Differential Equations / 7

Collect the two dependent variables in a column vector. The vector field and initial vector areThe given system of ordinary differential equations is then $Y^{\prime }=F(x,Y)$ with $Y(0)=B$. Both components of $F$ are continuous on all of $\mathbb{R}\times {\mathbb{R}}^2$, since their denominators are at least one.

Each denominator is at least one and positive, so each component of $F$ lies in $(0,1]$. Taking the maximum norm therefore givesThis is a global bound, so in particular the constant $M=1$ is valid throughout the specified cube.

For $a,b\ge 0$ and $c\ge 1$,Apply this to the first component with $a=|z_2|$, $b=|w_2|$, $c=x^2+1$. The reverse triangle inequality givesFor the second component use $a=|z_1|$, $b=|w_1|$, $c=1$ to obtain $|F_2(x,Z)-F_2(x,W)|\le |z_1-w_1|$. Taking the maximum yieldsThus $F$ is globally Lipschitz continuous in the dependent vector, uniformly in $x$, with Lipschitz constant $1$. Differentiability of the absolute-value function at zero is unnecessary.

A Lipschitz condition here is uniform in the independent variable: there is a finite constant $L\ge 0$ such that, whenever both $(x,Z)$ and $(x,W)$ belong to the cube,The same $L$ must work for every allowed $x,Z,W$. Part (c) proves this with $L=1$. This controls changes in the dependent variables; no Lipschitz condition in $x$ is required for the Picard-Lindelöf theorem.

The quantitative Picard-Lindelöf theorem applies on a rectangle $|x-x_0|\le a$, $\Vert Z-B{\Vert }_{\infty }\le b$. Suppose $F$ is continuous there, bounded in maximum norm by $M$, and uniformly Lipschitz continuous in $Z$. Then the initial value problem $Y^{\prime }=F(x,Y)$, $Y(x_0)=B$, has a unique solution onwith the usual interpretation $b/M=\infty$ if $M=0$. The solution remains in the dependent-variable ball. Its endpoints are included when $F$ is continuous on the closed rectangle, as here. Equivalently, existence first on every smaller interval and the derivative bound give extension to those endpoints.

For the given cube, $x_0=0$ and $a=b=1$. Thus the theorem gives $h=\min \{1,1/M\}$ once a bound $M$ and a uniform Lipschitz constant have been established. The theorem does not require $Lh<1$: convergence of Picard iteration can be proved by factorial estimates or by using successive shorter intervals.

The vector field is continuous on the cube. Part (b) gives $M=1$, and part (c) supplies a uniform Lipschitz constant $L=1$. In the quantitative Picard-Lindelöf theorem, the independent-variable half-width and the dependent-variable radius are both one, soThereforeThe global derivative bound also makes the cube condition explicit. Along any solution segment starting at the initial point,Thus it cannot leave the dependent-variable ball before $|x|=1$. Bounded derivatives give limiting endpoint values, and local existence extends to them because $F$ is defined continuously on all space. Uniform Lipschitz continuity ensures that all such local pieces agree on overlaps, proving uniqueness throughout the closed interval.