Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 6

Introduce the position and velocity variables $y_1=y$ and $y_2=y^{\prime }$. Then the required system of ordinary differential equations and initial condition areThus $f_1(x,\mathbf{y})=y_2$. Conversely, the first equation identifies $y_2=y_1^{\prime }$, and the second recovers the original second-order ordinary differential equation and both initial data.

Work with the L1 norm on ${\mathbb{R}}^2$. The scalar Lipschitz hypothesis impliesWe may take $L\ge 0$. The compact set $S$ and continuity give a finite bound $M={\max }_S\Vert \mathbf{f}{\Vert }_1$. ChooseLet $I=[a-\eta ,a+\eta ]$. The space $C_{\eta }$ of continuous maps $I\to B_k(\mathbf{b})$ is complete in the supremum norm: it is a closed subset of the Banach space $C(I,{\mathbb{R}}^2)$. Define the Picard operator on a closed ball byFor any $\mathbf{g}\in C_{\eta }$, the integral triangle inequality givesThe integrand is continuous, so the result is continuous and remains in the prescribed closed ball. Thus $\mathcal{P}$ is a self-map of the complete space. Moreover,It is a contraction mapping. The contraction mapping theorem gives a unique fixed point $\mathbf{y}$. By the fundamental theorem of calculus, the fixed-point identity makes $\mathbf{y}$ continuously differentiable and gives ${\mathbf{y}}^{\prime }=\mathbf{f}(x,\mathbf{y})$ and $\mathbf{y}(a)=\mathbf{b}$. Its first component is twice continuously differentiable, because its derivative is the second component.

Conversely, integrating any solution that remains in the ball makes it a fixed point, so it equals $\mathbf{y}$. A solution starting at $\mathbf{b}$ cannot first reach the boundary of this ball during $I$: before such an exit the same estimate gives distance at most $\eta M<k$, contradicting distance $k$ at the first exit. This proves existence and uniqueness on $|x-a|\le \eta$, not merely formal convergence of an iteration. It is the local Picard-Lindelöf theorem construction.

In position-velocity variables $(u,v)$, the scalar right-hand side isIts only relevant singularity is $u=0$. The initial state $(x,u,v)=(0,1,1)$ is separated from it. For an explicit application of part (a), take $h=1$ and $k=1/2$ in the L1 norm. On the resulting closed ball, $1/2\le u,v\le 3/2$, and $|x|\le 1$.

The partial derivatives obeyThe ball is convex, so integrating these partial derivatives along a segment between two states gives the uniform Lipschitz conditionContinuity holds on the whole slab, andThus $L=9$, $M\le 9$, and the choice $\boxed{\eta =1/20}$ satisfies the self-map and contraction bounds in part (a). This proves a unique solution on $[-1/20,1/20]$, with $y$ bounded away from zero there.

There is also a useful exact representation. Where $y>0$, set $q=y^{\prime }/y$, the logarithmic derivative. Differentiation cancels the quadratic term:Solving this first-order linear differential equation gives $q=e^{x^2}$, and a second integration yieldsThis function is positive and smooth for every real $x$, satisfies both initial data, and verifies the original ordinary differential equation by direct differentiation. It therefore extends the unique local solution to the entire real line. The logarithmic-derivative reduction of a second-order equation explains why this apparently nonlinear example is explicitly integrable by integration.