Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Differential Equations / 7

First prove the general uniqueness claim. The bound $Mh\le k$ prevents a solution from leaving the dependent-variable interval before $|x-a|=h$: up to any first exit, $|y(x)-b|\le M|x-a|\le k$. Let $y_1,y_2$ be two solutions with the same initial value. Their integral equations and the Lipschitz condition give, for $x\ge a$,Since their difference is bounded by $2k$, repeated substitution into this inequality proves, for every integer $n\ge 0$,The same argument with reversed limits works for $x\le a$. Letting $n\to \infty$ forces the difference to zero on the whole interval. This proves at most one solution without assuming that $Ah<1$.

For case (i), choose a rectangle around $(0,1)$ that stays in $y>0$. There $f_y=(2/3)x{y}^{-1/3}$ is bounded, giving a uniform Lipschitz constant by the mean value theorem. For example $h=k=1/2$ works: $M\le \frac{1}{2}(3/2{)}^{2/3}$ and $Mh<1/2$. Separate variables using $u=y^{1/3}$, obtaining $3u^{\prime }=x$. The initial condition gives $u=1+x^2/6$, henceIt remains positive for all real $x$, so the same local uniqueness argument continues along this solution.

No rectangle of positive width around $(0,0)$ has a uniform Lipschitz constant in $y$. For any nonzero $x$ in such a rectangle,The vanishing factor $x$ at the single initial point does not control neighbouring $x$-values. The failure of the hypothesis is accompanied by actual nonuniqueness: bothsatisfy the equation and have value zero at $x=0$. Direct differentiation gives $y_2^{\prime }=x^5/36=x(y_2{)}^{2/3}$, including the zero point by continuity.

The same quotient $|x||y{|}^{-1/3}$ is unbounded near $(1,0)$, so the uniform Lipschitz hypothesis again fails. Two explicit solutions through the initial point areFor the second, put $u=(x^2-1)/6$. Then $y_2^{\prime }=3u^2(x/3)=x{u}^2=x(y_2{)}^{2/3}$, and $y_2(1)=0$. The real power here means the square of the real cube root, so the identity also holds when $y_2$ is negative. Both functions are differentiable through their zeros. Thus uniqueness fails in this case as well.