Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2007 / 7

Use Picard iteration on $[a-h,a+h]$:By induction, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The Lipschitz condition gives, for $n\ge 0$,For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.

The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The Lipschitz condition also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit givesThe fundamental theorem of calculus makes $u$ a continuously differentiable solution with $u(a)=b$.

For uniqueness, let $u,v$ be two solutions and put $D_0=\sup |u-v|$ on the interval. Repeated use of the Lipschitz condition in their integral equations givesfor every $n$. Letting $n\to \infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore a unique solution exists on the full stated interval. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.

At zero the right side is continuous but fails every local Lipschitz condition in its second variable, since ${\delta }^{1/4}/\delta ={\delta }^{-3/4}\to \infty$. For every $c\ge 0$, the functionis continuously differentiable and satisfies the differential equation, including at the joining point where both one-sided derivatives are zero. Each has $u_c(0)=0$. The identically zero function is another possibility. Thus solutions exist but are not unique, even locally: choosing $c=0$ and taking the zero function already gives two different solutions on every neighborhood of zero. These are instances of waiting-time solutions of a sublinear Bernoulli equation.

Near $y=1$, the right side is continuously differentiable in $y$, with derivative $1/(4y^{3/4})$ bounded on, for example, $1/2\le y\le 3/2$. It therefore satisfies a local Lipschitz condition; it is also bounded on an appropriate bounded neighborhood. The existence and uniqueness result applies there. Separation of variables givesThe unique extension to the whole real axis is $u(x)=[\max \{1+3x/4,0\}{]}^{4/3}$. It joins zero with zero derivative at $x=-4/3$. To see uniqueness of this extension, $u^{\prime }\ge 0$ makes every solution nondecreasing. Before it reaches zero it cannot be positive, while a negative interval would have $u^{\prime }=0$ and could not continuously reach zero. Thus it must be zero for $x\le -4/3$; the positive part is fixed by the separated formula.

Translation in $x$ leaves this autonomous differential equation unchanged. For every $c\ge 1$, the same formulasatisfies $u_c(1)=0$, and the zero function does too. Thus solutions exist and are not unique, exactly as for the initial point $(0,0)$. The relevant failure is the Lipschitz condition at $y=0$; moving the initial value of $x$ does not remove it.