Intervals of Validity — Question 6

Question 6 – Solution

Strategy. Check continuation at the joins directly. The length of a solution’s interval and uniqueness of the solution are different questions.

Step 1: Verify all three pieces and the joins. For $x<a$, $$Y'_{a,b}=3(x-a)^2=3|(x-a)^3|^{2/3},$$ and the same computation holds with $b$ for $x>b$. On the middle piece both sides vanish.

At $x=a$, the values from both sides are $0$, and the one-sided derivatives tend to $0$. The derivative at the join itself is $0$ by the difference quotient. The same holds at $b$, including the case $a=b=0$. Thus $Y_{a,b}$ is continuously differentiable on $\mathbb R$, satisfies the equation everywhere, and has $Y_{a,b}(0)=0$.

Step 2: Exhibit distinct maximal solutions. Two examples are $y(x)=0$ for all $x$ and $y(x)=x^3=Y_{0,0}(x)$. Their maximal intervals are both $\boxed{\mathbb R}$: no larger real interval exists. The function $Y_{-1,1}$ is a third global solution.

Step 3: Separate a restriction from a maximal solution. On $(-1/2,1/2)$, $Y_{-1,1}=0$, but it extends as $Y_{-1,1}$ to the whole real line. Its restriction is therefore not maximal. That restriction can also be extended as the identically zero solution; an extension need not be unique here.

Step 4: Locate the failed uniqueness condition. The right-hand side is continuous, but it is not locally Lipschitz in $y$ at $0$: $$\frac{|3|y|^{2/3}-0|}{|y-0|}=3|y|^{-1/3}\longrightarrow\infty.$$ Consequently the usual local uniqueness hypothesis is unavailable at the initial point. The explicit examples prove nonuniqueness, even though every exhibited global solution has the same maximal interval $\mathbb R$.