Intervals of Validity — Question 9

Question 9 – Solution

Strategy. A formula obtained by integrating on either side of a jump need not be differentiable at the joining point.

Step 1: Solve on the initial component. For $x<0$, $y'=-1$, so the initial condition gives $$\boxed{y=-x-1}.$$ It solves the equation at every negative $x$. Its maximal classical interval containing $-1$ is $\boxed{(-\infty,0)}$, as the join test below shows.

Step 2: Compute the finite endpoint limits. As $x\uparrow 0$, $y(x)\to-1$ and $y'(x)\to-1$. Neither the solution nor its derivative blows up. The obstruction will be the change of slope imposed immediately to the right.

Step 3: Test the unique continuous gluing. On $x>0$, any solution is $x+C$. Continuity with the left branch forces $C=-1$ and the value $y(0)=-1$. The only continuous gluing is therefore $$Y(x)=|x|-1.$$ Its left difference quotient at $0$ is $-1$ and its right difference quotient is $1$. Consequently $Y'(0)$ does not exist. It is not a classical solution of the equation on any interval containing $0$.

Any larger interval extending the negative-half-line solution would include points on both sides of $0$ and would require this same gluing. Thus no classical extension is possible, proving the claimed maximality despite the finite limits.

Step 4: Identify the missing hypothesis. Here $f$ is defined everywhere but is discontinuous at $0$. A continuation result that requires continuity of the right-hand side near the endpoint cannot be applied. Domain membership and bounded values alone are insufficient when that hypothesis fails.

Changing $f(0)$ to any single real value cannot reconcile the unequal one-sided derivatives forced by $f=-1$ to the left and $f=1$ to the right. The same corner remains. The function $Y$ solves the equation away from the join, but the question requires the equation at every point of a classical interval.