Basic Concepts — Question 9

Question 9 – Solution

Strategy. Verify the join at the derivative order required by the equation, then use the family as a counterexample to an overgeneralized uniqueness claim.

Step 1: Verify away from the join. For $t<a$, both sides of the equation are zero. For $t>a$, $$y_a'=\frac{(t-a)^3}{36},\qquad y_a''=\frac{(t-a)^2}{12},\qquad \sqrt{|y_a|}=\sqrt{\frac{(t-a)^4}{144}}=\frac{(t-a)^2}{12}.$$ The coefficient 144 is essential to this identity.

Step 2: Verify classical regularity at the join. The right-hand limits of $y_a,y_a',y_a''$ as $t\downarrow a$ are all zero, matching the values on the left. The first and second derivatives at the join are also zero, as follows directly from their difference quotients. Hence $y_a$ is $C^2$ on $\mathbb R$.

At $t=a$, the equation reads $0=\sqrt{|0|}$, so it holds there, not merely away from the join. There is no hidden corner in position or velocity.

Step 3: Check the common data and distinctness. Since $a\ge 0$, every family member has $y_a(0)=y_a'(0)=0$, including $a=0$ by the join calculation. If $a_1<a_2$, choose $a_1<t<a_2$. Then $y_{a_1}(t)>0$ but $y_{a_2}(t)=0$, proving they are different solutions. The function $\boxed{y\equiv 0}$ is another solution, different from every finite-delay member.

Step 4: State exactly what has failed. The right-hand side $\sqrt{|y|}$ is continuous but depends nonlinearly on the unknown function. This is not an equation $y''+p(t)y'+q(t)y=g(t)$ with prescribed continuous coefficients, so that linear theorem cannot be invoked. The verified family proves that continuity alone does not ensure uniqueness for a general nonlinear second-order equation.

Two initial data determine one solution under appropriate hypotheses, not for every differential equation. This example exhibits infinitely many solutions; it does not claim to classify all solutions of the nonlinear equation.