Let \(y:[0,\infty)\to\mathbb R\) solve \[ y''=-|y|, \qquad y(0)=1, \qquad y'(0)=0. \] Prove that there is exactly one \(t>0\) such that \(y(t)=0\).
Solution
1. The initial-value problem has a unique solution on \([0,\infty)\).
Write the second-order equation as the first-order system \[ \begin{pmatrix}y\\v\end{pmatrix}' = \begin{pmatrix}v\\-\abs y\end{pmatrix}. \] The vector field \[ (y,v)\longmapsto(v,-\abs y) \] is globally Lipschitz on \(\RR^2\), since the absolute-value function is Lipschitz with constant \(1\). The standard existence-and-uniqueness theorem therefore gives a unique global solution for the prescribed initial data.
2. Define\[ \widetilde y(t) \coloneqq \begin{cases} \cos t,&0\le t\le\pi/2,\\ -\sinh(t-\pi/2),&t\ge\pi/2. \end{cases} \]Then \(\widetilde y\) is \(C^2\) on \([0,\infty)\).
At \(t=\pi/2\), both formulas give the value \(0\). Their first derivatives are \[ -\sin t \qquad\text{and}\qquad -\cosh(t-\pi/2), \] which both equal \(-1\) at \(t=\pi/2\). Their second derivatives are \[ -\cos t \qquad\text{and}\qquad -\sinh(t-\pi/2), \] which both equal \(0\) there. Thus the two pieces join with matching derivatives through order \(2\).
3. The function \(\widetilde y\) satisfies\[ \widetilde y''=-\abs{\widetilde y}, \qquad \widetilde y(0)=1, \qquad \widetilde y'(0)=0. \]
For \(0\le t\le\pi/2\), one has \(\widetilde y(t)=\cos t\ge0\), so \[ \widetilde y''(t)=-\cos t=-\abs{\widetilde y(t)}. \] For \(t\ge\pi/2\), one has \[ \widetilde y(t)=-\sinh(t-\pi/2)\le0, \] so \[ \widetilde y''(t) = -\sinh(t-\pi/2) = \widetilde y(t) = -\abs{\widetilde y(t)}. \] The initial conditions follow directly from the first formula.
4. The given solution is \(y=\widetilde y\).
Step 3 shows that \(\widetilde y\) solves the same initial-value problem as \(y\). By uniqueness from step 1, they coincide.
5. The equation \(y(t)=0\) for \(t>0\) has the unique solution\[ \boxed{t=\pi/2}. \]
By step 4, for \(0<t<\pi/2\) one has \[ y(t)=\cos t>0, \] while \(y(\pi/2)=0\). For \(t>\pi/2\), \[ y(t)=-\sinh(t-\pi/2)<0. \] Thus no other positive zero exists.
6. Q.E.D.
Step 5 is exactly the assertion.