Let \(y:\mathbb R\to\mathbb R\) solve \[ y''=-|y|, \qquad y(0)=1, \qquad y'(0)=0. \]
Show that \(y\) is even.
Show that \(y\) has exactly one zero on the positive real axis.
Solution
1. The initial-value problem\[ y''=-\abs{y}, \qquad y(0)=1, \qquad y'(0)=0 \]has at most one solution.
Introduce \[ v=y' \] and rewrite the equation as the first-order system \[ \begin{pmatrix} y\\ v \end{pmatrix}' = \begin{pmatrix} v\\ -\abs{y} \end{pmatrix}. \] The vector field \[ F(y,v) = (v,-\abs{y}) \] is globally Lipschitz because \[ \abs{\abs{y_1}-\abs{y_2}} \leq \abs{y_1-y_2}. \] The uniqueness theorem for ordinary differential equations therefore gives uniqueness for the stated initial data.
2. Define\[ \widetilde y(x)=y(-x). \]Then \(\widetilde y\) satisfies the same differential equation and initialconditions as \(y\).
Differentiating twice, \[ \widetilde y''(x) = y''(-x) = -\abs{y(-x)} = -\abs{\widetilde y(x)}. \] Moreover, \[ \widetilde y(0)=y(0)=1 \] and \[ \widetilde y'(0)=-y'(0)=0. \]
3. The function \(y\) is even:\[ \boxed{ y(-x)=y(x) } \]for every \(x\in\RR\).
By step 2, the functions \(y\) and \(\widetilde y\) solve the same initial-value problem. Uniqueness from step 1 gives \[ \widetilde y=y. \] This is exactly the displayed identity.
4. Define \(\phi:[0,\infty)\to\RR\) by\[ \phi(x) = \begin{cases} \cos x, & 0\leq x\leq\pi/2, \\ -\sinh(x-\pi/2), & x\geq\pi/2. \end{cases} \]Then \(\phi\) is \(C^2\) on \([0,\infty)\).
Each branch is smooth away from \(x=\pi/2\). At the joining point, \[ \cos(\pi/2)=0 = -\sinh0, \] so the values agree. The first derivatives are \[ -\sin x \] and \[ -\cosh(x-\pi/2), \] and both equal \(-1\) at \(x=\pi/2\). The second derivatives are \[ -\cos x \] and \[ -\sinh(x-\pi/2), \] and both equal \(0\) there. Hence the two branches join with matching first and second derivatives.
5. The function \(\phi\) satisfies\[ \phi''=-\abs{\phi} \]on \([0,\infty)\) and\[ \phi(0)=1, \qquad \phi'(0)=0. \]
For \[ 0\leq x\leq\pi/2, \] one has \(\cos x\geq0\), so \[ \phi''(x) = -\cos x = -\abs{\phi(x)}. \]
For \(x\geq\pi/2\), \[ \phi(x)=-\sinh(x-\pi/2)\leq0. \] Thus \[ \abs{\phi(x)} = \sinh(x-\pi/2), \] while \[ \phi''(x) = -\sinh(x-\pi/2) = -\abs{\phi(x)}. \] Step 4 handles the joining point. Finally, \[ \phi(0)=\cos0=1 \] and \[ \phi'(0)=-\sin0=0. \]
6. For every \(x\geq0\),\[ y(x)=\phi(x). \]
5. Uniqueness from step1 therefore gives equality on their common interval\[ [0,\infty). \]
7. The function \(y\) has exactly one zero on the positive real axis,namely\[ \boxed{ x=\frac\pi2. } \]
By step 6, \[ y(x)=\cos x \] for \(0\leq x\leq\pi/2\). Hence \[ y(x)>0 \] for \(0\leq x<\pi/2\), while \[ y(\pi/2)=0. \] For \(x>\pi/2\), step 6 gives \[ y(x) = -\sinh(x-\pi/2) < 0. \] Thus there is no other positive zero.
8. Q.E.D.
Step 3 proves part (1), and step 7 proves part (2).