Toronto Math Forum
MAT2442013S => MAT244 MathLectures => Reading Week Challenge => Topic started by: Victor Ivrii on February 16, 2013, 10:37:44 AM

Using Reading Week Bonus problem 2 consider $Q$ as a constant which is the maximum (minimum) of $q(x)$ on interval $[x_n,x_{n+1}]$ where $x_n$ and $x_{n+1}$ are two consecutive zeros of $y(x)$ and estimate $(x_{n+1}x_n)$ from above (from belowrespectively).

We proved in Problem 2 that if $y''+q(x)y=0$, $z''+Q(x)z=0$, $y(x_n)=z(x_n)=0$, $y(x_{n+1})=0$ and $Q(x)\ge q(x)$ on $[x_n,x_{n+1}]$ then either $z$ has a $0$ on $(x_n,x_{n+1})$ or $z(x_{n+1})=0$ and $Q=q$ on $[x_n,x_{n+1}]$.
For comparison we select constant coefficient equation $w''+c w=0$ with either $c=\max_{[x_n,x_{n+1}]} q(x)$ or $c=\min_{[x_n,x_{n+1}]} q(x)$. Recall that for $c\le 0$ solution does not oscillate and for $c>0$ the distance between consecutive $0$is exactly $\frac{\pi}{\sqrt{c}}$ (really, solution is $C\sin (\sqrt{c}(xx_n))$).
Therefore
* If $q(x)\le 0$ as $x\ge x_n$ then $y$ does not vanish on $(x_n,+\infty)$;
* If $c=\min_{[x_n,+\infty}) q(x)>0$ then $x_{n+1}x_n \le \frac{\pi}{\sqrt{c}}$;
* If $c=\max_{[x_n,+\infty}) q(x)>0$ then $x_{n+1}x_n \ge \frac{\pi}{\sqrt{c}}$.