Find the distance between two consecutive zeros of an…
Find the distance between two consecutive zeros of an arbitrary nontrivial solution of $y''+my=0, \qquad m=\mathrm{const}>0$. How many zeros can there be on an interval $(a,b)$?
Explanation of Method
Writing the general solution in amplitude–phase form shows that consecutive zeros of any nontrivial solution are spaced exactly π/√m apart, independent of the initial data.
Step 1: General solution
The characteristic equation $r^2+m=0$ gives $r=\pm i\sqrt{m}$, so every solution has the form
\[ y = C_1\cos(\sqrt{m}\,x) + C_2\sin(\sqrt{m}\,x), \]
and "nontrivial" means $(C_1,C_2)\neq(0,0)$.
Step 2: Amplitude–phase form
Write $C_1=A\cos\varphi_0$, $C_2=A\sin\varphi_0$ with $A=\sqrt{C_1^2+C_2^2}>0$ (positive precisely because $C_1,C_2$ are not both zero) and $\varphi_0=\arctan(C_2/C_1)$ (quadrant fixed by the signs of $C_1,C_2$). Then
\[ y = A\cos\varphi_0\cos(\sqrt{m}x) + A\sin\varphi_0\sin(\sqrt{m}x) = A\cos\bigl(\sqrt{m}\,x-\varphi_0\bigr). \]
Since $A\neq0$, the zeros of $y$ coincide exactly with the zeros of $\cos(\sqrt{m}x-\varphi_0)$.
Step 3: Locating the zeros
$\cos\theta=0$ exactly at $\theta=\dfrac{\pi}{2}+k\pi$, $k\in\mathbb{Z}$. Setting $\theta=\sqrt{m}x-\varphi_0$ and solving for $x$: \[x_k = \frac{\varphi_0}{\sqrt{m}} + \frac{\pi}{2\sqrt{m}} + \frac{k\pi}{\sqrt{m}}, \qquad k\in\mathbb{Z}.\]
Step 4: Distance between consecutive zeros
Subtracting consecutive values of $k$: $x_{k+1}-x_k = \frac{\pi}{\sqrt{m}}.$
This spacing is the same for every consecutive pair, and it does not depend on $C_1,C_2$ (equivalently, on the initial data) — only on $m$. It equals exactly half the period $T=2\pi/\sqrt{m}$ of the oscillation.
$d = \frac{\pi$.
Step 5: How many zeros can lie in $(a,b)$?
The full lattice of zeros $\{x_k\}$ is equally spaced with gap $d=\pi/\sqrt{m}$, but its exact position is shifted by the phase $\varphi_0$, which depends on the particular solution. For an interval of length $L=b-a$:
Maximum count.
Sliding the lattice so that a zero lands just inside $a$ packs in as many zeros as possible: at most $\left\lfloor \dfrac{L\sqrt{m}}{\pi}\right\rfloor+1$ zeros can occur in $(a,b)$.
Possible absence of zeros.
If $L
Guaranteed minimum.
Because consecutive zeros are always exactly $d$ apart, any interval of length strictly greater than $d=\pi/\sqrt m$ cannot fit between two consecutive zeros without containing at least one — so every such interval contains at least one zero of every nontrivial solution.
Step 6: Numerical illustration
Take $m=\pi^2$, so $\sqrt{m}=\pi$ and $d=\pi/\pi=1$. On the interval $(0,10)$ ($L=10$):
| Quantity | Value |
|---|---|
| Spacing $d=\pi/\sqrt m$ | $1$ |
| Period $T=2\pi/\sqrt m$ | $2$ |
| Interval length $L=b-a$ | $10$ |
| Maximum zeros in $(0,10)$ | $\lfloor 10\rfloor+1=11$ |
| Guaranteed minimum (any sub-interval of length $>1$) | $\geq 1$ |
The actual number of zeros of a given solution in $(0,10)$ depends on $\varphi_0$; e.g.\ for $y=\sin(\pi x)$ ($\varphi_0=\pi/2$) the zeros fall exactly at the integers $0,1,\dots,10$, so $10$ of them lie strictly inside the open interval $(0,10)$, one short of the theoretical maximum of $11$, which is attained only in the limit as a zero approaches the endpoint $a$ from inside.
Result
Final answer:
\[\boxed{\,d = \frac{\pi}{\sqrt{m}}\,}\]