I have to prove, that for $w(x)\geq 1$ every solution for differential equation $y''+w(x)y=0$ has infinity many zeros.

My idea is, that I first take $w(x)=1$, which has 2 linearly independent solutions $y=A\sin x+B\cos y$ Booth $\sin x$ and $\cos x$ have infinity many zeros, which is true for any of their linear combinations.

So I have proven that for $w=1$. For $w(x) > 1$ I can use Sturm–Picone comparison theorem, so between every 2 zeros of solution for $w=1$, there is at least one solution of every solution for $w>1$, therefore it is at least the same amount minus one of them as for $w=1$.

Does this proof make any sense, or what are its limits/problems? Is there any other way to prove that?

  • 2
    $\begingroup$ If you have a problem with the strict inequality in the Sturm-Picone theorem, note that by assumption $w(x)\ge 1>\frac49$, so that you know with certainty that between any two roots of $\cos(\frac23x)$ there is at least one root of $y(x)$. $\endgroup$ Sep 5, 2020 at 18:43

4 Answers 4


I didn't know of the Sturm-Picone comparison theorem before, but after having looked it up, it totally makes sense to use it. In the notation of the Wikipedia article you will have $p_1=p_2=1,$ $q_1=1$ and $q_2=q.$ The proof can hardly be simpler.


Here is a proof that does not rely on the Sturm–Picone comparison theorem. The most important thing about $w(x)$ is that it is bounded below by some $\epsilon>0$ (in your case $\epsilon=1$). Now, suppose that $y$ has a finite number of zeros and let $x_0$ be the largest zero plus $1$. Then for $x\geq x_0$, $y<0$ or $y>0$.

Let us consider the first case (the second follows in a similar manner). Since $y<0$ and $w(x)>\epsilon>0$, we know $x\geq x_0$ implies $y''>0$. However, we can say something further: that $y''>\delta>0$ (it is bounded away from zero by some $\delta$). To see this, note that for $z>x\geq x_0$ by the mean value theorem for integrals there exists $c\in (x,z)$ such that


Now, if $y'(x_1)>0$ for some $x_1>x_0$, let $\rho>0$ be defined such that $y'(x)\geq 0$ for $x\in [x_1-\rho,x_1+\rho]$. We can then restrict the interval of the integral such that there exists $c\in (x_1-\rho,x_1+\rho)$ such that

$$0>y(c)=\frac{1}{2\rho}\int_{x_1-\rho}^{x_1+\rho}y'(t)dt\geq \frac{1}{2\rho}\int_{x_1-\rho}^{x_1+\rho}0dt=0$$

a contradiction. Thus, $y'(x)\leq 0$ for all $x>x_0$. This implies $y(x)$ is decreasing on $[x_0,\infty)$ and therefore

$$y''(x)=-w(x)y(x)=w(x)|y(x)|>\epsilon |y(x_0)|>0$$

We now come to the conclusion of our proof. Using the fundamental theorem of calculus for $x>x_0$ we can write

$$y(x)=\int_{x_0}^x\left[ \int_{x_0}^zy''(t)dt+y'(x_0)\right]dz+y(x_0)$$

$$>\int_{x_0}^x\left[ \int_{x_0}^z\epsilon |y(x_0)|dt+y'(x_0)\right]dz+y(x_0)$$

$$=\frac{\epsilon |y(x_0)|}{2} x^2+(y'(x_0)-\epsilon|y(x_0)|)x+\frac{{x_0}^2\epsilon|y(x_0)|}{2}-x_0y'(x_0)+y(x_0)$$

Since this is a quadratic whose leading term is positive, we conclude $y(x)$ goes to infinity, a contradiction. As the case where $y>0$ is worked in the same manner (except with a sign change), we conclude $w(x)>\epsilon>0$ implies $y(x)$ has an infinite number of zeros.


First of all, yes you can simply use the Sturm-Picone theorem which however would require you to proof this theorem, or? Here is another proof not relying on it.

Suppose there are only a finite number of zeros $x_1,...,x_n$ in that order (i.e. $x_n$ is the largest zero) and furthermore assume WLOG $y(x)>0$ for all $x>x_n$ (otherwise consider the solution $-y(x)$). Using the equation $y''=-\omega \, y$, we can write down a formal solution $$y(x)=y(x_0) + y'(x_0) \, (x-x_0) - \int_{x_0}^x\omega(x') y(x') (x-x') \, {\rm d}x' \\ \leq y(x_0) + y'(x_0) \, (x-x_0) - \int_{x_0}^x y(x') (x-x') \, {\rm d}x' \\ =y(x_0) + y'(x_0) \, (x-x_0) - \frac{y(x_0) \, (x-x_0)^2}{2} - \frac{1}{2} \int_{x_0}^x y'(x') \, (x-x')^2 \, {\rm d}x' $$ where the third line follows after partial integration. $x_0>x_n$ is an arbitrary expansion point to our disposal. Let us distinguish two cases.

case 1: $y'(x) \geq 0$ for all $x\geq x_n$

In this case $x_0>x_n$ can be chosen arbitrarily and the third line for $y(x)$ allows the estimate $$y(x) \leq y(x_0) + y'(x_0) \, (x-x_0) - \frac{y(x_0) \, (x-x_0)^2}{2} $$ by our supposition $y(x)>0$ for $x\geq x_0$. However, the RHS decreases without bound for large $x$ leading to the desired contradiction.

case 2: $y'(x) < 0$ for at least some values of $x>x_n$

We can choose $x_0$ such that $y'(x_0)<0$. The supposition $y(x) > 0$ for all $x\geq x_0$ then gives $y(x) \leq y(x_0)+y'(x_0) \, (x-x_0)$, impossible for large $x$.


Nice question, this is my (intuitive) answer:

Since $w(x)$ is positive, then if $y$ is negative then $y''$ is necessarily positive, and vice versa. That is if the point $(x,y)$ is over the $x$-axis then the curve must be curved downwards until it meets the $x$-axis (that is a zero). Below the $x$-axis $y$ is negative so $y''$ is positive, that is the curve is bent upwards until it meet the $x$-axis again (that is another zero), and so on. At the zero one reads from the equation the $y'' = 0$. Then this process continues forever and so you have infinite number of zeros.

If $y = 0$ for all $x$ then $y'' = 0$ as well, and you have infinite number of zeros again.

Since $w(x) \geq 1$, then $y \rightarrow 0$ iff $y'' \rightarrow 0$

I assume that the domain of $y$ is $\mathbb{R}$ and that the siolutions are bounded.

  • $\begingroup$ $\frac1x-1$ in $2$ has $yy''<0$ but never zero after $2$ $\endgroup$
    – enzotib
    Sep 5, 2020 at 13:59
  • $\begingroup$ what do you mean with "in 2" ? $\endgroup$
    – Physor
    Sep 5, 2020 at 14:02
  • $\begingroup$ I mean for $x=2$ $\endgroup$
    – enzotib
    Sep 5, 2020 at 14:03
  • $\begingroup$ your $y$ is not bounded and not defined on all $\mathbb{R}$ $\endgroup$
    – Physor
    Sep 5, 2020 at 14:05
  • $\begingroup$ Ok, take $-\arctan x$ $\endgroup$
    – enzotib
    Sep 5, 2020 at 14:07

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