# Prove that, if $y''+w(x)y=0,$ then $y$ has infinitely many zeroes

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?

• 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 $y(x)$. Sep 5, 2020 at 18:43

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

$$\frac{1}{z-x}\int_x^zy'(t)dt=y(c)<0$$

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.

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