Prove that y(x) has finitely many positive zero

[drawar]

Homework Statement

Given $$y''+e^{-x}y=0. \qquad (*)$$ Let $y(x)$ be any nontrivial solution of $(*)$, show that y has finitely many positive zeros.
Hint: Consider $z''+\frac{C}{x^4}z=0$ where $C>0$ is sufficiently large, which has a solution $z(x)=x\sin \frac{\sqrt C}{x}$.

Homework Equations

1. (Sturm's Comparison Theorem)
Let y(x) and z(x) be nontrivial solutions of
$$y'' + q(x)y = 0$$
and
$$z'' + r(x)z = 0, $$
where q(x) and r(x) are positive functions such that q(x) > r(x). Then y(x) vanishes at least once between any two successive zeros of z(x).
2. Moreover, if there exists a constant $m>0$ such that $q(x) \geq m^2$ for all x, then y(x) has infinitely many zeros.

The Attempt at a Solution

My guess is $C$ must be big enough s.t $\frac{C}{x^4}>e^{-x}$, so we can make use of the theorem. I also think that a proof by contradiction (i.e. suppose y has infinitely many positive zeros) may be useful, but don't know how to proceed further from here.
Could anyone shed some light on it please?

 

[mfb]

Your attempt looks good so far.

"Then y(x) vanishes at least once between any two successive zeros of z(x). "

As you use q(x) > r(x), q(r) will be your C/x^4 and r(x) will be e^(-x).
Therefore, your contradiction starts with "assume $z'' + r(x)z = 0$ has infinitely many zeroes [...]"
And then use the known solution for the other case to get a contradiction.That is a nice theorem.

 

[haruspex]

Your attempt looks good so far.

"Then y(x) vanishes at least once between any two successive zeros of z(x). "

As you use q(x) > r(x), q(r) will be your C/x^4 and r(x) will be e^(-x).
Therefore, your contradiction starts with "assume $z'' + r(x)z = 0$ has infinitely many zeroes [...]"
And then use the known solution for the other case to get a contradiction.


That is a nice theorem.

Right, but noting that y and z in the problem statement have reversed roles from the y and z in the theorem statement.
Also, it doesn't quite get you there because sin(1/x) has infinitely many positive zeroes. You might need to split the range of x into (0, ε) and [ε, ∞).

 

[drawar]

Thank you for the help, but doesn't sin(1/x) have infinitely many zeros in [ε, ∞)?

 

[haruspex]

Thank you for the help, but doesn't sin(1/x) have infinitely many zeros in [ε, ∞)?

I don't think so. Can you find any in (1/π, ∞)?

 

[drawar]

Oh that's enlightening, thank you!