airy equation vanish infinitely many times Hello I'm studying Airy's equations. In particular I'm interested in the following istance of the equation $$v''(x)+xv(x)=0.\tag{1}$$
I'm asked to prove that $v$ vanishes infinitely many times on the positive $x$-axis and at most one time on the negative $x$-axis.
How do I answer this question?
I've tried some manipulations, especially connections with the Riccati form. Substituting $$u=\frac{v'}{v}\tag{2}$$ one arrives to the formula $$u'+u^2+x=0,\tag{3}$$ however I cannot see if this helps.
Does anybody have any suggestion?
Thanks
-Guido-
 A: It is convenient to apply the Sturm comparison theorem here.
For the negative part of the axis consider $v''=0$ as a comparison equation and a particular solution $v=1$. Now suppose a solution of $v''+xv=0$, $x<0$ vanishes more than once and derive a contradiction.
On the positive semi-axis a definite conclusion can be made for the case $x>1$. Use $v''+v=0$ (which is known to have oscillating solutions) as a comparison equation in this case.
A: Ok, we have $v''(x) = -x v(x).$
I am no expert at this, and you have not provided any initial conditions,
but suppose $v(0)>0$, and $v'(0)>0.$
Thus, a bit to the right of $x=0$, we have the signs $(y,y',y'') = (+,+,-)$.
So, using the relation above that the derivative is decreasing, we will have at some $x_1$ that $(y(x_1),y'(x_1),y''(x_1)) = (+,0,-)$ and a moment later $(+,-,-)$.
As $x$ and $y$ is still positive, $y''$ and thus $y'$ will continue to decrease,
so eventually, at some $x_2$ we have  $(y(x_2),y'(x_2),y''(x_2)) = (0,-,-)$ and a moment later, we will have signs $(-,-,+)$. Forward a bit, and we will then have
an $x_3$ with $(y(x_1),y'(x_1),y''(x_1)) = (-,0,+)$.
You now see the pattern, and the function will continue to oscillate in this fashion. With a similar argument, you can examine the behaviour when $x<0.$
