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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-

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2 Answers 2

up vote 1 down vote accepted

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.

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nice. Didn't know about this one. Can we solve the exercise without using that theorem? –  guido giuliani Oct 29 '12 at 9:55
    
yes, you can use the so-called Pruefer transform to cast the equation in the polar coordinates and then try to prove that the polar angle grows infinitely wit $x$. You would be basically following the lines of proof of the oscillation theorem then. –  Valentin Oct 29 '12 at 12:38

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 $x<0.$

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