Solving Airy's Equation and Applying the Sturm Comparison Theorem

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Homework Statement



(a) By using a suitable transformation, show that the normal form of the DE [tex]y'' - 2y' + (x+1)y = 0\;\;\;\;\;(*)[/tex] is Airy's equation [tex]u'' + xu = 0.[/tex]
(b) State the Sturm comparison theorem for zeros of 2 second order linear DEs in normal form.

(c) By comparing with the DE [tex]v'' + v = 0[/tex] prove that every solution y(x) of (*) has infinitely many positive zeros.

The Attempt at a Solution



I've done (a). For (b), the theorem is:
[PLAIN]http://img101.imageshack.us/img101/1237/sturmh.png

I'm not sure how to proceed with (c).
 
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It's pretty straightforward, isn't it? The two functions to be compared are q(x)= x and r(x)= 1. It is certainly true that x> 1 for all x in, say, [itex][2, \infty)[/itex]. Now, how many times do solutions of y''+ y= 0 vanish in that interval?
 
HallsofIvy said:
It's pretty straightforward, isn't it? The two functions to be compared are q(x)= x and r(x)= 1. It is certainly true that x> 1 for all x in, say, [itex][2, \infty)[/itex]. Now, how many times do solutions of y''+ y= 0 vanish in that interval?

The general solution of [tex]v'' + v = 0[/tex] is [tex]v = A\sin x + B\cos x[/tex]

It has successive zeros at [tex]x = n\pi - \frac{\pi}{4}[/tex] where [tex]n\in\mathbb{Z}.[/tex]

Does this prove that every solution y of (*) has infinitely many positive zeros?

(By the comparison theorem, any solution of [tex]u'' + xu=0[/tex] and therefore of (*) has a solution in between those successive zeros. Since there are infinitely many positive zeros, there are infinitely many positive zeros in (*))
 
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I've finally got hold of the 'solution' but does using an example of a solution like it does with [tex]u=\sin x[/tex] prove that the equation has infinitely many (positive) zeroes for all solutions?

[PLAIN]http://img94.imageshack.us/img94/9296/sturmd.jpg
 
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