Solving Airy's Equation and Applying the Sturm Comparison Theorem

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Homework Help Overview

The discussion revolves around solving Airy's equation and applying the Sturm comparison theorem to analyze the zeros of second-order linear differential equations. The original poster presents a differential equation and seeks to transform it into Airy's equation while also exploring the implications of the Sturm comparison theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the given differential equation into Airy's equation and the application of the Sturm comparison theorem. There are inquiries about the implications of comparing the functions q(x) = x and r(x) = 1, particularly regarding the zeros of the solutions. Some participants question whether using specific examples of solutions can substantiate claims about the existence of infinitely many positive zeros for all solutions.

Discussion Status

The discussion is active, with participants sharing their thoughts on the transformation and comparison processes. Some guidance has been offered regarding the comparison of functions and the nature of their zeros, but there is no explicit consensus on the conclusions drawn from these comparisons.

Contextual Notes

Participants are navigating through the requirements of the homework statement, particularly focusing on the transformation and the application of the Sturm comparison theorem. There are indications of uncertainty regarding the completeness of their arguments and the validity of using specific examples to prove broader claims.

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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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