I would just like to clarify a fw things I am not really understanding about Sturm-Liouville forms and eigenvalue problems:

I have the practice question:

$x^2y''+xy'+\lambda y = 0$

with boundary conditions $y(1)=0$, and $y(b)=0$

In part (a) I put the equation into SL form: $\frac{d}{dx}(x\frac{dy}{dx})+\frac{\lambda}{x}y=0$, no problem here.

part (b) says show that $\lambda \ge 0$ (hint use the Rayleigh Quotient)

From my understanding the Rayleigh Quotient is some complicated integral divided by an integral that describes a relationship between an eigenvalue and its eigenfunction.

But in the solutions provided it simply says multiply the SL form by $y$ and then integrate from $1$ to $b$.

Two questions: 1. Why is this using the Rayleigh quotient? I am struggling to see the connection here.

  1. The solution shows:

$$ \lambda \int_1^b \frac{y^2}{x}dx + \int_1^b y\frac{d}{dx}(x\frac{dy}{dx})dx = 0 $$ and then simply concludes with:

$$ \lambda \int_1^b \frac{y^2}{x}dx + xy\frac{dy}{dx}|_1^b = \int_1^b x(\frac{dy}{dx})^2dx $$

With the far left term and the RHS $\ge 0$, and the middle = 0. I am struggling to see how the jump was made. I know the middle term is 0 by the boundary conditions, but what about the other two integrals? how can we simply conclude that they are greater than 0?

Many thanks!

  • $\begingroup$ actually Rayleigh Q. provides a general description for eigenvalues.its applicable to compact operators as well. in practice most of the time cannot be used unless maybe for the largest eigenvalue. $\endgroup$ – BigM Sep 27 '14 at 4:04
  • $\begingroup$ meant to say the buggest and smallest.you might be able to use it in here.if could show the smallest is non-negative. $\endgroup$ – BigM Sep 27 '14 at 4:06
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    $\begingroup$ About that part of your question "RHS$\geq 0$ " . the middle term is zero because of initial condition. Those two integrals are non-negative since integrands are non-negative. Then its clear from here that $\lambda\geq0$ $\endgroup$ – BigM Sep 27 '14 at 4:18

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