What are the eigenfunctions for the ODE y′′−2xy′+2αy=0?

[peripatein]

Hi,

Homework Statement

I have the following ODE:
y′′−2xy′+2αy=0
I'd like to determine the first three eigenfunctions.

Homework Equations

The Attempt at a Solution

The solution y(x) may be recursively represented as:
a_n+2=a_n(2n−2α)/[(n+2)(n+1)]
I have found the eigenvalues to be −2α, however I find the manner whereby the eigenfunctions are determined to be rather perplexing. I'd sincerely appreciate an explanation. For instance, I know that for α=0, a₂=a₀(0−0)/2, but why would that entail y₀(x)=a₀? I mean, how was that derived?

 

[pasmith]

Hi,

Homework Statement

I have the following ODE:
y′′−2xy′+2αy=0
I'd like to determine the first three eigenfunctions.

Homework Equations

The Attempt at a Solution

The solution y(x) may be recursively represented as:
a_n+2=a_n(2n−2α)/[(n+2)(n+1)]
I have found the eigenvalues to be −2α, however I find the manner whereby the eigenfunctions are determined to be rather perplexing. I'd sincerely appreciate an explanation. For instance, I know that for α=0, a₂=a₀(0−0)/2, but why would that entail y₀(x)=a₀? I mean, how was that derived?

Whenever you have a recurrence of the form
<br /> a_{n+2} = F(n)a_n<br />
then you know immediately that if a_n = 0 or F(n) = 0 then a_{n + 2m} = 0 for all m \geq 0.

Starting from a_0 will give you the even terms a_{2n}, but those terms don't affect the odd terms a_{2n+1}, which are obtained by starting from a_1. Here we have a second-order linear ODE, so we expect that for each \alpha there will be two linearly independent solutions. We can define these to be E_\alpha(x) which is obtained by taking a_0 = 1 and a_1 = 0 and O_\alpha(x) which is obtained by taking a_0 = 0 and a_1 = 1.

Thus for \alpha = 0 we have E_0(x) = 1 and O_0(x) = x + \frac13 x^3 + \dots, and the general solution will be cE_0(x) + dO_0(x). In fact we can solve the ODE analytically when \alpha = 0 to find that
<br /> O_0(x) = \int_0^x \exp(u^2)\,\mathrm{d}u.<br />

You can see that if \alpha is an even positive integer then the even series terminates, and if \alpha is an odd positive integer then the odd series terminates, and if \alpha is not a positive integer then neither series terminates.

 

[peripatein]

I am sorry, but I am not really following. I pretty much lost you at "thus...". In any case, I happen to know that the first three eigenfunctions are: a₀, a₁x and a₀(1-2x²). If you could explain to me how these were obtained I'd be grateful.

 

[vela]

When $\alpha=0$, if you write out the series solution, you get
$$y = a_0(1) + a_1\left(x + \frac{x^3}{3} + \cdots\right).$$ This is of the form $y = a_0 y_1 + a_1 y_2$ where $a_0$ and $a_1$ are arbitrary constants and $y_1=1$ and $y_2=x+\cdots$ are the two linearly independent solutions.

For this problem, you're apparently looking for polynomial solutions. For $\alpha=0$, you can do this be setting $a_1=0$ to get rid of the infinite series.

It turns out, you can obtain polynomial solutions for only certain values of $\alpha$. You should be able to deduce what these values are from the recurrence relation.