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
$$a_{n+2} = F(n)a_n$$
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
$$O_0(x) = \int_0^x \exp(u^2)\,\mathrm{d}u.$$

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.