Second order diff. eq. Frobenius

Telemachus

Hi there. I have this exercise, which says:

Demonstrate that:

[tex]xy''+(1-x)y'+\lambda y=0[/tex]

has a polynomial solution for some λ values.
Indicate the orthogonality relation between polynomials, the fundamental interval, and the weight function.

So I thought I should solve this using Frobenius method. I have one singular point at x=0, which is regular. I assumed a solution of the form:
[tex]y(x)=\sum_0^{\infty}a_n x^{n+r}[/tex]

And then replacing in the diff. eq. I get:
[tex]\sum_0^{\infty}a_n (n+r)(n+r-1) x^{n+r-1}+\sum_0^{\infty}a_n (n+r)x^{n+r-1}-\sum_0^{\infty}(n+r)a_n x^{n+r}+\lambda \sum_0^{\infty}a_n x^{n+r}=0[/tex]
[tex]\sum_0^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]
[tex]a_0r^2x^{r-1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]

Therefore r=0.

Then replacing r=0, and changing the index for the first summation, with m=n-1, n=m+1:
[tex]\sum_0^{\infty}a_{m+1} (m+1)^2 x^{m}-\sum_0^{\infty}a_n (n-\lambda) x^{n}=0[/tex]
And now calling m=n
[tex]\sum_0^{\infty}x^m \left ( a_{m+1} (m+1)^2 x^{m}-a_m (m-\lambda) \right )=0[/tex]
So I have the recurrence relation:
[tex]a_{m+1}=\frac{a_m(m-\lambda)}{(m+1)^2}[/tex]

Trying some terms:
[tex]a_1=-a_0\lambda[/tex]
[tex]a_2=\frac{a_1(1-\lambda)}{2^2}=-\frac{a_0\lambda(1-\lambda)}{2^2}[/tex]
[tex]a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)}{2^23^2}[/tex]
[tex]a_4=\frac{a_3(3-\lambda)}{4^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)(3-\lambda)}{2^23^24^2}[/tex]

I'm not sure what this gives, I tried this:
[tex]a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}[/tex]
This is wrong, because the factorial in the numerator is only defined for positive values of (n-1-λ), and if n=1 I get (-\lambda)!, which wouldn't work for a_1, unless λ=0, which gives the trivial solution. But I think it works for n>1.

chiro

Hey Telemachus.

You can define the factorial for negative values, but the values can not be integers: if this holds then the factorial function does extend to the negative real line (minus the integers). Just in case you need more details:

http://en.wikipedia.org/wiki/Gamma_function

Telemachus

Thank you chiro. Do you think that what I did is ok?

I should take the diff. eq. into the self-adjoint form to get the weight function. About the fundamental interval, I think I should look at the convergence radius for the solution, right?

Telemachus

Ok. I worked this in a different fashion:

[tex]a_1=-a_0\lambda[/tex]
[tex]a_2=\frac{a_1(1-\lambda)}{2^2}=\frac{a_0\lambda(\lambda-1)}{2^2}[/tex]
[tex]a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(\lambda-1)(\lambda-2)}{2^23^2}[/tex]
[tex]a_4=\frac{a_3(3-\lambda)}{4^2}=\frac{a_0\lambda(\lambda-1)(\lambda-2)(\lambda-3)}{2^23^24^2}[/tex]

And now I called:
[tex]a_n=a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}[/tex]

Then λ-n can't be a negative integer, and the polynomials would be given by:
[tex]\sum_0^{\infty}a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}x^n[/tex]
Anyway, I think the an are wrong again, because if I take n=1 I get [tex]a_1=-a_0 \Gamma(\lambda-1)[/tex] which doesn't fit.

There is another solution, it is given by using the Frobenius theorem, and it involves a logarithm, but I think it isn't needed.

I actually think that I didn't have to get this explicit solution. To demonstrate what the problem asks I think I should take the equation to the self adjoint form.
[tex]xy''+(1-x)y'+\lambda y=0\rightarrow y''+(\frac{1}{x}-1)y'+\frac{\lambda}{x}y=0[/tex]

Multiplying by [tex]r(x)=e^{\ln (x) -x}[/tex]
I get:
[tex]\frac{d}{dx}\left ( e^{\ln (x) -x}\frac{dy}{dx} \right) +\lambda\frac{e^{\ln (x) -x}}{x}y=0[/tex]
This is the self adjoint form for my differential equation. Then the weight function is given by: [tex]p(x)=\frac{e^{\ln (x) -x}}{x}[/tex]

I don't know how to get the fundamental interval.

By the way, should I post this in homework and coursework questions? if it is so, please move it, and I'm sorry.

Telemachus

Ok. It's solved.

HallsofIvy

The original problem was show that the equation "has a polynomial solution for some λ values." So you really just need to show that for some [itex]\lambda[/itex], The coefficients are eventually 0.

Telemachus

Yes, but for which λ? besides, the coefficients doesn't seem that easy to get. I actually couldn't. I used some theorems on the sturm liouville theory to solve this, I didn't get the coefficients explicitly. I've tried, but I couldn't find the coefficients. I would like to find the right expression for the a_n in the recurrence relation, but it doesn't seem to be that easy.