Second order diff. eq. Frobenius

[Telemachus]

Hi there. I have this exercise, which says:

Demonstrate that:

xy''+(1-x)y'+\lambda y=0

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:
y(x)=\sum_0^{\infty}a_n x^{n+r}

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

Therefore r=0.

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

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

I'm not sure what this gives, I tried this:
a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}
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]

Hi there. I have this exercise, which says:

Demonstrate that:

xy''+(1-x)y'+\lambda y=0

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:
y(x)=\sum_0^{\infty}a_n x^{n+r}

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

Therefore r=0.

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

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

I'm not sure what this gives, I tried this:
a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}
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.

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:

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

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

Then λ-n can't be a negative integer, and the polynomials would be given by:
\sum_0^{\infty}a_0\frac{(-1)^n\Gamma(\lambda-n)}{(n!)^2}x^n
Anyway, I think the an are wrong again, because if I take n=1 I get a_1=-a_0 \Gamma(\lambda-1) 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.
xy''+(1-x)y'+\lambda y=0\rightarrow y''+(\frac{1}{x}-1)y'+\frac{\lambda}{x}y=0

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

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 \lambda, 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.