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Second order diff. eq. Frobenius 
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#1
Apr112, 06:39 PM

P: 552

Hi there. I have this exercise, which says:
Demonstrate that: [tex]xy''+(1x)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+r1) x^{n+r1}+\sum_0^{\infty}a_n (n+r)x^{n+r1}\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+r1}\sum_0^{\infty}a_n (n+r\lambda) x^{n+r}=0[/tex] [tex]a_0r^2x^{r1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r1}\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=n1, 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(n1\lambda)!}{(n!)^2}[/tex] This is wrong, because the factorial in the numerator is only defined for positive values of (n1λ), 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. 


#2
Apr112, 07:12 PM

P: 4,578

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 


#3
Apr112, 08:17 PM

P: 552

Thank you chiro. Do you think that what I did is ok?
I should take the diff. eq. into the selfadjoint form to get the weight function. About the fundamental interval, I think I should look at the convergence radius for the solution, right? 


#4
Apr212, 10:06 AM

P: 552

Second order diff. eq. Frobenius
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(\lambda1)}{2^2}[/tex] [tex]a_3=\frac{a_2(2\lambda)}{3^2}=\frac{a_0\lambda(\lambda1)(\lambda2)}{2^23^2}[/tex] [tex]a_4=\frac{a_3(3\lambda)}{4^2}=\frac{a_0\lambda(\lambda1)(\lambda2)(\lambda3)}{2^23^24^2}[/tex] And now I called: [tex]a_n=a_0\frac{(1)^n\Gamma(\lambdan)}{(n!)^2}[/tex] Then λn can't be a negative integer, and the polynomials would be given by: [tex]\sum_0^{\infty}a_0\frac{(1)^n\Gamma(\lambdan)}{(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(\lambda1)[/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''+(1x)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. 


#5
Apr212, 11:13 AM

P: 552

Ok. It's solved.



#6
Apr212, 03:38 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

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.



#7
Apr212, 06:45 PM

P: 552

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.



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