- #1
omyojj
- 37
- 0
Hello,
Let us assume we have a differential equation
[tex] ( \frac{d^2}{dx^2}+ \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 ) y = 0[/tex]
where
[tex]\Theta_n = (1-\mu^2)^{1/(n+1)}[/tex]
is a function of [tex]\mu[/tex] which is a function of [tex]x[/tex] :
[tex] \frac{d\mu}{dx} = \Theta_n^n = (1-\mu^2)^{n/(n+1)} [/tex]
In terms of [tex]\mu[/tex] and [tex]\frac{d}{dx} = \frac{d\mu}{dx}\frac{d}{d\mu}[/tex], the above equation becomes
[tex]\left( \Theta_n^n\frac{d}{d\mu}\Theta_n^n\frac{d}{d\mu} + \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 \right)y = 0[/tex]
In the limit of [tex]n \rightarrow \infty [/tex], we have
[tex]\lim_{n\rightarrow \infty} \Theta_n^n = 1- \mu^2 [/tex]
[tex]\left( (1-\mu^2)\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu} + 2(1-\mu^2) - \nu^2 \right) y = 0 [/tex]
and this is just associated Legendre equation.
[tex]y^{\prime\prime} - \frac{2\mu}{1-\mu^2}y^{\prime} + \left[ \frac{a(a+1)}{1-\mu^2}-\frac{\nu^2}{(1-x^2)^2}\right]y = 0[/tex]
with a=1.
(The solutions go like this
[tex] P_{1}^{\pm\nu} \propto ( 1+ \mu )^{\pm \nu/2} \cdot ( 1 - \mu )^{\mp \nu/2} \cdot (\nu - \mu) [/tex]
)
I think that solutions should show qualitatively similar behavior to that obtained from the associated Legendre's equation, But coudn't make any progress
Can I solve this equation utilizing a commonly-used technique like the Frobenius method for any rational number n > 0?
Can't I just have the form that the solutions might have? (like power series)
Thank you..
p.s. I should further note that the differential relation b/w x and mu can be directly integrated to give (x = 0 when mu = 0)
[tex] x = \mu {}_2 F_1 (1/2,\,\, n/(n+1);\,\,3/2;\,\,\mu^2) [/tex]
In the limiting case (n -> ∞ )
[tex] x = \mu {}_2 F_1 (1/2,\,\,1;\,\,3/2;\,\,\mu^2) = \tanh^{-1}\mu[/tex]
Let us assume we have a differential equation
[tex] ( \frac{d^2}{dx^2}+ \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 ) y = 0[/tex]
where
[tex]\Theta_n = (1-\mu^2)^{1/(n+1)}[/tex]
is a function of [tex]\mu[/tex] which is a function of [tex]x[/tex] :
[tex] \frac{d\mu}{dx} = \Theta_n^n = (1-\mu^2)^{n/(n+1)} [/tex]
In terms of [tex]\mu[/tex] and [tex]\frac{d}{dx} = \frac{d\mu}{dx}\frac{d}{d\mu}[/tex], the above equation becomes
[tex]\left( \Theta_n^n\frac{d}{d\mu}\Theta_n^n\frac{d}{d\mu} + \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 \right)y = 0[/tex]
In the limit of [tex]n \rightarrow \infty [/tex], we have
[tex]\lim_{n\rightarrow \infty} \Theta_n^n = 1- \mu^2 [/tex]
[tex]\left( (1-\mu^2)\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu} + 2(1-\mu^2) - \nu^2 \right) y = 0 [/tex]
and this is just associated Legendre equation.
[tex]y^{\prime\prime} - \frac{2\mu}{1-\mu^2}y^{\prime} + \left[ \frac{a(a+1)}{1-\mu^2}-\frac{\nu^2}{(1-x^2)^2}\right]y = 0[/tex]
with a=1.
(The solutions go like this
[tex] P_{1}^{\pm\nu} \propto ( 1+ \mu )^{\pm \nu/2} \cdot ( 1 - \mu )^{\mp \nu/2} \cdot (\nu - \mu) [/tex]
)
I think that solutions should show qualitatively similar behavior to that obtained from the associated Legendre's equation, But coudn't make any progress
Can I solve this equation utilizing a commonly-used technique like the Frobenius method for any rational number n > 0?
Can't I just have the form that the solutions might have? (like power series)
Thank you..
p.s. I should further note that the differential relation b/w x and mu can be directly integrated to give (x = 0 when mu = 0)
[tex] x = \mu {}_2 F_1 (1/2,\,\, n/(n+1);\,\,3/2;\,\,\mu^2) [/tex]
In the limiting case (n -> ∞ )
[tex] x = \mu {}_2 F_1 (1/2,\,\,1;\,\,3/2;\,\,\mu^2) = \tanh^{-1}\mu[/tex]
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