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How to solve this 2nd order Ordinary Differential Equation

0

$begingroup$

I was reading this, and wasn't able to solve equation (2.34). The equation is:

$$Big[nu^2 + fracrho^2 -1rho^2 partial_rho(rho^2 (rho^2 -1)partial_rho) Big]f(rho) = 0,$$

where $rho$'s range is $(1,infty)$.

I tried solutions of the form $f(rho) = fracg(rho)rho$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth fracx2right)^inu g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$fracd^2g_1dx^2 + [coth x -2inu, textcosech, x]fracdg_1dx-2g_1=0$$

I don't know how to proceed from here. I tried out the solutions using Mathematica also, but that didn't help. How do I solve the same? Thanks.

ordinary-differential-equations

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

$endgroup$

add a comment | 

0

$begingroup$

I was reading this, and wasn't able to solve equation (2.34). The equation is:

$$Big[nu^2 + fracrho^2 -1rho^2 partial_rho(rho^2 (rho^2 -1)partial_rho) Big]f(rho) = 0,$$

where $rho$'s range is $(1,infty)$.

I tried solutions of the form $f(rho) = fracg(rho)rho$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth fracx2right)^inu g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$fracd^2g_1dx^2 + [coth x -2inu, textcosech, x]fracdg_1dx-2g_1=0$$

I don't know how to proceed from here. I tried out the solutions using Mathematica also, but that didn't help. How do I solve the same? Thanks.

ordinary-differential-equations

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

$endgroup$

add a comment | 

$begingroup$

I was reading this, and wasn't able to solve equation (2.34). The equation is:

$$Big[nu^2 + fracrho^2 -1rho^2 partial_rho(rho^2 (rho^2 -1)partial_rho) Big]f(rho) = 0,$$

where $rho$'s range is $(1,infty)$.

I tried solutions of the form $f(rho) = fracg(rho)rho$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth fracx2right)^inu g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$fracd^2g_1dx^2 + [coth x -2inu, textcosech, x]fracdg_1dx-2g_1=0$$

I don't know how to proceed from here. I tried out the solutions using Mathematica also, but that didn't help. How do I solve the same? Thanks.

ordinary-differential-equations

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

$endgroup$

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

asked Mar 25 at 9:51

Bruce LeeBruce Lee

187

1 Answer
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1

$begingroup$

Writing $f(rho) = fracg(rho)rho$ is a good idea, you then get
$$
(1-rho^2)^2 g'' -2 rho (1-rho^2) g' + (2(1-rho^2) + nu^2) g = 0. tag*
$$
This is a form of the (associated) Legendre equation, which has solutions given by the associated Legendre functions $P_1^i nu(rho)$, $Q_1^inu(rho)$. In this case, these take a relatively simple form in $rho$; the general solution to $(*)$ is given by
$$
g(rho) = c_1 G(rho) + c_2 G(-rho),
$$
with
$$
G(rho) = (rho - i nu) left(frac1+rho1-rhoright)^fracinu2.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

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1

$begingroup$

Writing $f(rho) = fracg(rho)rho$ is a good idea, you then get
$$
(1-rho^2)^2 g'' -2 rho (1-rho^2) g' + (2(1-rho^2) + nu^2) g = 0. tag*
$$
This is a form of the (associated) Legendre equation, which has solutions given by the associated Legendre functions $P_1^i nu(rho)$, $Q_1^inu(rho)$. In this case, these take a relatively simple form in $rho$; the general solution to $(*)$ is given by
$$
g(rho) = c_1 G(rho) + c_2 G(-rho),
$$
with
$$
G(rho) = (rho - i nu) left(frac1+rho1-rhoright)^fracinu2.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

$endgroup$

add a comment | 

1

$begingroup$

Writing $f(rho) = fracg(rho)rho$ is a good idea, you then get
$$
(1-rho^2)^2 g'' -2 rho (1-rho^2) g' + (2(1-rho^2) + nu^2) g = 0. tag*
$$
This is a form of the (associated) Legendre equation, which has solutions given by the associated Legendre functions $P_1^i nu(rho)$, $Q_1^inu(rho)$. In this case, these take a relatively simple form in $rho$; the general solution to $(*)$ is given by
$$
g(rho) = c_1 G(rho) + c_2 G(-rho),
$$
with
$$
G(rho) = (rho - i nu) left(frac1+rho1-rhoright)^fracinu2.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

$endgroup$

add a comment | 

$begingroup$

Writing $f(rho) = fracg(rho)rho$ is a good idea, you then get
$$
(1-rho^2)^2 g'' -2 rho (1-rho^2) g' + (2(1-rho^2) + nu^2) g = 0. tag*
$$
This is a form of the (associated) Legendre equation, which has solutions given by the associated Legendre functions $P_1^i nu(rho)$, $Q_1^inu(rho)$. In this case, these take a relatively simple form in $rho$; the general solution to $(*)$ is given by
$$
g(rho) = c_1 G(rho) + c_2 G(-rho),
$$
with
$$
G(rho) = (rho - i nu) left(frac1+rho1-rhoright)^fracinu2.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

$endgroup$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1312921