Dtjytyk

How to compare two different files line by line in unix?

How to play a character with a disability or mental disorder without being offensive?

Did Deadpool rescue all of the X-Force?

Do any jurisdictions seriously consider reclassifying social media websites as publishers?

Chinese Seal on silk painting - what does it mean?

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

How do I use the new nonlinear finite element in Mathematica 12 for this equation?

Can the Great Weapon Master feat's damage bonus and accuracy penalty apply to attacks from the Spiritual Weapon spell?

Why should I vote and accept answers?

What is this clumpy 20-30cm high yellow-flowered plant?

If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?

Is there hard evidence that the grant peer review system performs significantly better than random?

How does Python know the values already stored in its memory?

Disembodied hand growing fangs

Morning, Afternoon, Night Kanji

How to write this math term? with cases it isn't working

How much damage would a cupful of neutron star matter do to the Earth?

Why weren't discrete x86 CPUs ever used in game hardware?

Is there any word for a place full of confusion?

Significance of Cersei's obsession with elephants?

Is grep documentation about ignoring case wrong, since it doesn't ignore case in filenames?

Time to Settle Down!

Converted a Scalar function to a TVF function for parallel execution-Still running in Serial mode

How to install press fit bottom bracket into new frame

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:

$$Big[nu^2 + frac{rho^2 -1}{rho^2} partial_{rho}(rho^2 (rho^2 -1)partial_{rho}) Big]f(rho) = 0,$$

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

I tried solutions of the form $f(rho) = frac{g(rho)}{rho}$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth {frac{x}{2}}right)^{inu} g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$frac{d^2g_1}{dx^2} + [coth x -2inu, text{cosech}, x]frac{dg_1}{dx}-2g_1=0$$

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

207

$endgroup$

add a comment | 

0

$begingroup$

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

$$Big[nu^2 + frac{rho^2 -1}{rho^2} partial_{rho}(rho^2 (rho^2 -1)partial_{rho}) Big]f(rho) = 0,$$

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

I tried solutions of the form $f(rho) = frac{g(rho)}{rho}$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth {frac{x}{2}}right)^{inu} g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$frac{d^2g_1}{dx^2} + [coth x -2inu, text{cosech}, x]frac{dg_1}{dx}-2g_1=0$$

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

207

$endgroup$

add a comment | 

$begingroup$

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

$$Big[nu^2 + frac{rho^2 -1}{rho^2} partial_{rho}(rho^2 (rho^2 -1)partial_{rho}) Big]f(rho) = 0,$$

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

I tried solutions of the form $f(rho) = frac{g(rho)}{rho}$, and further $rho = cosh[x]$. Then in the asymptotic limit $x to 0$, the solution goes like
$$g(cosh x) = left(coth {frac{x}{2}}right)^{inu} g_1(cosh x) $$

The differential equation for $g_1$ becomes then
$$frac{d^2g_1}{dx^2} + [coth x -2inu, text{cosech}, x]frac{dg_1}{dx}-2g_1=0$$

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

207

$endgroup$

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

207

share|cite|improve this question

asked Mar 25 at 9:51

Bruce LeeBruce Lee

207

asked Mar 25 at 9:51

Bruce LeeBruce Lee

207

asked Mar 25 at 9:51

Bruce LeeBruce Lee

207

1 Answer
1

active

oldest

votes

1

$begingroup$

Writing $f(rho) = frac{g(rho)}{rho}$ 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
$$
G(rho) = (rho - i nu) left(frac{1+rho}{1-rho}right)^{frac{inu}{2}}.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161581%2fhow-to-solve-this-2nd-order-ordinary-differential-equation%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Writing $f(rho) = frac{g(rho)}{rho}$ 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
$$
G(rho) = (rho - i nu) left(frac{1+rho}{1-rho}right)^{frac{inu}{2}}.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

$endgroup$

add a comment | 

1

$begingroup$

Writing $f(rho) = frac{g(rho)}{rho}$ 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
$$
G(rho) = (rho - i nu) left(frac{1+rho}{1-rho}right)^{frac{inu}{2}}.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

$endgroup$

add a comment | 

$begingroup$

Writing $f(rho) = frac{g(rho)}{rho}$ 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
$$
G(rho) = (rho - i nu) left(frac{1+rho}{1-rho}right)^{frac{inu}{2}}.
$$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

$endgroup$

share|cite|improve this answer

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921

answered Mar 25 at 10:42

Frits VeermanFrits Veerman

7,1462921