Transform differential equation into hypergeometric differential equationSolutions in terms of the...
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Transform differential equation into hypergeometric differential equation
Solutions in terms of the hypergeometric functionsCatagorising a Differential Equationsecond order differential equation to hypergeometric equationCan this differential equation be transformed into an hypergeometric equation?Differential equations - Hypergeometric functionInitial conditions on hypergeometric equationRelationship between the Picard-Fuchs differential equation and the hypergeometric differential equationHow to relate the solutions to a Fuchsian type differential equation to the solutions to the hypergeometric differential equation?Transforming second order differential equation into the form of a hypergeometric differential equationSolutions of hypergeometric equation when $cinmathbb{Z}$How do I prove that any second-order equation with three regular singular points can be transformed into a hypergeometric equation?
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I would like to know if this differential equation can be transformed into the hypergeometric differential equation
$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$
ordinary-differential-equations hypergeometric-function
asked Mar 1 at 15:39
J.DoeJ.Doe
62
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J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I would like to know if this differential equation can be transformed into the hypergeometric differential equation
$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$
ordinary-differential-equations hypergeometric-function
asked Mar 1 at 15:39
J.DoeJ.Doe
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
2
$begingroup$
Without $=$ in it, this is not an equation.
$endgroup$
– JJacquelin
Mar 1 at 15:51
$begingroup$
Sorry i missed the =0
$endgroup$
– J.Doe
Mar 2 at 11:47
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For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
$endgroup$
– doraemonpaul
Mar 3 at 15:41
add a comment |
$begingroup$
I would like to know if this differential equation can be transformed into the hypergeometric differential equation
$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$
ordinary-differential-equations hypergeometric-function
asked Mar 1 at 15:39
J.DoeJ.Doe
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I would like to know if this differential equation can be transformed into the hypergeometric differential equation
$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$
ordinary-differential-equations hypergeometric-function
ordinary-differential-equations hypergeometric-function
asked Mar 1 at 15:39
J.DoeJ.Doe
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 1 at 15:39
J.DoeJ.Doe
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Mar 2 at 11:45
J.Doe
asked Mar 1 at 15:39
J.DoeJ.Doe
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 1 at 15:39
J.DoeJ.Doe
62
asked Mar 1 at 15:39
J.DoeJ.Doe
62
62
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J.Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Without $=$ in it, this is not an equation.
$endgroup$
– JJacquelin
Mar 1 at 15:51
$begingroup$
Sorry i missed the =0
$endgroup$
– J.Doe
Mar 2 at 11:47
$begingroup$
For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
$endgroup$
– doraemonpaul
Mar 3 at 15:41
add a comment |
2
$begingroup$
Without $=$ in it, this is not an equation.
$endgroup$
– JJacquelin
Mar 1 at 15:51
$begingroup$
Sorry i missed the =0
$endgroup$
– J.Doe
Mar 2 at 11:47
$begingroup$
For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
$endgroup$
– doraemonpaul
Mar 3 at 15:41
2
2
$begingroup$
Without $=$ in it, this is not an equation.
$endgroup$
– JJacquelin
Mar 1 at 15:51
$begingroup$
Without $=$ in it, this is not an equation.
$endgroup$
– JJacquelin
Mar 1 at 15:51
$begingroup$
Sorry i missed the =0
$endgroup$
– J.Doe
Mar 2 at 11:47
$begingroup$
Sorry i missed the =0
$endgroup$
– J.Doe
Mar 2 at 11:47
$begingroup$
For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
$endgroup$
– doraemonpaul
Mar 3 at 15:41
$begingroup$
For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
$endgroup$
– doraemonpaul
Mar 3 at 15:41
add a comment |
2 Answers
2
active
oldest
votes
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Hint:
$4(u-1)u((u-1)uvarphi1''(u)+(u-2)varphi1'(u))+varphi1(u)((u-1)uomega^2-u(u+4)+8)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+dfrac{(omega^2-1)u(u-1)-5u+8}{4u(u-1)}varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{5}{4(u-1)}+dfrac{2}{u(u-1)}right)varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{2}{u}+dfrac{3}{4(u-1)}right)varphi1(u)=0$
$varphi1''(u)+dfrac{u-2}{u(u-1)}varphi1'(u)+left(dfrac{omega^2-1}{4u(u-1)}-dfrac{2}{u^2(u-1)}+dfrac{3}{4u(u-1)^2}right)varphi1(u)=0$
$varphi1''(u)+left(dfrac{2}{u}-dfrac{1}{u-1}right)varphi1'(u)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)varphi1(u)=0$
Let $varphi1=u^a(u-1)^bv$ ,
Then $dfrac{dvarphi1}{du}=u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)v$
$dfrac{d^2varphi1}{du^2}=u^a(u-1)^bdfrac{d^2v}{du^2}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v=u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v$
$therefore u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)u^a(u-1)^bv=0$
$dfrac{d^2v}{du^2}+2left(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{dv}{du}+left(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2a}{u}+dfrac{2b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)dfrac{dv}{du}+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{a}{u}+dfrac{b}{u-1}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2a}{u^2}-dfrac{a-2b}{u(u-1)}+dfrac{b}{(u-1)^2}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a^2+a+2}{u^2}+dfrac{8ab-4a+8b+omega^2-12}{4u(u-1)}+dfrac{4b^2+3}{4(u-1)^2}right)v=0$
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
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$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
add a comment |
$begingroup$
$$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$$
HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind :
$$varphi(u)=u^a(u-1)^bF(u)$$
where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
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– J.Doe
2 days ago
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Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
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– J.Doe
2 days ago
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$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
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– JJacquelin
2 days ago
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In this case $varphi1(u)$ is a scalar field an has that kind of expansion
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– J.Doe
21 hours ago
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Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
$endgroup$
– JJacquelin
19 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
Hint:
$4(u-1)u((u-1)uvarphi1''(u)+(u-2)varphi1'(u))+varphi1(u)((u-1)uomega^2-u(u+4)+8)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+dfrac{(omega^2-1)u(u-1)-5u+8}{4u(u-1)}varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{5}{4(u-1)}+dfrac{2}{u(u-1)}right)varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{2}{u}+dfrac{3}{4(u-1)}right)varphi1(u)=0$
$varphi1''(u)+dfrac{u-2}{u(u-1)}varphi1'(u)+left(dfrac{omega^2-1}{4u(u-1)}-dfrac{2}{u^2(u-1)}+dfrac{3}{4u(u-1)^2}right)varphi1(u)=0$
$varphi1''(u)+left(dfrac{2}{u}-dfrac{1}{u-1}right)varphi1'(u)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)varphi1(u)=0$
Let $varphi1=u^a(u-1)^bv$ ,
Then $dfrac{dvarphi1}{du}=u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)v$
$dfrac{d^2varphi1}{du^2}=u^a(u-1)^bdfrac{d^2v}{du^2}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v=u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v$
$therefore u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)u^a(u-1)^bv=0$
$dfrac{d^2v}{du^2}+2left(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{dv}{du}+left(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2a}{u}+dfrac{2b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)dfrac{dv}{du}+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{a}{u}+dfrac{b}{u-1}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2a}{u^2}-dfrac{a-2b}{u(u-1)}+dfrac{b}{(u-1)^2}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a^2+a+2}{u^2}+dfrac{8ab-4a+8b+omega^2-12}{4u(u-1)}+dfrac{4b^2+3}{4(u-1)^2}right)v=0$
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
add a comment |
$begingroup$
Hint:
$4(u-1)u((u-1)uvarphi1''(u)+(u-2)varphi1'(u))+varphi1(u)((u-1)uomega^2-u(u+4)+8)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+dfrac{(omega^2-1)u(u-1)-5u+8}{4u(u-1)}varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{5}{4(u-1)}+dfrac{2}{u(u-1)}right)varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{2}{u}+dfrac{3}{4(u-1)}right)varphi1(u)=0$
$varphi1''(u)+dfrac{u-2}{u(u-1)}varphi1'(u)+left(dfrac{omega^2-1}{4u(u-1)}-dfrac{2}{u^2(u-1)}+dfrac{3}{4u(u-1)^2}right)varphi1(u)=0$
$varphi1''(u)+left(dfrac{2}{u}-dfrac{1}{u-1}right)varphi1'(u)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)varphi1(u)=0$
Let $varphi1=u^a(u-1)^bv$ ,
Then $dfrac{dvarphi1}{du}=u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)v$
$dfrac{d^2varphi1}{du^2}=u^a(u-1)^bdfrac{d^2v}{du^2}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v=u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v$
$therefore u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)u^a(u-1)^bv=0$
$dfrac{d^2v}{du^2}+2left(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{dv}{du}+left(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2a}{u}+dfrac{2b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)dfrac{dv}{du}+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{a}{u}+dfrac{b}{u-1}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2a}{u^2}-dfrac{a-2b}{u(u-1)}+dfrac{b}{(u-1)^2}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a^2+a+2}{u^2}+dfrac{8ab-4a+8b+omega^2-12}{4u(u-1)}+dfrac{4b^2+3}{4(u-1)^2}right)v=0$
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
add a comment |
$begingroup$
Hint:
$4(u-1)u((u-1)uvarphi1''(u)+(u-2)varphi1'(u))+varphi1(u)((u-1)uomega^2-u(u+4)+8)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+dfrac{(omega^2-1)u(u-1)-5u+8}{4u(u-1)}varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{5}{4(u-1)}+dfrac{2}{u(u-1)}right)varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{2}{u}+dfrac{3}{4(u-1)}right)varphi1(u)=0$
$varphi1''(u)+dfrac{u-2}{u(u-1)}varphi1'(u)+left(dfrac{omega^2-1}{4u(u-1)}-dfrac{2}{u^2(u-1)}+dfrac{3}{4u(u-1)^2}right)varphi1(u)=0$
$varphi1''(u)+left(dfrac{2}{u}-dfrac{1}{u-1}right)varphi1'(u)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)varphi1(u)=0$
Let $varphi1=u^a(u-1)^bv$ ,
Then $dfrac{dvarphi1}{du}=u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)v$
$dfrac{d^2varphi1}{du^2}=u^a(u-1)^bdfrac{d^2v}{du^2}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v=u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v$
$therefore u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)u^a(u-1)^bv=0$
$dfrac{d^2v}{du^2}+2left(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{dv}{du}+left(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2a}{u}+dfrac{2b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)dfrac{dv}{du}+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{a}{u}+dfrac{b}{u-1}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2a}{u^2}-dfrac{a-2b}{u(u-1)}+dfrac{b}{(u-1)^2}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a^2+a+2}{u^2}+dfrac{8ab-4a+8b+omega^2-12}{4u(u-1)}+dfrac{4b^2+3}{4(u-1)^2}right)v=0$
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
Hint:
$4(u-1)u((u-1)uvarphi1''(u)+(u-2)varphi1'(u))+varphi1(u)((u-1)uomega^2-u(u+4)+8)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+dfrac{(omega^2-1)u(u-1)-5u+8}{4u(u-1)}varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{5}{4(u-1)}+dfrac{2}{u(u-1)}right)varphi1(u)=0$
$u(u-1)varphi1''(u)+(u-2)varphi1'(u)+left(dfrac{omega^2-1}{4}-dfrac{2}{u}+dfrac{3}{4(u-1)}right)varphi1(u)=0$
$varphi1''(u)+dfrac{u-2}{u(u-1)}varphi1'(u)+left(dfrac{omega^2-1}{4u(u-1)}-dfrac{2}{u^2(u-1)}+dfrac{3}{4u(u-1)^2}right)varphi1(u)=0$
$varphi1''(u)+left(dfrac{2}{u}-dfrac{1}{u-1}right)varphi1'(u)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)varphi1(u)=0$
Let $varphi1=u^a(u-1)^bv$ ,
Then $dfrac{dvarphi1}{du}=u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)v$
$dfrac{d^2varphi1}{du^2}=u^a(u-1)^bdfrac{d^2v}{du^2}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v=u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v$
$therefore u^a(u-1)^bdfrac{d^2v}{du^2}+2u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(u^a(u-1)^bdfrac{dv}{du}+u^a(u-1)^bleft(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)u^a(u-1)^bv=0$
$dfrac{d^2v}{du^2}+2left(dfrac{a}{u}+dfrac{b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{dv}{du}+left(dfrac{a}{u}+dfrac{b}{u-1}right)vright)+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2a}{u}+dfrac{2b}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2}{u}-dfrac{1}{u-1}right)dfrac{dv}{du}+left(dfrac{2}{u}-dfrac{1}{u-1}right)left(dfrac{a}{u}+dfrac{b}{u-1}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a(a-1)}{u^2}+dfrac{2ab}{u(u-1)}+dfrac{b(b-1)}{(u-1)^2}right)v+left(dfrac{2a}{u^2}-dfrac{a-2b}{u(u-1)}+dfrac{b}{(u-1)^2}right)v+left(dfrac{2}{u^2}+dfrac{omega^2-12}{4u(u-1)}+dfrac{3}{4(u-1)^2}right)v=0$
$dfrac{d^2v}{du^2}+left(dfrac{2(a+1)}{u}+dfrac{2b-1}{u-1}right)dfrac{dv}{du}+left(dfrac{a^2+a+2}{u^2}+dfrac{8ab-4a+8b+omega^2-12}{4u(u-1)}+dfrac{4b^2+3}{4(u-1)^2}right)v=0$
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
edited 14 hours ago
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
answered 2 days ago
doraemonpauldoraemonpaul
12.7k31660
12.7k31660
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
add a comment |
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
Good job. It can be factorized with $u^a(u-1)^b$. After that the point is to determine $a$ and $b$.
$endgroup$
– JJacquelin
2 days ago
add a comment |
$begingroup$
$$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$$
HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind :
$$varphi(u)=u^a(u-1)^bF(u)$$
where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
$endgroup$
– J.Doe
2 days ago
$begingroup$
Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
$endgroup$
– J.Doe
2 days ago
$begingroup$
$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
In this case $varphi1(u)$ is a scalar field an has that kind of expansion
$endgroup$
– J.Doe
21 hours ago
$begingroup$
Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
$endgroup$
– JJacquelin
19 hours ago
add a comment |
$begingroup$
$$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$$
HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind :
$$varphi(u)=u^a(u-1)^bF(u)$$
where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
$endgroup$
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
$endgroup$
– J.Doe
2 days ago
$begingroup$
Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
$endgroup$
– J.Doe
2 days ago
$begingroup$
$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
In this case $varphi1(u)$ is a scalar field an has that kind of expansion
$endgroup$
– J.Doe
21 hours ago
$begingroup$
Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
$endgroup$
– JJacquelin
19 hours ago
add a comment |
$begingroup$
$$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$$
HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind :
$$varphi(u)=u^a(u-1)^bF(u)$$
where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
$endgroup$
$$ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0$$
HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind :
$$varphi(u)=u^a(u-1)^bF(u)$$
where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
edited 2 days ago
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
answered Mar 2 at 13:20
JJacquelinJJacquelin
44.4k21854
44.4k21854
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
$endgroup$
– J.Doe
2 days ago
$begingroup$
Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
$endgroup$
– J.Doe
2 days ago
$begingroup$
$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
In this case $varphi1(u)$ is a scalar field an has that kind of expansion
$endgroup$
– J.Doe
21 hours ago
$begingroup$
Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
$endgroup$
– JJacquelin
19 hours ago
add a comment |
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
$endgroup$
– J.Doe
2 days ago
$begingroup$
Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
$endgroup$
– J.Doe
2 days ago
$begingroup$
$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
$endgroup$
– JJacquelin
2 days ago
$begingroup$
In this case $varphi1(u)$ is a scalar field an has that kind of expansion
$endgroup$
– J.Doe
21 hours ago
$begingroup$
Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
$endgroup$
– JJacquelin
19 hours ago
$begingroup$
The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
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– J.Doe
2 days ago
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The equation is right. Some other ways to express it $ 4 (u-1) u left((u-1) u text{$varphi $1}''(u)+(u-2) text{$varphi $1}'(u)right)+text{$varphi $1}(u) left((u-1) u omega ^2-u (u+4)+8right)=0 $ $ 4 u^4 text{$varphi $1}''(u)-8 u^3 text{$varphi $1}''(u)+4 u^3 text{$varphi $1}'(u)+4 u^2 text{$varphi $1}''(u)-12 u^2 text{$varphi $1}'(u)+u^2 omega ^2 text{$varphi $1}(u)-u^2 text{$varphi $1}(u)+8 u text{$varphi $1}'(u)-u omega ^2 text{$varphi $1}(u)-4 u text{$varphi $1}(u)+8 text{$varphi $1}(u)=0 $
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– J.Doe
2 days ago
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Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
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– J.Doe
2 days ago
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Other thing i can use and may be useful is the expansion of $varphi$: $ varphi1(u)=sqrt{1-u} y(u)$
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– J.Doe
2 days ago
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$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
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– JJacquelin
2 days ago
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$varphi1(u)=sqrt{1-u}y(u)$ is a too restrictive form for the change of function. This is a particular case of $varphi1(u)=u^a(1-u)^bF(u)$ with $a=0$ and $b=1/2$. This is much too specific to expect transform the original equation into an hypergeometric ODE.
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– JJacquelin
2 days ago
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In this case $varphi1(u)$ is a scalar field an has that kind of expansion
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– J.Doe
21 hours ago
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In this case $varphi1(u)$ is a scalar field an has that kind of expansion
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– J.Doe
21 hours ago
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Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
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– JJacquelin
19 hours ago
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Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck.
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– JJacquelin
19 hours ago
add a comment |
J.Doe is a new contributor. Be nice, and check out our Code of Conduct.
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Without $=$ in it, this is not an equation.
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– JJacquelin
Mar 1 at 15:51
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Sorry i missed the =0
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– J.Doe
Mar 2 at 11:47
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For some advanced cases, we may need some advanced approach for example shown in math.stackexchange.com/questions/2377289/….
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– doraemonpaul
Mar 3 at 15:41