1

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 $1}(u) \left((u-1) u \omega ^2-u (u+4)+8\right)=0$

J.Doe
  • 11

2 Answers2

1

$$ 4 (u-1) u \left((u-1) u \text{$\varphi $1}''(u)+(u-2) \text{$\varphi $1}'(u)\right)+\text{$\varphi $1}(u) \left((u-1) u \omega ^2-u (u+4)+8\right)=0$$ 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.

JJacquelin
  • 68,401
  • 4
  • 40
  • 91
  • 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)+8\right)=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 $ – J.Doe Mar 05 '19 at 12:32
  • Other thing i can use and may be useful is the expansion of $\varphi$: $ \varphi1(u)=\sqrt{1-u} y(u)$ – J.Doe Mar 05 '19 at 12:40
  • $\varphi1(u)=\sqrt{1-u}y(u)$ 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. – JJacquelin Mar 05 '19 at 14:10
  • In this case $\varphi1(u)$ is a scalar field an has that kind of expansion – J.Doe Mar 07 '19 at 09:46
  • Well. I give you an hint. Follow it as you like. If you have a better method, that up to you. Good luck. – JJacquelin Mar 07 '19 at 11:28
  • That is only the asymptotic expansion for a scalar field at the lowest order in anti De Sitter spacetime. – J.Doe Mar 08 '19 at 12:33
1

Hint:

$4(u-1)u((u-1)u\varphi1''(u)+(u-2)\varphi1'(u))+\varphi1(u)((u-1)u\omega^2-u(u+4)+8)=0$

$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\dfrac{(\omega^2-1)u(u-1)-5u+8}{4u(u-1)}\varphi1(u)=0$

$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\left(\dfrac{\omega^2-1}{4}-\dfrac{5}{4(u-1)}+\dfrac{2}{u(u-1)}\right)\varphi1(u)=0$

$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\left(\dfrac{\omega^2-1}{4}-\dfrac{2}{u}+\dfrac{3}{4(u-1)}\right)\varphi1(u)=0$

$\varphi1''(u)+\dfrac{u-2}{u(u-1)}\varphi1'(u)+\left(\dfrac{\omega^2-1}{4u(u-1)}-\dfrac{2}{u^2(u-1)}+\dfrac{3}{4u(u-1)^2}\right)\varphi1(u)=0$

$\varphi1''(u)+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\varphi1'(u)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)\varphi1(u)=0$

Let $\varphi1=u^a(u-1)^bv$ ,

Then $\dfrac{d\varphi1}{du}=u^a(u-1)^b\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v$

$\dfrac{d^2\varphi1}{du^2}=u^a(u-1)^b\dfrac{d^2v}{du^2}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v=u^a(u-1)^b\dfrac{d^2v}{du^2}+2u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v$

$\therefore u^a(u-1)^b\dfrac{d^2v}{du^2}+2u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(u^a(u-1)^b\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v\right)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)u^a(u-1)^bv=0$

$\dfrac{d^2v}{du^2}+2\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(\dfrac{dv}{du}+\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v\right)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$

$\dfrac{d^2v}{du^2}+\left(\dfrac{2a}{u}+\dfrac{2b}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$

$\dfrac{d^2v}{du^2}+\left(\dfrac{2(a+1)}{u}+\dfrac{2b-1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2a}{u^2}-\dfrac{a-2b}{u(u-1)}+\dfrac{b}{(u-1)^2}\right)v+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$

$\dfrac{d^2v}{du^2}+\left(\dfrac{2(a+1)}{u}+\dfrac{2b-1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a^2+a+2}{u^2}+\dfrac{8ab-4a+8b+\omega^2-12}{4u(u-1)}+\dfrac{4b^2+3}{4(u-1)^2}\right)v=0$

Choose $a^2+a+2=0$ and $4b^2+3=0$ , and the ODE reduces to Gaussian hypergeometric equation.

doraemonpaul
  • 16,488