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consider the following second order ordinary differential equation:

$$\sinh^2(x)u''(x)+\cosh(x)\sinh(x)u'(x)+(\lambda-\sinh^2(x))u(x)=0,$$

where $\lambda$ is some constant ($\lambda\in\mathbb{C}$, or $\lambda\in\mathbb{R}$ if you want).

I'm wondering how to find the solutions to this equation. I tried to find solutions by power series expansion but the calculations become very complicated. Does anyone know a solution to this equation? Maybe by a tricky transformation…

Any help will be very much appreciated.

Best wishes

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  • $\begingroup$ Try let $t=\sinh(x)$ $\endgroup$ – doraemonpaul Feb 23 '15 at 15:34
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Hint:

Let $r=\sinh(x)$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dr}\dfrac{dr}{dx}=\cosh(x)\dfrac{du}{dr}$

$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(\cosh(x)\dfrac{du}{dr}\right)=\cosh(x)\dfrac{d}{dx}\left(\dfrac{du}{dr}\right)+\sinh(x)\dfrac{du}{dr}=\cosh(x)\dfrac{d}{dr}\left(\dfrac{du}{dr}\right)\dfrac{dr}{dx}+\sinh(x)\dfrac{du}{dr}=\cosh(x)\dfrac{d^2u}{dr^2}\cosh(x)+\sinh(x)\dfrac{du}{dr}=\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)\dfrac{du}{dr}$

$\therefore\sinh^2(x)\left(\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)\dfrac{du}{dr}\right)+\cosh^2(x)\sinh(x)\dfrac{du}{dr}+(\lambda-\sinh^2(x))u=0$

$\sinh^2(x)\cosh^2(x)\dfrac{d^2u}{dr^2}+\sinh(x)(\sinh^2(x)+\cosh^2(x))\dfrac{du}{dr}+(\lambda-\sinh^2(x))u=0$

$r^2(r^2+1)\dfrac{d^2u}{dr^2}+r(2r^2+1)\dfrac{du}{dr}+(\lambda-r^2)u=0$

Let $s=r^2$ ,

Then $\dfrac{du}{dr}=\dfrac{du}{ds}\dfrac{ds}{dr}=2r\dfrac{du}{ds}$

$\dfrac{d^2u}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{du}{ds}\right)=2r\dfrac{d}{dr}\left(\dfrac{du}{ds}\right)+2\dfrac{du}{ds}=2r\dfrac{d}{ds}\left(\dfrac{du}{ds}\right)\dfrac{ds}{dr}+2\dfrac{du}{ds}=2r\dfrac{d^2u}{ds^2}2r+2\dfrac{du}{ds}=4r^2\dfrac{d^2u}{ds^2}+2\dfrac{du}{ds}$

$\therefore r^2(r^2+1)\left(4r^2\dfrac{d^2u}{ds^2}+2\dfrac{du}{ds}\right)+2r^2(2r^2+1)\dfrac{du}{ds}+(\lambda-r^2)u=0$

$4r^4(r^2+1)\dfrac{d^2u}{ds^2}+2r^2(3r^2+2)\dfrac{du}{ds}+(\lambda-r^2)u=0$

$4s^2(s+1)\dfrac{d^2u}{ds^2}+2s(3s+2)\dfrac{du}{ds}+(\lambda-s)u=0$

$\dfrac{d^2u}{ds^2}+\dfrac{3s+2}{2s(s+1)}\dfrac{du}{ds}+\dfrac{\lambda-s}{4s^2(s+1)}u=0$

$\dfrac{d^2u}{ds^2}+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{du}{ds}+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)u=0$

Let $u=s^av$ ,

Then $\dfrac{du}{ds}=s^a\dfrac{dv}{ds}+as^{a-1}v$

$\dfrac{d^2u}{ds^2}=s^a\dfrac{d^2v}{ds^2}+as^{a-1}\dfrac{dv}{ds}+as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v=s^a\dfrac{d^2v}{ds^2}+2as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v$

$\therefore s^a\dfrac{d^2v}{ds^2}+2as^{a-1}\dfrac{dv}{ds}+a(a-1)s^{a-2}v+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\left(s^a\dfrac{dv}{ds}+as^{a-1}v\right)+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)s^av=0$

$\dfrac{d^2v}{ds^2}+\dfrac{2a}{s}\dfrac{dv}{ds}+\dfrac{a(a-1)}{s^2}v+\left(\dfrac{1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\left(\dfrac{a}{s^2}+\dfrac{a}{2s(s+1)}\right)v+\left(\dfrac{\lambda}{4s^2}-\dfrac{\lambda+1}{4s(s+1)}\right)v=0$

$\dfrac{d^2v}{ds^2}+\left(\dfrac{2a+1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\left(\dfrac{4a^2+\lambda}{4s^2}+\dfrac{2a-\lambda-1}{4s(s+1)}\right)v=0$

Choose $4a^2+\lambda=0$ , i.e. $a=\pm i\dfrac{\sqrt\lambda}{2}$ , the ODE becomes

$\dfrac{d^2v}{ds^2}+\left(\dfrac{\pm i\sqrt\lambda+1}{s}+\dfrac{1}{2(s+1)}\right)\dfrac{dv}{ds}+\dfrac{\pm i\sqrt\lambda-\lambda-1}{4s(s+1)}v=0$

Which reduces to Gaussian hypergeometric equation.

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