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How do I solve the following differential equation?

$\partial^{2}_{x} \left[x^{2}p\right] + \partial_{x} \left[\left(x-1\right)p\right] = 0$

I tried a Fourier transform which leads to

$\left[k^{2}\partial^{2}_{k} + k\left(\partial_{k}+i\right)\right]\tilde{p} = 0$

where $\tilde{p}$ is the Fourier transform of $p$ but that doesn't really help.

Any ideas?

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It's like a Cauchy Euler problem, modulo that pesky $i$ term. I think you might be able to solve via series method on the Fourier transform. –  James S. Cook Oct 12 '12 at 10:50
    
Expand it first. –  doraemonpaul Oct 12 '12 at 13:58
    
Hm, if I expand $p$ into a power series, I get a problem with the radius of convergence. –  Aton Oct 12 '12 at 15:28
    
Would this question be suitable for mathoverflow? –  Aton Oct 15 '12 at 9:22
    
Okay, technically the above differential equation has an irregular singularity at x=0. That is why e.g. the Frobenius method does not work. Any ideas how to proceed from here? –  Aton Oct 15 '12 at 13:29
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1 Answer

$\partial_x^2[x^2p]+\partial_x[(x-1)p]=0$

$\partial_x[x^2p_x+2xp]+(x-1)p_x+p=0$

$x^2p_{xx}+2xp_x+2xp_x+2p+(x-1)p_x+p=0$

$x^2p_{xx}+(5x-1)p_x+3p=0$

$x^2\dfrac{d^2p}{dx^2}+(5x-1)\dfrac{dp}{dx}+3p=0$

This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0217.pdf

Let $\begin{cases}x=\dfrac{1}{u}\\p=u^3e^uy\end{cases}$ ,

Then $\dfrac{dp}{dx}=\dfrac{\dfrac{dp}{du}}{\dfrac{dx}{du}}=\dfrac{u^3e^u\dfrac{dy}{du}+(u^3+3u^2)e^uy}{-\dfrac{1}{u^2}}=-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy$

$\dfrac{d^2p}{dx^2}=\dfrac{d}{dx}\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)=\dfrac{\dfrac{d}{du}\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)}{\dfrac{dx}{du}}=\dfrac{-u^5e^u\dfrac{d^2y}{du^2}-(2u^5+8u^4)e^u\dfrac{dy}{du}-(u^5+8u^4+12u^3)e^uy}{-\dfrac{1}{u^2}}=u^7e^u\dfrac{d^2y}{du^2}+2(u^7+3u^6)e^u\dfrac{dy}{du}+(u^7+8u^6+12u^5)e^uy$

$\therefore\dfrac{1}{u^2}\left(u^7e^u\dfrac{d^2y}{du^2}+(2u^7+8u^6)e^u\dfrac{dy}{du}+(u^7+8u^6+12u^5)e^uy\right)+\left(\dfrac{5}{u}-1\right)\left(-u^5e^u\dfrac{dy}{du}-(u^5+3u^4)e^uy\right)+3u^3e^uy=0$

$u^5e^u\dfrac{d^2y}{du^2}+(2u^5+8u^4)e^u\dfrac{dy}{du}+(u^5+8u^4+12u^3)e^uy-5u^4e^u\dfrac{dy}{du}-(5u^4+15u^3)e^uy+u^5e^u\dfrac{dy}{du}+(u^5+3u^4)e^uy+3u^3e^uy=0$

$u^5e^u\dfrac{d^2y}{du^2}+(3u^5+3u^4)e^u\dfrac{dy}{du}+(2u^5+6u^4)e^uy=0$

$u\dfrac{d^2y}{du^2}+(3u+3)\dfrac{dy}{du}+(2u+6)y=0$

Obviously, $y=e^{-2u}$ is a particular solution.

Let $y=e^{-2u}v$ ,

Then $\dfrac{dy}{du}=e^{-2u}\dfrac{dv}{du}-2e^{-2u}v$

$\dfrac{d^2y}{du^2}=e^{-2u}\dfrac{d^2v}{du^2}-4e^{-2u}\dfrac{dv}{du}+4e^{-2u}v$

$\therefore u\left(e^{-2u}\dfrac{d^2v}{du^2}-4e^{-2u}\dfrac{dv}{du}+4e^{-2u}v\right)+(3u+3)\left(e^{-2u}\dfrac{dv}{du}-2e^{-2u}v\right)+(2u+6)e^{-2u}v=0$

$u\left(\dfrac{d^2v}{du^2}-4\dfrac{dv}{du}+4v\right)+(3u+3)\left(\dfrac{dv}{du}-2v\right)+(2u+6)v=0$

$u\dfrac{d^2v}{du^2}-4u\dfrac{dv}{du}+4uv+(3u+3)\dfrac{dv}{du}-(6u+6)v+(2u+6)v=0$

$u\dfrac{d^2v}{du^2}-(u-3)\dfrac{dv}{du}=0$

$u\dfrac{d^2v}{du^2}=(u-3)\dfrac{dv}{du}$

$\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}=1-\dfrac{3}{u}$

$\int\dfrac{\dfrac{d^2v}{du^2}}{\dfrac{dv}{du}}du=\int\left(1-\dfrac{3}{u}\right)du$

$\ln\dfrac{dv}{du}=u-3\ln u+c_1$

$\dfrac{dv}{du}=\dfrac{c_2e^u}{u^3}$

$v=\int\dfrac{c_2e^u}{u^3}du$

$y=e^{-2u}\int\dfrac{c_2e^u}{u^3}du$

$y=e^{-\frac{2}{x}}\int c_2x^3e^{\frac{1}{x}}d\left(\dfrac{1}{x}\right)$

$y=e^{-\frac{2}{x}}\int C_2xe^{\frac{1}{x}}dx$

$y=e^{-\frac{2}{x}}(C_2\int_k^xxe^{\frac{1}{x}}dx+C_1)$

$y=C_1e^{-\frac{2}{x}}+C_2e^{-\frac{2}{x}}\int_k^xxe^{\frac{1}{x}}dx$

$p=\dfrac{C_1e^{-\frac{1}{x}}}{x^3}+\dfrac{C_2e^{-\frac{1}{x}}}{x^3}\int_k^xxe^{\frac{1}{x}}dx$

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