$a^2x^2\frac{d^2y}{dx^2}+ax\frac{dy}{dx}-xy=0$

Question

Is there a solution for this ODE? If $a=1$, then this ODE reduces to the Modified Bessel Differential equation for which there is a solution in terms of special functions.

Answer 1

Let $x = t^k$ we have $$ \frac{d}{dx} = \frac{1}{kt^{k-1}}\frac{d}{dt}\\ \frac{d^2}{dx^2}= \frac{1}{k^2}t^{2 - 2k}\frac{d^2}{dt^2} + \frac{1 -k}{k^2}t^{1 - 2k}\frac{d}{dt} $$ we can sub into original equation $$ a^2t^{2k}\frac{1}{k^2}t^{2 - 2k}\frac{d^2y}{dt^2} + a^2t^{2k}\frac{1 -k}{k^2}t^{1 - 2k}\frac{dy}{dt} + at^k\frac{1}{kt^{k-1}}\frac{dy}{dt} - t^{k}y = 0 $$ we can simplify and collect terms $$ \frac{a^2}{k^2}t^2\ddot{y} + t\left[a^2\frac{1-k}{k^2} + \frac{a}{k}\right]\dot{y} - t^ky = 0 $$ if we set $k = 2$ we are closer to the Bessel ODE $$ t^2\ddot{y} + t\left[\frac{2}{a} - 1\right]\dot{y} - \frac{4}{a^2}t^2y = 0 $$ then we can use $y = t^mu(t)$ we have $$ \dot{y} = mt^{m-1}u + t^m\dot{u}\\ \ddot{y} = m(m-1)t^{m-2}u + 2mt^{m-1}\dot{u} + t^m\ddot{u} $$ we then arrive at $$ t^{m+2}\ddot{u} + \left[2m + \frac{2}{a} - 1\right]t^{m+1}\dot{u} + \left[m(m-1) + m\left(\frac{2}{a} - 1\right) - \frac{4}{a^2}t^2\right]t^mu $$ or $$ t^2\ddot{u} + \lambda_1 t\dot{u} + \left[\lambda_2 - \lambda_3t^2\right]u = 0 $$ where $$ \begin{eqnarray} \lambda_1 &=& 2m + \frac{2}{a} - 1\\ \lambda_2 &=& m(m-1) + m\left(\frac{2}{a} - 1\right) \\ \lambda_3 &=& \frac{4}{a^2} \end{eqnarray} $$ If we let $$ \lambda_1 = 2m + \frac{2}{a} - 1 = 1 \implies m = 1 - \frac{1}{a} $$ then we have $$ \lambda_2 = -\left(1-\frac{1}{a}\right)^2 = -\alpha^2 $$ $$ t^2\ddot{u} + t\dot{u} + \left[ - \lambda_3t^2 - \alpha^2\right]u = $$ now if we do a final mapping $t \to c\tau$ we have $$ \tau^2\frac{d^2u}{d\tau^2} + \tau\frac{du}{d\tau} + \left(-\lambda_3c^2\tau^2- \alpha^2\right)u = 0 $$ setting $c = \sqrt{\frac{1}{-\lambda_3}}$ we then have $$ \tau^2\frac{d^2u}{d\tau^2} + \tau\frac{du}{d\tau} + \left(\tau^2- \alpha^2\right)u = 0 $$ which is the bessel equation.

Answer 2

One can guesstimate the necessary substitution by the characteristic properties of the basis solutions of the modified Bessel equation:

• The roots of the indicial equation, or the characteristic equation for the Euler-Cauchy terms, have to be symmetric to zero. Here we have $0=a^2r(r-1)+ar=ar(ar-a+1)$. To shift the solutions into symmetry one has to change the dependent variable $y$ to $x^{-r/2}y(x)$ with the (in general) non-zero root $r=1-\frac1a$.

• The asymptotic behavior at infinity is in first approximation $s\mapsto e^{\pm s}$. Here however we get $x\mapsto e^{\pm\frac2{a}\sqrt{x}}$. So replace $x$ with the exponent $s=\frac2{a}\sqrt{x}\iff x=(as/2)^2$.

Together both transformations give $u(s)=s^{-r}y(a^2s^2/4)$, this function should satisfy a Bessel equation: \begin{align} u'(s)&=-rs^{-r-1}y(a^2s^2/4)+\frac12a^2s^{-r+1}y'(a^2s^2/4)\\ u''(s)&=r(r+1)s^{-r-2}y(x)-\frac12a^2(2r-1)s^ry'(x)+\frac14a^4s^{-r+2}y''(x) \\ &=s^{-r-2}\left(r(r+1)y(x)-2(2r-1)xy'(x)+4x^2y''(x)\right) \\\hline s^2u''(s)+su'(s)&=s^{-r}\left(r^2y(x)-4(r-1)xy'(x)+4x^2y''(x)\right) \\ &=s^{-r}\left(r^2y(x)+\frac4{a^2}xy(x)\right) \\ &=(r^2+s^2)u(s) \end{align}