For the differential equation \(x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=0\) Which of the following statements is true ?
\(x = 0\) is a regular singular point, while the point at infinity is an irregular singular point
This is Bessel's equation of order \(n\). Dividing by \(x^{2}\) puts it in the standard form \(y''+P(x)y'+Q(x)y=0\) with \(P(x)=\frac{1}{x}\) and \(Q(x)=1-\frac{n^{2}}{x^{2}}\).
At \(x=0\) both \(P\) and \(Q\) blow up, so the origin is not an ordinary point. However \(xP(x)=1\) and \(x^{2}Q(x)=x^{2}-n^{2}\) are analytic there, so \(x=0\) is a regular singular point.
To study the point at infinity put \(x=1/t\). Then \(\frac{dy}{dx}=-t^{2}\frac{dy}{dt}\) and \(\frac{d^{2}y}{dx^{2}}=t^{4}\frac{d^{2}y}{dt^{2}}+2t^{3}\frac{dy}{dt}\).
Substituting, \(x^{2}y_{xx}=t^{2}y_{tt}+2ty_{t}\) and \(xy_{x}=-ty_{t}\), so the equation becomes \(t^{2}y_{tt}+ty_{t}+\left(\frac{1}{t^{2}}-n^{2}\right)y=0\), that is \(y_{tt}+\frac{1}{t}y_{t}+\left(\frac{1}{t^{4}}-\frac{n^{2}}{t^{2}}\right)y=0\).
At \(t=0\) we get \(t\cdot\frac{1}{t}=1\) (analytic) but \(t^{2}\left(\frac{1}{t^{4}}-\frac{n^{2}}{t^{2}}\right)=\frac{1}{t^{2}}-n^{2}\), which diverges. So the point at infinity is an irregular singular point.
Check: for large \(x\) the Bessel functions behave like \(\cos(x-\phi)/\sqrt{x}\), an oscillation that cannot be written as a Frobenius power series in \(1/x\), which is exactly the behaviour of an irregular singular point.
Hence, \(x = 0\) is a regular singular point, while the point at infinity is an irregular singular point.
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