Context
This question is related to the unsolved questions in [2].
I want to solve the Helmholtz equation in spherical coordinates by separation of variables. It is typical to see the differential equation written in terms of an integer $n$ as, $$x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + \left(x^2 - n(n + 1)\right) y = 0.$$ Now, I know that,
``The two linearly independent solutions to this equation are called the spherical Bessel functions, and are related to the ordinary Bessel functions $J_n$ and $Y_n$ by \begin{align} j_n(x) &= \sqrt{\frac{\pi}{2x}} J_{n+\frac{1}{2}}(x), \\ y_n(x) &= \sqrt{\frac{\pi}{2x}} Y_{n+\frac{1}{2}}(x). [1]'' \end{align}
Question
I want to solve the Helmholtz equation in spherical coordinates too. However, I want to solve the equation terms of an unknown number $\lambda$ that is not necessarily of integer (or half-integer value). What are eigenfunctions and eigenvalue of the Equation 1? $$x^2 \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} + \left(x^2 - \lambda\right) y = 0.\tag{1}$$
My second attempt
In my second attempt, I use the integral form of the bessel-function of the first king. This seems be bare better fruit.
$$J_\alpha(x) = \frac{1}{\pi} \int_0^\pi \cos(\alpha\tau - x \sin\tau)\,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty e^{-x \sinh t - \alpha t} \, dt.$$
$$j_\alpha(x) = c\,x^{-1/2} \left[\int_0^\pi \cos(\alpha\tau - x \sin\tau)\,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty e^{-x \sinh t - \alpha t} \, dt.\right]$$ I take the partial for the first time with the result being \begin{align*} \frac{\partial{j_\alpha(x)} }{\partial{x}} &= -\frac{c}{2}\, x^{-3/2} \left[\int_0^\pi \cos(\alpha\tau - x \sin\tau)\,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty e^{-x \sinh t - \alpha t} \, dt.\right] \\ &+ c\,x^{-1/2} \left[ \int_0^\pi \sin\tau\, \sin(\alpha\tau - x \sin\tau) \,d\tau + \frac{\sin \alpha\pi}{\pi} \int_0^\infty \,\sinh(t) \, e^{-x \sinh( t) - \alpha t} \, dt\right]. \end{align*} I take the partial for a second time with the result being \begin{align*} \frac{\partial^2{j_\alpha(x)} }{\partial{x}^2} &= -\frac{c}{2}\, x^{-3/2} \left[ \int_0^\pi \sin(\tau)\,\sin(\alpha\tau - x \sin\tau) \,d\tau + \frac{\sin \alpha\pi}{\pi} \, \int_0^\infty \sinh(t)\,e^{-x \sinh t - \alpha t} \, dt. \right] \\ &+ \frac{3\,c}{4}\, x^{-5/2} \left[ \int_0^\pi \cos(\alpha\tau - x \sin\tau) \,d\tau - \frac{\sin \alpha\pi}{\pi} \, \int_0^\infty e^{-x \sinh t - \alpha t} \, dt. \right] \\ &+ c\,x^{-1/2} \left[ - \int_0^\pi \sin^2\tau\, \cos(\alpha\tau - x \sin\tau) \,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty \,\sinh^2(t) \, e^{-x \sinh( t) - \alpha t} \,dt \right]. \\ &- \frac{c}{2}\,x^{-3/2} \left[ \int_0^\pi \sin\tau\, \sin(\alpha\tau - x \sin\tau) \,d\tau + \frac{\sin \alpha\pi}{\pi} \int_0^\infty \,\sinh(t) \, e^{-x \sinh( t) - \alpha t} \,dt \right]. \end{align*}
\begin{align*} 0 &= x^2\, \frac{\partial^2{ u{(x)} }}{\partial{x}^2} + 2\,x \, \frac{\partial{ u{(x)} }}{\partial{x}} + \left[ x^2 + \lambda\right]\,u {(x)} \end{align*} Thus, upon re-organizing some more allows me to rewrite as \begin{align*} 0 &= x^{+3/2} \left[ \int_0^\pi \cos^2(\tau ) \, \cos(\alpha\tau - x \sin\tau) \,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty \, \cosh^2(t) \, e^{-x \sinh( t) - \alpha t} \,dt \right]. \\ &+ x^{1/2} \left[ \int_0^\pi \sin(\tau)\,\sin(\alpha\tau - x \sin\tau) \,d\tau + \frac{\sin \alpha\pi}{\pi} \, \int_0^\infty \sinh(t)\,e^{-x \sinh t - \alpha t} \, dt. \right] \\ &+ \left[\lambda - \frac{1 }{4}\right]\, x^{-1/2} \left[ \int_0^\pi \cos(\alpha\tau - x \sin\tau) \,d\tau - \frac{\sin \alpha\pi}{\pi} \, \int_0^\infty e^{-x \sinh t - \alpha t} \, dt. \right] \end{align*} It is likely that if I integrate by parts twice sequentially, and the second integral by parts once, then I will have solved this problem. But I am out of time for now.
My first attempt
Going off of what I have understood from above, I try the following test function for $y$: $$y = c\,x^{-1/2}\, J_{n+\frac{1}{2}}(x) $$
\begin{align*} 0 &= x^2\, \frac{\partial^2{ \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] }}{\partial{x}^2} \\ &+ 2\,x \, \frac{\partial{ \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] }}{\partial{x}} \\ &+ \left[ x^2 + \lambda\right]\,c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha} \end{align*} Continuing \begin{align*} 0 &= +\frac{-1}{2}x^2\, \frac{\partial { \left[c\,x^{-3/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] }}{\partial{x} } \\ &+ x^2\, \frac{\partial { \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha-1}\right] }}{\partial{x} } \\ &+ \frac{-1}{2}\,2\,x \, \left[c\,x^{-3/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ 2\,x \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(2m+\alpha)}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha-1}\right] \\ &+ \left[ x^2 + \lambda\right]\,c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha} \end{align*} Contiuing \begin{align*} 0 &= \frac{-1}{2}\frac{-3}{2}x^2\, \left[c\,x^{-5/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ \frac{-1}{2}x^2\, \left[c\,x^{-3/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ x^2\,\frac{-1}{2} \left[c\,x^{-3/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha-1}\right] \\ &+ x^2\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha-1}{2}\,\frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha-1}\right] \\ &+ \frac{-1}{2}\,2\,x \, \left[c\,x^{-3/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ 2\,x \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(2m+\alpha)}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha-1}\right] \\ &+ \left[ x^2 + \lambda\right]\,\left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \end{align*} Simplifying: \begin{align*} 0 &= \frac{ 3}{4} \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ \frac{-1}{2}x\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ - \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ 2\,x \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha-1}{2}\,\frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha }\right] \\ &+ \frac{-1}{2}\,2 \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ 4\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(2m+\alpha)}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha }\right] \\ &+ \left[ x^2 + \lambda\right]\,\left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \end{align*} Simplifying more: \begin{align*} 0 &= \left[ x^2 + \lambda - \frac{ 1}{4} \right]\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ \left[3- \frac{ 1}{2}x\right]\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha}{2}\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ 2\,x \, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{2m+\alpha-1}{2}\,\frac{2m+\alpha}{2}\,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha }\right] \end{align*} Simplifying more more: \begin{align*} 0 &= \left[ x^2 + \lambda - \frac{ 1}{4} \right]\, \left[c\,x^{-1/2}\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ \left[ \frac{4- x}{4}\right]\, \left[c\,x^{-1/2}\sum_{m=0}^\infty (2m+\alpha)\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}\right] \\ &+ \frac{1}{2}\,x \, \left[c\,x^{-1/2}\sum_{m=0}^\infty (2m+\alpha ) \, (2m+\alpha) \,\frac{(-1)^m}{m! \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha }\right] \end{align*}
I have this $x^2$ term that I would like to cancel out. If I could do that, then I would have the eigenvalues. I have tried a few approaches without avail. I am not even sure if this effort would bare fruit.
Looking for a bit of help.
Bibliography
[1] Wikipedia contributors. (2021, February 7). Bessel function. In Wikipedia, The Free Encyclopedia. Retrieved 23:05, February 7, 2021, from https://en.wikipedia.org/w/index.php?title=Bessel_function&oldid=1005472771