Preface
This is a can you please check my work post. Maybe these are frowned upon. However, I have been working on this for a couple of days now. I can use some help.
Context
I am solving a physics problem on the subject of spherical multipole radiation. The solutions are of interest harmonic and tend to satisfy the Helmholtz equation in spherical coordinates.
As part of this problem I obtain the equation $$ f_{\pm}(\mathbf{r},t) = \exp{\left(\pm i k\,r \right)} \, \cos(\theta). \tag{1} $$ I wish to expand (1) using elementary solutions of the Helmholtz equation. Some scant background can be found in [1].
I write \begin{align} \exp{\left(\pm i k\,r \right)} \, \cos(\theta) = \sum_{\ell=0}^\infty\sum_{m=-\ell}^\ell \left[A_{\ell_\pm}\,j_{\ell}{(k\,r)} + B_{\ell_\pm}\,n_{\ell}{(k\,r)} \right]Y_{\ell}^m(\theta,\varphi),\qquad \tag{2} \end{align} where $j_\ell$ and $n_\ell$ are spherical Bessel functions of the first and second kind, respectively, and $Y_\ell^m$ is are the spherical harmonics.
Question
How can I determine the values of the coefficients?
My attempt
Well, I can rewrite the left-hand side in terms of one, and only one, spherical harmonic. I find that \begin{align} \exp{\left(\pm i k\,r \right)} \, 2\,\sqrt{\frac{\pi}{3}}\, Y_1^0(\theta,\varphi) = \sum_\ell^\infty\sum_{m=-\ell}^\ell \left[A_{\ell_\pm}\,j_{\ell}{(k\,r)} + B_{\ell_\pm}\,n_{\ell}{(k\,r)} \right]Y_{\ell}^m(\theta,\varphi), \end{align} Thus, the problem is substantially simplified: \begin{align} \exp{\left(\pm i k\,r \right)} \, 2\,\sqrt{\frac{\pi}{3}} = A_{1_\pm}\,j_{1}{(k\,r)} + B_{1_\pm}\,n_{1}{(k\,r)} . \end{align}
Moving on, I turn to Sturm-Liouvill theory. The Bessel's have an orthogonality condition and a normaliztion condition. From [2], I know that ``the spherical Bessel functions the orthogonality relation is: $$\int_0^\infty x^2 j_\alpha(ux) j_\alpha(vx) \,dx = \frac{\pi}{2u^2} \delta(u - v)$$ for $\alpha > −1$''. This is relevant to my solution. I write \begin{align} A_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) = 2\,\sqrt{\frac{\pi}{3}} \, \int_{0}^\infty r^2 \exp{\left(\pm i k\,r \right)} \, j_{1}{(k^\prime\,r)} \,dr, \\ B_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) = 2\,\sqrt{\frac{\pi}{3}} \, \int_{0}^\infty r^2 \exp{\left(\pm i k\,r \right)} \, n_{1}{(k^\prime\,r)} \,dr . \end{align} This seems all right to me. Next, I am aware that these particular spherical Bessel functions can be rewritten in terms of circular functions. I find that \begin{align} A_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= 2\,\sqrt{\frac{\pi}{3}} \, \int_{0}^\infty r^2 \exp{\left(\pm i k\,r \right)} \, \left[ \frac{\sin(k^\prime\,r)}{(k^\prime\,r)^2} - \frac{\cos(k^\prime\,r)}{(k^\prime\,r) } \right] \,dr, \\ B_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= 2\,\sqrt{\frac{\pi}{3}} \, \int_{0}^\infty r^2 \exp{\left(\pm i k\,r \right)} \, \left[ - \frac{\cos(k^\prime\,r)}{(k^\prime\,r)^2} - \frac{\sin(k^\prime\,r)}{(k^\prime\,r) } \right] \,dr . \end{align} Upon evaluation, I obtain that \begin{align} A_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= \frac{2\,{k^\prime}^3-\lim\limits_{r\rightarrow \infty }{\mathrm{e}}^{\pm i\,r\,k}\,\left( \left[ r \, {k^\prime}^2 \left( {k^\prime}^2 - k^2 \right) \pm i\,k\left( k^2 - 3\, {k^\prime}^2 \right) \right] \sin\left(r\,k^\prime\right) + \left[ 2\,{k^\prime}^3 \pm i\,r\,k\,{k^\prime}\left( {k^\prime}^2 - \,k^2 \right) \right] \cos\left(r\,k^\prime\right) \right) }{{k^\prime}^2\,{\left(k^2-{k^\prime}^2\right)}^2} \\ B_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= \frac{ \pm i\,k\left(3\,{k^\prime }^2 -k^2 \right) - \lim\limits_{r\rightarrow \infty }{e}^{\pm i\,k\,r}\,\left( \left[ 2\,{k^\prime }^3 \pm i\,k\,k^\prime\,r \left( {k^\prime }^2 - k^2 \right) \right] \sin\left(k^\prime \,r\right) + \left[ {k^\prime }^2\,r \left( k^2 - {k^\prime }^2 \right) \pm i\,k \left( 3 \,{k^\prime }^2 - k^2 \right) \right] \cos\left(k^\prime \,r\right) \right) } { {k^\prime }^2\,{\left(k^2-{k^\prime }^2\right)}^2} \end{align}
Geez! Assuming this is even correct, it appears that I have two options. I am not sure what to do first. So, first I will take the limit as $r\to\infty$ and then integrate over $k^\prime$. Second, I will do it the other way around.
Option 1 \begin{align} A_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= \frac{2\,{k^\prime}^2-\lim\limits_{r\rightarrow \infty } \pm i\, e^{\pm i\,r\,k}\,r\,\left( {k^\prime}^2 - \,k^2 \right) \,\left( k \, \cos\left(r\,k^\prime\right) \mp i\, k^\prime \, \sin\left(r\,k^\prime\right) \right) }{{k^\prime} \,{\left(k^2-{k^\prime}^2\right)}^2} \\ B_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) &= \frac{ \pm i\,k\left(3\,{k^\prime }^2 -k^2 \right) - \lim\limits_{r\rightarrow \infty } {k^\prime }\,r \left( k^2 - {k^\prime }^2 \right) {e}^{\pm i\,k\,r}\,\left( k^\prime \, \cos\left(k^\prime \,r\right) \pm i\,k \sin\left(k^\prime \,r\right) \right) } { {k^\prime }^2\,{\left(k^2-{k^\prime }^2\right)}^2} \end{align}
Next, I integrate \begin{align} \int_0^\infty A_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) \,dk^\prime &= \int_0^\infty \frac{2\,{k^\prime}^2-\lim\limits_{r\rightarrow \infty } \pm i\, e^{\pm i\,r\,k}\,r\,\left( {k^\prime}^2 - \,k^2 \right) \,\left( k \, \cos\left(r\,k^\prime\right) \mp i\, k^\prime \, \sin\left(r\,k^\prime\right) \right) }{{k^\prime} \,{\left(k^2-{k^\prime}^2\right)}^2} \,dk^\prime \\ \int_0^\infty B_{1_\pm}\,\frac{\pi}{2\,k^2} \delta(k - k^\prime) \,dk^\prime &= \int_0^\infty \frac{ \pm i\,k\left(3\,{k^\prime }^2 -k^2 \right) - \lim\limits_{r\rightarrow \infty } {k^\prime }\,r \left( k^2 - {k^\prime }^2 \right) {e}^{\pm i\,k\,r}\,\left( k^\prime \, \cos\left(k^\prime \,r\right) \pm i\,k \sin\left(k^\prime \,r\right) \right) } { {k^\prime }^2\,{\left(k^2-{k^\prime }^2\right)}^2} \,dk^\prime. \end{align} I obtain \begin{align} A_{1_\pm} &= 0 \\ B_{1_\pm} &= 0. \end{align}
Ugh! This can not be right.
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