Question Number 220480 by SdC355 last updated on 13/May/25

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teach}\:\mathrm{me}\:\mathrm{about} \\ $$$$\mathrm{Weber}\:\mathrm{function}\:\boldsymbol{\mathrm{E}}_{\nu} \left({z}\right)\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\:\boldsymbol{\mathrm{J}}_{\nu} \left({z}\right)?? \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Consider}\:{n}-\mathrm{dimensional}\:\mathrm{Euclidean}\:\mathrm{Space} \\ $$$$\mathrm{and}\:\mathrm{function}\:{f}\:,\:{f};\mathbb{R}^{{n}} \rightarrow\mathbb{R} \\ $$$$\mathrm{Helmholt}{z}\:\mathrm{Equation}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\left(\bigtriangledown^{\mathrm{2}} +{k}^{\mathrm{2}} \right){f}=\mathrm{0}\:\mathrm{and}\:\mathrm{in}\:\:\mathrm{2}-\mathrm{dimensional}\:\mathrm{Solution}\:\mathrm{is} \\ $$$${f}\left({r},\theta\right)=\underset{\ell=\mathrm{0}} {\overset{\infty} {\sum}}\left({a}_{\ell} ^{\:} \mathrm{cos}\left(\ell\theta\right)+{b}_{\ell} \mathrm{sin}\left(\ell\theta\right)\right)\left({c}_{\ell} ^{\:} {J}_{\ell} \left({kr}\right)+{d}_{\ell} {Y}_{\ell} \left({kr}\right)\right) \\ $$$$\: \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{variables}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{functions} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{comes}\:\mathrm{out}\:\mathrm{as}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{for}\:\mathrm{solution} \\ $$$$\mathrm{But}, \\ $$$$\mathrm{When}\:\mathrm{Bessel}\:\mathrm{Equation}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{other}\:\mathrm{word} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=\frac{\left({x}−\nu\right)\mathrm{sin}\left(\nu\pi\right)}{\pi} \\ $$$$\left(\mathrm{A}\right) \\ $$$$\mathrm{and} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=−\frac{{x}+\nu+\left({x}−\nu\right)\mathrm{sin}\left(\pi\nu\right)}{\pi} \\ $$$$\left(\mathrm{B}\right)\: \\ $$$$\mathrm{and}\:\mathrm{Each}\:\mathrm{Solution}\:\mathrm{as}\:\mathrm{Follows} \\ $$$$\mathrm{Solution}\:\left(\mathrm{A}\right)\:\left\{\mathrm{Weber}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{E}}_{\nu} \left({x}\right) \\ $$$$\mathrm{Solution}\:\left(\mathrm{B}\right)\:\left\{\mathrm{Anger}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{J}}_{\nu} \left({x}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{know}\:\mathrm{how}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{works}\:\mathrm{aka}\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{How}\:\mathrm{these}\:\mathrm{two}\:\mathrm{functions} \\ $$$$\left(\mathrm{each}\:\mathrm{Weber}\:\mathrm{function}\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\right)\:\mathrm{work}… \\ $$$$\:\mathrm{I}'\mathrm{d}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{its}\:\mathrm{for}\:\mathrm{or}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{just}\:\mathrm{a}\:\mathrm{nonlinear}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{thats}\:\mathrm{been} \\ $$$$\mathrm{create}\:\mathrm{by}\:\mathrm{these}\:\mathrm{weirdo}\:\mathrm{mathematicians}\:\mathrm{for}\:\mathrm{their} \\ $$$$\mathrm{intellectual}\:\mathrm{play}??? \\ $$
Answered by MrGaster last updated on 13/May/25

Commented by MrGaster last updated on 13/May/25
\[
\begin{aligned}
&\text{Weber function}\mathbf{E}{\nu}(z)\text{and Anger function}\mathbf{J}{\nu}(z)\text{are defined as particular solutions of non-homogeneous Bessel equations:}\\
&1.\text{Weber function}\mathbf{E}{\nu}(z):\\
&\mathbf{E}{\nu}(z)\triangleq\frac{1}{\pi}\int{0}^{\pi}\cos\left(\nu\theta-z\sin\theta\right)\,d\theta-\frac{\sin(\nu\pi)}{\pi}\int{0}^{\infty}e^{-z\sinh t-\nu t}\,dt\\
&\text{It satisfies the non-homogeneous Bessel equation:}\\
&x^{2}y''+x y'+(x^{2}-\nu^{2})y=\frac{(x-\nu)\sin(\nu\pi)}{\pi}\\
&2.\text{Anger function}\mathbf{J}{\nu}(z):\\
&\mathbf{J}{\nu}(z)\triangleq\frac{1}{\pi}\int{0}^{\pi}\cos\left(\nu\theta-z\sin\theta\right)\,d\theta\\
&\text{It satisfies the non-homogeneous Bessel equation:}\\
&x^{2}y''+x y'+(x^{2}-\nu^{2})y=-\frac{x+\nu+(x-\nu)\sin(\pi\nu)}{\pi}\\
&\text{Verification that}\mathbf{E}{\nu}(z)\text{satisfies equation(A):}\\
&x^{2}\mathbf{E}{\nu}''+x\mathbf{E}{\nu}'+(x^{2}-\nu^{2})\mathbf{E}{\nu}=\frac{(x-\nu)\sin(\nu\pi)}{\pi}\\
&\text{By differentiating the integral representation of}\mathbf{E}{\nu}(z)\text{and using the properties of Bessel equations,the non-homogeneous term can be obtained.}\\
&\text{Application background:}\\
&\text{-Weber function:Used in electromagnetic problems with cylindrical symmetry and non-homogeneous conditions.}\\
&\text{-Anger function:Appears in non-homogeneous solutions of wave equations,such as acoustic scattering.}\\
&\text{Structure of the general solution:}\\
&\text{For equations(A)and(B),the general solution is the sum of the homogeneous solution(containing}J{\nu}\text{and}Y{\nu}\text{)and the particular solution(}\mathbf{E}{\nu}\text{or}\mathbf{J}{\nu}\text{):}\\
&\text{Solution(A)}=C{1}J{\nu}(x)+C{2}Y{\nu}(x)+\mathbf{E}{\nu}(x)\\
&\text{Solution(B)}=C{1}J{\nu}(x)+C{2}Y{\nu}(x)+\mathbf{J}{\nu}(x)\\
&\text{Integral representation and convergence:}\\
&\mathbf{J}{\nu}(z)=\frac{1}{\pi}\int{0}^{\pi}\cos(\nu\theta-z\sin\theta)\,d\theta\quad\text{(absolutely convergent)}\\
&\mathbf{E}{\nu}(z)=\mathbf{J}{\nu}(z)-\frac{\sin(\nu\pi)}{\pi}\int{0}^{\infty}e^{-z\sinh t-\nu t}\,dt\quad\text{(conditionally convergent,requires}\Re(z)>0\text{)}\\
&\text{Series expansion(when}\nu\notin\mathbb{Z}\text{):}\\
&\mathbf{J}{\nu}(z)=\sum{k=0}^{\infty}\frac{(-1)^{k}}{k!}\left(\frac{z}{2}\right)^{2k}\frac{\sin\left((\nu-k)\pi\right)}{(\nu-k)\pi}\\
&\mathbf{E}{\nu}(z)=\sum{k=0}^{\infty}\frac{(-1)^{k}}{k!}\left(\frac{z}{2}\right)^{2k}\frac{\cos\left((\nu-k)\pi\right)-1}{(\nu-k)\pi}\\
&\text{Asymptotic behavior(as}z\to\infty\text{):}\\
&\mathbf{J}{\nu}(z)\sim\sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\nu\pi}{2}-\frac{\pi}{4}\right)\\
&\mathbf{E}{\nu}(z)\sim\sqrt{\frac{2}{\pi z}}\sin\left(z-\frac{\nu\pi}{2}-\frac{\pi}{4}\right)\\
&\text{Orthogonality relation(under a specific weight function):}\\
&\int{0}^{\infty}\mathbf{E}{\nu}(x)\mathbf{J}{\mu}(x)\frac{dx}{x}=\delta{\nu\mu}\cdot\frac{\pi}{2\sin(\nu\pi)}\\
&\text{Differential relations:}\\
&\frac{d}{dz}\mathbf{J}{\nu}(z)=\frac{\nu}{z}\mathbf{J}{\nu}(z)-\mathbf{J}{\nu+1}(z)\\
&\frac{d}{dz}\mathbf{E}{\nu}(z)=\frac{\nu}{z}\mathbf{E}{\nu}(z)-\mathbf{E}{\nu+1}(z)+\frac{\sin(\nu\pi)}{\pi z}\\
&\text{Special values:}\\
&\text{When}\nu\in\mathbb{Z}\text{,}\mathbf{J}{\nu}(z)=J{\nu}(z)\text{,and}\mathbf{E}{\nu}(z)\text{degenerates into a combination of second-order Bessel functions.}\\
&\text{Boundary condition applications:}\\
&\text{In the Helmholtz equation,the non-homogeneous term induced by the source is represented by}\mathbf{E}{\nu}\text{and}\mathbf{J}{\nu}\text{,whose physical meaning is the radiation field or bound state correction.}\\
&\boxed{\text{The structure of the solution and the definition of the functions are as described above.}}
\end{aligned}
\]
Commented by SdC355 last updated on 13/May/25

$$\mathrm{Wow}……\mathrm{Perfect}….\mathrm{Thx}\:\mathrm{a}\:\mathrm{lot} \\ $$