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Has anyone an idea how to find the Green's function of the operator $$ \widehat{{\mathcal{O}}}=-\frac{1}{R} \partial_z^2 - \partial_R \frac{1}{R} \partial_R? $$

UPDATE1: For my own record (and anyone else who is interested) here is what one may call a solution or the way how I found it. I was first looking for solutions of the homogenous equation using a separation ansatz $f_R(R)f_z(z)$. The general solution then found this way is $$ G(R,z) = \int_{0}^\infty c_k \, R \, J_1\left(kR\right) \, e^{-k|z|} \, {\rm d}k $$ because it should be invariant as $z\rightarrow -z$. I'm not quite sure why one has to discard the second solution $Y_1$ as $R Y_1$ is finite at the origin. However it does not seem to yield the correct result as the integral over the surface after using Gauss theorem seems to give a zero result as the integrand is odd.

In analogy to the cylindrical laplacian whose Greens function $G\sim \frac{1}{\sqrt{R^2 + z^2}}$ can be found as a linear combination $\int_{0}^\infty J_0\left(kR\right) e^{-k|z|} \, {\rm d}k$ I expected to get the solution by plugging in $c_k=c$ where c is a dimensionless constant. Note that $\widehat{\cal O}$ has dimension $L^{-3}$ and therefore $$\widehat{\cal O} G = \frac{\delta(R)\delta(\varphi)\delta(z)}{R}$$ should be dimensionless when integrated over the entire volume. This implies that $G$ should be dimensionless too. Assuming $c_k$ does not depend on any other hidden parameters and $k$ has dimension $L^{-1}$ the only viable combination is $c_k=c$ being a dimensionless constant.

The solution of the integral then is $$G(R,z) = c\left(1-\frac{|z|}{\sqrt{R^2+z^2}}\right)$$. However as a Greens function should vanish at $\infty$ this result can not be quite right, because it does vanish as $|z| \rightarrow \infty$, but not as $R \rightarrow \infty$. Discarding the first constant term (which is irrelevant for the PDE) gives the opposite discrepancy: It vanishes as $R \rightarrow \infty$, but not as $|z| \rightarrow \infty$. This combined with the fact, that $$\int_V \underbrace{\widehat{\cal O} G}_{\equiv \, {\rm div} \vec{F}} \, {\rm d}V = \int_A \vec{F}\cdot {\rm d}\vec{A}$$ depends on the precise integration region $(\sim \ln (zR))$ led me to conclusion that probably this operator does not have a Greens function. Does anyone know if there is an existence theorem to any arbitrary Operator? Has it something to do with a PDE of second order has to have dimension $L^{-2}$ ?? See below:

Since in my case I want to solve the equation $\widehat{\cal O}\Psi(R,z) = \omega_{\varphi}(R,z)$ this is equivalent to solving $R\widehat{\cal O}\Psi(R,z) = R\,\omega_{\varphi}(R,z)$ and therefore I'm looking for a Greensfunction $G$ to the operator $R{\widehat{\cal O}}$ which has dimension $L^{-2}$ same as for the laplacian. Thus the Greensfunction $G(R,z)$ as the $\frac{1}{r}$ solution to the laplacian has to have dimension $L^{-1}$. Following the argument similar to above I conclude $c_k=ck$ again with a dimensionless constant $c$ and therefore $$G(R,z)=\frac{c\,R^2}{\left(R^2+z^2\right)^{\frac{3}{2}}}$$. This function now indeed has all the properties it needs to have and in particular $$\int_V \underbrace{R\widehat{\cal O} G}_{\equiv \, {\rm div} \vec{F}} \, {\rm d}V = \int_A \vec{F}\cdot {\rm d}\vec{A}$$ is independent on the integration region!

By choosing the tube of radius $R>0$ from $z=-\infty..\infty$ I readily obtain $$\int_A \vec{F}\cdot {\rm d}\vec{A} = 2\pi \int_{-\infty}^{\infty} R F_R(R,z) \, {\rm d}z = 8\pi c \stackrel{!}{=} 1$$ fixing $c$.

$F_R$ is determined by the equation $$\frac{1}{R} \partial_R R F_R = -R\partial_R \frac{1}{R} \partial_R G = \partial_z^2 G$$ i.e. $$ RF_R(R,z) - R_0F_R(R_0,z) = \int_{R_0}^{R} R' \, \partial_z^2 G(R',z) \, {\rm d}R' \\ \Longrightarrow \qquad F_R(R,z) = \frac{3cR^3}{\left(R^2 + z^2\right)^{\frac{5}{2}}} $$.

UPDATE2: In continuing I noticed that above function is not really good when it comes to finding an inhomogenous solution since that would need to be of the form $$ \Psi(R,z) = \int_V {\rm d}V' \, G\left(R,z,R',z'\right) \, R'\omega_{\varphi}\left(R',z'\right) $$ but $G(R-R',z-z')$ from above is not a solution to the Operator $R\widehat{\cal O}$. :-(

${\rm d}V'$ is an appropriate Volume element (not nececarily the cylinder or??)

In which cases can this function be found just by replacing $R \rightarrow R- R'$ and $z \rightarrow z-z'$ ???

So I'm basically starting all over again and this time I try to construct it as $$ G(R,z,R',z') = \int_{0}^\infty c_k \, RJ_1\left(kR\right) \, R'J_1\left(kR'\right) \, e^{-k|z-z'|} \, {\rm d}k $$ The idea now is to choose $c_k$ such that the integral over the parameterspace $(R',z')$ with Gauss-Theorem yields a result independent on $(R,z)$. I'm not sure though (as hinted above) if I need to consider in fact a full cylindrical geometry such that $$ R\widehat{\cal O} \, G = \frac{\delta(R-R') \delta(z-z') \delta(\phi-\phi')}{R} $$ with physical meaning or if any Greensfunction in an arbitrary parameterspace will suffice i.e. choosing the following equation $$ R\widehat{\cal O} \, G = \delta(R-R') \delta(z-z') $$ so that $G$ now is dimensionless ($R\widehat{\cal O}$ has dimension $L^{-2}$ as does the RHS). If I try to construct it using the latter equation, then since $G$ is dimensionless the most simple ansatz (with powers of $k$) must be $c_k=ck$ ($k$ has dimension $L^{-1}$) again with $c$ a dimensionless constant. The integral is actually calculatable by the formula $$ \int_{0}^\infty e^{-kz} \, J_\mu(kR) \, J_\mu(kR') \, {\rm d}k = \frac{1}{\pi \sqrt{RR'}} \, Q_{\mu-\frac{1}{2}} \left( \frac{z^2 + R^2 + R'^2}{2RR'} \right) $$ The additional $k$ in the integrand is obtained by deriving with respect to $z$. However things are getting rather complicated so I will stay with the integral to calculate the flux through the surface. In a 2d co-ordinate system I will take $R'$ as x-coordinate and $z'$ as y. The only way I could solve the Flux through the surfaces was to choose the boundary as the line (from $R'=0..\infty$) at $z'=z+\epsilon$ and $z'=z-\epsilon$ parallel to the R' axis, and the line from $z'=z-\epsilon$ to $z'=z+\epsilon$ at $R'=0$ and $R'=\infty$ respectively. Thus we need $$ F_{z'} = - \partial_{z'} G = -\int_{0}^\infty c k \begin{cases} -k \quad &z'>z \\ k \quad &z'<z \end{cases} \bigg\} \, RJ_1\left(kR\right) \, R'J_1\left(kR'\right) \, e^{-k|z-z'|} \, {\rm d}k $$ and the flux through the area at $z'=z\pm\epsilon$ is $$ \int_{z'=z\pm\epsilon} \vec{F}\cdot {\rm d}\vec{A} = \int_{0}^\infty {\rm d}R' \, \left( F_{z+\epsilon} - F_{z-\epsilon} \right) $$ The integral over $R'$ gives a factor of $\frac{2}{k^2}$ and thus $$ \int_{z'=z\pm\epsilon} \vec{F}\cdot {\rm d}\vec{A} = 2c \int_{0}^\infty RJ_1\left(kR\right) \, e^{-k|z-z'|} \, {\rm d}k \stackrel{|z-z'|=\epsilon}{=} 2c\left(1-\frac{\epsilon}{\sqrt{\epsilon^2 + R^2}}\right) $$

The integral over the surface in positive $R'-$direction (from $z'=z-\epsilon..z+\epsilon$) needs $$ F_{R'} - F_{R'_0} = -\int_{R'_0}^{R'} R'' \partial_{R''} \frac{1}{R''} \partial_{R''} G(R,z,R'',z') \, {\rm d}R'' \\ = \int_{R'_0}^{R'} \partial_{z'}^2 G(R,z,R'',z') \, {\rm d}R'' \\ = \int_{0}^\infty ck \, RJ_1\left(kR\right) \, \partial_{z'}^2 e^{-k|z-z'|} \, {\rm d}k \int_{R'_0}^{R'} R''J_1\left(kR''\right) \, {\rm d}R'' $$ and thus for $R'=\infty$ and $R_0'=0$ $$ \int_{z-\epsilon}^{z+\epsilon} F_{R'}(R,z,R',z') \, {\rm d}z' \\ = \int_{z-\epsilon}^{z+\epsilon} F_{0}(R,z,0,z') \, {\rm d}z' + \int_{0}^\infty ck \, RJ_1\left(kR\right) \, \left[ \partial_{z'} e^{-k|z-z'|} \right]_{z'=z-\epsilon}^{z'=z+\epsilon} \, {\rm d}k \int_{0}^{\infty} R''J_1\left(kR''\right) \, {\rm d}R'' \\ = \int_{z-\epsilon}^{z+\epsilon} F_{0}(R,z,0,z') \, {\rm d}z' + \int_{0}^\infty ck \, RJ_1\left(kR\right) \, \left[ -2k \, e^{-k\epsilon} \right] \, {\rm d}k \, \frac{1}{k^2} \\ = \int_{z-\epsilon}^{z+\epsilon} F_{0}(R,z,0,z') \, {\rm d}z' - 2c \int_{0}^\infty RJ_1\left(kR\right) \, e^{-k\epsilon} \, {\rm d}k \\ = \int_{z-\epsilon}^{z+\epsilon} F_{0}(R,z,0,z') \, {\rm d}z' - 2c \left( 1 - \frac{\epsilon}{\sqrt{\epsilon^2+R^2}} \right).$$

Conversely the opposite part in negative $R'$-direction at $R'=0$ gives $$-\int_{z-\epsilon}^{z+\epsilon} F_{0}(R,z,0,z') \, {\rm d}z'$$ and therefore after adding up all the parts we have $$ \int_A \vec{F} \cdot {\rm d}\vec{A} = 0 $$ implying this specific $G(R,z,R',z')$ being no Greens-function?

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  • $\begingroup$ What does the -3 on the left mean? $\endgroup$ – Diger Jun 2 '18 at 15:18
  • $\begingroup$ It means people dislike your questions. It would help to include more context, i.e your motivation, where this problem came from, and your attempts to solve it. $\endgroup$ – Dylan Jun 3 '18 at 1:44
  • $\begingroup$ When trying to solve the viscous equation $\nabla p = - \nabla \times \vec{\omega}$ for cylindrical coordinates in a tube it arises in terms of an inhomogenous equation for the streamfunction $\Psi$ and the inhomogeneity $\omega_\varphi$ which is known. The equation then sounds $\widehat{{\mathcal{O}}} \Psi = \omega_\varphi$. Thus I need to find a Greens-function right? What I have so far is from using separation of Variables i.e. $G \sim \int_{0}^\infty {\rm d}k \, e^{-kz} \, R \, J_1(kR) = \frac{\sqrt{R^2+z^2}-z}{\sqrt{R^2+z^2}}$. But is this the Greens function up to normalization? $\endgroup$ – Diger Jun 4 '18 at 0:13
  • $\begingroup$ When doing the completely analogous procedure for the cylindrical laplacian this would indeed give the Greens-function $G \sim \int_{0}^\infty {\rm d} k \, e^{-kz} \, J_0(kR) = \frac{1}{\sqrt{R^2 + z^2}}$ $\endgroup$ – Diger Jun 4 '18 at 0:20
  • $\begingroup$ However: Being a Greens-function implies that it should have poles somewhere (right?) which the function above doesn't have. It's series abobut z=0 is $1-\frac{z}{R} + {\cal O}(z^3)$ and about R=0 is $\frac{R^2}{2z^2} + {\cal O}(R^4)$. Also I expect the Greens function to solve $\widehat{{\mathcal{O}}}G = \frac{\delta (R) \delta (\varphi) \delta(z) }{R}$ such that $\int_V {\rm d}^3x \, \underbrace{\widehat{{\mathcal{O}}}G}_{\equiv {\rm div} \vec{F}} = \int_A {\rm d}\vec{A} \cdot \vec{F} = c$ where c is a constant. $\endgroup$ – Diger Jun 4 '18 at 0:40

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