Usbrth

Has anyone an idea how to find the Green's function of the operator
$$
widehatmathcalO=-frac1R partial_z^2 - partial_R frac1R partial_R?
$$

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
$$
G(R,z) = int_0^infty c_k , R , J_1left(kRright) , e^-k , rm dk
$$
because it should be invariant as $zrightarrow -z$.
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 $Gsim frac1sqrtR^2 + z^2$ can be found as a linear combination $int_0^infty J_0left(kRright) e^-k , rm dk$ I expected to get the solution by plugging in $c_k=c$ where c is a dimensionless constant. Note that $widehatcal O$ has dimension $L^-3$ and therefore $$widehatcal O G = fracdelta(R)delta(varphi)delta(z)R$$ 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 $$G(R,z) = cleft(1-fracsqrtR^2+z^2right)$$. 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 $$int_V underbracewidehatcal O G_equiv , rm div vecF , rm dV = int_A vecFcdot rm dvecA$$ 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 $widehatcal OPsi(R,z) = omega_varphi(R,z)$ this is equivalent to solving $Rwidehatcal OPsi(R,z) = R,omega_varphi(R,z)$ and therefore I'm looking for a Greensfunction $G$ to the operator $Rwidehatcal O$ which has dimension $L^-2$ same as for the laplacian. Thus the Greensfunction $G(R,z)$ as the $frac1r$ 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 $$G(R,z)=fracc,R^2left(R^2+z^2right)^frac32$$. This function now indeed has all the properties it needs to have and in particular $$int_V underbraceRwidehatcal O G_equiv , rm div vecF , rm dV = int_A vecFcdot rm dvecA$$ is independent on the integration region!

By choosing the tube of radius $R>0$ from $z=-infty..infty$ I readily obtain $$int_A vecFcdot rm dvecA = 2pi int_-infty^infty R F_R(R,z) , rm dz = 8pi c stackrel!= 1$$ fixing $c$.

$F_R$ is determined by the equation $$frac1R partial_R R F_R = -Rpartial_R frac1R partial_R G = partial_z^2 G$$ i.e.
$$
RF_R(R,z) - R_0F_R(R_0,z) = int_R_0^R R' , partial_z^2 G(R',z) , rm dR' \
Longrightarrow qquad F_R(R,z) = frac3cR^3left(R^2 + z^2right)^frac52
$$.

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
$$
Psi(R,z) = int_V rm dV' , Gleft(R,z,R',z'right) , R'omega_varphileft(R',z'right)
$$
but $G(R-R',z-z')$ from above is not a solution to the Operator $Rwidehatcal O$. :-(

$rm dV'$ 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
$$
G(R,z,R',z') = int_0^infty c_k , RJ_1left(kRright) , R'J_1left(kR'right) , e^-k , rm dk
$$
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
$$
Rwidehatcal O , G = fracdelta(R-R') delta(z-z') delta(phi-phi')R
$$
with physical meaning or if any Greensfunction in an arbitrary parameterspace will suffice i.e. choosing the following equation
$$
Rwidehatcal O , G = delta(R-R') delta(z-z')
$$
so that $G$ now is dimensionless ($Rwidehatcal O$ 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
$$
int_0^infty e^-kz , J_mu(kR) , J_mu(kR') , rm dk = frac1pi sqrtRR' , Q_mu-frac12 left( fracz^2 + R^2 + R'^22RR' right)
$$
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
$$
F_z' = - partial_z' G = -int_0^infty c k begincases -k quad &z'>z \ k quad &z'<z endcases bigg} , RJ_1left(kRright) , R'J_1left(kR'right) , e^-k , rm dk
$$
and the flux through the area at $z'=zpmepsilon$ is
$$
int_z'=zpmepsilon vecFcdot rm dvecA = int_0^infty rm dR' , left( F_z+epsilon - F_z-epsilon right)
$$
The integral over $R'$ gives a factor of $frac2k^2$ and thus
$$
int_z'=zpmepsilon vecFcdot rm dvecA = 2c int_0^infty RJ_1left(kRright) , e^-k , rm dk stackrelz-z'= 2cleft(1-fracepsilonsqrtepsilon^2 + R^2right)
$$

The integral over the surface in positive $R'-$direction (from $z'=z-epsilon..z+epsilon$) needs
$$ F_R' - F_R'_0 = -int_R'_0^R' R'' partial_R'' frac1R'' partial_R'' G(R,z,R'',z') , rm dR'' \ = int_R'_0^R' partial_z'^2 G(R,z,R'',z') , rm dR'' \
= int_0^infty ck , RJ_1left(kRright) , partial_z'^2 e^-k , rm dk int_R'_0^R' R''J_1left(kR''right) , rm dR'' $$
and thus for $R'=infty$ and $R_0'=0$
$$
int_z-epsilon^z+epsilon F_R'(R,z,R',z') , rm dz' \
= int_z-epsilon^z+epsilon F_0(R,z,0,z') , rm dz' + int_0^infty ck , RJ_1left(kRright) , left[ partial_z' e^-k right]_z'=z-epsilon^z'=z+epsilon , rm dk int_0^infty R''J_1left(kR''right) , rm dR'' \
= int_z-epsilon^z+epsilon F_0(R,z,0,z') , rm dz' + int_0^infty ck , RJ_1left(kRright) , left[ -2k , e^-kepsilon right] , rm dk , frac1k^2 \
= int_z-epsilon^z+epsilon F_0(R,z,0,z') , rm dz' - 2c int_0^infty RJ_1left(kRright) , e^-kepsilon , rm dk \
= int_z-epsilon^z+epsilon F_0(R,z,0,z') , rm dz' - 2c left( 1 - fracepsilonsqrtepsilon^2+R^2 right).$$

Conversely the opposite part in negative $R'$-direction at $R'=0$ gives
$$-int_z-epsilon^z+epsilon F_0(R,z,0,z') , rm dz'$$
and therefore after adding up all the parts we have
$$
int_A vecF cdot rm dvecA = 0
$$
implying this specific $G(R,z,R',z')$ being no Greens-function?