ilie cylinder problems

A note on Laplace’s equation inside a cylinder SILVANAILIE& DAVIDJ. JEFFREY Department of Applied Mathematics, The University of Western Ontario, London, Ontario, Canada N6A5B7 Abstract Two difficulties connected with the solution of Laplace’s equation around an ob- ject inside an infinite circular cylinder are resolved. One difficulty is the non- convergence of Fourier transforms used, in earlier publications, to obtain the gen- eralsolution,andtheseconddifficultyconcernstheexistenceofapparentlydiffer- ent expressions for the solution. By using a Green’s function problem as an easily analyzed model problem, we show that, in general, Fourier transforms along the cylinderaxisexistonlyinthesenseofgeneralizedfunctions,butwheninterpreted assuch,theyleadtocorrectsolutions. Wedemonstratetheequivalenceofthecor- rected solution to a different general solution, also previously published, but we point out that the two solutions have different numerical properties. Keywords Laplace equation; Fourier Transforms; Green’s function; Sphere in Cylinder; General- ized functions; 1 Introduction This paper addresses the methods used to study the electric field around a cavity in a wire, or the fluid motion around a drop inside a pipe, or similar problems in which an object is placed inside an infinite cylinder. Several papers have presented solutions for the field around a spherical cavity or drop in a cylinder, the most recent paper be- ing Linton [1] and the earliest being Knight [2]. These works overlooked the fact that their integral transforms do not converge in all cases. This non-convergence can be demonstrated without solving the full problem of a sphere inside a cylinder, because the convergence problem is already present in the simplest problem that can be posed: theGreen’sfunctionforLaplace’sequationinacylinderwithNeumannboundarycon- ditions, which in physical terms means the electric field created by a point charge, or the ideal flow from a point source. This paper uses the derivation of the Green’s func- tion as a model problem, which allows us to pinpoint the difficulty and its resolution, withoutthedistractionsofthemorecomplicatedfullproblem(ofafinite-sizedbodyin a cylinder). 1 Having shown the existence of a convergence problem in the Knight—Linton ap- proach, we give a remedy. We must bear in mind, when considering possible reme- dies, that the Green’s function problem used here is a model problem, and any method proposed must generalize back to the problems originally considered by Knight and Linton. We show that the Knight—Linton approach can be repaired using general- ized functions, and then their method remains viable. We also point out an alternative method, outlined by Morse & Feshbach [3] for one simple set of problems. We show that the Morse—Feshbach method gives a solution equivalent to the Fourier transform method, but that the numerical properties of the two solutions are different. Morse— Feshbachhasbetterpropertiesforlarge zandKnight—Lintonforsmall z. Itshouldbe realized, however, that the Morse—Feshbach method has not been tried on the spheri- cal cavity problem, but only on the model problem given here. The problem for the Green’s function is as follows. We scale cylindrical coordi- nates (r, θ, z )so that the boundary conditions are imposed on r= 1. The Green’s function satisfies ∇2G=−4πδ(x) (1) and the Neumann boundary condition ∂G/∂r = 0onr= 1. This is our model prob- lem, and we wish to solve it in a way that illuminates the Knight—Linton approach. Jumping ahead to the solution, given below in equation (11), we shall see that asymp- totically G∼ −2|z|for large zowing to the Neumann boundary condition. In section 2 we consider the consequencesof this. 2 Fourier transform method Knight[2]andotherseffectivelytakeaFouriertransformof(1)withrespectto z. Since we have already stated that G∼ − 2|z|+o(1)forz→ ∞, a Fourier transform does not exist in the ordinary sense. In looking for a response to this difficulty, we must not be misled by the simplicity of the problem (1). It is tempting to consider deriving equationsfor G+ 2|z|,aquantitywhosetransformwouldexist. However,forthemore difficult problems considered by Knight [2] and Linton [1], the asymptotic behaviour of the solution is one of the main goals of the calculation. Therefore, although refor- mulating the problem in terms of convergent integrals would be a possibility in this model problem, it is a solution that does not generalize to harder problems. We can, however, continue to use Knight’s method, provided we are later willing to interpret the integrals as generalized functions. It is convenient to separate thesingularity in Gby writing G= (r2+z2)−1/2+ϕ , (2) and considering the problem for ϕ, which is ∇2ϕ= 0 , (3) ∂ϕ ∂r= (1 + z2)−3/2on r= 1. (4) As with G, the asymptotic behaviour of ϕwill be −2|z|asz→ ∞. We note from (3), (4) that this problem is obviously symmetric in z; however, we do not take advantage 2 of this symmetry to reformulate the problem for two reasons. First, the papers we are commenting on did not do it, and second, we wish to consider a method that would apply to non-symmetric situations. Also the difficulties we address are present even if one restricts the problem to z≥0. If¯ϕis the Fourier transform of ϕwith respect to z, it satisfies a modified Bessel equation ∂2¯ϕ ∂r2+1 r∂¯ϕ ∂r−t2¯ϕ= 0, whose independent solutions are I0(tr)andK0(tr). Since K0is singular at r= 0, it is rejected, and the solution, symmetric with respect to zis ϕ(r, z) =/integraldisplay∞ 0g(t)I0(rt) cos zt dt , (5) withg(t)to be determined from the boundary condition. By differentiating (see [4]) /integraldisplay∞ 0K0(rt) cos zt dt =π 2(r2+z2)−1/2, (6) we deduce that (4) is apparently satisfied by setting g(t) = (2 /π)K1(t)/I1(t).We combine (6) and (2) to write theGreen’s function finally as G(r, z) =2 π/integraldisplay∞ 0/parenleftbigg K0(rt) +K1(t) I1(t)I0(rt)/parenrightbigg coszt dt . (7) The problem, now, is to rewrite (7) as a convergent integral, because for small t, the integrand has the expansion K1(t) I1(t)I0(rt)→2 t2+O(1),fort→0, (8) and therefore the integral does not converge. However, the theory of generalized func- tions allows us to write [5] /integraldisplay∞ 01 t2costz dt =−π 2|z|. (9) This result could also be obtained using the concept of Hadamard’s finite part (see [6]). The connexion between the theory of generalized functions (distributions) and Hadamard’s finite part is described in [5, 6]. Hence (7) can be rewritten as G(r, z) =−2|z|+2 π/integraldisplay∞ 0/parenleftbigg K0(rt) +K1(t) I1(t)I0(rt)−2 t2/parenrightbigg coszt dt . (10) Thus the approach of Knight andLinton can be used with this re-interpretation. 3 3 Morse & Feshbach’s solution A different solution for the Green’s function (2) is given in Morse & Feshbach [3]. Theydescribetheproblemasflowin z >0whenfluidentersthecylinderfromasmall hole in a wall at z= 0. Using separation of variables, they obtain a solution in terms of Bessel functions J0(βkr), with βkdefined by J1(βk) = 0. In present notation, we normalize their problem by setting flux/unit area = 1and obtain G(r, z) =−2z+∞/summationdisplay k=12 βkJ0(βk)2e−βkzJ0(βkr). (11) Before showing the equivalence of (11) and (7), we comment that both solutions canbeunderstoodintermsofthetechniqueofseparationofvariables. Whenseparating Laplace’s equation in cylindrical coordinates, one can take the constant of separation as positive, in which case we are led to (11), or negative, in which case we obtain (7). Introductorycoursesonpartialdifferentialequationstypicallyexploreonlyonechoice for the constant of separation. We explicitly convert solution (7) to (11) by expanding the integrand in (10) as a Dini-Bessel series, valid for 0< r < 1, K0(rt) +K1(t) I1(t)I0(rt)−2 t2=∞/summationdisplay k=12 J2 0(βk)1 t2+β2 kJ0(βkr), where the coefficients were derived using the formula [4] /integraldisplay1 0rKn(tr)Jn(λr)dr=1 t2+λ2/bracketleftbigg/parenleftbiggλ t/parenrightbiggn +λJn+1(λ)Kn(t)−tJn(λ)Kn+1(t)/bracketrightbigg , which is valid for n >−1. Using this expansion in (10), we can integrate term by term, using the formula [4] /integraldisplay∞ 0costz t2+β2 kdt=π 2βke−βk|z|, and conclude that the two expressions are equivalent. 4 Numerical properties of the solutions Although the two expressions are equivalent, they have different (numerical) conver- gence properties: (7) converges slowly for large z, while (11) converges slowly for small z. In (11), the exponential terms e−βk|z|will all tend to 1asz→0, and the ex- pansionwillbeslowlyconvergentwhen zissmall. Ontheotherhand,the cos(zt)fac- tor in (7) will cause numerical difficulties for large z, because it will oscillaterapidly. 4 5 Dipole on axis Inmanyapplications,thedominantresponseofasphereorasimilarobjectinacylinder will be as a dipole rather than as a pole [1]. We note here that convergence problems, similar to those observed in the case of the pole, persist in the dipole case. By dif- ferentiating (7) with respect to z, we obtain the potential φof a unit dipole in a form equivalent to Linton’s φ(r, z) =−2 π/integraldisplay∞ 0t/parenleftbigg K0(rt) +K1(t) I1(t)I0(rt)/parenrightbigg sinzt dt . (12) Theintegrandisasymptotically t(K0(rt) +K1(t)/I1(t)I0(rt))∼2/tandtheintegral again converges only in the sense of generalized functions. Either by separating the singular behaviour in (12) or by differentiating (11), we see that φ(r, z)∼ −2 sgn z+o(1)asz→ ∞ . 6 Conclusions We have considered two general methods for solving Laplace’s equation around an object in a cylinder. We have shown that a re-interpretation of the transforms used by KnightandLintonallowstheirmethodtobeusedreliably. Further,weshowedthatthe different methods offer different numerical properties, although the Morse—Feshbach method has only been applied to the problems described here, and not yet to more complicated situations. 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