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Astronomy & Astrophysicsmanuscript no. 7132 c© ESO 2008February 4, 2008

A new equation for the mid-plane potential of power law disksJean-Marc Hure1,2 and Franck Hersant1

1 Universite Bordeaux 1/CNRS/OASU/UMR 5804/L3AB, 2 rue de l’Observatoire, F-33270 Floirac,e-mail:jean-marc.hure@obs.u-bordeaux1.fr franck.hersant@obs.u-bordeaux1.fr

2 LUTh/Observatoire de Paris-Meudon-Nancay, Place Jules Janssen, F-92195 Meudon Cedex

Received ???/ Accepted ???

ABSTRACT

Aims. We show that the gravitational potentialψ in the plane of an axisymmetrical flat disk where the surface density varies as apowers of the radiusR obeys an inhomogeneous first-order Ordinary Differential Equation (ODE) solvable by standard techniques.Methods. The exact derivative of the midplane potentiel in its integral form is found to be algebrically linked to the potential itself.Results. The ODE reads

dψdR− (1+ s)

ψ

R= Λ(R),

whereΛ is fully analytical. The potential being exactly known at the origin R = 0 for any indexs (and at infinity as well), thesearch for solutions consists of a Two-point Boundary ValueProblem (TBVP) with Dirichlet conditions. The computatingtime isthen linear with the number of grid points, instead of quadratic from direct summation methods. Complex mass distributions whichcan be decomposed into a mixture of power law surface densityprofiles are easily accessible through the superposition principle.Conclusions. This ODE definitively takes the place of the untractable bidimensional Poisson equation for planar calculations. It opensnew horizons to investigate various aspects related to self-gravity in astrophysical disks (force calculations, stability analysis, etc.).

Key words. Gravitation — Methods : analytical — Accretion, accretion disks

1. Introduction

The Poisson equation plays a major role in a large variety ofastrophysical problems where the dynamics of particles andflu-ids is influenced by gravitation. Like most Partial DifferentialEquations (PDEs), it is difficult to solve accurately in space andparticularly inside matter due to its tridimensional nature and be-cause it needs proper boundary conditions which are not easilyaccessible in the vicinity of systems. There are many situationswhere potential values or forces are required in a single planewhere matter is gathered through a flat disk. In such cases, thePoisson equation in cylindrical coordinates (R, φ, z)

(

R−1∂RR∂R + R−2∂2φφ + ∂

2zz

)

ψ = 4πGρ(R, φ, z),

whereρ is the mass density andG is the gravitation constant,is of no use without i) the extension of the computational gridalong thez-direction perpendicular to the disk plane and ii) theadditional knowledge of the gravity vertical gradient in the planeof the disk — the most critical point. Unfortunately, the quantity∂2

zzψ is not available in a simple manner. This is the reason whyψ is always determined through direct summation techniques.

In this letter, we report the discovery of a new local equationsatisfyied by the midplane potentialψ in flat axially symmetri-cal disks provided the surface densityΣ =

∫ ∞−∞ ρdz varies as a

power law of the cylindrical radiusR. As shown, it is an inhom-geneous first-order Ordinary Differential Equation (ODE) sub-ject to perfectly known Dirichlet boundary conditions. In somesense, this new equation circumvents the∂2

zzψ-problem quotedabove by converting the genuine PDE into an ODEfor planarcalculations. Also, it is solvable by standard algorithms with

Send offprint requests to: Jean-Marc Hure

very low computational cost scaling linearly with the numberN of grid points (instead ofN2 with summation methods). Aspower laws are often met in astrophysical disks (e.g. Pringle,1981; Andrews & Williams, 2005), this ODE seems very wellsuited to investigate all these cases where disk gravity is tobe acounted for. More generally, this represents a significativeprogress in potential theory with applications in other fields ofphysics.

2. Basic theoretical considerations

The potentialψ due to a flat axially symmetric disk in its planeat a cylindrical distanceR from the center is (e.g. Durand, 1953)

ψ(R) = −2G∫ aout

ain

√

aRΣ(a)mK(m)da, (1)

whereK is the complete elliptic integral of the first kind,m isthe modulus with

m =2√

aRa + R

and 0≤ m ≤ 1, (2)

ain ≥ 0 is the inner edge andaout > ain is the outer edge. Weconsider surface density profiles in the disk of the form

Σ(a) = Σout

(

aaout

)s

, (3)

whereΣout is the outer edge value ands is a real exponent. Thischoice is natural. Power laws with negative exponents are com-monly met in disk theories (e.g. Pringle, 1981) and observa-tions as well (e.g. Andrews & Williams, 2005). These can also

2 Jean-Marc Hure and Franck Hersant: A new equation for the mid-plane potential of power law disks

be combined to describe other kinds of mass distributions (seeSect. 8). The total disk mass is then

Md = 2π∫ aout

ain

Σ(a)ada = 2πΣouta2outξ(∆), (4)

where1

ξ(∆) =1− ∆2+s

2+ s(5)

A key quantity is the potential at the centerψ(0) ≡ ψc whichcan easily be deduced from Eq.(1) sinceK(0) = π

2 . We have

ψc = −2πG∫ aout

ain

Σ(a)da = −2πGΣoutaoutχ(∆), (6)

where (see note 1)

χ(∆) =1− ∆1+s

1+ s(7)

and∆ = ain/aout is the axis ratio. Note that casess = {−2,−1} arenot compatible with a disk filling the center (i.e. withain = 0).

3. Towards the ODE

In order to obtain the ODE, we divide theR-axis into 3 domains.For R ≤ ain (i.e. inside the central hole), we setv = R/a ≤ 1 andchange the modulus of the complete elliptic integral accordingto the identity (Gradshteyn & Ryzhik, 1965)

K(

2√

x1+ x

)

= (1+ x)K(x), wherex ≤ 1. (8)

SinceΣ(a) = Σoutvsoutv−s, Eq.(1) also reads

ψ(R) = 4GRΣoutvsout

∫ vout

vin

K(v)dv

vs+2, (9)

wherevin = R/ain andvout = R/aout. The potential depends onfive quantities: four parametersain, aout, s andΣout and a singlespace variableR. The radial gradient of the potential (i.e. theopposite of the acceleration) is then given by an exact derivative

dψdR= (1+ s)

ψ

R+ 4GRΣoutv

sout

ddR

{∫ vout

vin

K(v)dv

vs+2

}

. (10)

It is not necessary to perform the integration, as using

ddx

∫ y2(x)

y1(x)f (y)dy = f (y2)

dy2

dx− f (y1)

dy1

dx, (11)

we directly get

dψdR=(1+ s)

ψ

R(12)

+ 4GRΣoutvsout

K(vout)

v2+sout

dvout

dR− K(vin)

v2+sin

dvin

dR

,

wheredv/dR = 1/a. Let ψ = ψ/ψc be the normalized potentialand = R/aout the normalized radial coordinate (it is equivalentto vout). Using Eq.(6) to eliminateΣout, Eq.(12) takes the form

dψd= (1+ s)

ψ

+ S left(), (13)

1 Note that limn→01−∆n

n = − ln∆ for any∆ , 0.

where

S left() = − 2πχ(∆)

[

K() −K(

∆

)

∆1+s]

. (14)

We see that the normalized potential obeys an inhomogeneousfirst-order Ordinary Differential Equation (ODE), withS left assecond member ands and∆ as parameters.

ForR ≥ aout (i.e. outside to the disk), we proceed in a similarmanner withu = a/R ≤ 1. Using Eq.(8), Eq.(1) becomes

ψ(R) = −4GR∫ uout

uin

Σ(a)K(u)udu, (15)

whereuin = ain/R anduout = aout/R. Taking the exact derivativeof ψ and using normalized quantities, we find

dψd= (1+ s)

ψ

+ S right(), (16)

where

S right() = − 2πχ(∆)2

[

K(

1

)

−K(

∆

)

∆s+2

]

. (17)

This is the same ODE as Eq.(13), but with a different secondmember.

Finally, for ain ≤ R ≤ aout (i.e. within the disk), we use boththe variablev andu. The potential reads

ψ(R) = −4GRΣoutvsout

{∫ 1

uin

K(u)u1+sdu (18)

−∫ vout

1K(v)

dvv2+s

}

,

and so

dψdR= (1+ s)

ψ

R+ 4GΣout

[

K(vout)vout

−K(uin)u2in

]

. (19)

Multiplying this expression byaout/ψc and using Eq.(6) again toeliminateΣout, we obtain the expression

dψd= (1+ s)

ψ

+ S inside(). (20)

We recognize the same ODE as above, but with yet another sec-ond member, namely

S inside() = − 2πχ(∆)2

[

K() −K(

∆

)

∆s+2

]

. (21)

4. Asymptotic properties

It is interesting to verify that the differential equation possessesthe right properties both at small and at large distances. A secondorder expansion of the termS left() as/∆ → 0 for ain > 0shows that (Gradshteyn & Ryzhik, 1965)

lim→0

S left() = −(1+ s) −

(

1− ∆s−1)

4χ(∆)2. (22)

As a consequence, we get the approximation

d(ψ − 1)

d− (1+ s)(ψ − 1) ≈ −

2(

1− ∆s−1)

4χ(∆), (23)

Jean-Marc Hure and Franck Hersant: A new equation for the mid-plane potential of power law disks 3

which admits a trivial solution, namely

ψ() ≈ 1+2

4χ(∆)

(

1− ∆s−1

s − 1

)

. (24)

As expected, the potential is quadratic withclose to the cen-ter, implying the linearity of the gravitational acceleration with. This approximate expression is fully compatible with Eq.(9)provided the complete elliptic integral is preliminarily second-order expanded over the modulus asR → 0. Note that the ODEis not defined at the center. This is however not a big problemsinceψ = 1 anddψ/d = 0 at = 0 by symmetry.

Conversely, forR ≫ aout, the potential is expected to mergewith that of a central point mass with massMd. We havem→ 0as → ∞, henceχ(∆)2S right() → ∆s+2 − 1. From Eqs.(4)and (6), we then find

RdψdR≈ s

(

ψ +GMd

R

)

+ ψ + 2GMd

R, (25)

which must be fulfilled for anys. This has a double implication:ψ→ −GMd

R and−dψ/dR→ −GMd

R2 asR→ ∞. This is correct.

5. The ODE in compact form

As shown, at any point in the disk plane,ψ satisfies the ODE

dψd− (1+ s)

ψ

= S (), (26)

where the second member is a piece-wise defined function

S () =

S left(), for 0 ≤ ≤ ∆,S inside(), for ∆ ≤ ≤ 1,S right(), for ≥ 1.

(27)

These three expressions are apparently different. However,they can be converted back into a single function by changingthe modulus of the functionK using Eq.(8). From Eq.(2), wehave

min =2√∆

∆ +, and mout =

2√

1+, (28)

and so we get the expression

S () =K(min)min∆

s+ 32 −K(mout)mout

πχ(∆)√3

. (29)

valid for any ∈ [0,∞[. One thus recognizes the expres-sion given in the abstract, withΛ() = ψcS ()/a2. This ODEcan obviously be rewritten in different equivalent forms. Withcaution, it can even be deduced by direct calculus (see theAppendix). Note that Eq.(26) is compatible with the famous so-lution of Mestel (1963) where the square of the disk rotationalvelocity −dlnψ is uniform for s = −1 provided the disk hasinfinite mass/extension (i.e.ain = 0 andaout = ∞).

6. Gravitational forces and zero-gravity point

Eq.(26) can also be regarded as an algebraic relationship be-tween the potentialψ and the radial acceleration of gravitygR = −dψ/dR. This is particularly attractive from a dynamicalpoint of view since the forceFR = µgR undergone by a particle

with massµ moving inside the disk plane is directly accessiblefrom the potential (instead of its gradient), by the formula

FR(R) = −µ ψc

aout

[

(1+ s)

ψ() − S ()

]

. (30)

There is always one point inside any disk where the potentialgoes through a maximum (and the total force vanishes). Leteq ∈ [∆, 1] be this equilibrium radius in units ofaout; we have

(1+ s)ψeq = −eqS (eq). (31)

7. The ODE, in practice

As it is well known, ODEs are easier to solve than PDEs. Here,ψ is exactly known in two places in the disk plane: at the cen-ter whereψ = 1 by construction, and at infinity whereψ = 0by nature. Thus, the problem is intrinsically a two-point bound-ary value problem (TBVP) with exactly known Dirichlet condi-tions. The presence of the second memberS presents no specialdifficulty since special functions can be determined at computerprecision from numerical libraries. Obviously, any finite size do-main [ left ≥ 0, right < ∞] can also be considered, provided thenormalized potential is accurately determined at the left boundψleft = ψ(left) and right boundψright = ψ(right) of the interval.For instance, on a regular-grid made ofN + 1 mesh points,each with coordinate i = left + iℓ whereℓ is the mesh size(i.e. Nℓ = right −left) andi = 0, . . . ,N, then the ODE can bediscretized as follows

ψi+1 − 2(2+ s)ℓψi

i− ψi−1 = 2ℓS i, for i = 1,N − 1, (32)

at second order, withψ0 ≡ ψleft andψN ≡ ψright. TheseN−1 alge-braic expressions can be put in matrix form, and rapidly solvedfor ψi by standard techniques (e.g. with the Thomas algorithm).

If a potential valueψleft is known at a “starting” radius left(possibly the center), solving the ODE amounts to an InitialValue Problem (IVP) since Eq.(26) also writes

ψ() = ψleft +

∫

left

[

(1+ s)ψ(′)′

+ S (′)

]

d′. (33)

A 1st-order implicit scheme (slightly better than an explicitscheme) appropriate fors < 0 would then give at i+1 the value

ψi+1 = −i+1

[

ψi + (i+1 −i)S i+1

]

si+1 − (1+ s)i, (34)

and so the solution can be carefully propagated up toright.We immediately see two major advantages of solving the

ODE with these methods: first, the computing time is expectedto be proportional to the numberN of grid points (instead ofN2 with direct summation methods) and second, the method si-multaneously supplies the potential and the accceleration. Let usgive a simple example. Figure 1a displays the potentialψi(i)when the solution of the IVP is propagated from the center(0, ψ0) ≡ (0, 1) through a disk using Eq.(34) with the fol-lowing parameters:N = 1000 andN = 1.1 for the grid,and ain = 0.1, aout = 1 (axis ratio∆ = 0.1) and s = −1.5for the disk. Figure 1b displays the corresponding accelerationg = −dψ/d obtained from Eq.(26). We have compared thesedata with the 16-digit reference values obtained from the split-ting method by Hure & Pierens (2005). The relative errors are ofthe order 3× 10−4 on average, which is already remarkably low.

4 Jean-Marc Hure and Franck Hersant: A new equation for the mid-plane potential of power law disks

0 0,5 1normalized radius ϖ=R/aout

0,5

1

norm

aliz

ed p

oten

tial ψ

/ψc

reference valuesolution of the ODE

0 0,5 1normalized radius ϖ=R/aout

-2

0

2

norm

aliz

ed a

ccel

erat

ion

-dϖψ

/ψcreconstructed

exponential disk

reconstructedexponential disk

s=-1.5

s=-1.5

Fig. 1. Normalized potentialψ obtained by solving the ODE(left), and acceleration of gravity deduced from Eq.(26)(right)(see text for the disk parameters and setup). Reference values arecomputed from the splitting method by Hure & Pierens (2005).The relative errors, less than 0.1%, are not visible on the graphs.

Better results can be obtained with other schemes. A crude com-parison of computing times shows that, at a given precision level,potential values and accelerations are determined together muchmore rapidly from the ODE (by a factor∼ 40 in the presentexample) than from direct summation. This is far from beingnegligeable, especially if the potential is to be determined manytimes.

8. Complex distributions

The linearity of the Poisson equation enables the description ofcomplex distributions of matter from individual systems throughthe superposition principle. This is particularly attractive sincepower laws represent probably the most natural (if not the sim-plest) set of basis functions (in particular, a basis for polynomi-als). So, a composite system made up ofJ co-planar and co-axialdisks, each with inner edgeain, j, outer edgeaout, j, axis ratio∆ j,outer surface densityΣout, j and power indexs j, generates at dis-tanceR from the center a total potential

ψtot(R) =J

∑

j=1

ψ j( j)ψcj , (35)

where j = R/aout, j andψcj = −2πGΣout, jaout, jχ(∆ j) accordingto Eq.(6). The problem then involvesJ independant ODEs whichcan be rapidly solved, possibly in parallel. This is illustrated inFig. 1 which shows the potential (same disk as before) for anexponential profileΣ(a) ∝ exp(−a/aout) reconstructed from itsTaylor expansion (a series of power laws). Here,ψtot has beendetermined from the TBVP and space mapping, withJ = 20terms (sufficient to perfectly match the exponential in the range[0, 1] with 16 digits).

9. Concluding remarks

We have demonstrated that the classical Poisson equation canadvantageously be replaced by an ODE when potential valuesare required in the plane of an axially symmetrical flat finitesizedisk where the surface density is a power law of the radius. Some

properties of the ODE have been discussed and a simple exampleof numerical integration through one of the most basic schemeshas been given. As quoted, the combination of power law pro-files make the description of a wide variety of matter distribu-tions possible. This work can be analyzed in much more detail(like the numerical implementation of the ODE) and extendedinseveral ways. For instance, it would be very interesting to con-sider other density profiles, the off-plane case as well as the fullplanar case by releasing theφ-invariance. These questions willbe tackled in forthcoming papers.

Acknowledgements. F. Hersant was supported by a CNRS fellowship which isgratefully ackonwledged. We thank J. Braine and the anonymous referee forvaluable comments.

ReferencesAndrews, S. M. & Williams, J. P. 2005, ApJ, 631, 1134Durand, E. 1953, Electrostatique. Vol. I. Les distributions. (Ed. Masson)Gradshteyn, I. S. & Ryzhik, I. M. 1965, Table of integrals, series and products

(New York: Academic Press, 1965, 4th ed., edited by Geronimus, Yu.V. (4thed.); Tseytlin, M.Yu. (4th ed.))

Hure, J.-M. & Pierens, A. 2005, ApJ, 624, 289Mestel, L. 1963, MNRAS, 126, 553Pringle, J. E. 1981, ARA&A, 19, 137

Appendix A: Direct derivation of the ODE

Equation (26) can be derived directly from Eq.(1) by integratingthe kernel over the modulusm. By reversing Eq.(2), we find

aR=

(

1± m′

m

)2

≡ u(m), (A.1)

wherem′ =√

1− m2 is the complementary modulus. The sur-face densityΣ(a) is then an explicit function ofm andR with

Σ(a) ≡ Σ(m,R) = Σoutus(m)u−s

out, (A.2)

as well as the exact derivative(

dadm

)

R

=2Ru(m)

m

[

1+ u(m)1− u(m)

]

. (A.3)

Under these conditions, the midplane potential writes

ψ(R) = −2GΣoutRu−sout

∫ mout

min

us+ 12 (m)

(

1R

dadm

)

K(m)mdm. (A.4)

Note that the term1R

dadm (and subsequently the whole inte-

grand) is a function ofm only. AsRu−sout ∝ R1+s, we have

dψdR= (1+ s)

ψ

R(A.5)

− 2GΣoutRu−sout

ddR

{∫ mout

min

us+ 12 (m)

(

1R

dadm

)

K(m)mdm

}

Using identity (8), we find

dψdR= (1+ s)

ψ

R− 2GΣoutu

−sout

us+ 1

2out

(

dadm

)

aout

K(mout)moutdmout

dR

−us+ 1

2

in

(

dadm

)

ain

K(min)mindmin

dR

. (A.6)

Jean-Marc Hure and Franck Hersant: A new equation for the mid-plane potential of power law disks 5

Since(

dmdR

)

a

=m2R

[

u(m) − 1u(m) + 1

]

, (A.7)

the above expression can be rearranged into

dψdR= (1+ s)

ψ

R(A.8)

+ 2GΣoutu32out

[

K(mout)mout− ∆s+ 32 K(min)min

]

, (A.9)

which is equivalent to Eqs.(26) and (29).