2 cantarella, deturck, gluck and teytel...2 cantarella, deturck, gluck and teytel 1. Introduction...

2 cantarella, deturck, gluck and teytel

1. Introduction

The helicity of a smooth vector field defined on a domain in 3-space is the standardmeasure of the extent to which the field lines wrap and coil around one another. Itplays important roles in fluid mechanics, magnetohydrodynamics, and plasma physics.

The Biot-Savart operator BS starts with a divergence-free vector field V definedon and tangent to the boundary of a domain Ω in 3-space, regards it as a distributionof electric current, and computes its magnetic field. Restricting the magnetic field tothe given domain, we modify it by subtracting a gradient vector field so as to keep itdivergence-free while making it tangent to the boundary of the domain. The resultingoperator BS ′, when extended to the L2 completion of this family of vector fields, iscompact and self-adjoint, and thus has an eigenvalue of largest absolute value, whosecorresponding eigenfields are smooth by elliptic regularity.

The helicity of the vector field V is the L2 inner product of V with its image underthe modified Biot-Savart operator BS ′. Therefore, the eigenfields corresponding tothe eigenvalues of BS ′ having largest absolute value necessarily maximize the absolutevalue of helicity among all divergence-free fields tangent to the boundary of Ω havingthe same energy (square of L2 norm).

The curl operator, acting on divergence-free vector fields tangent to the boundaryof Ω, is a left inverse to the modified Biot-Savart operator BS ′. Therefore, everyeigenfield of BS ′ is also an eigenfield of curl (with reciprocal eigenvalues), but notnecessarily vice versa.

In [CDG1], we considered the manifold M = D2(a) S1, where D2(a) is the diskof radius a in the Euclidean plane, and S1 is a circle of any length. Although thisis not a domain in 3-space, we can still define the modified Biot-Savart operator onit and calculate its eigenvalues and eigenfields. (We could also consider the infiniteright circular cylinder of radius a, with periodicity conditions along the axis of thecylinder corresponding to the length of the S1 factor.) We found that the largesteigenvalue of the modified Biot-Savart operator on M is λ = a/2.405 . . ., where thedenominator is the first positive zero of the Bessel function J0(x). The correspondingeigenfield is given in cylindrical coordinates by:

V (r, θ, z) = J1(r/λ)φ + J0(r/λ)z

where φ and z are unit vector fields in the φ and z directions, respectively. This vectorfield was discovered by Lundquist [Lu] in 1951 in his study of force-free magneticfields on a cylindrical domain; its invariant surfaces are shown in the third of thethree pictures on page one.

Our goal in this paper is to study the eigenvalue problem for the modified Biot-Savart operator acting on divergence-free vector fields defined on and tangent tothe boundary of a round ball in R3, or a spherical shell (the domain between twoconcentric spheres in R3). In particular, we have the following:

biot-savart operator on spherically symmetric domains 3

Theorem A. For the ball B3(b) of radius b, the eigenvalue of the modified Biot-Savart operator having largest absolute value is λ = b/4.4934 . . ., where the denom-inator is the first positive solution of the equation x = tan x. It is an eigenvalue ofmultiplicity three, and its eigenfields are all images under rotations of R3 of constantmultiples of the one given in spherical coordinates by:

V (r, θ, φ) = u(r, θ)r + v(r, θ)θ+ w(r, θ)φ

where r, θ and φ are unit vector fields in the r, θ and φ directions, respectively, and

u(r, θ) =2λ

r2

(λ

rsin(r/λ) − cos(r/λ)

)cos θ

v(r, θ) = −1

r

(λ

rcos(r/λ) − λ2

r2sin(r/λ) + sin(r/λ)

)sin θ

w(r, θ) =1

r

(λ

rsin(r/λ) − cos(r/λ)

)sin θ.

This vector field was discovered by Woltjer [W1] in 1958, who believed it to bethe helicity maximizer, and used it to model the magnetic field in the Crab Nebula.A picture of it is given at the beginning of this paper. Its integral curves fill up afamily of concentric “tori”, with a “core” closed orbit, and a special orbit that beginsat the south pole of the bounding sphere at time −1, proceeds vertically up the zaxis and reaches the north pole at time +1. Orbits on the bounding sphere start atthe north pole at time −1 and proceed down lines of longitude to the south pole attime +1.

Theorem B. For the spherical shell B3(a, b) of inner radius a and outer radius b,the eigenvalue of the modified Biot-Savart operator having largest absolute value is

1/λ(1)1 , where λ(1)

1 is the smallest of the infinite sequence of positive numbers xk thatsatisfy

J 32(ax)Y 3

2(bx)− J 3

2(bx)Y 3

2(ax) = 0

(as a approaches zero, this reduces to the equation x = tan x of Theorem A). It is aneigenvalue of multiplicity three, and its eigenfields are all images under rotations ofR3 of constant multiples of the one given in spherical coordinates by:

V (r, θ, φ) = u(r, θ)r + v(r, θ)θ+ w(r, θ)φ

whereu(r, θ) = r−3=2

(c1J 3

2(λ

(1)1 r) + c2Y 3

2(λ

(1)1 r)

)cos θ

v(r, θ) = − 1

2r

∂

∂r

(pr(c1J 3

2(λ

(1)1 r) + c2Y 3

2(λ

(1)1 r)

))sin θ

w(r, θ) =λ(1)

1

2pr

(c1J 3

2(λ

(1)1 r) + c2Y 3

2(λ

(1)1 r)

)sin θ


This vector field is qualitatively like the one on the ball, having a family of con-centric tori as invariant surfaces, and exceptional orbits on both the inner and outerspherical boundaries. This is pictured as the center of the three figures at the begin-ning of the paper.

An outline of this paper is as follows. Section 2 contains a discussion of helicity anda rough upper bound based on geometric estimates, valid on any domain. The HodgeDecomposition Theorem in section 3 is a crucial ingredient in the reduction of theproblem of maximizing helicity to that of finding the eigenvalue and eigenfields of curlhaving minimum absolute value (and that satisfy appropriate boundary conditions).This reduction is carried out in section 4. Section 5 contains formulas for the curland other expressions in spherical coordinates that will be used in the remainder ofthe paper.

The real work begins in section 6, where we calculate all axially symmetric eigen-fields of the curl operator on the ball and on spherical shells. It is true that everynonzero eigenvalue of curl has an axially symmetric eigenfield, but we calculate theother eigenfields as well. In section 7, we show that the radial component of anyeigenfield of curl must satisfy an elliptic boundary value problem whose solutionsare expressible in terms of eigenfunctions of the Laplace operator. Based on thisobservation, the other components of curl eigenfields are calculated in section 8.

In section 9, we explain why our set of eigenfields is complete. Then we identify theeigenvalue of smallest absolute value in section 10, and show that among sphericallysymmetric domains, the ball has the smallest such eigenvalue. Since the maximumratio of helicity to energy is the reciprocal of this eigenvalue, it shows that the balladmits the maximum helicity for given energy among spherically symmetric domainsof a fixed volume. Finally, in section 11, we examine some of the other eigenvaluesand eigenfields.

2. Helicity of vector elds.

Let Ω be a compact domain in 3-space with smooth boundary ∂Ω; “smooth” forus always means of class C∞. We allow both Ω and ∂Ω to be disconnected.

Let VF(Ω) be the set of all smooth vector fields V on Ω. Then VF(Ω) is aninfinite-dimensional vector space, on which we use the L2 inner product hV,W i =∫Ω V W d(vol).

The helicity H(V ) of the vector field V on Ω, defined by the formula

H(V ) =1

4π

∫Ω×Ω

V (x) V (y) x− yjx− yj3 d(volx)d(voly),

was introduced by Woltjer [W2] in 1958 and named by Moffatt [M] in 1969.

To help understand this formula, think of V as a distribution of electric current,and use the Biot-Savart law to compute its magnetic field:

BS(V )(y) =1

4π

∫ΩV (x) y − x

jy − xj3 d(volx).


Although the magnetic field BS(V ) is well-defined throughout all of 3-space, we willrestrict it to the domain Ω and thus view the Biot-Savart law as providing an operator

BS: VF(Ω)! VF(Ω).

The relation between helicity and the Biot-Savart operator is as follows:

H(V ) =1

4π

∫Ω×Ω

V (x) V (y) x− yjx− yj3 d(volx)d(voly)

=∫

ΩV (y)

[1

4π

∫ΩV (x) y − x

jy − xj3 d(volx)

]d(voly)

=∫

ΩV (y) BS(V )(y) d(voly)

=∫

ΩV BS(V ) d(vol),

so the helicity of V is just the L2 inner product of V and BS(V ),

H(V ) = hV,BS(V )i.

Consider the subspace K(Ω) of VF(Ω) consisting of all divergence-free vector fieldson Ω which are tangent to ∂Ω:

K(Ω) = fV 2 V F (Ω) : r V = 0, V n = 0g,

where n is the unit outward normal vector field to ∂Ω. These vector fields are the fluidanalogues of knots and links, and are the main focus of our attention. For simplicity,we call them fluid knots.

The following result, extracted from [CDG1], provides a bound on the helicity ofany vector field V that depends only on the energy of V and the volume of Ω.

Theorem C. Let V be a smooth vector field in 3-space, defined on the compactdomain Ω with smooth boundary. Then the helicity H(V ) of V is bounded by

jH(V )j R(Ω)E(V ),

where R(Ω) is the radius of a round ball having the same volume as Ω and E(V ) isthe engergy of V , defined as E(V ) =

∫Ω V V d(vol).

This upper bound is not sharp, but it is the right order of magnitude. In particular,Theorem A provides an example of a vector field on the round ball with helicity greaterthan one-fifth of the asserted upper bound.

3. The Hodge decomposition theorem

As before, let Ω be a compact domain with smooth boundary in 3-space, and letVF(Ω) be the set of all smooth vector fields on Ω. We regard VF(Ω) as an infinite-dimensional vector space endowed with the L2 inner product.


The following is arguably the single most useful expression of the interplay betweenthe topology of Ω, the inner product structure on VF(Ω) and the traditional calculusof vector fields. We will use this result a number of times in the sections to come.The reader can find a detailed treatment and proof of this theorem in [CDG2].

Hodge Decomposition Theorem. We have the direct sum decomposition of VF(Ω)into five mutually orthogonal subspaces:

VF(Ω) = FK HK CG HG GG

withker curl = HK CG HG GG

image grad = CG HG GGimage curl = FK HK CG

ker div = FK HK CG HG

where

FK = Fluxless Knots = fr V = 0, V n = 0, all interior fluxes = 0gHK = Harmonic Knots = fr V = 0, r V = 0, V n = 0gCG = Curly Gradients = fV = rϕ, r V = 0, all boundary fluxes = 0gHG = Harmonic Gradients = fV = rϕ, r V = 0, ϕ locally constant on ∂ΩgGG = Grounded Gradients = fV = rϕ, ϕj@Ω = 0gFurthermore:

HK = H1(Ω; R) = H2(Ω, ∂Ω; R)

is finite-dimensional, and the dimension of HK is equal to the (total) genus of ∂Ω.

Also,HG = H2(Ω; R) = H1(Ω, ∂Ω; R)

is finite-dimensional, and the dimension of HG is equal to the number of connectedcomponents of ∂Ω minus the number of connected components of Ω.

The outward pointing unit vector field orthogonal to ∂Ω is denoted by n, so thecondition V n = 0 indicates that the vector field V is tangent to the boundary of Ω.

We say all interior fluxes = 0 for the vector field V if the flux of V through everysmooth surface Σ in Ω with ∂Σ ∂Ω vanishes. For fluid knots, that is, for divergence-free vector fields tangent to the boundary of Ω, the value of this flux depends onlyon the homology class of Σ in the relative homology group H2(Ω, ∂Ω; Z).

A more detailed explanation of the Hodge decomposition theorem and its proofmay be found in our papers [CDG2] and [CDGT2].

It is worth noting that for the ball B3(b) of radius b in R3, both spaces HK andHG are zero. On the spherical shell B3(a, b) between spheres (centered at the origin)of radii a and b, we still have HK = 0, but the space HG of harmonic gradients is


one-dimensional and spanned by the vector field (1/r2)r, which is the gradient of theharmonic function −1/r.

4. Spectral Methods.

Start with a vector field V 2 K(Ω), thus V is divergence-free and tangent to theboundary of Ω. Recall the definition of the modified Biot-Savart operator BS ′ givenin the introduction. We start with V , compute its magnetic field BS(V ) and restrictthis to Ω. Then we refer to the Hodge Decomposition Theorem and subtract anappropriate gradient vector field from BS(V ) so that the resulting vector field lies inK(Ω). To say it another way, we take the L2 orthogonal projection of BS(V ) backto K(Ω). In this way we define the modified Biot-Savart operator

BS ′:K(Ω)! K(Ω).

Just as the Biot-Savart operator is related to helicity by the formula

H(V ) = hV,BS(V )i

for any V 2 VF(Ω), so the modified Biot-Savart operator is related to helicity by theformula

H(V ) = hV,BS ′(V )ifor any V 2 K(Ω). The second formula follows from the first, since BS ′(V ) differsfrom BS(V ) by a gradient vector field, which is L2-orthogonal to V if V 2 K(Ω).

It is useful to have a clear picture of the image of the Biot-Savart operator. Wewill say that a vector field V 2 K(Ω) satisfies Ampere’s law if∫

CV ds = 0

for all closed curves C on ∂Ω which bound in the complement of Ω.

We refer to [CDG1] and [CDG3] for the proofs of the following three theorems:

Theorem D. The image of the modified Biot-Savart operator consists of those vectorfields V 2 K(Ω) which satisfy Ampere’s law.

Theorem E. The ordinary and modified Biot-Savart operators BS and BS ′ arebounded operators, and hence they extend to bounded operators on the L2 completionsof their domains. There they are both compact and self-adjoint.

Since we are restricting our attention to vector fields which are divergence-free andtangent to the boundary of their domain, we will be using the modified Biot-Savartoperator exclusively.


We apply the spectral theorem to BS ′. The eigenfields corresponding to the eigen-values λ(Ω) of maximum absolute value are the vector fields in K(Ω) with maximumabsolute helicity for given energy. They satisfy

H(V ) = λ(Ω)E(V ),

while all vector fields in K(Ω) satisfy

jH(V )j jλ(Ω)jE(V ).

Theorem F. The equation r BS(V ) = V holds in Ω if and only if V 2 K(Ω),that is, if and only if V is divergence-free and tangent to the boundary of Ω.

Since the modified Biot-Savart operator BS ′ is only defined when V 2 K(Ω), andsince BS(V ) and BS ′(V ) differ by a gradient vector field, we have

r BS ′(V ) = r BS(V ) = V.

Thus the curl operator is a left inverse to BS ′, and so the eigenvalue problem forBS ′ can be converted to an eigenvalue problem for curl on the image of BS ′, whichmeans to a system of partial differential equations. In fact, we see from Theorems Dand F that finding all of the eigenvalues and eigenfields of BS ′ is equivalent to findingall of the nonzero eigenvalues and eigenfields of curl that are tangent to the boundaryof Ω and also satisfy Ampere’s law. It is precisely this task that we will carry out onballs and spherical shells.

Remark: Even though we extended BS ′ to the L2 completion of K(Ω) in orderto apply the spectral theorem, the eigenfields will be smooth vector fields in K(Ω).This follows, thanks to elliptic regularity, because on divergence-free vector fields, thesquare of the curl is the negative of the Laplacian.

5. The curl operator in spherical coordinates.

We are searching for eigenvalues and eigenfields of the modified Biot-Savart op-erator on spherically symmetric domains. Since all of these arise as eigenvalues andeigenfields of the curl operator, we begin by writing the curl operator in sphericalcoordinates.

Throughout this paper, we take r, θ and φ to be the standard spherical coordinates,so we have

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ .

We let r =∂

∂r, θ =

1

r

∂

∂θand φ =

1

r sin θ

∂

∂φbe unit vector fields in the r, θ and φ

directions respectively.


We will always consider a vector field V (r, θ, φ) with components

V (r, θ, φ) = u(r, θ, φ)r+ v(r, θ, φ)θ+ w(r, θ, φ)φ.

For such a vector field, we have that

r V =1

r sin θ

(∂

∂θ(sin θ w)− ∂v

∂φ

)r +

1

r

(1

sin θ

∂u

∂φ− ∂

∂r(rw)

)θ

+1

r

(∂

∂r(rv)− ∂u

∂θ

)φ .

The eigenvalue equation r V = λV thus reduces to a system of three partialdifferential equations:

1

r sin θ

(∂


∂φ

)= λu

1

r

(1

sin θ

∂u

∂φ− ∂

∂r(rw)

)= λv

1

r

(∂

∂r(rv)− ∂u

∂θ

)= λw .

In case λ = 0, there is a fourth equation we must take into account, namelyrV = 0.In spherical coordinates this is:

1

r2

∂

∂r(r2u) +

1

r sin θ

∂

∂θ(sin θ v) +

1

r sin θ

∂w

∂φ= 0.

We do not need to consider the r V = 0 equation when λ 6= 0 because it is impliedby taking the divergence of both sides of r V = λV and using the fact that thedivergence of a curl is zero.

Since we are interested in eigenfields of the modified Biot-Savart operator, wemust restrict our attention to divergence-free vector fields that are tangent to theboundary of Ω and that satisfy Ampere’s law. The divergence-free condition onV will be guaranteed by the system of partial differential equations given above.The conditions of tangency to the boundary and Ampere’s law will translate intoboundary conditions on the components of V . For general domains, these can bequite complicated, but since our choices of Ω are spherically symmetric, we get thecondition:

u = 0 on ∂Ω

in order that V be tangent to ∂Ω. The Amperian condition will arise in the axiallysymmetric case discussed briefly in the next section.

Because the square of the curl is the negative of the Laplacian for divergence-free vector fields, we will need the formula for the Laplacian of a scalar function inspherical coordinates. It is:

∆f =1

r2

∂

∂r

(r2∂f

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂f

∂θ

)+

1

r2 sin2 θ

∂2f

∂φ2.


It will also be helpful to have the formula for the gradient of a scalar function inspherical coordinates:

rf =∂f

∂rr +

1

r

∂f

∂θθ +

1

r sin θ

∂f

∂φφ .

We need two more preliminary formulas. These concern the standard round metricinduced on the sphere S2(r) of radius r in Euclidean space. Because the Euclideanmetric in R3 is given in spherical coordinates by

ds2 = dr2 + r2dθ2 + r2 sin2 θdφ2 ,

it is easy to see that the metric induced on S2(r) is

ds2r = r2(dθ2 + sin2 θdφ2) ,

where r is taken to be a constant. The Laplacian taken with respect to this metricof a function g defined only on S2(r) is then

∆r(g) =1

r2

(1

sin θ

∂

∂θ

(sin θ

∂g

∂θ

)+

1

sin2 θ

∂2g

∂φ2

).

The gradient of a function f defined only on S2(r) taken with respect to this metricis

rrf =1

r

∂f

∂θθ +

1

r sin θ

∂f

∂φφ.

6. Axially symmetric solutions

In this section, we find solutions of r V = λV on B3(b) and on B3(a, b) thatare invariant under rotations around the z-axis. We do this for two reasons. First,we will show in subsequent sections that all the eigenvalues of the curl operator onour domains have multiplicity at least 3. It is the case that for each eigenvalue, atleast one eigenfield is invariant under rotations around the z-axis. Thus, we willdiscover all of the eigenvalues in this section, if not all of the eigenfields. Second, thecomputations of the axially symmetric eigenfields are much easier than the generalcase, and serve to motivate some of the choices we will make in sections 7 and 8.

So we assume in this section that

V = u(r, θ)r + v(r, θ)θ+ w(r, θ)φ.

In other words, V is independent of φ. Under this assumption, the formula for rVsimplifies to

r V =1

r sin θ

∂

∂θ(sin θ w)r − 1

r

∂

∂r(rw)θ +

1

r

(∂

∂r(rv)− ∂u

∂θ

)φ.


The equation r V = λV becomes the system of equations:

1

r sin θ

∂

∂θ(sin θ w) = λu

−1

r

∂

∂r(rw) = λv

1

r

(∂

∂r(rv)− ∂u

∂θ

)= λw.

Assume λ 6= 0. Then the first two of these equations give us u and v in terms offirst derivatives of w. If we substitute these into the third equation, we will get asecond-order partial differential equation for w in the following manner. We rewritethe second equation as

rv = −1

λ

∂

∂r(rw).

Therefore∂

∂r(rv) = −1

λ

∂2

∂r2(rw).

If we solve for u in the first equation and differentiate with respect to θ, we get that

∂u

∂θ=

1

λr

∂2w

∂θ2− 1

λr sin2 θw +

cos θ

λr sin θ

∂w

∂θ.

We substitute these into the third equation, then multiply by λr sin θ to obtain thefollowing equation:

∂2

∂r2(r sin θ w)− 1

r2

cos θ

sinθ

∂

∂θ(r sin θ w) +

1

r2

∂2

∂θ2(r sin θ w) = −λ2r sin θ w.

This gives us an equation for f(r, θ) = r sin θ w(r, θ):

∂2f

∂r2− 1

r2

cos θ

sin θ

∂f

∂θ+

1

r2

∂2f

∂θ2= −λ2f.

We also have boundary conditions for f , which come from the fact that we want Vto satisfy Ampere’s law. This means that the line integral of V around any curveon the boundary of our domain should vanish, since all curves on the boundary ofeither B3(b) or B3(a, b) bound in the exterior of the domain. If we take our curve tobe a line of latitude (so r and θ are constant and φ varies from 0 to 2π), we see thatthe integral of V dx around the boundary is equal to 2πrf(r, θ). Therefore we musthave f(r, θ) = 0 on the boundary of our domain. (It is also possible to show thatf(r, θ) = 0 on the boundary without appealing to Ampere’s law, and simply to usethe condition that the vector field be tangent to the boundary.)

We can solve the equation for f by separation of variables. If we assume thatf(r, θ) = R(r) Θ(θ), substitute into the partial differential equation and rearrange,we get the equations:

r2(R′′ + λ2R)

R=−Θ′′ + cot θ Θ′

Θ= K.


This separates into a pair of equations for R and Θ that are related to Bessel’sequation and Legendre’s equation, respectively.

The Θ equation, sin θ Θ′′ − cos θ Θ′ + K sin θ Θ = 0, upon thinking of Θ as afunction of x = cos θ (we will write Q(x) = Q(cos θ) = Θ(θ), becomes

(1− x2)Q′′(x) +KQ(x) = 0.

This is related to the Legendre equation

(1− x2)P ′′n − 2xP ′n + n(n + 1)Pn = 0

satisfied by the Legendre polynomial Pn of degree n. In fact, we can calculate that ifn = −1

2 1

2

p1 + 4K (so that K = n(n+ 1)), then the function Q(x) = (1−x2)P ′n(x)

will satisfy our differential equation. Since the only bounded solutions of Legendre’sequation (and hence of ours) occur when K = n(n + 1), n 2 f1, 2, . . .g (see [TS,p. 691]), we have in hand all of the solutions of the equation for Θ, namely

Θ(θ) = sin θ P 1n(cos θ),

where P 1n is the associated Legendre function P 1

n(x) =p

1− x2P ′n(x) (see [TS, p. 698]).(In fact, we could have Θ(θ) = A+B cos θ for K = 0, but this possibility is eliminatedby the fact that we must have Θ(0) = Θ(π) = 0 because this is true for f(r, θ) by itsdefinition as r sin θ w).

Now we know that K = n(n+ 1) and we can turn to the equation for R(r), whichnow reads:

r2R′′ + (λ2r2 − n(n+ 1))R = 0

together with the boundary conditions that R(0) is bounded and R(b) = 0 if ourdomain is the ball B3(b), or the conditions R(a) = R(b) = 0 for B3(a, b).

As is well-known, the R equation is related to Bessel’s equation. We obtain it, andhence its solution, by setting α = 1

2, β = λ, γ = 1 and ν = n+ 1

2in equation (5.4.11)

on page 106 of [Le]. This tells us that

R(r) = r1=2(AJn+1

2(λr) +BYn+1

2(λr)

)where J and Y are Bessel functions of the first and second kind. Consequently, wecan write the general separable solution of the partial differential equation for f as:

f(r, θ) = r1=2(AJn+1

2(λr) +BYn+1

2(λr)

)sin θ P 1

n(cos θ)

for n 2 f1, 2, 3, . . .g. We divide this by r sin θ to get

w(r, θ) = r−1=2(AJn+1

2(λr) +BYn+1

2(λr)

)P 1n(cos θ).

The boundary conditions will determine which values of λ occur as eigenvalues. Inthe case of the ball B3(b), the requirement that w be continuous at the origin impliesthat there are no Bessel functions of the second kind in the solution, thus on the ballwe have:

w(r, θ) = r−1=2Jn+12(λr)P 1

n (cos θ).


We also need this to be zero for r = b, that is, Jn+12(λb) = 0. This will be true for an

infinite sequence of values of λ, which we will call λ(n)k . These are the eigenvalues we

are seeking.

In the case of B3(a, b), for any positive value of λ, we determine A and B (upto a constant multiple) by requiring that

AJn+12(λa) +BYn+1

2(λa) = 0.

Then, as in the case of the ball, there will be an infinite sequence of λ(n)k for which

A

(n)k

Jn+12(λ(n)

k b) +B

(n)k

Yn+12(λ(n)

k b) = 0.

These λ(n)k are the eigenvalues we are seeking. Except for possible fortuitous coinci-

dences of zeros of different Bessel functions, there is only one linearly independenteigenfunction for each eigenvalue. This is in contrast with the general result presented

in the next two sections, where for each eigenvalue λ(n)k there are 2n + 1 linearly in-

dependent eigenfields.

With all the solutions w(r, θ) in hand, we now calculate the u and v part of

V = ur+vθ+wφ. This will be immediate from the r and θ components ofrV = λVthat we had calculated in the axially symmetric case:

1

r sin θ

∂

∂θ(sin θ w) = λu

−1

r

∂

∂r(rw) = λv.

For example, the first of these equations yields:

u =1

λr sin θ

∂

∂θ

(r−1=2Jn+1

2(λr) sin θ P 1

n(cos θ))

= − 1

λr3=2Jn+1

2(λr)

1

sin θ

∂

∂θ

(sin θ

∂

∂θ(Pn(cos θ))

)

=n(n + 1)

λr3=2J(λr)Pn(cos θ)

where we have used the definition of P 1n(z) = (1− z2)

12d

dzPn(z), the chain rule, and

the fact that Pn(cos θ) satisfies the differential equation

1

sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+ n(n+ 1)u = 0

(equation (4.3.10) on page 48 of [Le]).

If we multiply the three functions u, v and w through by n(n+1)

, we get that our


eigenfields are V = ur + vθ+ wφ, where

u = r−3=2Jn+12(λr)Pn(cos θ)

v =1

n(n+ 1)r

∂

∂r

(prJn+1

2(λr)

) ∂

∂θPn(cos θ)

w = − λ

n(n+ 1)prJn+1

2(λr)

∂

∂θPn(cos θ)

where λ is one of the λ(n)k we found above, and Jn+1

2should be replaced by the

appropriate linear combination of Jn+12

and Yn+12

in the case of B3(a, b). This solution

is easily seen to agree with the m = 0 case of the solution we will obtain at the endof section 8 in the general (non-axially-symmetric) case.

We defer further exploration of specific solutions until sections 10 and 11.

7. r component of the general solution

Now we turn to the situation where we want to solve the equationrV = λV onthe spherically symmetric domain Ω, and the components of V are allowed to varywith all three spherical coordinate variables. So we are looking for three functionsu(r, θ, φ), v(r, θ, φ) and w(r, θ, φ) that satisfy the system of three partial differentialequations:

1

r sin θ

(∂


∂φ

)= λu

1

r

(1

sin θ

∂u

∂φ− ∂

∂r(rw)

)= λv

1

r

(∂

∂r(rv)− ∂u

∂θ

)= λw.

We also want V to be tangent to the boundary of Ω. As in section 6, this translatesinto boundary conditions on u as follows: If Ω is the ball B3(b), then we requireu(0, θ, φ) to be bounded and u(b, θ, φ) = 0. If Ω is the spherical shell B3(a, b), thenwe require that u(a, θ, φ) = 0 and u(b, θ, φ) = 0.

We will also need to check that any solutions we find of the above boundary-valueproblem are Amperian (the line integral of V dx over any curve on ∂Ω must vanish),but we will wait to address this until after we find our solutions.

In contrast to our approach in the axially symmetric case, where we extracteda single second-order equation for w from the first order system r V = λV , inthis section we will concentrate on u, the r component of V . Our motivation comesfrom the fact that for divergence-free vector fields, we have rr V = −∆V . Inrectangular coordinates, ∆V is simply the coordinate-wise Laplacian of V . But inspherical coordinates, this is no longer the case. We concentrate on u because it isthe “most rectangular” of the spherical coordinates (in that the integral curves of rare straight lines) and extract our second-order PDE for u from the r component ofrr V = λ2V .


Proposition 1. Let Ω be the ball B3(b) or the spherical shell B3(a, b). If u(r, θ, φ)is the r component of an eigenfield V 2 K(Ω) of r V = λV on Ω, then u mustsatisfy the equation

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

together with the boundary conditions u(0, θ, φ) is bounded and u(b, θ, φ) = 0 if Ω =B3(b), or u(a, θ, φ) = 0 and u(b, θ, φ) = 0 if Ω = B3(a, b).

Proof. The proposition follows from two observations about eigenfields of the curloperator. If r V = λV for V 2 K(Ω), then ∆V = −λ2V and r V = 0. And for

any vector field V = ur+ vθ+wφ that satisfies ∆V = −λ2V and r V = 0, we havethat

∆(ru) = −λ2(ru).

To see this, define the vector field R = xi + yj + zk = rr and suppose that inrectangular coordinates V = ai + bj + ck. Then R V = ru and so

∆(ru) = ∆(R V ) = ∆(xa+ yb+ zc)

= x∆a+ y∆b+ z∆c+ 2(∂a

∂x+∂b

∂y+∂c

∂z)

= R ∆V + 2r V = −λ2R V= −λ2(ru).

We use the formulas at the end of section 5 to calculate that ∆r = 2/r and of courserr = r; therefore

∆(ru) = r∆u+ 2(rr) (ru) + u∆r = r∆u+ 2∂u

∂r+

2

ru = −λ2ru.

Divide both sides by r to obtain the equation in the proposition. The boundaryconditions follow easily from the fact that u = V n on the boundary.

This completes the proof of Proposition 1. Our task now is to solve the boundary-value problem for u.

Proposition 2. On the ball B3(b), the eigenvalues of the equation

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

are given by the set fλ(n)k g for n = 1, 2, 3, . . . and k = 1, 2, 3, . . ., where the product

bλ(n)k is the kth (positive) zero of the Bessel function Jn+1

2(x). Each eigenvalue λ(n)

k

has multiplicity 2n+ 1 with eigenfunctions given by:

u(r, θ, φ) = r−3=2Jn+12(λ(n)

k r)Pmn (cos θ)

cosmφ

sinmφ

where m runs from 0 to n 2 f1, 2, 3, . . .g (and of course we can only use the cosinesolution for m = 0).


(see [Le, p. 134]), we see that the solution for n = 0, which is r−3=2J1=2(λr), will beunbounded as r! 0. This yields the conclusion of Proposition 2.

The situation for the spherical shell B3(a, b) is a little more complicated. For everychoice of n 2 f0, 1, 2, . . .g and each λ > 0, there are constants c1 and c2 (unique upto scalar multiple of the vector [c1, c2]) such that:

R(a) = c1Jn+12(λa) + c2Yn+1

2(λa) = 0.

This determines R(r) up to a constant multiple. For most values of λ, we will have

R(b) 6= 0. But there will be an infinite sequence of λ, which we label λ(n)k as above,

such thatR(b) = c1Jn+1

2(λ

(n)k b) + c2Yn+1

2(λ

(n)k b) = 0.

These values of λ(n)k will be our eigenvalues for curl corresponding to the fixed value

of n. As in the case of the ball, the multiplicity of λ(n)k is 2n+1. Also, as in the case of

the ball, we may not choose n = 0, although the reason for this is more complicated,and we will discuss it at the end of the next section. Our r components for the regionbetween two concentric spheres are thus

u(r, θ, φ) = r−3=2(c(n)

1k Jn+12(λ(n)

k r) + c(n)2k Yn+1

2(λ(n)

k r))Pmn (cos θ)

cosmφ

sinmφ

where n 2 f1, 2, 3, . . .g, k 2 f1, 2, 3, . . .g and m runs from 0 to n as before, provingProposition 3.

8. An ordinary dierential system for the θ and φ components

Now we turn to finding the θ and φ components of our solution to r V = λV .We assume that we have in hand a function u(r, θ, φ) that satisfies the equation

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

of Proposition 1. Our first goal is to prove that given any function u, there is atmost one pair of functions v and w so that the vector V = ur + vθ + wφ satisfiesr V = λV (and r V = 0 in the case where λ = 0).

Proposition 4. Let Ω = B3(a, b) (where a is allowed to be 0), and suppose that thefunction u(r, θ, φ) is given. If v1(r, θ, φ), w1(r, θ, φ) and v2(r, θ, φ), w2(r, θ, φ) are two

pairs of functions such that both V1 = ur+ v1θ+w1φ and V2 = ur+ v2θ+w2φ satisfyr V1 = λV1 and r V2 = λV2 (and r V1 = r V2 = 0 in case λ = 0), then itmust be the case that v1 = v2 and w1 = w2.

Proof. Set f = v1 − v2 and g = w1 − w2. Then the vector field Q = fθ + gφ issmooth (analytic) on the interior of Ω and satisfies rQ = λQ and r Q = 0. Ther component of the equation rQ = 0 is

1

r sin θ

(∂

∂θ(sin θ g)− ∂f

∂φ

)= 0 ,


and

r Q =1

r sin θ

(∂

∂θ(sin θ f) +

∂g

∂φ

)= 0.

The first of these equations implies that

∂2

∂φ∂θ(sin θ g) =

∂2f

∂φ2

and the second implies that

∂2

∂φ∂θ(sin θ g) = − ∂

∂θ

(sin θ

∂

∂θ(sin θ f)

).

Together, these last two equations imply that

∂

∂θ

(sin θ

∂

∂θ(sin θ f)

)+

1

sin θ

∂2

∂φ2(sin θ f) = 0.

We divide this equation by r2 sin θ and recognize it (using the formula from section 5for the Laplacian on the sphere) as the condition ∆r(sin θ f) = 0 for each value ofr. Thus, the restriction of sin θ f to any sphere centered at the origin is a harmonicfunction, and is thus constant on the sphere. But f must be bounded and continuous,and so sin θ f is zero at the north and south poles of the spheres. Therefore f isidentically zero. A similar argument shows that g must be identically zero as well,proving the proposition.

Now we know that for each eigenfunction u of the equation

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

there can be at most one eigenfield V of r V = λV having u as its r component.We now show that all of the eigenfunctions found in Propositions 2 and 3 give rise tovector fields. To save notation, we will write J(λr) instead of c1J(λr) + c2Y(λr) inthe statement and proof of the following proposition. Also, where there are functionsone atop the other and plus and minus signs one atop the other, only two solutionsare represented: one must choose all of the top choices or all of the bottom ones.

Proposition 5. Suppose n 2 f1, 2, 3, . . .g. Let u be one of the 2n+1 eigenfunctionsof

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

on B3(a, b) (where a is allowed to be zero in the case of the ball) corresponding to the

eigenvalue λ = λ(n)k of Proposition 2 or 3, with u(a, θ, φ) = u(b, θ, φ) = 0 (or u(0, θ, φ)

is bounded in the case of the ball). Set ν = n+ 12, choose m 2 f0, 1, . . . , ng, and write

u = (ν2 − 14)r−3=2J(λr)P

mn (cos θ)

cosmφ

sinmφ

.


Then for

v =1

r

∂

∂r

(prJ(λr)

) ∂

∂θPmn (cos θ)

cosmφ

sinmφ

λmp

rJ(λr)

Pmn (cos θ)

sin θ

sinmφ

cosmφ

and

w = mr

∂

∂r

(prJ(λr)

) Pmn (cos θ)

sin θ

sinmφ

cosmφ

− λp

rJ(λr)

∂

∂θPmn (cos θ)

cosmφ

sinmφ

the vector field V = ur+vθ+wφ satisfies rV = λV and V n = 0 on the boundaryof the domain.

Proof. One obvious way to prove this proposition is to check that V as specifiedactually solves the equation rV = λV . On the other hand, this would give no clueas to how v and w were calculated. So we give an indication of how we solved for vand w. To begin, we assume that we have u(r, θ, φ) as in the proposition, and set

f =1

sin θ

∂u

∂φand g =

∂u

∂θ.

Then the θ and φ components of r V = λV become

1

r

(f − ∂

∂r(rw)

)= λv and

1

r

(∂

∂r(rv)− g

)= λw.

So we let p = rv and q = rw and note that these equations can be considered to bea system of two ordinary differential equations in p and q with r as the independentvariable, and parameters that depend on θ and φ. The equations are

p′ − λq = g and q′ + λp = f.

We need to find a particular solution of this constant-coefficient system of ODEs, andwe proceed to do so by a variation on the method of undetermined coefficients.

To begin, we let J stand for any solution of the Bessel equation of order ν (in-cluding the linear combination of J and Y that we need in the solution for B3(a, b)),and then we can write

u(r, θ, φ) = Q(θ, φ)r−3=2J(λr),

where ν = n+ 12

for some n 2 f1, 2, 3, . . .g. We also write

f =1

sin θ

∂u

∂φ= S(θ, φ)r−3=2J(λr)

and

g =∂u

∂θ= R(θ, φ)r−3=2J(λr),

where R and S are known functions that can be specified later.


We will make use of two well-known identities for Bessel functions (see [Le, pp. 100,103 and 105]), namely

J ′(ζ) =ν

ζJ(ζ)− J+1(ζ)

andJ(ζ)

ζ=

1

2ν(J−1(ζ) + J+1(ζ)).

The second one immediately yields:

J(λr)

r=

λ

2ν(J−1(λr) + J+1(λr)),

and so the first gives:

d

dr

(r1=2J(λr)

)=

1

2prJ(λr) +

νprJ(λr) −

prλJ+1(λr)

=pr

((ν + 1

2)λ

2ν(J−1(λr) + J+1(λr)) − λJ+1(λr)

)

=λpr

2ν

((1

2+ ν)J−1(λr) + (1

2− ν)J+1(λr)

).

We can also use the identities to rewrite our expressions for f and g:

r−3=2J(λr) = r−1=2J(λr)

r=r1=2λ

2ν

(J−1(λr)

r+J+1(λr)

r

)

=λ2r1=2

4

(J−2(λr)

ν(ν − 1)+

2J(λr)

(ν − 1)(ν + 1)+J+2(λr)

ν(ν + 1)

),

and f is S times this expression, g is R times it.

The point of using all these identities is to note that all derivatives of r1=2J(λr)can be expressed as a linear combination of terms of the same sort, for various valuesof τ . This is the basis of our method of undetermined coefficients.

We make the assumption that

p =∑

ar1=2J(λr) and q =

∑

br1=2J(λr)

where τ ranges over half-integers (i.e., numbers of the form k + 12, where k is an

integer). If we substitute this assumption into the equation p′− λq = g, we get, afterusing our Bessel function identities and shifting some indices,

∑

λr1=2

(2τ + 3

4(τ + 1)a+1 −

2τ − 3

4(τ − 1)a−1 − b

)J (λr)

=λ2Rr1=2

4

(J−2(λr)

ν(ν − 1)+

2J(λr)

(ν − 1)(ν + 1)+J+2(λr)

ν(ν + 1)

),


and ∑

λr1=2

(2τ + 3

4(τ + 1)b+1 −

2τ − 3

4(τ − 1)b−1 + a

)J(λr)

=λ2Sr1=2

4

(J−2(λr)

ν(ν − 1)+

2J(λr)

(ν − 1)(ν + 1)+J+2(λr)

ν(ν + 1)

).

This gives a recurrence relation for the coefficients a and b , if we can figure out howto get the recurrence started. Note that the first value of τ for which there is not azero term on the right side of the equations is τ = ν − 2, and this pair of equationswould allow us to solve for a−1 and b−1. Thus, the most optimistic assumption wecan make, which is supported by the fact that our solutions in the axial case can beexpressed as finite sums of the above form, is that a = b = 0 for all τ ν − 2. Sowe assume this and begin solving for the subsequent a and b .

With our assumption, the coefficients of r1=2J−2(λr), which come from the termsfor τ = ν − 2 yield the equations:

λ(2ν − 1)

4(ν − 1)a−1 =

λ2R

4ν(ν − 1)and

λ(2ν − 1)

4(ν − 1)b−1 =

λ2S

4ν(ν − 1)

which imply that

a−1 =λR

ν(2ν − 1)and b−1 =

λS

ν(2ν − 1)

(here we can see why we must exclude the case when ν = 1/2, i.e., when n = 0, whichwe treat separately below).

We continue following the recurrence: for τ = ν − 1, we obtain

a =4λS

(2ν − 1)(2ν + 1)and b =

−4λR

(2ν − 1)(2ν + 1),

and then in a miracle of factoring and cancellation, for τ = ν we obtain:

a+1 =−λR

ν(2ν + 1)and b+1 =

−λSν(2ν + 1)

.

We then check that the next two pairs a+2, b+2 and a+3, b+3 are zero, so all sub-sequent a and b are zero.

We have thus calculated finite sums for p and q:

p =pr

(λR

ν(2ν − 1)J−1(λr) +

4λS

(2ν − 1)(2ν + 1)J(λr) −

λR

ν(2ν + 1)J+1(λr)

)

q =pr

(λS

ν(2ν − 1)J−1(λr) − 4λR

(2ν − 1)(2ν + 1)J(λr) −

λS

ν(2ν + 1)J+1(λr)

).

We can use our Bessel function identities to combine the first and last terms of each


of the above expressions, yielding:

p =R

ν2 − 14

d

dr

(prJ(λr)

)+λSpr

ν2 − 14

J(λr)

q =S

ν2 − 14

d

dr

(prJ(λr)

)− λR

pr

ν2 − 14

J(λr).

We recall that v = p/r and w = q/r, and use the definitions of R and S to obtainthe formulas in the proposition.

Finally, we note that we do not have to worry about any other solutions of thesystem of ODEs from which we calculated p and q, because the solutions we havealready found are smooth, and by Proposition 4, they are the only possible smoothsolutions. We could also calculate the other solutions explicitly and check that theyare not smooth (along the z-axis). This completes the proof of Proposition 5.

We have thus calculated all the solutions ofrV = λV with the possible exceptionof solutions that come from Proposition 3 for n = 0 on the domain B3(a, b). We noweliminate this possibility.

Proposition 6. There are no nonzero solutions of r V = λV with r V = 0having r component equal to a function of r alone.

This will deal with the n = 0 case because the eigenvalues λ(0)k occur with multi-

plicity one, and because the Legendre function P 00 (cos θ) is a constant as is cosmφ

for m = 0.

Proof. To begin the proof of Proposition 6, we return to the system of ODEs thatwe derived at the beginning of the proof of Proposition 5 for the θ and φ componentsof V = ur+ vθ+wφ, which are implied by the θ and φ components of r V = λV .Since u is a function of r alone, these components yield the system

∂(rw)

∂r= −λrv

∂(rv)

∂r= λrw.

These imply that

v =c1(θ, φ)

rsin(λr) − c2(θ, φ)

rcos(λr)

w =c1(θ, φ)

rcos(λr) +

c2(θ, φ)

rsin(λr)

where c1 and c2 are smooth functions on the (unit) sphere. The r component ofr V = λV multiplied by r2 now becomes the equation:

cos(λr)

(1

sin θ

∂

∂θ(sin θ c1) +

1

sin θ

∂c2

∂φ

)

+ sin(λr)

(1

sin θ

∂

∂θ(sin θ c2)−

1

sin θ

∂c1

∂φ

)= λr2u(r).


By the discussion preceding the theorem, to prove the Proposition it is sufficientto exhibit a scalar function whose gradient is equal to the restriction of V to (acomponent of) ∂Ω. Since the proof is the same for either component of ∂B3(a, b), werestrict our attention to the sphere S2(b) of radius b. There, we have that J(λb) = 0and all of the r terms that survive (i.e., those with derivatives of Bessel functions) areevaluated at b, and thus are constant on S2(b). In fact, if we set C to be the value ofr−1(d/dr)(

prJ(λr)) at r = b, then we see that the restriction of V to S2(b) is equal

to v0θ + w0φ, where

v0 = C∂

∂θ(Pm

n (cos θ))

cosmφ

sinmφ

w0 = mCPmn (cos θ)

sin θ

sinmφ

cosmφ

.

But according to the formula given at the end of section 5 for the gradient of afunction on S2(r), this vector field is the gradient on S2(r) of the function

f(θ, φ) = bCPmn (cos θ)

cosmφ

sinmφ

which is a smooth function, being the restriction to S2(b) of a harmonic homogeneouspolynomial on R3 of degree n. This observation completes the proof of Proposition 7.

9. Completeness and comparison with Chandrasekhar-Kendall solutions

In [CK], Chandrasekhar and Kendall gave a method for constructing solutionsof r V = λV from scalar eigenfunctions of the Laplace operator (solutions of∆u = −λ2u) and simple (usually parallel) vector fields. This technique was reviewedin [CDG1], where we showed that it can be used to construct all of the eigenfields ofr V = λV that are divergence free and tangent to the boundary of the flat solidtorus.

The vector fields constructed in the preceding section constitute all (or at least abasis for, since all of the eigenvalues are multiple) of the eigenfields of r V = λVthat are divergence free and tangent to the boundary of the ball B3(b) or the domainB3(a, b) between two concentric spheres. This follows from the fact that we havecomplete sets of solutions of the Dirichlet problem for the equation

∆u+2

r

∂u

∂r+

2

r2u = −λ2u

for the r components of the eigenfields on both domains, and from Propositions 4 and6, which show that for each solution u of this equation, either there is a unique pairof functions v and w that complete u to a solution V = ur+ vθ+wφ of rV = λVor else there is none.

It is also the case that all of these vector fields were found by Chandrasekhar andKendall in [CK], where there was no discussion of their completeness as solutionsof our boundary-value problem. We have written our solutions so that they can be


immediately recognized as the same as those of equations (12), (13) and (14) of [CK],although there is a misprint in equation (13) there.

10. The largest eigenvalue of the Biot-Savart operator

As noted above, the divergence-free vector fields of a given energy that are tangentto the boundary of the domain Ω and have maximum helicity are the eigenfields ofthe modified Biot-Savart operator corresponding to the eigenvalue of largest absolutevalue. These eigenfields are the Amperian eigenfields of the curl operator correspond-ing to the nonzero eigenvalue of smallest absolute value (because the eigenvalues arereciprocals).

Thus, to find the helicity maximizers on our domains B3(b) and B3(a, b), we haveto find the smallest eigenvalues of r V = λV . Since we have in our possessionall of the solutions of this equation, the task is not too difficult. We will prove thefollowing:

Proposition 8. The smallest non-zero eigenvalue of rV = λV on B3(b) and on

B3(a, b) is λ(1)1 , in the notation of Propositions 2 and 3.

Proof. Since the eigenvalues certainly satisfy

λ(n)1 < λ(n)

2 < λ(n)3 <

(since these are successive zeroes of some combination of Bessel functions), in orderto prove the proposition, we need only show that

λ(1)1 < λ(2)

1 < λ(3)1 < ,

i.e., that λn11 < λn2

1 if n1 < n2.

Recall from Propositions 2 and 3 and their proofs that the eigenvalues λ(n)k are the

values of λ for which the expression

Sn(r) = AJn+12(λr) +BYn+1

2(λr)

is nonzero but vanishes for both r = a and r = b, where A and B are chosen so thatSn(a) = 0. For the ball, a = 0, which implies that B = 0 but we have Sn(0) = 0 inthis case as well, since the function u(r, θ, φ) is smooth and has an additional factorof r−3=2. Being a linear combination of Bessel functions of order n+ 1

2with argument

λr, the function Sn satisfies the differential equation

r2d2Sndr2

+ rdSndr

+(λ2r2 − (n+

1

2)2)Sn = 0

(see [St], p. 404).

We shall use Sturm’s comparison theorem (see [S]). Suppose f(x) is a nonzerofunction that satisfies f ′′ + P (x)f = 0 with f(a) = 0 and g(x) is a nonzero functionthat satisfies g′′ + Q(x)g = 0 with g(a) = 0 (or we could assume f ′(a) = g′(a) = 0).


Sturm’s comparison theorem tells us that if P (x) Q(x) for all x, then on anyinterval containing a and no (other) zeroes of f(x), there can be no (other) zeroes ofg(x). Moreover, the “next” zero of f(x) must be closer to a than that of g(x).

In order to apply Sturm’s theorem to our problem, we need to make a change ofvariables to eliminate the first-order term in the differential equation. If we defineF (r) =

pr Sn (r), then F will satisfy the equation

d2F

dr2+

(λ2 − n(n+ 1)

r2

)F = 0.

Suppose F (1)1 is the solution of our equation that satisfies F (1)

1 (x) > 0 for all x between

a and b, and F (1)1 (b) = 0. So F (1)

1 satisfies

d2F (1)1

dr2+(

(λ(1)1 )2 − 2

r2

)F (1)

1 = 0.

For any λ < λ(1)1 , and any n > 1, we will certainly have

λ2 − n(n+ 1)

r2< (λ

(1)1 )2 − 2

r2.

Sturm’s theorem tells us that no nonzero F corresponding to such values of λ and ncan satisfy F (a) = F (b) = 0. Therefore all of the other eigenvalues of the boundary-

value problem at hand must be larger than λ(1)1 . Repeated application of the argument

will prove the stronger inequality λn11 < λn2

1 if n1 < n2. This proves Proposition 8.

With the help of Maple, we calculate the following table of values of λ(1)1 for

B3(a, b), where a = (b3 − 1)1=3 (so all of the domains have the same volume as theunit ball):

a b λ(1)1 a b λ(1)

1

0 1 4.493409458 1.334200824 1.5 18.97447753.1442730320 1.001 4.568933565 1.912931183 2 36.08896092.2470322424 1.005 4.799466836 4.986630952 5 234.9901755.3117589910 1.01 5.033680133 9.996665555 10 942.1635951.5401839794 1.05 6.423856305 99.99996667 100 94257.20533.6917396417 1.1 7.858576168 999.9999997 1000 9424777.95775

All of the eigenfields corresponding to the first eigenvalue on spherically symmetric

domains are axially symmetric. The multiplicity of the λ(1)1 is three, corresponding to

three dimensions’ worth of choices for the axis of symmetry. Maple plots of the threecomponent functions (u, v and w in V = ur+ vθ+wφ) corresponding to b = 1.02 ona constant-φ section of the domain B3(a, b) are as follows:


This yields the picture in the introduction of the vector field with maximal helicityon this spherical shell (recall that the eigenvalues of the Biot-Savart operator arereciprocals of those of the curl operator, and the eigenfields are the same). Theintersections in the invariant surfaces under the flow of the vector field with a constant-φ section of the domain are shown in the first plot in the next section.

11. Other eigenelds of the Biot-Savart operator

It is interesting to examine the nature of the eigenfields of the Biot-Savart operator

corresponding to the other eigenvalues λ(n)k of the Biot-Savart operator. We will look

at the axially symmetric (m = 0) solutions corresponding to other eigenvalues. Thetopology of these solutions is essentially the same whether the domain is B3(b) orB3(a, b), so we will exhibit them only for the ball. Each is made up of several familiesof invariant tori (nk of them), half of which cycle in one direction around the z-axisand the other half cycle the other way. We look at plots of the intersections withthe φ = 0 half-plane of the surfaces that are invariant under the flow defined by the

eigenfields corresponding to λ(1)1 , λ

(2)1 , λ

(3)1 , λ

(1)2 , λ

(1)3 and λ

(2)2 below. Here are the first

three plots:


For the ball B3(1):

n = 1 n = 2 n = 3 n = 4 n = 5

k = 1 4.49341 5.76346 6.98792 8.18256 9.35581k = 2 7.72525 9.09501 10.4171 11.7949k = 3 10.9041 12.3229 13.6980 15.0397k = 4 14.0662 15.5146 16.9236k = 5 17.2208 18.6890 20.1218k = 6 20.3713 21.8539 23.3042

For the shell B3(0.312, 1.01):

n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

k = 1 5.03368 5.92554 6.99240 8.12397 9.26946 10.4104k = 2 9.32168 9.93620 10.7867k = 3 13.7236k = 4 18.1695

For the shell B3(0.692, 1.1):

n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 11 n = 12

k = 1 7.85858 8.17541 8.62800 9.19535 9.85647 10.5924 14.9312 15.8707k = 2 15.4746

For the shell B3(1.913, 2):

n = 1 n = 2 n = 3 n = 20 n = 100 n = 121 n = 122 n = 125

k = 1 36.0890 36.1034 36.1251 37.5717 62.7739 71.8208 72.2631 73.5952k = 2 72.1671 72.1743


REFERENCES

[CDG1] J. Cantarella, D. DeTurck and H. Gluck, Upper bounds for the writhingof knots and the helicity of vector fields, preprint, March 1997.

[CDG2] J. Cantarella, D. DeTurck and H. Gluck, Hodge decomposition of vectorfields on bounded domains in 3-space, preprint, December 1997.

[CDG3] J. Cantarella, D. DeTurck and H. Gluck, The Biot-Savart operator forapplication to knot theory, fluid mechanics and plasma physics, preprint, December1997.

[CDGT1] J. Cantarella, D. DeTurck, H. Gluck and M. Teytel, Isoperimetric prob-lems for the helicity of vector fields and the Biot-Savart and curl operators, preprint,November 1998.

[CDGT2] J. Cantarella, D. DeTurck, H. Gluck and M. Teytel, Influence of geom-etry and topology on helicity, preprint, November 1998.

[CK] S. Chandrasekhar and P.C. Kendall, On Force-Free Magnetic Fields, Astro-physical Jornal 126 (1957) 457-460.

[Le] N.N. Lebedev, Special Functions and their Applications, Dover Publications,1972.

[Lu] S. Lundquist, Magneto-hydrostatic fields, Archiv Fysik 2, number 35, (1951)361-365.

[M] H.K. Moffatt, The degree of knottedness of tangled vortex lines, J. Fluid Mech.35 (1969) 117-129 and 159, 359-378.

[S] G.F. Simmons, Differential Equations with Applications and Historical Notes,second edition, McGraw-Hill, 1991.

[St] J.A. Stratton, Electromagnetic Theory, first edition, McGraw-Hill, 1941.

[TS] A.N. Tikhonov and A.A. Samarskii, Equations of Mathematical Physics, Perg-amon Press, 1963.

[W1] L. Woltjer, The Crab Nebula, Bull. Astr. Netherlands 14 (1958) 39-80.

[W2] L. Woltjer, A theorem on force-free magnetic fields, Proc. Nat. Acad. Sci.USA 44 (1958) 489-491.

Department of MathematicsUniversity of PennsylvaniaPhiladelphia, PA 19104-6395

[email protected]

[email protected]@[email protected]

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