Twist maps as energy minimisers in homotopy classes: symmetrisation and the coarea formula
Charles Morris Ali Taheri









   








Twist maps as energy minimisers in homotopy classes: symmetrisation and the coarea formula
    C. Morris, A. Taheri^†
    
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Let = [a, b] = {x: a<|x|<b}⊂^n with 0<a<b<∞ fixed be an open annulus and consider the energy 
functional,

    𝔽 [u; ] = 1/2∫_|∇ u|^2/|u|^2  dx,

over the space of admissible incompressible Sobolev maps 

    𝒜_ϕ() = {  u ∈ W^1,2(, ^n) : ∇ u = 1  a.e. in  and u|_∂ = ϕ},

where ϕ is the identity map of . Motivated by the earlier works <cit.> in this paper we examine 
the twist maps as extremisers of 𝔽 over 𝒜_ϕ() and investigate their minimality properties by 
invoking the coarea formula and a symmetrisation argument. In the case n=2 where 𝒜_ϕ() 
is a union of infinitely many disjoint homotopy classes we establish the minimality of these extremising twists in their respective 
homotopy classes a result that then leads to the latter twists being L^1-local minimisers of 𝔽 in 𝒜_ϕ(). 
We discuss variants and extensions to higher dimensions as well as to related energy functionals. 













§ INTRODUCTION AND PRELIMINARIES
 




Let = [a, b] = {(x_1, ..., x_n) : a < |x| < b} with 0<a<b<∞ fixed be an open annulus in ^n and consider the energy 
functional

    𝔽[u; ] = 1/2∫_|∇ u|^2/|u|^2   dx,

over the space of incompressible Sobolev maps, 

    𝒜_ϕ () = {  u ∈ W^1,2(,^n) : ∇ u = 1  a.e. in  and u|_∂ = ϕ}.

Here and in future ϕ denotes the identity map of  and so the last condition in (<ref>) means that 
u ≡ x on ∂ in the sense of traces. 



By a twist map u on ⊂^n we mean a continuous self-map of  onto itself which agrees with the identity map ϕ 
on the boundary ∂ and has the specific spherical polar representation (see <cit.>-<cit.> for background and further results) 

    u : ( r, θ) ↦(r,Q(r)θ),      x ∈.

Here r = |x| lies in [a, b] and θ = x/|x| sits on 𝕊^n-1 with Q ∈ C([a,b],  SO(n)) satisfying Q(a)=Q(b)=I. 
Therefore Q forms a closed loop in SO(n) based at I and for this in sequel we refer to Q as the twist loop associated with 
u. Also note that (<ref>) in cartesian form can be written as

    u: x ↦ Q(r) x = r Q(r) θ,      x ∈.

Next subject to a differentiability assumption on the twist loop Q it can be verified that u ∈𝒜_ϕ() with its 𝔽 
energy simplifying to  

    𝔽[Q(r) x; ]    = 1/2∫_|∇ u|^2/|u|^2 dx = 1/2∫_|∇ Q(r)x|^2/|x|^2  dx 
       = n/2∫_dx/|x|^2 + ω_n/2∫_a^b Q̇^2  r^n-1  dr,
 
where the last equality uses |∇ [Q(r)x]|^2 = n + r^2 |Q̇θ|^2. Now as the primary task here is to search for extremising twist 
maps we first look at the Euler-Lagrange equation associated with the loop energy 𝔼=𝔼[Q] defined by the last integral 
in (<ref>) over the loop space {Q ∈ W^1,2([a, b];  SO(n)) : Q(a)=Q(b)=I}. Indeed this can be shown to take the form 
(see below for justification)

    d/dr[ ( r^n-1Q̇)Q^t ] =0,
 
with solutions  

Q(r) =  exp [-β(r) A] P,

where P ∈ SO(n), A ∈^n × n is skew-symmetric and β=β(|x|) is described for a ≤ r ≤ b by

    β(r) = ln 1/r    n=2,
    
    r^2-n/(n-2)    n≥ 3.

Now to justify (<ref>) fix Q∈ W^1,2([a,b],  SO(n)) and for F ∈ W^1,2_0([a,b], ^n × n) 
set H= (F-F^t)Q and Q_ϵ = Q + ϵ H. Then Q_ϵ^tQ_ϵ = I + ϵ^2 H^tH and 


    d/dϵ∫_a^b 2^-1|Q̇_ϵ|^2 r^n-1  dr |_ϵ = 0   = ∫_a^b ⟨Q̇ , (Ḟ - Ḟ^t )Q + (F - F^t )Q̇⟩ r^n-1  dr 
       = ∫_a^b ⟨Q̇ , (Ḟ - Ḟ^t )Q ⟩ r^n-1  dr 
       = ∫_a^b ⟨d/dr (r^n-1Q̇Q^t) , (F - F^t ) ⟩ dr = 0,

and so the arbitrariness of F with an orthogonality argument gives (<ref>). 



Returning to (<ref>) it is not difficult to see that the Euler-Lagrange equation associated with 𝔽 
over 𝒜_ϕ() is given by the system (cf. Section <ref>)

    ∇ u^2/u^4 u + div {∇ u/u^2 - p(x)  cof ∇ u }  = 0,      u=(u_1, ..., u_n),

where p=p(x) is a suitable Lagrange multiplier. Here a further analysis reveals that out of the solutions Q=Q(r) to 
(<ref>) just described only those twist loops in the form    

    Q(r) = R  diag[ R[g](r), ...,  R[g](r)]R^t,     R∈𝐒𝐎(n),

when n is even and Q(r) ≡ I (a ≤ r ≤ b) when n is odd can grant extremising twist maps u for the original energy 
(<ref>). For clarification R[g] denotes the SO(2) matrix of rotation by angle g:

    R [g] = [ [  cos g  sin g; -sin g  cos g;        ]].

Indeed direct computations give the angle of rotation g=g(r) to be   

    g(r) = 2π k log(r/a)/log(b/a) + 2π m,      k, m ∈,

when n=2 and 

    g(r) = 2π k (r/a)^2-n-1/(b/a)^2-n-1 + 2π m,       k, m ∈,

when n ≥ 4 even. (See also <cit.>, <cit.>, <cit.> for complementing and further results.)



Our point of departure is (<ref>)-(<ref>) and the aim is to study the minimising properties the twist maps calculated 
above. Of particular interest is the case n=2 where the space 𝒜_ϕ() admits multiple homotopy classes (A_k : k ∈). 
Here direct minimisation of the energy over these classes gives rise to a scale of associated minimisers (u_k). Using a symmetrisation 
argument and the coarea formula we show that the twist maps u_k with twist angle g as presented in 
(<ref>) are indeed energy minimisers in A_k and as a result also L^1 local minimisers of 𝔽 over 
𝒜_ϕ(). We discuss variants and extensions including a larger scale of energies where similar techniques can be applied 
to establish minimality properties in homotopy classes. 





§ THE HOMOTOPY STRUCTURE OF THE SPACE OF SELF-MAPS 𝔄=𝔄()
 




The rich homotopy structure of the space of continuous self-maps of the annulus =[a, b] ⊂^n will prove 
useful later on in constructing local energy minimisers. For this reason here we give a quick outline of the main tools and results and refer the 
reader to <cit.> for further details and proofs. To  this end set 
𝔄=𝔄 ()= { f ∈ C(, ) : f =ϕ}



equipped with the uniform topology. A pair of maps f_0, f_1 ∈𝔄 are homotopic iff there exists 
H ∈ C([0, 1] ×; ) such that, firstly, H(0, x) = f_0 (x) for all x∈, 
secondly, H(1, x) = f_1 (x) for all x∈ and finally H(t, x) = x for all t ∈ [0,1], x ∈∂. 
The equivalence class consisting of all g ∈𝔄 homotopic to a given f ∈𝔄 is referred to as the homotopy 
class of f and is denoted by [f]. Now the homotopy classes {[f] : f ∈𝔄} can be characterised as follows depending 
on whether n=2 or n ≥ 3. 








  * (n=2)  Using polar coordinates, for f ∈𝔄 and for θ∈ [0, 2 π] (fixed), the ^1-valued curve 
γ_θ defined by 

    γ_θ:  [a, b] →^1 ⊂^2,     γ_θ: r ↦  f|f|^-1(r, θ),
 
has a well-defined index or winding number about the origin. Furthermore, due to continuity of f, this index is independent of the 
particular choice of θ∈ [0, 2 π]. This assignment of an integer (or index) to a map f ∈𝔄 will be denoted by 

    f ↦ deg(f|f|^-1).
 
Note firstly that this integer also agrees with the Brouwer degree of the map resulting from identifying ^1 ≅ [a, b]/{a, b},  
justified as a result of γ_θ(a)=γ_θ(b) and secondly that for a differentiable curve (taking advantage of the 
embedding ^1 ⊂) we have the explicit formulation

    deg (f|f|^-1) = 1/2 π i∫_γdz/z.




  (n=2). The map deg: { [f] : f ∈𝔄}→ 
is bijective. Moreover, for any pair of maps f_0, f_1 ∈𝔄, we have

    [f_0]=[f_1]  deg(f_0 |f_0|^-1) =  deg(f_1|f_1|^-1).
 





  * (n ≥ 3) Using the identification ≅ [a,b] ×^m where for ease of notation we have set m=n-1 it is plain 
that for f ∈𝔄 the map 
[Here as usual ϕ denotes the identity map of the m-sphere and C_ϕ(^m, ^m) 
is the path-connected component of C(^m, ^m) containing ϕ.]

    ω: [a, b] → C_ϕ(^m, ^m),     ω: r ↦ f|f|^-1 (r, ·),

uniquely defines an element of the fundamental group π_1[ C_ϕ(^m, ^m)]. 
By considering the action of SO(n) on ^m – viewed as its group of 
orientation preserving isometries, i.e., through the assignment, 

    E: ξ∈ SO(n) ↦ω∈ C(^m, ^m),
 
where 

    ω(x) =  E[ξ](x) = ξ x,      x ∈𝕊^m,
 
it can be proved that the latter assignment induces a group isomorphism on the level of the fundamental groups, namely, 

    E^⋆: π_1[ SO(n), I_n] ≅π_1 [ C_ϕ(^m, ^m), ϕ] ≅_2.

Thus, summarising, we are naturally lead to the assignment of an integer mod 2 to any f ∈𝔄 which will 
be denoted by

    f ↦ deg_2 (f|f|^-1) ∈_2.




 (n ≥ 3) The degree mod 2 map deg_2: { [f] : f ∈𝔄}→_2 
is bijective. Moreover, for a pair of maps f_0, f_1 ∈𝔄, we have 

    [f_0]=[f_1]  deg_2(f_0 |f_0|^-1) =  deg_2(f_1|f_1|^-1).
 










§ A COUNTABLE FAMILY OF L^1 LOCAL MINIMISERS OF 𝔽 WHEN N=2





When n=2 by Lebesgue monotonicity and degree theory (see <cit.> as well as <cit.>,<cit.>,<cit.>,<cit.>) 
every map u in 𝒜 = 𝒜_ϕ() has a representative (again denoted u) 
in 𝔄. As a consequence we can introduce the components – hereafter called the 
homotopy classes, 

    A_k := {u ∈𝒜:  deg (u|u|^-1) = k },      k ∈.

Evidently A_k are pairwise disjoint and their union (over all k ∈) gives 𝒜. Furthermore it can be seen 
without difficulty that each A_k is W^1,2-sequentially weakly closed and that for u ∈ A_k and s>0 there 
exists δ=δ(u, s)>0 with








    {v : 𝔽[v]< s}∩𝔹^L^1_δ(u) ⊂ A_k.

Here 𝔹^L^1_δ(u) = {v ∈𝒜 : ||v-u||_L^1 < δ}, that is, the L^1-ball in 𝒜 centred at u. 
Indeed for the sequential weak closedness fix k and pick (u_j : j ≥ 1) ⊂ A_k so that u_j  u in W^1,2. 
Then by a classical result of Y. Reshetnyak 

    ∇ u_j ∇ u
 
(as measures) and so u ∈𝒜 while u_j → u uniformly on  gives by Proposition <ref> 
that u ∈ A_k. For the second assertion arguing indirectly and assuming 
the contrary there exist u ∈ A_k, s>0 and (v_j : j ≥ 1) in 𝒜 such that 𝔽[v_j; ] < s 
and ||v_j - u||_L^1→ 0 
  






while v_j ∉ A_k. However by passing to a 
subsequence (not re-labeled) v_j  u in W^1,2(, ^2) and as above 
v_j → u uniformly on . Hence again by Proposition <ref>, v_j ∈ A_k for large enough j 
which is a contradiction. □





Now in view of the sequential weak lower semicontinuity of 𝔽 in 𝒜 (see below) an application 
of the direct methods of the calculus of variations leads to the following existence and multiplicity result.



  (Local minimisers) Let = [a, b] ⊂^2 and for k ∈ consider the homotopy classes A_k as defined by 
(<ref>). Then there exists u= u (x; k) ∈ A_k such that

    𝔽[u; ] = inf_ v ∈ A_k𝔽[v; ].

Furthermore for each such minimiser u there exists δ=δ(u)>0 such that

    𝔽[u; ] ≤𝔽[v; ],

for all v ∈𝒜_ϕ() satisfying ||u-v||_L^1<δ.




Fix k and pick (v_j) ⊂ A_k an infimizing sequence: 𝔽[v_j] ↓ L:= inf_A_k𝔽[·]. Then as L<∞ and 
a ≤ |v(x)| ≤ b for v ∈𝒜 it follows that by passing to a subsequence (not re-labeled) v_j  u in W^1,2(, ^2) and uniformly in 
 where by the above discussion u ∈ A_k. Now  

    | ∫_|∇ v_k|^2/|v_k|^2 - ∫_|∇ v_k|^2/|u|^2|    ≤∫_ |∇ v_k|^2 ( ||v_k|^2-|u|^2|/|u|^2 |v_k|^2)  
       ≤sup_||v_k|^2-|u|^2|/|u|^2 |v_k|^2∫_ |∇ v_k|^2 → 0

as k ↗∞ together with 

    ∫_|∇ u|^2/|u|^2≤lim∫_|∇ v_k|^2/|u|^2

gives the desired lower semicontinuity of the 𝔽 energy on 𝒜_ϕ() as claimed, i.e., 

    𝔽[u; ]=∫_|∇ u|^2/2|u|^2≤lim∫_|∇ v_k|^2/2|v_k|^2 = lim 𝔽[v_k; ].

As a result L ≤𝔽[u] ≤lim inf𝔽[v_j] ≤ L and so u is a minimiser as required. 
To justify the second assertion fix k ∈ and u as above and with s=1+𝔽[u] pick δ>0 as in the discussion prior to the theorem. 
Then any v ∈𝒜 satisfying ||u-v||_L^1 < δ also satisfies (<ref>) [otherwise 
𝔽[v]<𝔽[u] < s implying that v ∈ A_k and hence in view of u being a minimiser, 𝔽[v] ≥𝔽[u] 
which is a contradiction.] □





§ TWIST MAPS AND THE EULER-LAGRANGE EQUATION ASSOCIATED WITH 𝔽
 




The purpose of this section is to formally derive the Euler-Lagrange equation associated with 𝔽 
over 𝒜_ϕ(). Note that 𝔽[u] can in principle be infinite if |u| is too small or 
zero, however, for twist maps or more generally L^n-integrable maps in 𝒜_ϕ, |u| is bounded away from zero  
as u is a self-map of  onto itself. Moreover ∇ u is 
L^1-integrable for the latter maps but not in general for maps u of Sobolev class W^1,2 (with n ≥ 3).




Now the derivation uses the Lagrange multiplier method and proceeds formally by considering the unstrained functional

    𝕂[u; 𝕏] = ∫_[ ∇ u^2/2u^2 - p(x) (∇ u -1 )  ]   dx,
 
for suitable p=p(x) where evidently 𝕂[u; ]=𝔽[u; ] when u ∈𝒜_ϕ(). 
We can calculate the first variation of this energy by setting 
d/dε 𝕂[u_ε; ] |_ε=0=0, where u ∈𝒜_ϕ() is sufficiently 
regular and satisfies |u| ≥ c>0 in , u_ε=u+εφ for all φ∈ C^∞_c(,^n) and 
ε∈ sufficiently small, hence obtaining,

    0 =   d/d ε∫_[ ∇ u_ε^2/2u_ε^2 
    - p(x) (∇ u_ε -1 )  ]   dx |_ε=0
    
    =    ∫_{∑_i,j=1^n [ 1/u^2∂ u_i/∂ x_j - p(x) [ cof∇ u]_ij] 
    ∂φ_i/∂ x_j - ∑_i=1^n ∇ u^2/u^4u_i φ_i } dx  
    
    =    ∫_{ - ∑_i,j=1^n ∂/∂ x_j[ 1/u^2∂ u_i/∂ x_j 
    - p(x) [ cof∇ u]_ij] 
    φ_i  - ∑_i=1^n  ∇ u^2/u^4 u_i φ_i } dx 
    
    =    ∫_ - ∑_i=1^n{∇ u^2/u^4 u_i + ∑_j=1^n ∂/∂ x_j[ 1/u^2∂ u_i/∂ x_j - p(x) [ cof∇ u]_ij]    }φ_i   dx.

As this is true for every compactly supported φ as above an application of the fundamental lemma 
of the calculus of variation results in the Euler-Lagrange system for u=(u_1, ..., u_n) in : 

    ∇ u^2/u^4 u + div {∇ u/u^2 - p(x)  cof ∇ u }  = 0,

where the divergence operator is taken row-wise. Proceeding further an application of the Piola identity on the 
cofactor term gives (with 1 ≤ i ≤ n)

    ∇ u^2/u^4 u_i + ∑_j=1^n {∂/∂ x_j( 1/u^2∂ u_i/∂ x_j) - [ cof ∇ u]_ij∂ p/∂ x_j} = 0.

Next expanding the differentiation further allows us to write

    0     = ∇ u^2/u^4 u_i + ∑_j=1^n {1/u^2∂^2 u_i /∂ x_j^2 
    - 2/u^4∑_k=1^n ∂ u_i/∂ x_j∂ u_k /∂ x_j u_k  
    - [ cof ∇ u]_ij∂ p/∂ x_j}
        = ∇ u^2/u^4 u_i + ∑_j=1^n {1/u^2∂^2 u_i /∂ x_j^2 
    - 2/u^4∂ u_i/∂ x_j [∇ u^tu]_j - [ cof ∇ u]_ij∂ p/∂ x_j}.




Finally transferring back into vector notation and invoking the incompressibilty condition ∇ u =1 it follows in turn that 

    Δ u/u^2 + ∇ u^2/u^4 u - 2/u^4∇ u (∇ u)^tu = ( cof ∇ u) ∇ p,

and subsequently 

    (∇ u)^t/u^2[ Δ u + ∇ u^2/u^2 u - 2/u^2∇ u (∇ u)^tu ] = ∇ p.

Thus the Euler-Lagrange system (<ref>) is equivalent to (<ref>) that in particular asks for the nonlinear term on the left of (<ref>) to be a gradient field 
in . Recall from earlier discussion that restricting 𝔽 to the class of twist maps results in the Euler-Lagrange equation (<ref>) 
where the solution Q=Q(r) as explicitly computed is the twist loop associated with the map 

    u: (r, θ) ↦ (r, Q(r) θ),      x ∈,
  
with Q(r)=exp [- β(r)A] P, P ∈𝐒𝐎(n), A ∈^n × n skew-symmetric (A^t=-A). 
The boundary condition u=ϕ on ∂ gives,
[The function β=β(r) was introduced earlier in Section <ref>.] 

    exp[-β(a) A] P=I,     exp[-β(b) A] P=I.

Therefore it must be that P= exp[β(a) A] and exp ([β(b)-β(a)] A)=I. Now as A lies in 𝔰𝔬(n) it must be conjugate 
to a matrix S in the Lie algebra of the standard maximal torus of orthogonal 2-plane rotations in 𝐒𝐎(n). This means that 
there exists R ∈𝐒𝐎(n) such that A = RSR^T for some S as described and so S∈ (β(b)-β(a))^-1𝕃 
where 𝕃={ T∈𝔱: exp(T)=I }, that is, 
𝕃 is the lattice in the Lie subalgebra 𝔱⊂𝔰𝔬(n) consisting of matrices sent by the 
exponential map to the identity I of SO(n). Hence Q(r)=R exp (-[β(r)-β(a)] S) R^t. 
Next upon noting that the derivatives of β=β(r) are given by

    β̇ (r)    = -1/r^n-1 ,     β̈ (r) = n-1/r^n,

we can write

    Q̇ =  A Q/r^n-1,     Q̈ = A^2Q/r^2n-2-(n-1)AQ/r^n.




Now, moving forward, a set of straightforward calculations show that for a twist map u with a twice continuously differentiable twist loop Q=Q(r) we have the differential relations 

    (∇ u)^t    = Q^t + r θ⊗Q̇θ, 
    ∇ u^2    = tr [(∇ u)^t (∇ u)] = n + r^2 Q̇θ^2,

and likewise

    Δ u = [ (n+1)Q̇ + r Q̈]θ.




Thus for the particular choice of a twist map with twist loop arising from a solution to (<ref>) the above quantities can be explicitly described by 
the relations

    (∇ u)^t    = Q^t + r^2-nθ⊗ AQθ, 
    ∇ u^2    = tr [ (Q^t + r^2-nθ⊗ AQθ) (Q + r^2-n AQ θ⊗θ) ] 
       = n + AQθ^2/r^2(n-2),

and likewise 

    Δ u    = [ (n+1) AQ/r^n-1 + r( A^2 Q/r^2n-2 -(n-1) AQ/r^n)   ] θ
        = [ 2AQ/r^n-1  + A^2Q/r^2n-3]θ.

For the ease of notation we shall hereafter write ω = Qθ. Proceeding now with the calculations and using (<ref>)-(<ref>) we have 

    Δ u + ∇ u^2/u^2u     = [ 2A/r^n-1  + A^2/r^2n-3  
    +  1/r(n + Aω^2/r^2n-4) I ]ω

and in a similar way 

    ∇ u (∇ u)^t/u^2u     = [  Q + r^2-n Aω⊗θ] 
    [  Q^t + r^2-nθ⊗ Aω] ω/r
        = [ I + Aω⊗θ Q^t +  Qθ⊗ Aω/r^n-2  
    + Aω⊗ Aω/r^2n-4] ω/r
        = [ I + Aω⊗ω +  ω⊗ Aω/r^n-2  
    + Aω⊗ Aω/r^2n-4] ω/r. 
    
        =  1/r (I + r^2-n A) ω,

where the last identity here uses (x⊗ y)z = ⟨ y,z⟩ x and ⟨ω,ω⟩ = ⟨ Qθ,Qθ⟩=1 along with 
⟨ Aω,ω⟩ =0 for skew-symmetric A. Hence putting together (<ref>) and (<ref>) gives

    Δ u  + ∇ u^2/u^2u - 2 ∇ u (∇ u)^t/u^2u = [  A^2+ Aω^2 I/r^2n-2  
    +  (n-2) I ] ω/r,

which when combined with (<ref>) results in

    (<ref>)   =(∇ u)^t/u^2[ Δ u  + ∇ u^2/u^2u - 2 ∇ u (∇ u)^t/u^2u ] 
       = 1/r^2[  Q^t + θ⊗ Aω/r^n-2] 
    [  A^2 + Aω^2 I/r^2n-4  +  (n-2) I ] ω/r
       =  Q^t [  A^2+ Aω^2 I/r^2n-1  
    +  n-2/r^3 I ]ω + (θ⊗ Aω )A^2 ω/r^3n-3.

Noting (θ⊗ Aω )A^2 ω = ⟨ Aω,A^2ω⟩θ = ⟨μ,Aμ⟩ = 0 
with A skew-symmetric and μ = Aω the last set of equations give

    (<ref>)   = (∇ u)^t/u^2[ Δ u  + ∇ u^2/u^2u - 2 ∇ u (∇ u)^t/u^2u ] 
       = Q^t [  A^2+ Aω^2 I/r^2n-2  +  n-2/r^2 I ] ω/r =  I.




Therefore to see if (<ref>) admits twist solutions it suffices to verify if the quantity described by (<ref>) is a gradient field in . Towards this end 
recall that here we have Q(r) = exp(-β(r)A)P where as seen P=exp(β(a)A). Thus a basic 
calculation gives

    Aω^2    = AQθ^2 =  θ^t Q^t A^t A Q θ = - θ^t Q^t A^2 Q θ
        =  - θ^t P^t exp(β(r)A) A^2 exp(-β(r)A)P θ
       = - θ P^t A^2 Pθ = θ P^t A^t A Pθ
       = AP θ^2,

and likewise by substitution we have

    Q^tA^2 ω = P^t A^2 P θ.
 



Hence using the above we can proceed by writing the Euler-Lagrange equation (<ref>) upon substitution as,

    I   = (∇ u)^t/u^2[ Δ u  + ∇ u^2/u^2u - 2 ∇ u (∇ u)^t/u^2u ] 
       = P^t ( A^2+ AP θ^2I ) P θ/r^2n-1  
    +  (n-2) θ/r^3.

Now as for a fixed skew-symmetric matrix B by basic differentiation we have 

    ∇( By^2 ) = -2B^2y,     ∇y^2n = 2ny^2n-2y,

it is evident that we can write 

    - ∇( By^2/2ny^2n) = B^2y/ny^2n + By^2y/y^2n+2.

In particular with B=P^tAP being skew-symmetric, (<ref>) can be written in the form

    I   = (∇ u)^t/u^2[ Δ u  + ∇ u^2/u^2u - 2 ∇ u (∇ u)^t/u^2u ] 
       = - ∇( P^tAP x^2/2nx^2n) + (n-1) P^tA^2 P x/ nx^2n  +  
    - (n-2) ∇1/|x|.

Therefore it is plain that (<ref>) is a gradient field in  provided that the term on the right and subsequently the middle 
term, that is, the expression 

    (n-1)P^tA^2 P x/n x^2n
 
is a gradient field in . By direct calculations (cf. <cit.>) this is seen to be the case iff all the eigenvalues of the skew-symmetric matrix A 
are equal. (Note that in odd dimensions this requirement leads to A=0.) As a result here (<ref>) would be a gradient (indeed ∇ p) 
and so the Euler-Lagrange system (<ref>) is satisfied by the twist u. 




Now using the representation A= RSR^t for some S ∈ (β(b)-β(a))^-1𝕃 and writing S = λ J where, J is the n × n 
block diagonal matrix: J=0 when n is odd and J=  diag( A_1, ⋯,  A_n/2) when n is even, i.e., 

    J = [[   𝐀_1     0     ⋯     0;     0   𝐀_2     ⋯     0;     ⋮           ⋱     ⋮;     0     0     ⋯ 𝐀_n/2;       ]] 
         A_j= [  0 -1;  1  0 ]


it is required that λ (β(b)-β(a)) J ∈𝕃. But invoking the lattice structure of 𝕃 this can happen iff


    λ = 2π k/β(b)-β(a),      k ∈,
 
and thus 

    u(x) = Rexp( -2kπβ(r)-β(a)/β(b)-β(a) J ) R^tx.

Noticing that here we have

    β(r)-β(a)/β(b)-β(a)   = log(r/a)/log(b/a),

for n=2 and 

    β(r)-β(a)/β(b)-β(a) = (r/a)^2-n-1/(b/a)^2-n - 1 ,

for even n ≥ 4 respectively we obtain the representation  

    u(x)    = Rexp( -g(r) J ) R^tx = exp( -g(r) A ) x,

where we have set A= R J R^t and the angle of twist function g=g(r) is given by (<ref>) for n=2 and 
(<ref>) for even n ≥ 4 respectively. 



For odd n ≥ 3 as shown A=0 and so the only twist solution to (<ref>) is the trivial solution u ≡ x. 






§ SYMMETRISATION AS A MEANS OF ENERGY REDUCTION ON 𝒜_Φ() WHEN N=2
 





Recall that the space of admissible maps 𝒜_ϕ() consists of maps u ∈ W^1,2(, ^2) satisfying the incompressibility condition 
∇ u = 1 a.e. in  and u|_∂ = ϕ. Also as mentioned earlier due to a Lebesgue-type monotonicity every such map is 
continuous on the closed annulus  and using degree theory the image of the closed annulus is again the closed annulus itself; hence, the 
"embedding" 

    𝒜_ϕ() = ⋃_k ∈ A_k ⊂𝔄(),
 
where the components A_k here are as defined by (<ref>). For the sake of future calculations it is useful to write (<ref>) as 

    deg (u|u|^-1) = 1/2π∫_a^b u × u_r/u^2  dr = k ∈,
 
where x=r. (Note that we adopt the convention that in two dimensions the cross product is a scalar and not a vector.) When u is a twist map, 
specifically, u=Q[g] x the integral reduces to g(b)-g(a) = 2π k where as before g=g(r) is the angle of rotation function. 
We now proceed by reformulating the 𝔽 energy of an admissible map u ∈𝒜_ϕ() in a more suggestive way. Indeed 
switching to polar co-ordinates it is seen that 

    ∇ u^2  = u_r^2 + 1/r^2u_θ^2
 
where 

    |u_r|^2 = (u · u_r)^2 + (u × u_r)^2/|u|^2,      
    |u_θ|^2 = (u · u_θ)^2 + (u × u_θ)^2/|u|^2.

Next we note that 

    (|u|_r)^2  = (u · u_r)^2/|u|^2,      (|u|_θ)^2  = (u · u_θ)^2/|u|^2.
 
Hence the gradient term on the left in (<ref>) can be expressed as  

    ∇ u^2     = (u · u_r)^2 + (u × u_r)^2/u^2 + (u · u_θ)^2 + (u × u_θ)^2 /r^2u^2
        = ∇u^2 + (u × u_r)^2/u^2 + (u × u_θ)^2 /r^2u^2.
 
From this we therefore obtain the 𝔽 energy as

    𝔽 [u; ]    = 1/2∫_∇ u^2/u^2   dx 
    = 1/2∫_0^2π∫_a^b ∇ u^2/u^2  r   dr dθ
       = 1/2∫^2π_0 ∫_a^b [ ∇u^2/u^2 +  (u × u_r)^2/u^4 + 
    (u × u_θ)^2 /r^2u^4] r   dr dθ.

Let us first state the following useful identity that will be employed in obtaining a fragment of the lower bound on the energy: For u ∈𝒜_ϕ() and 
a.e. r∈[a,b],

    ∫_0^2πu(r,θ)× u_θ(r,θ)/u^2  dθ = 2π.

The proof of this identity is postponed until later on in Section <ref> (cf. Proposition <ref>). 
Now assuming this for the moment an application of Jensen's inequality gives, again for a.e. r∈[a,b],

    1/2π∫_0^2π(u× u_θ)^2 /u^4  dθ   ≥(1/2π∫_0^2πu(r,θ)× u_θ(r,θ)/u^2  dθ)^2=1.

Hence it is plain that 

    ∫^2π_0 ∫_a^b (u × u_θ)^2 /r^2u^4 r   dr dθ≥ 2πln(b/a).

Therefore we have the following lower bound on the 𝔽 energy of an admissible map u:

    𝔽 [u; ]    ≥πln(b/a) + 1/2∫^2π_0 ∫_a^b [ ∇u^2/u^2 
    + (u × u_r)^2/u^4] r   dr dθ.

Interestingly here we have equality only for twist maps and so outside this class the inequality is strict (for more on questions of uniqueness 
see <cit.>). The next task is to show that by using a basic "symmetrisation" in 𝒜_ϕ() we can reduce the energy 
which will then be the main ingredient in the proof of the result. 



 (Symmetrisation) Let u ∈𝒜_ϕ() be an admissible map and associated with u define the angle of rotation 
function g=g(r) by setting 

    g(r) = 1/2π∫_a^r ∫_0^2πu × u_r/u^2  dθ dr,      a ≤  r ≤ b.

Then the twist map defined by u̅ (x) = Q[g] x with Q=Q[g]= R[g] has a smaller 𝔽 energy than the original map u, that is, 

    𝔽[u̅; ] ≤𝔽[u; ].
 
Furthermore if u ∈ A_k then the symmetrised twist map u̅ satisfies u̅∈ A_k. Thus the homotopy classes A_k are invariant under 
symmetrisation. 





Clearly the symmetrised twist map u̅ is in the same homotopy class as u since by definition g ∈ W^1,2[a, b] satisfies 
g(a)=0 and 

    g(b) = 1/2π∫_a^b ∫_0^2πu × u_r/u^2  dθ dr = 2 π k.
 
Therefore u̅∈ A_k as a result of (<ref>). Next the 𝔽 energy of u̅ satisfies the bound

    𝔽 [u̅; ] - 2πlog(b/a)    = π∫_a^b |Q̇(r)|^2 r   dr 
       = π∫_a^b |ġ(r)|^2 r   dr 
       = π∫_a^b [1/2π∫_0^2πu × u_r/u^2 dθ]^2 r   dr  
       ≤1/2∫_0^2π∫_a^b (u × u_r)^2/u^4 r   drdθ

where the last line is a result of Jensen's inequality. Therefore by referring to (<ref>) all that 
is left is to justify the inequality

    2πlog(b/a) ≤∫_∇u^2/u^2  dx.
 



Towards this end we use the isoperimetric inequality in the context of sets of finite perimeter and the coarea formula in the context 
of Sobolev spaces: For real-valued f and non-negative Borel g: 

    ∫_ g(x) |∇ f|   dx = ∫_∫_{ f=t } g(x)   dℋ^1(x)   dt.

(See, e.g., <cit.>.) Then upon taking f = |u| ∈ W^1,2() ∩ C() and g = 1/ |u|^2 this gives 

    ∫_∇u/u^2  dx    = ∫_a^b ( ∫_{u = t } dℋ^1 ) dt/t^2 
    = ∫_a^b ℋ^1({u=t })  dt/t^2.




Now since the level sets E_t = {x ∈ : u(x)≤ t } and F_t={x ∈ : x≤ t } enclose the same area due to 
∇ u = 1 a.e. (we can consider u as extended by identity inside {|x|<a}) an application of the isoperimetric 
inequality gives 2π t = ℋ^1({x=t }) = ℋ^1(∂^⋆ F_t) 
≤ℋ^1(∂^⋆ E_t) = ℋ^1({u=t }) for a.e. t∈ [a,b] (cf., e.g., 
<cit.>). Thus substituting in (<ref>) results in the lower bound

    ∫_∇u/u^2  dx     = ∫_a^b ℋ^1({u=t })  dt/t^2
       ≥∫_a^b ℋ^1({x=t })  dt/t^2 = ∫_a^b 2π t  dt/t^2 = 2πlog (b/a).
 



Finally we arrive at the conclusion by noting that u and ϕ have the same distribution function, that is, again as a result of the pointwise 
constraint ∇ u =1 a.e. in : 
α_u(t) = |{ x ∈ : u(x)≥ t } | = | { x ∈ : x≥ t } | = α_ϕ(t) and therefore

    ∫_dx/u^2 = ∫_a^∞ -2 α_u(t)  dt/t^3 + /a^2 
    = ∫_a^∞ -2 α_ϕ(t)  dt/t^3 + /a^2
    = ∫_dx/x^2 = 2πlog(b/a).




Now putting all the above together, a final application of Hölder inequality gives,

    (2πlog(b/a) )^2    ≤( ∫_∇u/u^2  dx )^2 
       ≤∫_∇u^2/u^2  dx ×∫_dx/u^2
       = 2 πlog(b/a)∫_∇u^2/u^2  dx
 
and thus eventually we have 

    2πlog(b/a) ≤∫_∇u^2/u^2  dx
 
and so the conclusion follows. 







§ THE 𝔽 ENERGY AND CONNECTION WITH THE DISTORTION FUNCTION





In this section we delve into the relationship between the energy functional 𝔽 in (<ref>) and the notions 
of distortion function and energy of geometric function theory. In particular we show that in two dimensions twist maps have minimum 
distortion among all incompressible Sobolev homeomorphisms of the annulus with identity boundary values in any given homotopy 
class. To fix notation and terminology let U, V ⊂^n be open sets and 
f ∈ W_loc^1,1(U, V). Then f is said to have finite outer distortion iff 
there exists measurable function K=K(x) with 1≤ K(x)<∞ such that 

    ∇ f(x)^n ≤ n^n/2 K(x) ∇ f (x).

The smallest such K is called the outer distortion of f and denoted by K_O(x, f). Note that here |A| = √( tr A^tA) is 
the Hilbert-Schmidt norm of the n × n matrix A. Naturally 1 ≤ K_O (x, f) < ∞ and it measures the deviation of f from 
being conformal. We also speak of the inner distortion function K=K_I(x, f) defined by the quotient 

    K_I(x,f) = n^-n/2 cof ∇ f^n/ ( cof ∇ f),

when ∇ f (x) ≠ 0 and K_I(x, f)=1 otherwise. We define the distortion energy associated to the inner distortion 
K_I(x, f) (<ref>) by the integral 

    𝕎[f; U]  = ∫_UK_I(x,f)/x^n  dx.

Related energies and more have been considered in <cit.> with close links to the work in <cit.>. 
The connection between the 𝔽 energy and the distortion energy 𝕎 (<ref>) is implicit in the following result 
of T. Iwaniec, G. Martin, J. Onninen and K. Astala <cit.>. (See also <cit.>, <cit.> and <cit.>.)  



 
Suppose f ∈ W^1,n_loc (, ) is a homeomorphism with finite outer distortion. Assume K_I is L^1-integrable over . 
Then the inverse map h=f^-1: → lies in the Sobolev space W^1,n(, ). 
Furthermore  

    n^-n/2∫_∇ h(y)^n/h(y)^n  dy = ∫_K(x,f)/x^n  dx.





The first assertion is Theorem 10.4 pp. 22 of <cit.>. For the second assertion using definitions we have

    n^n/2 K_I (x, f) = | cof ∇ f|^n/ ( cof ∇ f) = |(∇ f)^-1|^n ∇ f 
    =  |∇ h(f)|^n ∇ f,

and the conclusion follows upon dividing by |x|^n and integrating using the area formula. □






Now let 𝒜^n_ϕ() ={u ∈ W^1,n(; ^n) : ∇ u =1 }. 
As in Section <ref> each u in 𝒜^n_ϕ() admits a representative in 𝔄(). Now restricting to 
homeomorphisms u ∈𝒜^n_ϕ() the above theorem gives v=u^-1∈𝒜^n_ϕ() and so by the 
incompressibility constraint

    ∫_K_I(x,u)/x^n  dx 
    = n^-n/2∫_ cof ∇ u^n/x^n dx 
    = n^-n/2∫_∇ v^n/v^n dx.
  



In the planar case this allows us to relate the distortion energy of a homeomorphism u, say, in A_k to the 
𝔽 energy of the inverse map v=u^-1 in A_-k through

    𝔽[v;] = 1/2∫_∇ v(y)^2/v(y)^2  dy 
    =  1/2∫_∇ u(x)^2/x^2  dx = ∫_K_I(x,u)/x^2  dx = 𝕎[u;].




Therefore by showing that twist maps minimise the 𝔽 energy within their homotopy classes we have implicitly shown 
that twist maps minimise the distortion energy within their respective homotopy classes of homeomorphisms 
(as the inverse of a twist map is a twist map in opposite direction and clearly twist maps are homeomorphisms of annuli 
onto themselves). 



 The distortion energy 𝕎 has a minimiser u=u(x; k) (k∈) among all 
homeomorphisms within A_k. The minimiser is a twist map of the form u =Q[g] x 
where g(r) = 2π k ln(r/a)/ln(b/a).
In particular the minimum energy is given by,

    𝕎[u;] = ∫_K_I(x, u)/|x|^2  dx = 2πln(b/a) + 4π^3 k^2/ln(b/a).





That u=u_k minimises 𝕎 amongst homeomorphisms in A_k is a result of u_-k= (u_k)^-1 minimising 
𝔽 over A_-k (Proposition <ref>) and (<ref>). Indeed arguing indirectly assume there is a 
homeomorphism v∈ A_k: 𝕎[v; ] < 𝕎[u_k; ]. Then by (<ref>), 
𝔽[v^-1; ]<𝔽[u_-k; ] and this is a contradiction as 
v^-1, u_-k∈ A_-k while 𝔽[u_-k;] = inf_A_-k𝔽. We are thus left with the 
calculation of the 𝕎 energy of u=u_k. To this end put v=u^-1: 

    𝔽[v; ] =1/2∫_|∇ v|^2/|v|^2  dx    = 2πln (b/a) + π∫_a^b rġ_-k(r)^2   dr 
       = 2πln(b/a) + 4π^3 k^2/ln(b/a),

and so a further reference to (<ref>) completes the proof. □





In the higher dimensions, i.e. n>2, from Theorem <ref> we can get an analogous identity to (<ref>) for homeomorphisms 
u in 𝒜^n_ϕ. Indeed, with v=u^-1 

    n 𝔽[v; ] =  ∫_∇ v(y)^n/v(y)^n  dy 
     =  ∫_ cof ∇ u(x)^n/x^n  dx = n^n/2𝕎[u; ],

where the energies 𝔽=𝔽_n and 𝕎 are given by,

    𝔽[v; ] = 1/n∫_∇ v(y)^n/v(y)^n  dy,     𝕎[u; ]= ∫_K_I(x, u)/|x|^n  dx.




Therefore again to find minimisers of 𝕎 among homeomorphisms in u∈𝒜^n_ϕ() one can follows the lead of n=2 and 
consider the energy 𝔽 over 𝒜^n_ϕ(). It is straightforward to see that we have equality in (<ref>) for twist 
maps u∈𝒜^n_ϕ() with the distortion energy of u= Q(r)x given by

    𝕎[u;] = n^-n/2∫_(n x^-2 + Q̇θ^2)^n/2  dx.

Now restricting to the particular case of the twist map being (cf. <cit.>)

    u(x) = exp( -g(r) J ) x,      x ∈,
 
with J is as in Section <ref> and g∈ W^1,n([a,b]) the angle of rotation describing the twist. The corresponding distortion energy is

    𝕎[u;]    = n^-n/2∫_(n x^-2 + ġ^2)^n/2  dx 
       = ω_n n^-n/2∫_a^b (n r^-2 + ġ^2)^n/2 r^n-1  dr .




Note that in higher dimensions (i.e., n≥ 3) as discussed earlier there are only two homotopy classes in 𝒜^n_ϕ(). 
A twist map u=Q(r)x lies in the non-trivial homotopy class of 𝒜^n_ϕ() iff the twist loop 
Q=Q(r) ∈([a,b], 𝐒𝐎(n)) based at I lifts to a non-closed path R=R(r) ∈([a,b],𝐒𝐩𝐢𝐧(n)) connecting ± 1 
in 𝐒𝐩𝐢𝐧(n) (see <cit.> for more.)
[Note that 𝐒𝐩𝐢𝐧(n) is the universal cover of 𝐒𝐎(n) and {± 1}⊂ Spin(n) is the fibre 
over I under the covering map.] 



Likewise a twist u of the form (<ref>) lies in the non-trivial homotopy class of 𝒜^n_ϕ() iff the angle 
of rotation function g satisfies g(b)-g(a)=2π k for some k odd. When k is even the twist map u lies in the trivial homotopy 
class of 𝒜_ϕ(). 
[The identity boundary conditions on u dictates that the angle of rotation function must satisfy g(b)-g(a)=2π k for some k∈.]





§ OTHER VARIANTS OF THE DIRICHLET ENERGY
 





The goal of this section is to establish various energy bounds and identities, when n=2, by invoking the regularity and the measure preserving 
constraints satisfied by the elements of 𝒜_ϕ(). These inequalities will ultimately lead to useful results for extremisers 
and minimisers of variants of the Dirichlet energy in homotopy classes of 𝒜_ϕ(). We begin with the following identity.

For Φ∈ C^1[a,b] the integral identity 

    ∫_Φ(|u|)   dx = ∫_Φ(|x|)   dx

holds for all u ∈𝒜_ϕ().

 Denoting by α_u=α_u(t) the distribution function of |u| we can write using basic considerations and invoking the standard properties of distribution functions

    ∫_Φ(u)   dx     = ∫_∫_a^uΦ̇(t)   dt dx + Φ(a)  
    = ∫_a^b Φ̇(t) α_u(t)   dt + Φ(a) 
        = ∫_a^b Φ̇(t) α_x(t)   dt + Φ(a)  
    = ∫_∫_a^xΦ̇(t)   dt dx + Φ(a) 
       = ∫_Φ(x)   dx,

which is the required conclusion. 




We now collect a few more results which will be needed for the proof of our main theorem at the end of this section.



 
Let Φ∈ C^1[a,b] and pick u∈𝒜_ϕ(). Consider the continuous closed curve γ(θ) = u(r,θ) where 
a<r<b is fixed. Then the integral identity

    ∫_0^2πΦ(u)^2(u× u_θ)   dθ |^r_a = 2∫__r[uΦ(u)Φ̇(u)+Φ(u)^2 ]  dx,

holds for almost every r∈ (a,b), where _r=[a,r]={ x∈^2:a<x<r }.




We shall justify the assertion first when u is a sufficiently smooth diffeomorphism and then pass on to the general case by invoking a suitable 
approximation argument. Towards this end consider first the case where u is a smooth diffeomorphism with u ≡ x on ∂. [Here 
u need not satisfy the incompressibility condition in 𝒜_ϕ().] Let α denote 
the 1-form 

    α = Φ(x)^2 (x_1dx_2-x_2dx_1).
 
Then by a rudimentary calculation the pull-back of α under the C^∞ curve γ is given by 

    γ^*α = Φ(u)^2 (u× u_θ )   dθ.

Hence contour integration and basic considerations lead to the integral identity  

    ∫_γα = ∫_0^2πΦ(u)^2 (u× u_θ)   dθ |_r.




Note that the C^∞ curve γ here is diffeomorphic to ^1 and as such by the Jordan-Schöenflies theorem γ is the boundary 
of some bounded region C_γ⊂^2 diffeomorphic to the unit ball _1. In particular due to the boundary conditions on u we have that 
a < γ(θ)=u(r,θ) when a<r and so as a result C_a^γ = { x ∈^2: a<x<γ}⊂ 
with boundaries of ∂_a and γ. Hence application of Stoke's theorem gives

    ∫_C_a^γ dα = ∫_∂ C_a^γα 
    = ∫_γα - ∫_|x|=aα 
    = ∫_0^2πΦ(u)^2 u× u_θ  dθ  |_a^r,

for all r∈[a,b]. Next again by a rudimentary calculation we obtain that the pull-back of dα is,

    dα = 2 ∇ u [uΦ(u)Φ̇(u)+Φ(u)^2 ]  dx_1 ∧ d x_2.

Hence from (<ref>) and (<ref>) it follows that for all r∈[a,b],

    2∫__r[uΦ(u)Φ̇(u)+Φ(u)^2 ]  ∇ u   dx 
       = ∫_0^2πΦ(u)^2 u× u_θ  dθ |_a^r.

Now pick an arbitrary u in 𝒜_ϕ(). By approximation, e.g., using Theorem 1.1 in <cit.> there is a sequence 
of C^∞ diffeomorphisms (v^k) so that v^k-u ∈ W^1,2_0(,^2) with v_k → u uniformly on  
and strongly in W^1,2. Hence,

    f_k:= Φ(v^k)^2 v^k × v^k_θ/|x|→Φ(u)^2 u × u_θ/|x| =:f,

a.e. in . Note that f_k≤ c v^k_θ, f≤ c u_θ for some c>0 and so f_k,f ∈ L^2() 
and by virtue of v_k → u in W^1,2 and dominated convergence, for each r∈(a,b) and 0<δ<b-r, we have

    ∫_r^r+δ∫_0^2πΦ(v^k)^2(v^k× v^k_θ)   dθ dr →∫_r^r+δ∫_0^2πΦ(u)^2(u× u_θ)   dθ dr.

In a similar spirit we have (suppressing the arguments of Φ for brevity) 

    h_k:=[v^kΦΦ̇+Φ^2 ] ∇ v^k  →[uΦΦ̇+Φ^2 ]  ∇ u =:h,

a.e. in . Again since h_k≤ c v^k_x_1v^k_x_2, h≤ c u_x_1u_x_2 for some 
c>0 we have h_k,h∈ L^1() and so by dominated convergence

    ∫__r[v^kΦΦ̇+Φ^2 ] ∇ v^k   dx →∫__r[uΦΦ̇+Φ^2 ]  ∇ u   dx.

Now combining (<ref>) and (<ref>) together with the fact that 
u=v^k=ϕ on ∂ it follows that
 
    2∫_r^r+δ∫__r[v^kΦΦ̇+Φ^2 ] ∇ v^k   dx dr →∫_r^r+δ∫_0^2πΦ(u)^2(u× u_θ)   dθ  |_a^r dr.

Moreover (<ref>) and a final application of dominated convergence gives

    2 ∫_r^r+δ∫__r[uΦΦ̇+Φ^2 ]  ∇ u   dx dr 
    = ∫_r^r+δ∫_0^2πΦ(u)^2(u× u_θ)   dθ  |_a^r dr.

Therefore the result follows by recalling that ∇ u = 1 a.e. in  and applying the Lebsegue differentiation 
theorem, i.e., dividing by δ and letting δ↘ 0. □





Note that we can write the conclusion of the above proposition, namely, the integral identity (<ref>) in a shorter and somewhat more suggestive form 

    ∫_0^2πΦ(u)^2(u× u_θ)   dθ |_a^r = ∫__ru^-1Γ̇(u)  dx,
 	
where Γ (t) = t^2 Φ(t)^2 for Φ∈ C^1[a,b]. With this result and formulation at our disposal we can now prove the earlier 
relation (<ref>) as a specific proposition. 

Taking Φ(t)= 1/t in the above gives the integral identity 

    ∫_0^2πu(r,θ)× u_θ(r,θ)/u^2  dθ = 2π,      a.e.    r∈[a,b].



When Φ(t) = 1/t it can be easily seen that Γ̇(t)=0 and therefore Proposition <ref> gives that,

    ∫_0^2π(u× u_θ)/u^2  dθ |_a^r = 0,    a.e.   r∈ [a,b].

Then recalling that u(x)=x on ∂, i.e. when x=a, it can be seen that the integral over the inner boundary is 2π and 
hence from (<ref>) this implies that,

    ∫_0^2π(u× u_θ)/u^2  dθ   = 2π,    a.e.    r∈ [a,b],

which completes the proof.




 
Suppose Γ∈ C^2[a,b] such that Γ̇(t)/t is a monotone increasing function. Then for almost every r∈[a,b] we have that

    ∫_0^2πΓ(u)^2 (u(r,θ)× u_θ(r,θ))^2 /u^4  dθ≥ 2πΓ(r)^2.





 First we note that by Proposition <ref> we have for a.e. r ∈ [a, b] that 

    ∫_0^2πΓ (u) u× u_θ/u^2  dθ  |_a^r= ∫__ru^-1Γ̇(u)  dx.

As Γ̇(t)/t is monotone increasing and u and x share the same distribution function α_x(t) we have that, 

    ∫_^b_ru^-1Γ̇(u)   dx     = ∫_a^b d/d t(Γ̇(t)/t) 
    ∫_χ_{ x ∈_r^b : u(x)>t } dx   dt + _r^bΓ̇(a)/a
       ≤∫_a^b α_x(t) d/d t(Γ̇(t)/t) dt + _r^bΓ̇(a)/a
        = ∫__r^bx^-1Γ̇(x)   dx 
       = 2π[ Γ(b)-Γ(r)].

In the above _r^b ={ x ∈ : r< x<b }. Now since,


    2π[ Γ(b)-Γ(a)]= ∫_x^-1Γ̇(x) dx = ∫_u^-1Γ̇(u)  dx,

by Proposition <ref> we obtain upon using (<ref>) in (<ref>) that,

    ∫__ru^-1Γ̇(u)  dx ≥∫__rx^-1Γ̇(x)  dx = 2π[Γ(r)-Γ(a)].

Therefore,

    ∫_0^2πΓ (u) u× u_θ/u^2  dθ  |_a^r≥ 2π[Γ(r)-Γ(a)],

but from the identity boundary conditions on u we know that,

    ∫_0^2πΓ (u) u× u_θ/u^2  dθ  |_a = 2πΓ(a),

and therefore,

    ∫_0^2πΓ (u) u× u_θ/u^2  dθ  |_r ≥ 2 πΓ(r),      a.e.    r∈[a,b].

The result then follows from an application of Jensen's inequality. 




With the aid of these bounds we can now move on to the main goal of the section, namely, formulating and proving minimality for twist maps 
in homotopy classes of 𝒜_ϕ() for a larger class of energies than those considered earlier. 


 

Let =[a, b] ⊂^2 and let ℍ=ℍ[u; ] denote the energy functional,

    ℍ[u;]    = 1/2∫_Φ (u) [ ∇u^2  + 
    (u× u_θ)^2/r^2u^2] +(u× u_r)^2/u^4  dx,
 
where u lies in 𝒜_ϕ() and Φ (t) = t^-2Γ(t)^2 with Γ(t)∈ C^2[a,b] such that Γ̇(t)/t 
is monotone increasing. Then for any u ∈ A_k (k ∈) there exists a twist map 
u̅ = u̅_k=Q[g] x defined by the same symmetrisation as in (<ref>) such that,

    ℍ[u̅; ] ≤ℍ[u; ]

whilst u̅∈ A_k. 
[Note that taking Φ(t)= 1/t^2, i.e., Γ(t) = 1, gives ℍ=𝔽.]




 As the first step in the proof we wish to prove the inequality 

    ∫_Φ(x)   dx  = ∫_Φ(u̅(x)) ∇u̅(x)^2   dx ≤∫_Φ(u) ∇u(x)^2   dx.




In order to do this we again need to apply the isoperimetric inequality, the coarea formula for Sobolev functions as in the proof of Proposition <ref> 
and then the integral identity (<ref>). Thus we proceed by writing 

    ∫_Φ(u) ∇u(x)  dx    =  ∫_a^b Φ(t)ℋ^1(u=t)   dt 
       ≥∫_a^b Φ(t)ℋ^1(x=t)   dt 
       = ∫_Φ(x)   dx.
 
Therefore it follows from basic considerations that 

    (∫_Φ(x) dx)^2    ≤(∫_Φ(u) ∇u(x)  dx )^2 
       ≤∫_Φ(u) ∇u(x)^2 dx ∫_Φ(u)   dx 
        = ∫_Φ(u) ∇u(x)^2 dx ∫_Φ(x)   dx,

and so rearranging gives the desired inequality (<ref>), namely, 

    ∫_Φ(x) dx ≤∫_Φ(u) ∇u(x)^2   dx.

Next we proceed by writing  

    ∫_Φ(u)(u× u_θ)^2/r^2u^2  dx    = 
    ∫_a^b 1/r∫_0^2πΦ(u)(u× u_θ)^2/u^2  dθ dr.
 
As Φ(t) = t^-2Γ(t)^2 it follows upon noting Proposition <ref> that we have 

    1/2π∫_0^2πΦ(u)(u× u_θ)^2/u^2 dθ   = 1/2π∫_0^2πΓ(u)^2(u× u_θ)^2/u^4 dθ
       ≥Γ(r)^2 = Φ(r) r^2,    a.e.    r∈[a,b].

Hence by combining the above it follows that 

    ∫_Φ(u)(u× u_θ)^2/r^2u^2  dx  ≥ 2π∫_a^b Φ(r)r  dr 
    = ∫_Φ(u̅)(u̅×u̅_θ)^2/r^2u̅^2  dx.

Therefore with the above at our disposal all that remains is to use the inequality 

    ∫_(u× u_r)^2/u^4  dx ≥∫_(u̅×u̅_r)^2/u̅^4  dx,

whose proof proceeds similar to that of Proposition <ref> by using the 
same angle of rotation function (<ref>) in defining u̅. This therefore completes the proof. 






































































































§ MEASURE PRESERVING SELF-MAPS AND TWISTS ON SOLID TORI





In this section we propose and study extensions of twist maps to a larger class of domains. Recalling that an n-dimensions annulus takes the 
form = [a,b] ×^n-1 the natural extension here would be domains of the product type =^m ×^n-1 (with m ≥ 1, 
n ≥ 2) embedded in ^m+n. Twist maps in turn will be suitable measure preserving self-maps of  that agree with the 
identity map ϕ on the boundary ∂ (see <cit.>). To keep the discussion tractable we confine ourselves here to the 
case m+n=3. We proceed by first considering the solid torus T≅𝔹^2 ×^1 embedded in ^3 as (see Fig. 1):

    T = { x=(x_1, x_2, x_3) :   ( √(x_1^2+x_2^2) - ρ )^2 + x_3^2 = r^2 ,    0 ≤ r < 1 }.

Here T= T_ρ and the fixed parameter ρ is chosen ρ>1 to avoid self-intersection. Now let us set 
μ = √(x_1^2+x_2^2)- ρ. Then T above can be represented as

    0 ≤μ^2 + x_3^2  = r^2 < 1.
   
From now on (μ, x_3) is the preferred choice of co-ordinates for =𝔹_1^2  where 
={ (μ,x_3)∈^2 : μ^2+x_3^2 < 1 } is unit ball in the (μ, x_3) plane. In polar co-ordinates we have μ = r cosθ, 
x_3 = rsinθ and upon noting μ = √(x_1^2+x_2^2)- ρ we have (x_1, x_2) as the co-ordinates of a sphere of radius 
ρ + rcosθ:

    x_1 = (ρ + rcosθ) cosϕ,    x_2 = (ρ + rcosθ) sinϕ.

For our purposes in this section we shall write the above co-ordinate system in the following way,

    x_1 = (μ+ρ)cosϕ
    
    x_2 = (μ+ρ) sinϕ
    
    x_3 = x_3.

Now with the above notation in place we can define the desired twist maps on the solid torus 𝐓 as
[Note that the aim here is to seek non-trivial extremising twist maps for the energy functional 𝔽 over the admissible 
class of maps 𝒜_ϕ(𝐓).],

    u(x)     = Q(μ,x_3) x,      x ∈ T,

where the rotation matrix Q in SO(3) takes the explicit form 

    Q(μ,x_3) = [  cos g(μ,x_3) -sin g(μ,x_3)             0;  sin g(μ,x_3)  cos g(μ,x_3)             0;             0             0             1 ].




Here the function g=g(μ,x_3) defines the angle of rotation as in the case for the annulus, however, in this case g depends on the two variables 
(μ, x_3) and not just one r=|x| as is the case for the annulus. There are two main reasons for this choice of representation of a twist map for 
𝐓, which we describe below. 


  * Firstly in order to be consistent with twist maps for the annulus we require that the rotation matrix is an isometry of the boundary

    ∂ T = { x=(x_1, x_2, x_3) :   ( √(x_1^2+x_2^2) - ρ )^2 + x_3^2 = 1  },

with respect to the metric induced from its embedding in ^3. (This is similar to what was done earlier in the case of an annulus). Then with this in mind the isometries 
of ∂𝐓 are 𝐒𝐎(3) matrices of the form,

    Q = [  cosφ -sinφ     0;  sinφ  cosφ     0;     0     0     1 ].


  * Secondly we allow the angle of rotation function g here to depend on the two variables (μ, x_3) instead of one to incorporate all the "ball" 
variables in the product structure on T. Note that (μ, x_3) collapses into r=|x| in the annulus case as here one deals with an interval 
(a one dimensional ball). 

Now in preparation for the upcoming calculations let us denote y=[x_1,x_2,0]^t, ϑ = y/|y| and g_μ= ∂ g / ∂μ. 
Then it is easily seen that 

    ∇ u = Q + Q̇x ⊗∇ g,
    
    (∇ u)(∇ u)^t u = Qx + ⟨∇ g, x ⟩Q̇x,
    ∇ u^2  = 3 +  (μ+ρ)^2∇ g^2, 
    u^2 =  (μ+ρ)^2+x_3^2, 
    (∇ u)  = (  Q + Q̇x ⊗∇ g )  
                = 1 + ⟨ Q^tQ̇x, ∇ g ⟩ = 1.

Note that the last equality results from the fact that the product Q^tQ̇ is skew-symmetric and ∇ g = [(μ+ρ)^-1x_1 g_μ,(μ+ρ)^-1x_2 g_μ, g_x_3]^t. 
Therefore using the above it is seen that the energy for a twist maps is given by

    𝔽[u; 𝐓] = 1/2∫_ T|∇ u|^2/|u|^2  dx 
    = π∫_3+(μ+ρ)^2∇ g^2/(μ+ρ)^2+x_3^2 (μ+ρ)   dμ dx_3.

Here we are using the fact that a change in the co-ordinates (r,θ,ϕ)→ (μ,x_3,ϕ) results in a Jacobian factor of μ+ρ 
in the integral, i.e., 

    ∫_0^2π∫_0^2π∫_-1^1 r(ρ + rcosθ )   drdθ dϕ = ∫_0^2π∫_ (μ+ρ)   dμ dx_3dϕ.

Hence (<ref>) becomes,

    𝔽[u;𝐓]     = π∫_(μ+ρ)^3(g_μ^2+g_x_3^2)/(μ+ρ)^2+x_3^2  dμ d x_3 + 3/2∫_𝐓x^-2dx,
 
where here the additional absolute constant does not affect the variational structure of 𝔽. Now to derive the Euler-Lagrange equation 
associated to the energy integral on the right it suffices to take variations of g=g(μ, x_3) by some φ∈𝐂^∞_c(). This calculation leads to the following 
divergence form equation:

    ∂/∂μ( (μ+ρ)^3 g_μ/(μ+ρ)^2 + x_3^2)  +  
    ∂/∂ x_3( (μ+ρ)^3 g_x_3/(μ+ρ)^2 + x_3^2) = 0, 
     div(μ+ρ)^3 ∇ g/(μ+ρ)^2 + x_3^2 =0.




Evidently the identity boundary condition on u translates into g(z) = 2 k π for some fixed k ∈ and all z=(μ,x_3)∈∂. 
Now suppose g solves (<ref>). Then by an application of the divergence theorem 
it is seen that the only solution to this boundary value problem is the trivial one, namely, g(μ,x_3)=2π k for all (μ,x_3) ∈. 
Indeed

    0      = ∫_ div(  (μ+ρ)^3 ∇ g/(μ+ρ)^2 + x_3^2)   dμ dx_3 
        = ∫_0^2π(cosθ+ρ)^3/(cosθ+ρ)^2 + x_3^2∂ g/∂ r(1,θ)   dθ.
 
Now again as g solves (<ref>) an application of the divergence theorem also gives 

    ∫_(μ+ρ)^3(g_μ^2+g_x_3^2)/(μ+ρ)^2+x_3^2  dμ dx_3    = 
     ∫_[ (μ+ρ)^3(g_μ^2+g_x_3^2)/(μ+ρ)^2+x_3^2 + g   div(μ+ρ)^3∇ g/(μ+ρ)^2 + x_3^2]   dμ dx_3 
       =  ∫_0^2π  g(1,θ)(cosθ+ρ)^3/(cosθ+ρ)^2 + x_3^2∂ g/∂ r(1,θ)   dθ
       = 2π k  ∫_0^2π(cosθ+ρ)^3/(cosθ+ρ)^2 + x_3^2∂ g/∂ r(1,θ)   dθ = 0.

Note that in obtaining the last identity we have used (<ref>) combined with the boundary condition satisfied by g, namely, 
g(1,θ)=2π k for 0≤θ≤ 2π. Hence

    ∫_(μ+ρ)^3(g_μ^2+g_x_3^2)/(μ+ρ)^2+x_3^2  dμ dx_3 = 0.




Now since by assumption ρ>1 we have (ρ + μ) >0 as μ=rcosθ≤ r < 1 and so μ>-1. 
Hence (<ref>) gives ∇ g^2=0 and thus g(μ,x_3)=2π k; again by invoking the boundary condition on 
g. It therefore follows that here we have no non-trivial solutions. Interestingly note that this conclusion stems from one crucial 
difference between the annulus  and the solid torus 𝐓 in that  has two boundary components whilst 
𝐓 only has one. It was precisely this difference that turned crucial in the application of the divergence theorem.  



 There are no non-trivial twist solutions (<ref>) to the Euler-Lagrange equations 
associated with the energy functional 𝔽 on a solid torus T.





§ TWIST MAPS ON TORI WITH DISCONNECTED DOUBLE COMPONENT BOUNDARY




In contrast to what was seen above let us next move on to considering a "thickened" torus, that is, the domain obtained 
topologically by taking the product of a two-dimensional torus and an interval. Note that here the boundary of the resulting 
domain consists of two disjoint copies of the initial torus and is in particular not connected. Now for definiteness and to fix 
notation let us set 𝕋=𝕋_ρ to be (see Fig. 1)

    𝕋 = { x=(x_1, x_2, x_3)  : ( √(x_1^2+x_2^2) - ρ)^2 + x_3^2 = r^2 ,    a < r < 1 }.

Here 0<a<1<ρ are fixed and the aim is to seek non-trivial extremising twist maps for the energy functional 𝔽 
over the admissible class of maps 𝒜_ϕ(𝕋). Using the same co-ordinate system as in the earlier case we see 
that the (μ,x_3) are the co-ordinates of the two dimensional annulus centred at the origin. Additionally for similar reasons to that 
discussed earlier we define twist maps on 𝕋 as u(x) = Q(μ,x_3) x where the rotation matrix Q=Q(μ, x_3) 
in SO(3) is as (<ref>). 




A straightforward calculation shows that the 
energy of a twist maps is given by the integral 

    𝔽[u; 𝕋]    = 1/2∫_𝕋|∇ u|^2/|u|^2  dx 
    = π∫__1\_a3+(μ+ρ)^2∇ g^2/(μ+ρ)^2+x_3^2 (μ+ρ)   dμ d x_3 
        = π∫__1\_a(μ+ρ)^3(g_μ^2+g_x_3^2)/(μ+ρ)^2+x_3^2  dμ d x_3 + 3/2∫_𝕋x^-2 dx.

Similar to what was described earlier in obtaining the second equality we have used the integral identity

    ∫_0^2π∫_0^2π∫_a^1 r(ρ + rcos(θ))   drdθ dϕ = ∫_0^2π∫__1\_a (μ+ρ)   dμ d x_3 dϕ .





The Euler-Lagrange equation can be obtained in the standard way by taking variations φ∈𝐂^∞_c(_1\_a) where 
_1\_a={ (μ,x_3)∈^2 : a^2<μ^2+x_3^2 <1 }. This calculation again leads to the Euler-Lagrange equation given 
by (<ref>) where we assume









without loss of generality that the boundary condition on the rotation angle function g is set to g(μ,x_3)=0 for (μ,x_3)∈∂_a 
and g(μ,x_3)=2π k for (μ,x_3)∈∂_1 with k ∈. Therefore solutions to (<ref>) satisfy, 

    0 = ∫__1\_a div(μ+ρ)^3 ∇ g/(μ+ρ)^2 + x_3^2  dx 
    = ∫_∂_1(μ+ρ)^3 ∇ g· n/(μ+ρ)^2 + x_3^2 -  
    ∫_∂_a(μ+ρ)^3 ∇ g· n/(μ+ρ)^2 + x_3^2.

Subsequently 

    𝔽[u; 𝕋] - 3/2 ∫_𝕋x^-2 dx/π =   ∫__1\_a[ (μ+ρ)^3∇ g^2/(μ+ρ)^2 + x_3^2  
    + g  div( (μ+ρ)^3 ∇ g/(μ+ρ)^2 + x_3^2) ]   dx 
    
    =   ∫_∂_1g(μ+ρ)^3 ∇ g· n/(μ+ρ)^2 + x_3^2 dℋ^1
    -  ∫_∂_ag(μ+ρ)^3 ∇ g · n/(μ+ρ)^2 + x_3^2  dℋ^1.

Hence taking into account the boundary conditions, e.g. g=0 on ∂_a and g=2π k on ∂_1, we gain that if g is a solution of (<ref>) then,

    𝔽[u; 𝕋] - 3/2 ∫_𝕋x^-2 dx/π   = 2π k ∫_∂_1(μ+ρ)^3 ∇ g· n/(μ+ρ)^2 + x_3^2 dℋ^1, 
         k ∈.




Evidently (<ref>) with the stated boundary conditions has a unique solution. Indeed if g, g are two solutions to 
(<ref>) with g=0 on ∂_a and g=2π k on ∂_1 
then g=g-g solves (<ref>) with g=0 on ∂ [_1\_a]. Then by (<ref>)

    ∫__1\_a(μ+ρ)^3 ∇ g^2/(μ+ρ)^2 + x_3^2  dx = 0.
  



However in view of ρ>1 this gives ∇ g^2 ≡ 0 and so invoking the boundary conditions g=0, i.e., g= g. 
As existence follows from standard arguments it follows that (<ref>) has a unique smooth solution g=g(μ, x_3; k) for each k ∈. 





§ EULER-LAGRANGE ANALYSIS AND TWISTS AS CLASSICAL SOLUTIONS





The goal of this section is to examine the solution g=g(μ, x_3; k) to (<ref>) with the prescribed boundary conditions in relation to the 
Euler-Lagrange system (<ref>) associated with 𝔽 on 𝒜_ϕ(𝕋). To this end recall that the system takes the form 

    (∇ u)^t/u^2[ Δ u + ∇ u^2/u^2 u - 2/u^2∇ u (∇ u)^tu ]    = ∇ p.

For the ease of notation from now on we shall set ξ = μ + ρ. Hence using the identities 
(<ref>) we have  

    (∇ u)(∇ u)^t u = Qx + ⟨∇ g,x ⟩Q̇x,
    ∇ u^2 u^-2  = (3 + ξ^2 ∇ g^2) x^-2, 
    Δ u = 2 ξ^-1 g_ξQ̇x + Δ g Q̇x + ∇ g^2 Q̈x.

Therefore from (<ref>) and a basic calculation we obtain

    Δ u + ∇ u^2/u^2 u - 2/u^2∇ u (∇ u)^tu     = (2g_ξ/ξ + 
    Δ g -2 ⟨∇ g,x ⟩/x^2)Q̇x + ∇ g^2 Q̈x 
        + 1 + ξ^2 ∇ g^2/x^2 Qx.




Now since we have Δ g = Δ_ξ,x g  + g_ξ/ξ where the Δ_ξ,x denotes the Laplacian with respect to the ξ and x_3 
variables we can rewrite this as 

    Δ u + ∇ u^2/u^2 u - 2/u^2∇ u (∇ u)^tu =   (3g_ξ/ξ + 
    Δ_ξ,x g - 2 ⟨∇ g,x ⟩/x^2)Q̇x + ∇ g^2 Q̈x 
        + 1 + ξ^2 ∇ g^2/x^2 Qx.

Now upon recalling that the desired twist solution satisfies (<ref>) we have that

    div( ξ^3 ∇ g/ξ^2 + x_3^2)  = ξ^3 Δ_ξ,x g/ξ^2 + x_3^2 + 
    ( 3ξ^2/ξ^2 + x_3^2- 2ξ^4/(ξ^2+x_3^2)^2)g_ξ - 2ξ^3 x_3 g_x_3/(ξ^2 + x_3^2)^2 =0.




Thus dividing both sides by ξ^3/(ξ^2+x_3^2) and taking the negative terms to one side gives 

    Δ_ξ,xg + 3 g_ξ/ξ = 2( ξ g_ξ + x_3 g_x_3/x^2) = 2⟨∇_ξ,x g, z⟩/x^2,
 
where z=(ξ,x_3)^t and ∇_ξ,x denotes the gradient with respect to the (ξ, x_3) variable. Now since 
⟨∇ g, x ⟩ = ⟨∇_ξ,xg, z ⟩ we obtain,

    Δ_ξ,xg + 3 g_ξ/ξ  = 2⟨∇ g, x⟩/x^2,

and so as a result 

    Δ u + ∇ u^2/u^2 u - 2/u^2∇ u (∇ u)^tu     =  ∇ g^2 Q̈x + 
    1 + ξ^2 ∇ g^2/x^2 Qx.




Next referring to the definition of Q basic calculation gives Q̇ = J_1 Q and Q̈=-J_2Q, where

    J_1 = [  0 -1  0;  1  0  0;  0  0  0 ],   
     J_2 = [ 1 0 0; 0 1 0; 0 0 0 ].




Hence with the above notation the Euler-Lagrange associated with the twist u, satisfying (<ref>), simplifies to 

    (∇ u)^t/x^4 (I + ξ^2 ∇ g^2 I - x^2∇ g^2 J_2) Qx 
        = (I + ξ^2 ∇ g^2I - x^2∇ g^2 J_2) x/x^4
        = 1/x^4[ (1-x_3^2∇ g^2 ) x_1;  (1-x_3^2∇ g^2) x_2;   (1+ξ^2 ∇ g^2) x_3 ]
       = ∇( -1/2x^2) + ∇ g^2/x^4[ -x_3^2 x_1; -x_3^2 x_2;    ξ^2 x_3 ] =∇ p.




Considering the last line in the above equation it is plain that for u to grant solution to the Euler-Lagrange equation it must be that  

    -1/2∇ |x|^-2 + ∇ g^2/x^4 (-x_3^2 x_1, -x_3^2 x_2, ξ^2 x_3)^t =∇ p,

or equivalently that the second term on the left is a gradient. But for this to be the case the latter term must necessarily be curl-free 
and so this leads to the system of equations

    0     = ∂/∂ x_1( -x_3^2 ∇ g^2/x^4x_2 ) 
    - ∂/∂ x_2( -x_3^2 ∇ g^2/x^4x_1 ), 
    
    0     = ∂/∂ x_1( ξ^2 ∇ g^2/x^4x_3 ) 
    - ∂/∂ x_3( -x_3^2 ∇ g^2/x^4x_1 ),
    
    0     = ∂/∂ x_2( ξ^2 ∇ g^2/x^4x_3 ) 
    - ∂/∂ x_3( -x_3^2 ∇ g^2/x^4x_2 ).




It can be easily verified that equation (<ref>) is satisfied for any twist map since here we have 

    ∂/∂ x_1( -x_3^2 ∇ g^2/x^4x_2 ) 
    - ∂/∂ x_2( -x_3^2 ∇ g^2/x^4x_1 ) 
       = x_3^2 ∇ g^2[ ∂/∂ x_2x_1/x^4 
    - ∂/∂ x_1x_2/x^4] +  x_3^2/x^4[ x_1∂∇ g^2/∂ x_2 -  x_2∂∇ g^2/∂ x_1] 
       = x_3^2/x^4[ x_1 x_2/ξ∂∇ g^2/∂ξ 
    -  x_2x_1/ξ∂∇ g^2/∂ξ]=0.

We point out that the last line results upon noting the relations 

    ∂/∂ x_1 = x_1/ξ∂/∂ξ-x_2/ξ^2∂/∂ϕ,     ∂/∂ x_2 = x_2/ξ∂/∂ξ+x_1/ξ^2∂/∂ϕ.
 
Using this we can again see that we can write (<ref>) and (<ref>) as a single equation in the following way,

    ∂/∂ x_1( ξ^2 ∇ g^2/x^4x_3 ) 
       + ∂/∂ x_3( x_3^2 ∇ g^2/x^4x_1 ) = 
        = x_1 x_3/ξ∂/∂ξ( ξ^2 ∇ g^2/x^4) 
    + x_1∂/∂ x_3( x_3^2 ∇ g^2/x^4),
    ∂/∂ x_2( ξ^2 ∇ g^2/x^4x_3 ) 
       - ∂/∂ x_3( -x_3^2 ∇ g^2/x^4x_2 ) = 
        = x_2 x_3/ξ∂/∂ξ( ξ^2 ∇ g^2/x^4) 
    + x_2∂/∂ x_3( x_3^2 ∇ g^2/x^4).

Therefore it is apparent that (<ref>) and (<ref>) become

    x_3 ∂/∂ξ( ξ^2 ∇ g^2/x^4) + ξ∂/∂ x_3( x_3^2 ∇ g^2/x^4)=0.

Now since we have the identities

    ∂/∂ξ( ξ^2/x^4) = 2ξ (x_3^2 - ξ^2)/(ξ^2+x_3^2)^3 
    =- ξ/x_3∂/∂ x_3( x_3^2/x^4),

we obtain that (<ref>) simplifies further to 

    x_3ξ/x^4   [ ξ∂∇ g^2/∂ξ 
    x_3∂∇ g^2 /∂ x_3] = 0 ξ∂∇ g^2/∂ξ  + x_3∂∇ g^2 /∂ x_3 = 0.





Hence for a solution g to (<ref>) with the prescribed boundary conditions to furnish a solution 
to the Euler-Lagrange system (<ref>) associated with 𝔽 it is necessary for g to satisfy

    ξ∂∇ g^2/∂ξ  + x_3∂∇ g^2 /∂ x_3 = 0.




We now show that (<ref>) is also sufficient. Indeed assuming (<ref>) the desired conclusion will 
follow upon showing that (<ref>) holds. Towards this end set f to be the function,

    f(ξ,x_3) = -∫_0^ξx_3^2∇ g^2/(τ^2 +  x_3^2)^2τ dτ.

Then one can easily verify that,

    ∂ f/∂ x_1 = -x_3^2 ∇ g^2/x^4x_1,   ∂ f/∂ x_2 
    = -x_3^2 ∇ g^2/x^4x_2.

Furthermore using (<ref>) it is plain that 

    ∂ f/∂ x_3   = - ∫_0^ξ∂/∂ x_3x_3^2∇ g^2 τ/(τ^2 +  x_3^2)^2  dτ
        = ∫_0^ξ∂/∂ττ^2∇ g^2 x_3/(τ^2 +  x_3^2)^2  dτ 
    = ξ^2∇ g^2/x^4 x_3.

As a result ∇ f = ∇ g^2 x^-4 (-x_3^2 x_1, -x_3^2 x_2, ξ^2 x_3)^t.




 A twist map u with the corresponding angle of rotation function g=g(μ, x_3; k) satisfying (<ref>) 
and g=0 on ∂_a, g=2π k on ∂_1 (with k ∈)  is a solution to the Euler-Lagrange system 
(<ref>) associated with 𝔽 on 𝒜_ϕ(𝕋) iff it satisfies (<ref>).





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 † DEPARTMENT OF MATHEMATICS, UNIVERSITY 
OF SUSSEX, FALMER, BRIGHTON BN1 9RF, ENGLAND, UK.    


E-mail address: a.taheri@sussex.ac.uk



