5.1 The elliptic equilibrium of a real system

Transcription

1 ÆÇÆ ÄÁÆ Ê ÄÄÁÈÌÁ ÉÍÁÄÁ ÊÁ The elliptic equilibrium of a real system. Stable for the linear approximation but the flow around a center point may be changed by non linear terms. Problem: study the stability of the equilibrium for the nn linear system. Particularly interesting for physical applications. Formal first integrals, but generically non convergent (asymptotic series). Numerical comparison with the dynamics. Exponential stability. 5.1 The elliptic equilibrium of a real system Consider the real linear system ż = Az, z R n and let the eigenvalues be pure imaginary, i.e., iω 1,...,iω n, iω 1,..., iω n. Paradigmatic form: if A can be put in complex diagonal form (e.g., the eigenvalues are distinct) then the system may be reduced to the real form ( ) 0 I ż = JΩz, Ω = diag(ω 1,...,ω n,ω 1,...,ω n ), J =. I 0 i.e., An useful notation for z R n : ( ) ( ) x X(x,y) z =, V = ; y Y(x,y) ω ω ωn JΩ =. ω ω ωn (ẋ ) = JΩ ẏ ( ) x + y ( ) X1 (x,y) + Y 1 (x,y) ( ) X (x,y) Y (x,y)

2 5.1.1 The Hamiltonian case A canonical system with Hamiltonian (5.1) H(x,y) = H 0 (x,y)+h 1 (x,y)+h (x,y)+..., H 0 (x,y) = 1 n ω l (x l +yl), where ω 1,...,ω n are constants and H s (x,y) is a homogeneous polynomial of degree s+. The canonical vector field is ( ) ( H1 ) ( H ) x y y X H = JΩ y H1 x H x Hereafter we consider the simpler case H = H 0 +H 1, i.e., H = H 3 =... = 0. Physical models: the triangular Lagrangian equilibria of the problem of three bodies; the secular model for the planetary system (Lagrange); orbits of stars in a galaxy (Contopoulus); the discretized string of Lagrange; the Fermi Pasta Ulam model with its variazioni such as the Toda lattice and the Klein Gordon lattice; disordered lattices; The electromagnetic field; &c Diagonal form of the linear system The linear system takes a diagonal form in complex variables, by the linear canonical transformation ( ) z = Mζ, ζ = M 1 ξ z, ζ = C n, η M = 1 ( ) I ii, M 1 = 1 ( ) I ii. ii I ii I Eplicit form of the transformation and of its inverse x l = 1 (ξ l +iη l ), ξ l = 1 (x l iy l ), y l = i (ξ l iη l ), η l = i (x l +iy l ), l = 1,...,n. Non linear system in complex variables ζ = iωζ +M 1 V(z), z=åζ Hamiltonian in complex variables n H 0 = ω l I l, I l = iξ l η l. Eigenvalues in the Siegel domain: the system can not be conjugated to its linear part (except in very particular cases). Non linear elliptic equilibria 6

3 5.1.3 Loss of stability for the non linear equilibrium Formally, may give the system a normal form ẋ =Λx+Z(x) with the condition {Λx,Z} = 0. Revisit the normal form method by solving the homological equation with the normal form condition above. The system in normal form needs not be stable. Example 5.1: An unstable non linear system. Consider the system in R Transform to complex variables and remark it is in normal form. ẋ = ωy x(x +y ), ẏ = ωx y(x +y ). ξ = iωξ iξ η, η = iωη+iξη, Take Φ(x,y) = (x +y )/ as a Lyapounov function. Its time derivative is Φ = (x +y ) < 0. The system is asymptotically stable in the future, but not perpetually stable. Study the stability of the non linear system by looking for first integrals. 5. First integrals The linear system possesses n independent first integrals: the actions I l = 1 (x l +y l) = iξ l η l, l = 1,...,n. Problem: Find a first integral expanded as a formal power series Φ(x,y) = Φ 0 (x,y)+φ 1 (x,y)+φ (x,y)+... where Φ 0 (x,y) = I l is one of the actions, and Φ s (x,y) for s 1 is a homogeneous polynomial of degree s+. General case. Split equation ( ω +L V ) Φ = 0 as a recurrent system ω Φ 0 = 0, ω Φ s = Ψ s, Ψ s = L V1 Φ s 1... L Vs Φ 0, where ω = LÂΩz. Equation for Φ 0 is obviously solved by the actions I 1,...,I n. Problem: Find a solution of the equation Hamiltonian case. Split equation {H,Φ} = 0 as a recurrent system ω Φ = Ψ ω Φ 0 = 0, ω Φ s = Ψ s, Ψ s = {H 1,Φ 0 }+... {H 1,Φ s 1 }, where ω = {,H 0 } = LÂΩz. where Ψ(x,y) is a known homogeneous polynomial of degree s+. Non linear elliptic equilibria 63

4 5..1 Formal scheme Lemma 5.: The operator ω is diagonal over the basis of complex monomials ξ j η k, with eigenvalues i j k,ω. Proof. Just repeat the proof for a generic diagonal vector field Λz. In complex variables n ( ) ω f = iω l ξ l iω l η l f. ξ l η l For a monomial ξ j η k calculate ω ξ j η k = i j k,ω ξ j η k. Solution of the equation ω Φ = Ψ. In complex variables, write (in multiindex notation) Ψ = with known (complex) coefficients ψ j,k. The formal solution is ϕ = j,k ϕ j,k with j,k N 0 ψ j,k ξ j η k ψ j,k ϕ j,k = i j k,ω, Q.E.D. must avoid zero divisors!... BUT... again Resonances and diophantine frequencies To a real vector of frequencies ω we associate the resonance module M ω defined as M ω = { k Z n : k,ω = 0 }. A resonance module is a subgroup of Z n. The dimension dim M ω is often called the multiplicity of the resonance: dim M ω = 0 : non resonant frequencies; 0 < dim M ω < n 1 : partial resonance; dim M ω = n 1 : full resonance. The frequencies ω R n are said to be diophantine in case k,ω γ k τ for 0 k Z n, with γ > 0 and τ n 1. Let P s be the space of homogeneous polynomials of degree s in n variables. Define the kernel Nω s and the range Rs ω of the operator ω as Nω s = ω 1 ( ) ( {0}, R s ω = ω P s ). N s ω and Rs ω are both subspaces of Ps. Since ω may be diagonalized N s ω R s ω = {0}, N s ω R s ω = P s. Non linear elliptic equilibria 64

5 Lissajous figures: plane orbits of two linear oscillators with frequencies ω 1, ω. ω /ω 1 = 1: resonance 1 : 1. ω /ω 1 = : resonance : 1. ω /ω 1 = /3: resonance : 3. ω /ω 1 = 3/5: resonance 3 : 5. ω /ω 1 = : non resonance. ω /ω 1 = 5 1 : non resonance. In complex variables ξ, η N s ω contains all monomials ξj η k with j k M ω. If ω is non resonant then N s ω contains only monomials ξ k η k, which have even degree. Introduce the action angle coordinates the transformation is canonical (or symplectic). x j = I j cosϕ j, y j = I j sinϕ j. A monomial x j y k with j +k = s is transformed to x j y k = (j+k)/ I (j+k)/ m c m exp ( i m,ϕ ) with m Z n satisfying m = s, s +,...,s,s and complex coefficients c m = c m (the monomial is real). More generally: a homogeneous polynomial f P s is transformed as f = c j,k I j/ ( ) exp i k,ϕ k Z n j N n 0 with k = s, s+,...,s,s and c j,k = c j, k. If ω is non resonant then f N ω is a polynomial on the actions I 1,...,I n, independent of the angles ϕ 1,...,ϕ n. If ω is resonant then f N s ω depends on the actions I 1,...,I n and on combinations of the angles k,ϕ with k M ω. Non linear elliptic equilibria 65

6 Reformulating the diophantine condition of strong non resonance. Definition 5.3: A frequency vector ω R n is said to be diophantine of class (γ,τ) with γ > 0 and τ R in case k,ω γ k τ for all k Z n, k = 0. Denote by Γ γ,τ the set of diophantine frequencies of class (γ,τ). Proposition 5.4: Let τ > n. Then for every ball B R R n we have Vol ( B R \Γ γ,τ ) cγ where c is a positive constant depending on the radius R of the ball and on the dimension n. The proof is essentially repetition of the proof of proposition.9. Need only to estimate # { k Z n : k = s } < n 1 s n 1. Hint: pick any k = (k 1,...,k n ) N n 0. By a change of sign of its components may generate at most n 1 independent vectors (k 1,±k,...,±k n ) Z n Reversible systems The system (ẋ ) = JΩ ẏ ( x y ) + is reversible in case ( ) X1 (x,y) + Y 1 (x,y) ( ) X (x,y) Y (x,y) X s (x, y) = X s (x,y), Y s (x, y) = Y s (x,y). The canonical system with Hamiltonian H(x,y) = H 0 (x,y)+h 1 (x,y)+... is reversible in case H(x, y) = H(x,y). Parity of functions: a function f(x,y) has even parity in case f(x, y) = f(x,y), and has odd parity in case f(x, y) = f(x,y). Write f + (x,y) and f (x,y) for an even and an odd function, respectively. Any function f(x,y) may be written as For, just set f + (x,y) = 1 f(x,y) = f + (x,y)+f (x,y). ( f(x,y)+f(x, y)) ), f (x,y) = 1 ( f(x,y) f(x, y)) ). A polynomial f(x,y) is even (resp. odd) if every monomial has even (resp. odd) degree in y. The Hamiltonian of a reversible canonical system is an even function. If ω is non resonant then every f N ω is an even function. For a component of a vector field use a similar notation, e.g., X s,j, Y + s,j. Non linear elliptic equilibria 66

7 The derivative, the product and the Poisson bracket obey + + x +, y, {, } Lemma 5.5: Let the vector field V be reversible. Then L V f + is an odd function, and L V f is an even function. Proof. If the vector field is reversible then the Lie derivative of an even function is ( LV f +) n ( (x, y) = X f + l +Y + f + ) l (x, y) x l y l n ( = X f + l +Y + l ( )) f+ (x,y) = ( L V f +) (x,y). x l y l For an odd function a similar calculation gives ( LV f ) (x, y) = ( L V f ) (x,y). Q.E.D. Proposition 5.6: If the frequencies ω are non resonant and the system (ẋ ) ( ( ) ( ) x X1 (x,y) X (x,y) = JΩ ẏ y) Y 1 (x,y) Y (x,y) is reversible then there exist n independent formal first integrals of the form Φ(x,y) = Φ 0 +Φ 1 (x,y)+φ (x,y)+... with Φ 0 = I j for j = 1,...,n, which are even functions. Corollary 5.7: The first integrals are made unique by the condition Φ s (x,y) R s+ ω for all s > 0. Proof of proposition 5.6. Recall the equations: ω Φ 0 = 0, ω Φ s = L V1 Φ s 1... L Vs Φ 0. Must prove that L V1 Φ s 1 R s+ for all s > 0. Firstequation solvedby anyfunction Φ 0 N ω. Bynon resonance,φ 0 depends onlyon the actions; so Φ 0 must be even. By induction. If Φ 0,...,Φ s 1 are even, then every Lie derivative in the r.h.s. is odd, hence it belongs to R s+, due to the non resonance condition. The operator ω is uniquely inverted on R s+, thus giving Φ s R s+. AnarbitraryΦ s N s+,maybeadded,butitiseven.thereforeφ s mustbeeven,thuscompleting the induction. Q.E.D. Proof of corollary 5.7. Just solve the system keeping Φ s R ω for all s. Q.E.D. Non linear elliptic equilibria 67

8 Proposition 5.8: If the Hamiltonian (5.1) is reversible and the frequencies are non resonant then there exist n independent formal first integrals of the form Φ(x,y) = I l +Φ 1 (x,y)+φ (x,y)+..., for l = 1,...,n, which are even functions and are made unique by the condition Φ s (x,y) Rω s+ for all s > 0. Just a reformulation of proposition 5.6. Proposition 5.9: The n first integrals Φ (j) = I j +Φ (j) , j = 1,...,n are independent and in involution, i.e, {Φ (j),φ (l) } = 0 for j,l = 1,...,n. Proof. Let Υ = {Φ (j),φ (l) }. By Poisson s theorem, Υ is a first integral, and so it must be an even function. Since both Φ (j) and Φ (l) are even functions, then Υ must be and odd function. Conclude that Υ = 0. The system is formally integrable (Liouville, Arnold, Jost). The concept of complete stability of Birkhoff, generalizing the adiabatic invariance. Truncate the series at an arbitrary degree r, i.e., Φ (r) = I j +Φ 1 (x,y)+...+φ r (x,y). In a neighbourhood of radius of the origin we have Φ (r) = O( r+3 ). Q.E.D. Then Φ (r) (t) Φ (r) (0) ( ) 1 = O( 3) up to t = O r. Question: Can we improve the complete stability of Birkhoff? 5..4 Polynomial norm and technical estimates For a homogeneous polynomial f(x,y) = f j,k x j y k j+k =s the polynomial norm is defined as f = f j,k. j,k The definition of norm applies to both real and complex polynomials. The transformation to complex variables ξ, η and the inverse transformation change the norm of a homogeneous polynomial of degree s at most by a factor s/. Lemma 5.10: Let f(ξ,η) and g(ξ,η) be homogeneous polynomials of degree r and s, respectively. Then {f,g} rs f g. Proof. Write {f,g} = f jk g j k y n xj+j k+k j,k,j,k Use n (j lk l +j l k l) r n (k l +j l ) = rs and compute {f,g} n f jk g j k (j l k l +j lk l ) rs j,k,j,k rs f g. j l k l j l k l x l y l, ( j,k )( ) f jk g j k j,k Q.E.D. Non linear elliptic equilibria 68

9 Recall the homological equation ω Φ = Ψ. (5.) Ψ = ψ k ξ j η k solved by Φ = i k =s+ Define the sequence {α s } s>0 as By non resonance, α s > 0. α s = min 0< k s+ k,ω. k =s+ ψ k k j,ω ξj η k. Lemma 5.11: The solution (5.) of the homological equation ω Φ s = Ψ s satisfies Φ s 1 α s Ψ s. An obvious consequence of the form (5.) of the solution. Proposition 5.1: The formal first integrals satisfy Φs A s+1 B (s+1)! s α l with positive constants A, B that could be explicitly estimated., s 1, Proof. Starting with I l = 1 and H 1 = E, transform the recurrent equations for Φ into a recurrent estimate for the norms: Ψ 1 = {H 1,I l } gives Ψ 1 6E, Ψ s = {H 1,Φ s 1 } gives Ψ s 3E(s+1) Φ s 1, ω Φ s = Ψ s gives Φ s 3E(s+1) Φ s 1. α s The constants A, B are estimated by using the latter formula. Q.E.D Truncated first integrals Can not prove convergence with the estimates of proposition 5.1. The first integrals so constructed are generically non convergent (Siegel, 1941). There is numerical evidence that the estimate of proposition 5.1 is very close to optimal (Contopoulos, Efthymiopoulos and Giorgilli, ). There is numerical evidence that the first integrals truncated at a suitable order do reproduce very well the dynamics. Example 5.13: Truncated first integrals. Comparison of the level lines of the truncated integrals with the Poincaré sections of the orbits for the Hamiltonian H = ω1 (x 1 +y ω 1 )+ (x +y )+x 1 x 1 3 x3 with ω 1 = 1 and ω = ( 5 1)/ (the golden number). Non linear elliptic equilibria 69

10 Pick r 1, and consider the truncated first integral Φ(r) = Il + Φ Φr. By construction: Φ (r) = {H1, Φr }, Φ (r) Cr r+3, (r + )! Cr = 3EAr+1 B Qr αl In a polydisk = (x, y) Rn : x + y we have: (a) A deformation of the actions Il bounded by Φ(r) (x, y) Il < Dr 3. In first approximation the time evolution of Il (t) is quasi periodic, oscillating in a bounded interval of width OO 3 ). (b) A slow noise due to the evolution of Φ(r) bounded by Φ(r) (t) Φ(r) (0) t Cr r+3, The effect of the noise can not be predicted, and induces small deviations with respect to the quasiperiodic evolution. Non linear elliptic equilibria 70

11 5..6 Exponential estimates The estimate of Φ (r) obtained by proposition 5.1 depends on two relevant parameters, namely: (i) the radius of the neighbourhood, which must be chosen in view of the initial condition; (ii) the order r of truncation of the first integral, which is our choice. Want to remove the dependence on the user defined parameter r (which is unnatural in a physical model). Diophantine condition on the frequencies: for k 0 assume k,ω > γ k τ for some γ > 0 and τ > n 1. That is, replace α s with γs τ. In a polydisk get (forget unessential constants and simplify the formula, also setting a = τ +1) Φ(r) (x,y) (r!)a r 3 For given (the initial conditions) look for the best choice of r. In the right member write (r!) a r = r a (r 1)! a r 1. For increasing r there is a minimum for r = r opt = ( ) 1/a 1. Using Stirling s formula calculate for r = r opt (r opt!) a ropt ( ropt e ) aropt r opt exp [ a ( ) ] 1/a 1. Theorem 5.14: Consider the Hamiltonian (5.1). Assume that the frequencies ω satisfy the Diophantine condition. Then there exist positive constants C, T and such that for every < the following holds true: for every orbit with initial point (x 0,y 0 ) / we have I l (t) I l (0) < C 3 for t < T exp [ (τ +1) ( ) ] 1/(τ+1) 1. Hints for the proof. By the triangle s inequality, for any r > 0 get Il (t) I l (0) Il (t) Φ (r) (t) + Φ (r) (t) Φ (r) (0) + Φ (r) (0) I l (0). The first and third term are estimated by the deformation, which in turn is estimated to be O( 3). The second term is due to the noise, and its contribution increases as t Φ(r) (x,y). By optimizing the choice of r we get Φ(r) (x,y) exp ( (1/ ) (1/(τ+1))). If we allow the effect of the noise to become O( 3), i.e., comparable with the deformation, then we get the exponential estimate for the time. A comment by Littlewood: If not eternity, this is a considerable slice of it. Non linear elliptic equilibria 71

12 5..7 A note on the general case Consider the full expansion of the Hamiltonian H(x,y) = H 0 (x,y)+h 1 (x,y)+h (x,y)+..., H 0 (x,y) = 1 The equation for a first integral {H,Φ} = 0 splits as { ω Φ 0 = 0, ω Φ s = Ψ s, Ψ s = n ω l (x l +yl), {H 1,I l } for s = 1 {H 1,Φ s 1 }+...+{H s,i l } for s > 1 The formal construction goes the same way: just more Poisson brackets. The quantitative scheme is a little more elaborate, but still one ends up with the estimate of proposition 5.1 Φ (l) s A s+1 B (s+1)! s 1 α, s 1, l just with different values for the constants A and B. The time derivative of a truncated integral Φ (r) is calculated as an infinite series Φ (r) = s>rψ s, Ψ s = {H 1,Φ r }... {H s,i l }. The series is convergent: it is the Poisson bracket between a polynomial (finite degree r +) and an analytc function. The exponential estimate is still valid. 5.3 A note on the concept of stability Usually one refers to the theory of Lyapounov, in particular to stability of an equilibrium, but......for a pysical system that evolves equilibrium is an exceptional state. Need a more refined approach. An integrable Hamiltonian system is characterized by its actions (I, say) and by the angles (ϕ, say). The actions are constant (first integrals), and the angles evolve linearly in time (Kronecker flow). Adding a small perturbation (of size ε, say) generically destroys such a nice (but boring) picture (the theorem of Poincaré on non existence of first integrals). The problem of stability is reformulated as: Prove that the actions I satisfy an inequality such as I(t) I(0) ε a for t T(ε) with T(ε) large, in a sense to be made precise, for 0 < a 1. Example 5.15: The Solar System. The actions are the semimajor axes, the eccentricities and the inclinations of the orbits The angles are the mean anomaly (related to an area according to the second Kepler s law), the argument of the perihelion and the argument of the node. For the development of life it is essential that the semimajor axes, the eccentricities and the inclinations do not change too much for a substantial fraction of the life of the Solar System. Non linear elliptic equilibria 7

13 Different approaches to the problem of stability of perturbed, near to integrable systems. Let ε be the size of the perturbation. Want I(t) I(0) ε a for t T(ε). i. T(ε) 1/ε: the theory of adiabatic invariants. Essentially the theory of Lagrange for the Solar System. Played a major role in the development of Quantum Mechanics. ii. T(ε) 1/ε r with r > 1: the theory of complete stability. Introduced by Birkhoff (197). Refers to the dynamics around an equilibrium, as in the present lectures. Based on a bound of type C r r on the noise, but without the estimate C r (r!) a, which prevents the process of optimization of r. iii. T(ε) exp(1/ε a ) with 0 < a 1: the exponential stability. Proposed by Moser (1955) and Littlewood (1959) for an elliptic equilibrium. Developed in general form by Nekhoroshev (1977). iv. T(ε) exp ( exp(1/ε a ) ) with 0 < a 1: the superexponential stability. Investigated by Morbidelli and Giorgilli (1995). v. T(ε) = : the perpetual stability The dream of many mathematicians and astronomers of the XIX th century: to prove that the Newtonian model of the Solar System is integrable. The guiding idea of the theory of Lyapounov (189). 5.4 Use of computer algebra Want to obtain good stability estimates by using the truncated first integrals explicitly constructed via computer algebra Basic estimate Let R = (R 1,...,R n ) be fixed positive constants (radii). Define the domain (polydisk) R = { (x,y) R n : I l (x,y) R l, l = 1,...,n }. Take an initial point (x 0,y 0 ) 0R. Want φ t (x 0,y 0 ) for t τ( 0, ) for 0 <. Recall I(t) I(0) I(t) Φ (r) (t) + Φ (r) (t) Φ (r) (0) + I(0) Φ (r) (0), δ r ( ) = sup Φ (r) (x,y) I l (x,y). }{{}}{{} δ r( t) δ r( 0) Choose 0 and. Let (a label l is omitted for brevity) D r ( 0, ) = R ( 0 ) δ r ( ) δ r ( 0). Compatibility condition on 0, : want D r ( 0, ) 0. Straightforward estimate: Φ (r) (t) Φ (r) (0) < t sup Φl (x,y). (x,y) R The stability time is estimated to be not less than (5.3) τ r ( 0, ) = min,...,n The estimate depends on r, 0 and. sup (x,y) D r ( 0, ) Φ(r) (x,y) Non linear elliptic equilibria 73

14 5.4. Estimating the uniform norm of a function Recall Φl = {H 1,Φ r }. Need to estimate sup (x,y) R {H1,Φ r }(x,y), δr ( ) Let f(x,y) = j+k =s f j,kx j y k. Want to estimate f R = sup (x,y) R f(x,y). Naive estimate: f R s Easy to calculate, but can do better. j+k =s Estimate by trigonometric functions. We have f R j+k =s r s=1 sup Φs (x,y). f j,k R j1+k1 1 R jn+kn n. f j,k n sup x j l l yk l l. R In polar coordinates x l = cosϑ l, y l = sinϑ l we have Conclude sup x j l l yk l l = jl+k l sup cos j l ϑ l sin k l ϑ l Θ l jl,k l jl+k, Θ j,k = R ϑ T f R s j+k =s f j,k n R j l+k l Θ jl,k l. j j k k (j +k) j+k. Estimate via complex variables Rewrite the transformation to complex variables as x l = 1 (ξ l +iη l ), y l = 1 (ξ l iη l ), where ξ l = r l e iϑ l, η l = r l e iϑ l. In complex variables f(ξ,η) = with coefficients c j,k that may be calculated. For (x,y) R we have 0 r j < R j. Thus f s R s/ j+k =s j+k =s c j,k ξ j η k c j,k R j1+k1 1 R jn+kn n. Non linear elliptic equilibria 74

15 5.4.3 Optimizing the stability time The estimate (5.3) τ r ( 0, ) = min,...,n D r ( 0, ) sup Φ(r) (x,y) depends on 0, and r. In a practical problem 0 is determined by the initial conditions, but and r are our choice. For fixed 0, optimize and r so as to maximize the estimated stability time. I.e., look for an optimal estimate T ( 0) = max r supτ r ( 0, ). Recall (by analytic estimates) δ r ( ) C 3, {H1,Φ (r) } r! r+3. For fixed r the condition R > 0 gives an allowed interval, and the estimated time has a maximum T r ( 0) (depending only on 0 and r) at some point opt (r). For r increasing, T r ( 0) has a maximum for r opt. Performing numerically the process above one finds the optimal estimated stability time T ( 0) = T ropt ( 0). D r (ρ 0,ρ) Φ (r) τ r (ρ 0,ρ) T r (ρ 0 ) ρ ρ ρ opt (r) ρ Expected behaviour of the stability time lnt * (ρ 0 ) There is a decreasing sequence 1,, 3,... of steps at which the optimal order r increases. In the interval [ s+1, s] the estimated stability time T ( ) behaves as a power 1/ r (a straight segment in log log scale).... ρ 3 ρ ρ 1 lnρ 0 Non linear elliptic equilibria 75

16 Non linear elliptic equilibria 76