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The interplay of group and dynamical systems analysis: The case of spherically symmetric charged fluids in general relativity
International Journal of Non-Linear Mechanics 62 (2014) 58–72
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The interplay of group and dynamical systems analysis: The case of spherically symmetric charged fluids in general relativity P.M. Tchepmo Djomegni, K.S. Govinder n School of Mathematics, Statistics and Computer Science, University of KwaZulu–Natal, Private Bag X54001, Durban 4000, South Africa
art ic l e i nf o
a b s t r a c t
Article history: Received 18 February 2014 Received in revised form 27 February 2014 Accepted 28 February 2014 Available online 12 March 2014
We investigate the relationship between the Dynamical Systems analysis and the Lie Symmetry analysis of ordinary differential equations. We undertake this investigation by looking at a relativistic model of self-gravitating charged fluids. Specifically we look at the impact of specific parameters obtained from Lie Symmetries analysis on the qualitative behaviour of the model. Steady states, stability and possible bifurcations are explored. We show that, in some cases, the Lie analysis can help to simply the dynamical systems analysis. & 2014 Elsevier Ltd. All rights reserved.
Keywords: Symmetries Dynamical systems Emden–Fowler equation
1. Introduction Most of the real world problems (in biology, finance, economics, industry, etc.), and many fundamental laws of physics and chemistry are formulated in the form of differential (or difference) equations. Various methods for solving or analysing such equations have been developed. In the late 19th century, Marius Sophus Lie unified many of these methods by introducing the notion of (what has become known as) Lie groups [23]. The Lie theory of differential equations has been phenomenally successful in determining solutions to differential equations [3,21]. It is a useful tool that can be applied to either find solutions explicitly or it can be used to classify equations via equivalence transformations. A major hurdle has been the oftentimes tedious calculations involved in finding the symmetries. However, with the advent of very capable computer packages [11,6], this disadvantage has been overcome. Symmetries can be easily calculated for a variety of equations and then used to obtain solutions, if possible. Another very useful approach to differential equations is that of dynamical systems analysis [26] in which the long-term behaviour of a system is investigated by focusing on linearization around equilibrium points. This approach has its genesis in Newtonian mechanics, and emphasizes on qualitative rather than quantitative questions [2,17,22]. For example, it was eventually realized that equations describing the motion of the three-body problem (sun, earth and moon) were difficult to solve analytically [26]. Instead of focusing
n
Corresponding author. E-mail addresses: [email protected] (P.M. Tchepmo Djomegni), [email protected] (K.S. Govinder). http://dx.doi.org/10.1016/j.ijnonlinmec.2014.02.010 0020-7462/& 2014 Elsevier Ltd. All rights reserved.
only of the exact positions of the planets at all times, people looked at their stability. Thus, the qualitative analysis is helpful, particularly when the exact solution of an equation cannot be found (and also can give more useful information even when exact solutions exist). Though these two approaches look as they belong to different areas of mathematics, they have some structures in common. For instance, in a natural way, the equivariant bifurcation theory can be viewed as an application of Lie groups in symmetric systems [30]. Therefore, it is natural to consider the use of both approaches when analysing any differential equation of interest. In what follows, we present the results of applying both methods to an Emden–Fowler equation of index three. Such equations have a lengthy history [29,19,25,10,14] but, we believe, have not been approached in this two-fold manner. We will show how each method can be used to obtain interesting information about the behaviour of the solutions.
2. The model The metric governing the shear-free motion of a spherically symmetric perfect fluid can be written as 2
2
2
2
ds ¼ e2νðt;rÞ dt þe2λðt;rÞ ½dr þr 2 ðdθ2 þ sin θ dϕ2 Þ
ð1Þ
where ν and λ are the gravitational potentials. If we impose the presence of an electromagnetic field, the usual Einstein field equations must be modified and supplemented with Maxwell's equations. The resulting Einstein–Maxwell system is then given by λ2t 1 4λr E2 4 4λ ρ ¼ 3 2ν 2λ 2λrr þ λ2r þ ð2Þ r e e r e
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
p¼
1 1 2νr 2λr E2 þ þ 4 4λ ð 3λ2t 2λtt þ2νt λt Þ þ 2λ λ2r þ 2νr λr þ 2ν r r e e r e
ð3Þ
p¼
1 1 νr λr E2 2 2 þ þλ ð 3λ 2λ þ2ν λ Þ þ ν þ ν þ tt t t rr rr 4 4λ t r 2ν 2λ r r e e r e
ð4Þ ð5Þ
E ¼ r 2 eλ ν Φr
ð6Þ
Er ¼ sr 2 e3λ
ð7Þ
Here, ρ is the energy density, p is the isotropic pressure, s is the proper charge density of the fluid and we interpret E as the total charge contained within the sphere of radius r centred around the origin of the coordinate system. The electromagnetic field is present via Φr. Integrating (5) and combining (3) and (4) and integrating allow us to reduce the system to 4λr E2 4 4λ ρ ¼ 3e2h e 2λ 2λrr þ λ2r þ ð8Þ r r e " # 1 2λr E2 λ 2 3λ þ 2h e e λr þ 4 λ p¼ r λt e3λ r e
Eq. (18) is the fundamental non-linear ordinary differential equation which determines the behaviour of self-gravitating charged fluids in general relativity [15]. Once y is determined, Kweyama et al. [15] found that G¼a
0 ¼ νr λt λtr
ð9Þ
∂ ∂ þ ðby þ cÞ ∂x ∂y
a″ ¼ 2b
0
ð20Þ
b″ ¼ 2fc
ð21Þ
c″ ¼ 0
ð22Þ
0
af þ ð2a0 þ bÞf ¼ 3cg
ð23Þ
ag 0 þ ð2a0 þ 2bÞg ¼ 0:
ð24Þ
This system was reduced to 2b ¼ a0 þ α
λr 2E2 ¼ F~ 4 λ eλ λrr λ2r r r e
ð11Þ
E ¼ r 2 eλ ν Φr
ð12Þ
Er ¼ sr 2 e3λ ;
ð13Þ
where h ¼ hðtÞ and F~ ¼ F~ ðrÞ are arbitrary functions of integration. This means that we only need to solve (11) in order to obtain all the other unknown functions. Using the transformation [7,15] x ¼ r2
ð14Þ
y ¼ eλ
ð15Þ
f ðxÞ ¼
F~ 4r 2
ð16Þ
gðxÞ ¼
E2 2r 6
ð17Þ
c ¼ c0 þc1 x
ð26Þ
g ¼ g 2 a 3 exp
Z
α dx ; a
ð27Þ
Z Z Z α dx α dx f 2 3g 2 ca 3=2 exp dx ; f ¼ a 5=2 exp 2a a ð28Þ where α, c0, c1, f2 and g2 are arbitrary constants. Setting Z dx X¼ a Z Z Z b dx c b dx exp dx; Y ¼ y exp a a a
ð29Þ ð30Þ
(18) can be transformed into autonomous form. When c a0, a satisfied the non-linear fourth order equation Z 0 5a α α dx caa⁗ þ c c0 a a‴ ¼ 12g 2 c3 a 3 exp ; ð31Þ þ a 2 2 and (18) became α2 Y ¼ g 2 Y 3 þ f 2 Y 2 þ N: Y″ þ αY 0 þ M þ 4
ð32Þ
When c ¼0
we can rewrite (11) as y″ ¼ f ðxÞy2 þgðxÞy3 :
ð25Þ
ð10Þ
ð19Þ
is a point symmetry of (18), provided the functions a(x), b(x), and c (x) satisfied the following system of ordinary differential equations:
t
eν ¼ λt e h
59
ð18Þ
a ¼ a0 þ a1 x þ a2 x2 ;
ð33Þ
and (18) became Remark. (1) In this approach, our interest is only in point symmetries. We acknowledge that a variety of other symmetries exist, including potential [4], Lie–Bäcklund/generalized [1,21], Lambda [20,24,8] and non-local [9]. Expanding our study to those symmetries may indeed yield other useful results but they are outside the scope of our work. (2) In general, (18) does not admit any Lie point symmetries. As a result, it cannot be linearized under a point transformation. In what follows, we investigate under what conditions the equation does admit point symmetries. We first look at conditions under which it admits a single Lie point symmetry (and so is part of the equivalence class of autonomous second order differential equations). Thereafter, we investigate additional conditions under which the equation admits a second Lie point symmetry. This is important as the possession of two Lie point symmetries guarantees that the equation can be reduced to quadratures.
Y″ þ αY 0 þ βY ¼ g 2 Y 3 þ f 2 Y 2 ;
ð34Þ
where 1 1 β ¼ a21 þ a0 a2 þ α2 ; 4 4
ð35Þ
and M, N, a0, a1 and a2 are constants. The quantities M and N are constants of integration and are given by [15] M¼
1 1 aa″ a02 2f 2 I þ 3g 2 I 2 2 4
ð36Þ
and
Z α dx 1 1 ac0 a0 c þ αc N ¼ a 1=2 exp 2a 2 2 1 1 02 α2 2 3 I þ f 2 I 2g 2 I ; aa″ a þ 2 4 4
ð37Þ
60
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
where Z Z α dx dx: I¼ ca 3=2 exp 2a
α2 U0 ¼ M þ Y αU þ N: 4
ð38Þ
ð44Þ
Since N is constant, the system (44) is linear.
Since (19) can be transformed into G1 ¼
∂ ∂X
1. For M a α2 =4, ð4N=ð4M þα2 Þ; 0Þ is the only equilibrium point. At that point, the Jacobian matrix is given by 0 1 0 1 2 B C ð45Þ A ¼ @ M þα α A: 4
ð39Þ
via (29)–(30), Eqs. (32) and (34) admit only one symmetry. Kweyama et al. [15] then presented conditions under which Eqs. (32) and (34) could have an additional symmetry. Our intention is to investigate the impact of all the parameters and of those conditions on the qualitative behaviour of our underlying equations in the phase plane near the equilibrium points.
Thus τ ¼ α, Δ ¼ M þα2 =4. (a) If M o α2 =4, then Δ o 0. Therefore, the equilibrium point is a saddle point. (b) If α2 =4 o M, then Δ 4 0. Thus τ2 4Δ ¼ α2 4M α2 ¼ 4M. (i) When M 4 0, τ2 4Δ o0. (A) For α a 0, we have a spiral around the equilibrium point. The spiral is stable if α 4 0, and unstable if α o 0. The phase portrait of the solutions is shown in Fig. 1(a), (b). (B) For α ¼ 0, the only equilibrium point ðN=M; 0Þ is a centre, and the direction is given by the sign of U (see the beginning of the section). The phase portrait is shown in Fig. 1(c). (ii) When α2 =4 o M o 0, then τ2 4Δ 40, and α a 0. Therefore the equilibrium point is a node. It is a stable node if α 4 0 and an unstable node if α o 0. (iii) When M ¼0, then τ2 4Δ ¼ 0. (A) If α ¼ 0, then Δ ¼ 0. Thus we have a whole plane of equilibrium points. (B) If α a0, we have a double eigenvalue λ ¼ τ=2 ¼ α=2, with one associated eigenvector V ¼ ð 2=α; 1Þ. Thus the equilibrium point is a degenerate node. The node is stable if α 4 0, and unstable if α o 0. 2. For M ¼ α2 =4, (44) is linear and the analysis is trivial.
3. Non-zero c The first part of our analysis is focused on Eq. (32). If we set U ¼ Y 0;
ð40Þ
we can rewrite (32) as the two-dimensional non-linear system Y 0 ¼ U ¼ hðY; UÞ
α2 Y αU þN ¼ kðY; UÞ: U 0 ¼ g2 Y 3 þ f 2 Y 2 M þ 4
ð41Þ
The motion in the phase plane is given by the phase portrait of solutions that comes from system (41). The phase portrait is determined by the linear system analysis associated with (41) at the equilibrium points. The equilibrium points occur at points satisfying U ¼ 0;
ð42Þ
and
α2 g2 Y 3 þ f 2 Y 2 M þ Y þ N ¼ 0: 4
ð43Þ
Before proceeding with the general analysis, we observe that the function h is odd in U (i.e., hðY; UÞ ¼ hðY; UÞ), and when α ¼ 0 the function k is even in U (i.e., kðY; UÞ ¼ kðY; UÞ). Therefore the system is reversible when α ¼ 0 [26]. The direction of the vector fields is given by the sign of U in (41). Thus Y increases when U 4 0, and decreases when U o0.
3.2. Neutral perfect fluid Here we take g 2 ¼ 0 and f 2 a 0 [18]. Thus there is no charge within the sphere, but we do have matter.
3.1. Neutral dust Here we take g 2 ¼ 0 and f 2 ¼ 0 [18]. From (27) g ¼0, therefore there is no charge contained within the sphere (E ¼0 in (17)). The system (41) becomes
1. When α ¼ 0, the system (41) becomes Y0 ¼ U
0
U 0 ¼ f 2 Y 2 MY þ N:
Y ¼U
U
4
U
4
4
2
2
2
2
4
Y
4
2
U 2
1
2
2
2
4
4
ð46Þ
4
Y
1
1
2
3
Y
1
2
Fig. 1. Behaviour of the system in the neutral dust case when M 4 0. The system predicts a spiral (α ¼ 7 1, M ¼ 2, N ¼3), or a centre (α ¼ 0, M ¼ 1, N ¼ 1). (a) Stable spiral. (b) Unstable spiral and (c) Centre.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
There are two possible equilibrium points
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 0 M þ M 2 4f 2 N M M 2 4f 2 N @ A @ ðY 0 ; 0Þ ¼ ; 0 ; ðY 1 ; 0Þ ¼ ; 0A; 2f 2 2f 2
ð47Þ provided M 2 4f 2 N Z0. (a) If M 2 4f 2 N ¼ 0, i.e., N ¼ M 2 =4f 2 , then ðM=2f 2 ; 0Þ is the only equilibrium point, and at that point the Jacobian matrix is given by 0 1 A¼ : ð48Þ 0 0 Thus τ ¼ 0 and Δ ¼ 0. We are in a borderline case. We cannot extrapolate from the linear case to the fully non-linear case. However, system (46) yields Y 0 ðf 2 Y 2 MY þ NÞ ¼ UU 0 ;
ð49Þ
i.e., 0
2
0
0
0
f 2 Y Y MY Y þ NY ¼ UU :
ð50Þ
After integration (50) becomes 1 1 1 f Y 3 MY 2 þ NY þ K 0 ¼ U 2 ; 3 2 2 2
ð51Þ
where K0 is a constant. The phase portrait is shown in Fig. 2. (b) If M 2 4f 2 N 4 0, we have two equilibrium points ðY 0 ; 0Þ and ðY 1 ; 0Þ in (47). (I) At ðY 0 ; 0Þ, the Jacobian matrix is 0 1 0qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 A: ð52Þ A¼@ M 2 4f 2 N 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ 0 and Δ ¼ M 2 4f 2 N o 0. Therefore we have a saddle point. (II) At ðY 1 ; 0Þ, as Jacobian matrix we have 0 1 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 @ A: ð53Þ A¼ M 2 4f 2 N 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ 0 and Δ ¼ M 2 4f 2 N 4 0. Therefore the point is a linear centre. Since the system is reversible, then the equilibrium point is a non-linear centre. From (47) if f 2 4 0, the equilibrium points exist when N o M 2 =4f 2 , and then fuse together when N ¼ M 2 =4f 2 , and
disappear when N 4 M 2 =4f 2 . There has been a saddle-node bifurcation [27]. Similarly, if f 2 o 0, the equilibrium points exist when N 4 M 2 =4f 2 , and then fuse when N ¼ M 2 =4f 2 , and are destroyed when N o M 2 =4f 2 . Therefore we still have a saddlenode bifurcation, specifically a blue sky bifurcation. Here, N is the bifurcation parameter (and not f2 as it cannot be zero). We can visualize the bifurcation diagram as shown in Fig. 3. 2. If α a 0, we have the following equilibrium points: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 4M þ α2 þ 16M 2 64f 2 N þ 8Mα2 þ α4 ð54Þ ðY 0 ; 0Þ ¼ @ ; 0A 8f 2 and 0 ðY 1 ; 0Þ ¼ @
4M þ α2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 16M 2 64f 2 N þ 8Mα2 þ α4 ; 0 A; 8f 2
ð55Þ
provided 16M 2 64f 2 N þ8Mα2 þ α4 Z 0. (a) If we assume that 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, then: (I) At ðY 0 ; 0Þ, the Jacobian matrix is given by 0 1 0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ A: 1 ð56Þ A¼ 16M 2 64f 2 N þ 8Mα2 þ α4 α 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ α and Δ ¼ 14 16M 2 64f 2 N þ 8Mα2 þ α4 o 0. Therefore the point is a saddle point. (II) At ðY 1 ; 0Þ, the Jacobian matrix given by 0 1 0 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q A; ð57Þ A¼@ 1 16M 2 64f 2 N þ 8Mα2 þα4 α 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi yields τ ¼ α and Δ ¼ 14 16M 2 64f 2 N þ 8Mα2 þ α4 40. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (i) If τ2 4Δ ¼ α2 16M 2 64f 2 N þ8Mα2 þ α4 4 0, then the equilibrium point is a node. The node is stable when α 4 0, and unstable when α o 0. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (ii) If τ2 4Δ ¼ α2 16M 2 64f 2 N þ 8Mα2 þα4 o 0, we have a spiral. It is a stable spiral for α 4 0, and unstable spiral for α o 0. (b) If we assume that 16M 2 64f 2 N þ8Mα2 þ α4 ¼ 0, ðð4M þ α2 Þ=8f 2 ; 0Þ is the only equilibrium point, and at that point the Jacobian matrix is given by 0 1 A¼ : ð58Þ 0 α Thus τ ¼ α and Δ ¼ 0. We are in a borderline case. Since our system is non-linear and difficult to solve analytically, we have only plotted the full system numerically with specific values of parameters just to show the behaviour in the phase space in Fig. 4(a), (b) (we used the package Dynpac [5]).
U 2
1
1
61
1
2
3
Y
1
2 Fig. 2. Behaviour of the system in the neutral perfect fluid case where α ¼ 0 (f 2 ¼ 1, M ¼2, and N ¼1).
From (54) and (55), when α 4 0 we have a bifurcation (because when α o 0 all the equilibrium points are unstable, and there is no change of stability). Thus for f 2 4 0, the two equilibrium points exist when N o ð16M 2 þ 8Mα2 þα4 Þ=64f 2 , then fuse when N ¼ ð16M 2 þ8Mα2 þ α4 Þ=64f 2 , and disappear when N 4 ð16M 2 þ 8Mα2 þ α4 Þ=64f 2 . A saddle-node bifurcation occurred.
Likewise, for f 2 o 0 the opposite happens. We have a blue sky bifurcation, with N as the bifurcation parameter. The bifurcation diagram is shown in Fig. 5.
62
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
Y
Y Stable node
N
Saddle point
N
Stable spiral Saddle point
Fig. 3. Bifurcation diagram Y ¼ f ðNÞ (f 2 ¼ 7 1 and M¼ 2). (a) f 2 4 0: saddle-node bifurcation. (b) f 2 o 0: blue sky bifurcation.
2
U 3
U 3
2
2
1
1
1
1
2
Y
2
1
1
1
1
2
2
3 2
2
Y
3
Fig. 4. Behaviour of the system when 16M 64f 2 N þ 8Mα þ α ¼ 0 (the linear analysis classified as a borderline case). We have plotted using f 2 ¼ 1, M¼ 1, N ¼ 1, α ¼ 7 2. (a) α 4 0 and (b) α o 0. 2
4
We use this condition because δ could be zero. It is clear that any of these points Y0, Y1, Y2 could be zero.
3.3. Charged fluid We now insist on g 2 a0. Thus E a 0 and the matter is charged. Eq. (43) admits as solutions
2
2
Y0 ¼
f2 16f 2 þ 48g 2 M þ 12g 2 α2 δ ffiffiffi ; pffiffiffi þ þ p 3g 2 6 3 4g 2 δ 12 3 2g 2
ð59Þ
Y1 ¼
pffiffiffi pffiffiffi 2 f2 ð1 þ i 3Þð16f 2 þ 48g 2 M þ 12g 2 α2 Þ ð1 i 3Þδ ffiffiffi ffiffiffi p p ; 3 3 3g 2 12 4g 2 δ 24 2g 2
ð60Þ
pffiffiffi pffiffiffi 2 f ð1 i 3Þð16f 2 þ 48g 2 M þ 12g 2 α2 Þ ð1 þi 3Þδ p p ffiffiffi ffiffiffi Y2 ¼ 2 ; 3g 2 12 3 4g 2 δ 24 3 2g 2
ð61Þ
3
3
Y 00 ¼
ðf g 3 27g 52 NÞ1=3 f2 þ 2 2 ; 3g 2 3g 22
ð64Þ
Y 11 ¼
pffiffiffi 3 ð1 i 3Þðf 2 g 32 27g 52 NÞ1=3 f2 ; 3g 2 6g 22
ð65Þ
Y 22 ¼
pffiffiffi 3 ð1 þ i 3Þðf 2 g 32 27g 52 NÞ1=3 f2 3g 2 6g 22
ð66Þ
as solutions. Since we are interested in real values, the only possible equilibrium point is ðY 00 ; 0Þ. At that point, the Jacobian matrix given by
2
provided 16f 2 þ 48g 2 M þ 12g 2 α2 a 0 [12,13,28], with δ ¼ ð 128f 2 576f 2 g 2 M 1728g 22 N 144f 2 g 2 α2 þ δ1 Þ1=3 ;
2
1. If 16f 2 þ 48g 2 M þ 12g 2 α2 ¼ 0, i.e., M ¼ ð 4f 2 3g 2 α2 Þ=12g 2 , Eq. (43) admits
ð62Þ
and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 δ1 ¼ ð 128f 2 576f 2 g 2 M 1728g22 N 144f 2 g2 α2 Þ2 4ð16f 2 þ 48g2 M þ 12g 2 α2 Þ3 :
ð63Þ
0
0 2=3 5 B 3 3 A ¼ @ ðf 2 g 2 27g 2 NÞ 3 3g 2
1
1
C α A; 3
yields τ ¼ α and Δ ¼ ðf 2 g 32 27g 52 NÞ2=3 =3g 32 .
ð67Þ
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
63
Y
Y Stable node
N
Saddle point
N
Stable spiral Saddle point
Fig. 5. Bifurcation diagram Y ¼ f ðNÞ (f 2 ¼ 0:5, M ¼ 1, α ¼ 3 for figure (a), and M ¼ 1, f 2 ¼ 1 and α ¼ 2 for figure (b)). (a) f 2 40: saddle-node bifurcation and (b) f 2 o 0: blue sky bifurcation.
U 3
U 3
2
2
1
1
1
1
2
3
4
Y
1
1
1
1
2
2
3
3
2
3
4
Y
pffiffiffi 2 Fig. 6. Nodes predicted when 16f 2 þ 48g2 M þ 12g2 α2 ¼ 0 in the charged fluid (f 2 ¼ 3, g 2 ¼ 1, M¼ 0, N ¼ 2, α ¼ 7 2 3). (a) Stable node and (b) Unstable node.
3
(a) If f 2 27g 22 N a 0 and g 2 4 0, then Δ o 0. Therefore ðY 00 ; 0Þ is a saddle point. 3 (b) If f 2 27g 22 N a 0 and g 2 o 0, then Δ 40. If we assume that α ¼ 0, ðY 00 ; 0Þ is a non-linear centre (since the system is reversible). However, if α a 0, the point is stable when α 4 0, and unstable when α o 0. Note that 3
τ2 4Δ ¼ α2 þ
4ðf 2 g 32 27g 52 NÞ2=3 : 3g 32
ð68Þ
(i) If τ2 4Δ 4 0, the point is a node. (ii) If τ2 4Δ o 0, we have a spiral. (iii) If τ2 4Δ ¼ 0, the linear equilibrium point is a degenerate node. This is a borderline case and we have numerically solved the full system to see the effect on the behaviour, as shown in Fig. 6(a), (b). 3 (c) If f 2 27g 22 N ¼ 0, then Δ ¼ 0. We are also in a borderline case. The behaviour for different values of parameters is shown in Fig. 7(a), (b). 2 2. If 16f 2 þ48g 2 M þ 12g 2 α2 o 0, then δ in (62) is real. Therefore Y0 in (59) is real, whereas Y1 and Y2 in (60) and (61) are still complex. Thus ðY 0 ; 0Þ is the only equilibrium point. At that point, the Jacobian matrix is given by 0
0
2 B A ¼ @ 3g Y 2 þ2f Y M þ α 2 0 2 0 4
1
1
C α A:
ð69Þ
Therefore τ ¼ α and Δ ¼ 3g 2 Y 20 2f 2 Y 0 þ ðM þ α2 =4Þ (which cannot be zero). (a) If Δ o 0, ðY 0 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, the equilibrium point is a non-linear centre. However, for α a0, τ2 4Δ ¼ 12g 2 Y 20 þ 8f 2 Y 0 4M. The equilibrium point is stable when α 4 0, and unstable when α o 0. If τ2 4Δ 4 0, then ðY 0 ; 0Þ is a node. If τ2 4Δ o 0, then ðY 0 ; 0Þ is a spiral. If τ2 4Δ ¼ 0, the linear point is a degenerate node. As before, we solve the full non-linear system numerically to determine its behaviour (see Fig. 8). 2 3. If 16f 2 þ 48g 2 M þ 12g 2 α2 4 0, Y0, Y1 and Y2 in (59), (60), (61) can be real or complex, but we are only interested in real values. It is difficult to transform them into the form A þ iB (where A and B are real), because δ in (62) can also be complex, and not easily transformable in this form. We will restrict ourselves to the constraint obtained in [15], with g 2 40. It was shown that 2
M¼
f2 α2 ; 3g 2 36
3
N¼
f2 2α2 f 2 27g 22 27g 2
led to the additional symmetry ∂ α αf ∂ eðα=3ÞX Yþ 2 : G2 ¼ eðα=3ÞX ∂X 3 9g 2 ∂Y
ð70Þ
ð71Þ
64
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
and then integrate to obtain
By substituting M and N in (59)–(61) by (70), we have three equilibrium points ! pffiffiffi 2α f2 f2 ðY 0 ; 0Þ ¼ ; 0 ; ðY 1 ; 0Þ ¼ pffiffiffiffiffi ; 0 ; 3g 2 3g 2 3 g 2 ! pffiffiffi f 2α ð72Þ ðY 2 ; 0Þ ¼ 2 þ pffiffiffiffiffi ; 0 : 3g 2 3 g 2
(a) Assume that α a 0. (I) At ðY 0 ; 0Þ, the Jacobian matrix given by 0 1 0 1 2 A A ¼ @ 2α α 9
Δ¼
ð73Þ
2α2 : 9
ð74Þ
Since Δ 4 0, τ2 4Δ ¼ α2 8α2 =9 ¼ α2 =9 40. Therefore the eigenvalues are real and have the same sign, hence the equilibrium point is a node. If α 40, then τ o 0. Thus ðY 0 ; 0Þ is a stable node. If α o0, then τ 4 0. Thus ðY 0 ; 0Þ is an unstable node. (II) At ðY 1 ; 0Þ and ðY 2 ; 0Þ, we find as Jacobian matrix 0
0 A ¼ @ 4α2 9
1 α
1 2 g Y 4 þ f 3 Y 3 MY 2 þ 2NY þ 2K 1 ; 2 2 3
ð79Þ
where K1 is a constant. The behaviour is obtained by sketching the above implicit function (79) as shown in Fig. 9.
4. Zero c
yields τ ¼ α;
U2 ¼
1 A:
ð75Þ
The second part of our analysis concerns the non-linear second order equation (34) Y″ þ αY 0 þ βY ¼ g 2 Y 3 þ f 2 Y 2 : ð80Þ 1 2 Since β ¼ γ þ α , with γ an arbitrary constant, this equation is the 4 same as (32) with M ¼ γ and N ¼0. 4.1. Neutral perfect fluid Here we take g 2 ¼ 0. Whatever the value of f2 the condition N ¼0 (according to the previous section) does not affect the existence and the stability of the equilibrium point. Therefore we have the same behaviour of solutions in the phase space. However, there is no bifurcation (since N is constant). 4.2. Charged fluid Now we take g 2 a 0. Setting U ¼ Y 0;
Thus 2
4α Δ¼ : 9
ð76Þ
Since Δ o0, ðY 1 ; 0Þ and ðY 2 ; 0Þ are saddle points. (b) If α ¼ 0, ð f 2 =3g 2 ; 0Þ is the only equilibrium point, and at that point τ ¼ 0;
Δ ¼ 0:
ð77Þ
This is a borderline case. Here we can show the global behaviour of the full system by reducing (41) to the equation α2 Y αU þ N Y 0 ; ð78Þ UU 0 ¼ g 2 Y 3 þf 2 Y 2 M þ 4
3
2
ð81Þ
Eq. (80) becomes the two-dimensional non-linear system Y0 ¼ U U 0 ¼ g 2 Y 3 þf 2 Y 2 βY αU: The equilibrium points are solutions of the system U ¼ 0;
ð83Þ
g 2 Y 3 þ f 2 Y 2 βY ¼ 0:
ð84Þ
Eq. (84) admits as solutions Y 0 ¼ 0;
ð85Þ
U 3
U 3
2
2
1
1
1
1
2
3
Y
3
2
1
1
1
1
2
2
3 Fig. 7. Behaviour of the system when Δ ¼ 0 and
ð82Þ
2
3
Y
3 2 16f 2 þ 48g 2 M þ 12g 2 α2
¼ 0 in the charged fluid (f 2 ¼ 3, g 2 ¼ 1, M ¼ 4, N ¼ 1, α ¼ 7 2). (a) α 4 0. (b) α o 0.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
2
Fig. 8. Spirals node predicted when
U 3
U 3
2
2
1
1
1
1
2
3
Y
2
2
3
3
2 16f 2 þ 48g2 M þ 12g2 α2 o 0
1
τ ¼ α;
Y
Y2 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ f 2 þ 4g 2 β 2g 2
1
1
C C: α A
ð90Þ
and M and N given by 70
Therefore 2
;
ð86Þ
τ ¼ α;
Δ¼
f 2 þ 4g 2 β þf 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ 4g 2 β
2g 2
:
ð91Þ
In this case Δ a 0. :
ð87Þ 2
2
We have four possible cases: f 2 þ 4g 2 β o 0 and β a0, f 2 þ4g 2 β 4 0 2 2 and β a 0, f 2 þ 4g 2 β 4 0 and β ¼ 0, or f 2 þ4g 2 β ¼ 0. When α ¼ 0, the system is reversible, and the direction of the vector fields is given by the sign of U in (82). Thus Y increases when U 4 0, and decreases when U o 0. 2
ð89Þ
0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B 2 2 A¼B @ f 2 þ4g 2 β þ f 2 f 2 þ 4g 2 β 2g 2
3
2g 2
Δ ¼ β:
0
2
Y1 ¼
Y
(a) If β o0, then Δ o 0. Therefore the origin is a saddle point. (b) If β 4 0, for α ¼ 0 the origin is a non-linear centre (since the system is reversible). However for α a 0, note that τ2 4Δ ¼ α2 4β. The point is stable when α 4 0, and unstable when α o 0. If 0 o β o α2 =4, the origin is a node. If β 4α2 =4, we have a spiral. If β ¼ α2 =4, the origin is a linear degenerate node. As in the other cases, the behaviour of the full system (34) is obtained by a numerical simulation, as shown in Fig. 10 (a), (b). 2 2. If f 2 þ 4g 2 β 4 0 and β a0, we have two additional equilibrium points ðY 1 ; 0Þ and ðY 2 ; 0Þ given by (86) and (87). I. At ðY 1 ; 0Þ, the Jacobian matrix is
1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 f 2 þ 4g 2 β
3
Thus
1
2 16f 2 þ 48g2 M þ 12g2 α2 4 0,
2
by the simulated system (41) (f 2 ¼ 3, g2 ¼ 1, M ¼3, N ¼ 2, α ¼ 7 2). (a) Stable spiral. (b) Unstable spiral.
1
Fig. 9. Phase portrait for (f 2 ¼ 3, g 2 ¼ 1, α ¼ 0).
1 1
2
2
1
1
U 3
3
2
65
1. If f 2 þ4g 2 β o 0 and β a 0, there is only one equilibrium point, the origin. At the origin, the Jacobian matrix is given by ! 0 1 A¼ : ð88Þ β α
(a) If Δ o 0, then ðY 1 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, ðY 1 ; 0Þ is a centre. However, for α a 0, the point is stable when α 40, and unstable when α o 0. Note that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 f 2 þ 4g 2 βÞ 2ðf þ 4g β þ f 2 2 2 τ2 4Δ ¼ α2 þ : ð92Þ g2 (i) If τ2 4Δ 4 0, then ðY 2 ; 0Þ is a node. (ii) If τ2 4Δ o 0, then we have spiral node. (iii) If τ2 4Δ ¼ 0, then ðY 1 ; 0Þ is a linear degenerate node. We have solved the system numerically to get the behaviour of the non-linear analysis (see Fig. 11).
66
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
(II) At ð f 2 =g 2 ; 0Þ, the Jacobian matrix given by
(II) At ðY 2 ; 0Þ, the Jacobian matrix is 0
0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B 2 2 f þ 4g β f f 2 þ4g 2 β A¼B 2 2 @ 2
1
C C: α A
2g 2
Therefore 2
τ ¼ α;
Δ¼
f 2 þ4g 2 β f 2
0
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ 4g 2 β
2g 2
ð93Þ
1 0 1 B f2 C A¼@ 2 A α g2
ð94Þ
yields τ ¼ α and Δ ¼ f 2 =g 2 (which cannot be zero Þ. (a) If g 2 4 0, then Δ o0. Therefore we have a saddle point. (b) If g 2 o 0, then Δ 40. For α ¼ 0, we have a centre. 2 However, for α a0, note that τ2 4Δ ¼ α2 þ 4f 2 =g 2 . The point is stable when α 4 0, and unstable when α o 0. (i) If α2 4 4f 2 =g , then the point is a node.
1
2
:
In this case Δ a 0. (a) If Δ o 0, then ðY 2 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, ðY 2 ; 0Þ is a centre. However, for α a 0, the point is stable when α 4 0, and unstable when α o 0. Note that
τ2 4Δ ¼ α2 þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 f 2 þ 4g 2 β f 2 f 2 þ 4g 2 β g2
:
ð95Þ
(i) If τ2 4Δ 4 0, then ðY 1 ; 0Þ is a node. (ii) If τ2 4Δ o 0, then we have spiral node. (iii) If τ2 4Δ ¼ 0, then ðY 2 ; 0Þ is a linear degenerate node. We have solved the system numerically to get the behaviour of the non-linear analysis (see Fig. 11). 2 3. If f 2 þ 4g 2 β 4 0 and β ¼ 0, i.e., f 2 a 0, from (85)–(87), we have two equilibrium points, ð0; 0Þ and ð f 2 =g 2 ; 0Þ. (I) At ð0; 0Þ, the Jacobian matrix is given by 0 1 A¼ : ð96Þ 0 α
2
2
Thus τ ¼ α and Δ ¼ f 2 =4g 2 . (a) If g 2 4 0 and β a 0, then Δ o 0. Therefore the origin is a saddle point. (b) If g 2 o 0 and β a0, then Δ 4 0. For α ¼ 0, we have a centre. 2 However, for α a 0, note that τ2 4Δ ¼ α2 þf 2 =g 2 . In this case, the origin is stable when α 40, and unstable when α o0. 2 (i) If α2 4 f 2 =g 2 , then the origin is a node. 2 2 (ii) If α o f 2 =g 2 , we have a spiral.
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
Y
1.0
0.5
0.5
0.5
0.5
1.0
1.0
1.5 Fig. 10. Node at the origin in charged fluid when
2
(ii) If α2 o 4f 2 =g , we have a spiral. 2 2 (iii) If α2 ¼ 4f 2 =g , then the origin is a linear degenerate 2 2 node. As in the other cases, we simulated the full system numerically to get the behaviour as shown in Fig. 11. 2 2 4. If f 2 þ 4g 2 β ¼ 0, i.e., β ¼ f 2 =4g 2 , from (86)–(87) we can say that we have two equilibrium points ð0; 0Þ and ð f 2 =2g 2 ; 0Þ for β a 0, and only one, the origin, for β ¼ 0. (I) At f 2 =2g 2 ; 0 , the Jacobian matrix given by 0 1 A¼ ð98Þ 0 α yields τ ¼ α and Δ ¼ 0. This is a borderline case, so the possible phase portrait is obtained by doing a numerical simulation. (II) At the origin, the Jacobian matrix is given by 0 1 0 1 B f2 C ð99Þ A¼@ 2 A: α 4g 2
Thus τ ¼ α and Δ ¼ 0. We are in a borderline case. As previously done, we have simulated the full system numerically to show the possible behaviour at this point. The results are indicated in Fig. 12.
1.0
ð97Þ
1.0
Y
1.5 2 f 2 þ 4g2 β o 0
and β ¼ α =4 4 0. (a) Stable node. (b) Unstable node. 2
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72 2
(iii) If α2 ¼ f 2 =g 2 , then the origin is a linear degenerate node. Since we cannot extrapolate to the non-linear case, we just simulate the system (see Fig. 13). (c) If β ¼ 0, then f 2 ¼ 0 and Δ ¼ 0. The origin is the only equilibrium point. Since we are in a borderline case, we simulated the system numerically to show the possible behaviour as shown in Fig. 14.
From (85)–(87), a pitchfork bifurcation could occur, provided f 2 ¼ 0 and α 4 0 (all the equilibrium points are unstable when α o 0, and there is no change of stability). Thus for g 2 4 0, when β r0 the origin is the only equilibrium point, and is unstable. Two unstable equilibrium points appear in symmetrical pairs when β 4 0, and the origin becomes stable (from (89)). There is a supercritical pitchfork bifurcation. However, for g 2 o 0, when β o 0 there are three equilibrium points. The origin is unstable whereas the other two are stable (from (89), (91) and (94)). When β ¼ 0, the symmetrical pair points disappear but the origin remains unstable, and when β 4 0 the origin becomes stable. In this case there is a subcritical pitchfork bifurcation, with β as the bifurcation parameter (g2 cannot be a bifurcation parameter since it is different to zero). We can see the bifurcation diagrams in Fig. 15(a), (b).
67
additional symmetry G2 ¼ eðα=5Þ X
∂ 2α αM α3 ∂ þ eðα=5Þ X Y þ ; þ ∂X 5 5f 2 500f 2 ∂Y
ð100Þ
provided N¼
M 2 Mα2 49α4 þ þ : 4f 2 8f 2 40 000f 2
ð101Þ
From Section 3.2 this parameter value for N belongs to the region where 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, but it is not a critical value in terms of the behaviour of the system. The stability analysis shows two steady states: one at ðM=2f 2 þ 49α2 =200f 2 ; 0Þ which is a saddle point, and the other at ðM=2f 2 þ α2 =200f 2 ; 0Þ which is a node. The node is stable if α 4 0, and is unstable if α o0. Here, there is no bifurcation. We can see the behaviour of the system as shown in Fig. 16(a), (b). 2. (a) For Eq. (34), Kweyama et al. [15] found when f 2 a 0 and αa0
G2 ¼ eðα=3Þ X
∂ α 2α3 ∂ eðα=3Þ X Yþ ∂X 3 9f 2 ∂Y
ð102Þ
as an additional symmetry, provided the following conditions apply: β¼
5. Comparison with symmetry analysis
4α2 ; 9
2
g2 ¼
f2 : 2α2
ð103Þ
In the previous section we provided the general qualitative behaviour for (32) (when g 2 ¼ 0 and f 2 a 0) and for (34) (when g 2 a 0, and when g 2 ¼ 0 and f 2 a 0). For these cases Kweyama et al. [15] provided special values of some parameters which allowed for an additional symmetry. We consider the implication of those parameters on our qualitative analysis.
From Section 4.2 the parameters g2 and β in (103) are not special – they do not change the stability of the system, but 2 they correspond to the region where f 2 þ 4g 2 β 4 0 and β a0. In this region we have three equilibrium points: ð0; 0Þ and ð 4α2 =3f 2 ; 0Þ which are saddle points, and 2α2 =3f 2 ; 0 which is a node. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 17(a), (b)). (b) When f 2 ¼ 0 (g 2 a 0), Kweyama et al. [15] found
1. Consider Eq. (32) in the neutral perfect fluid case ðg 2 ¼ 0 and f 2 a 0Þ. Kweyama et al. [15] obtained (when α a 0Þ the
G2 ¼ eðα=3Þ X
U 3
U 3
2
2
1
1
1
1
2
3
Y
1
∂ α ∂ eðα=3Þ X Y ∂X 3 ∂Y
1
1
1
2
2
3
3
2
ð104Þ
3
Y
pffiffiffi Fig. 11. Linear degenerate nodes at the equilibrium points predict by (34) (f 2 ¼ 3, g 2 ¼ 1, β ¼ 2, α ¼ 7 8). (a) Stable node at ð0; 0Þ and (b) Unstable node at ð0; 0Þ.
68
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U
U
1.0
2
2
1
1
0.5
0.5
1.0
1.5
2.0
Y
1.0
0.5
0.5
1
1
2
2
1.0
1.5
2.0
Y
2
Fig. 12. Changes of the system when α2 ¼ 4f 2 =g2 (f 2 ¼ 1, g 2 ¼ 1, α ¼ 7 2). (a) α 40 and (b) α o 0.
1.0
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
Y
1.0
0.5
0.5
0.5
0.5
1.0
1.0
1.5
1.5
1.0
1.5
2.0
Y
Fig. 13. Degenerate node at the origin obtained after simulation of the system (f 2 ¼ 1, g2 ¼ 1, β ¼ 0:25, α ¼ 7 1). (a) Stable node and (b) Unstable node.
as an additional symmetry, provided β¼
2α2 : 9
ð105Þ 2
In this case β is also not special. Note that f 2 þ 4g 2 β ¼ 89g 2 α2 . 2 (i) If g 2 4 0 and α a 0, then f 2 þ 4g 2 β 4 0. Thus from pffiffiffi Section pffiffiffi 4.2 we p have three equilibrium points: ð0; 0Þ, ð 2α=3 g ; 0Þ ffiffiffi pffiffiffi and ð 2α=3 g ; 0Þ. The stability analysis shows that the origin is a node (because 0 o β o α2 =4Þ, whereas the two others points are saddle points. The node is stable when α 4 0, and unstable when α o 0. The behaviour is shown in Fig. 18(a), (b). 2 (ii) If g 2 o 0 and α a 0, then f 2 þ 4g 2 β o 0. Therefore, from Section 4.2 the only equilibrium point, the origin, is a
node (since 0 o β o α2 =4Þ. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 19(a), (b)). (iii) When α ¼ 0, the origin is again the only equilibrium point, and at that point, the determinant Δ ¼ 0. This is a borderline case. The behaviour has been obtained in Section 4.2 by simulating the system numerically, as shown in Fig. 19(c). 3. Consider Eq. (34) in the neutral perfect fluid case ðg 2 ¼ 0 and f 2 a 0Þ. Maharaj et al. [18] obtained an additional symmetry subject to the condition β¼ 7
6α2 : 25
ð106Þ
The analysis in this case corresponds to the one in Section 3.2 (with M ¼ β α2 =4 and N ¼ 0Þ. The value of β is not special.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
1.0
U 1.5
U 1.5
U 1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
0.5
1.0
1.5
Y 2.0
1.0
0.5
0.5
1.0
1.5
Y 2.0
1.0
0.5
0.5
0.5
0.5
0.5
1.0
1.0
1.0
1.5
1.5
1.5
69
1.0
1.5
Y 2.0
Fig. 14. Behaviour of the system (34) in the case where β ¼ 0 and f 2 ¼ 0. (a) α 4 0. (b) α o 0 and (c) α ¼ 0.
Y
Y stable unstable
stableβ
unstable
unstable
β
unstable
stable
β
stable
Fig. 15. Bifurcation diagram in charged fluid (g2 ¼ 7 1). (a) Supercritical bifurcation and (b) Subcritical bifurcation.
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
Y
0.5
0.5
0.5
0.5
1.0
1.0
1.5
1.0
1.5
Y
1.5 2
Fig. 16. Behaviour of (32) in the neutral perfect fluid case when α a 0 and N ¼ M =4f 2 þ Mα2 =8f 2 þ 49α4 =40 000f 2 (f 2 ¼ 18, α ¼ 7 10, M ¼ 2). (a) Stable node, saddle point and (b) Saddle point, unstable node.
(a) When α ¼ 0, then β ¼ 0. Thus from Section 3.2 the origin is the only equilibrium point (since M 2 4f 2 N ¼ 0Þ, and at that point the trace τ ¼ 0 and the determinant Δ ¼ 0. We simulate the system numerically to obtain the behaviour as shown in Fig. 20(a). (b) When α a0, consider β ¼ 6α2 =25. Since 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, from Section 3.2 we have two equilibrium points, ð0; 0Þ and ð6α2 =25f 2 ; 0Þ. The stability analysis showed that the origin is a node, and ð6α2 =25f 2 ; 0Þ is a saddle point. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 21(a), (b)).
However, for β ¼ 6α2 =25, our equilibrium points reduce to ð0; 0Þ which is a saddle point, and ð 6α2 =25f 2 ; 0Þ which is a node (see Fig. 21(a), (b)). We observe that the system behaves differently depending on the sign of β in (106). Note that, in all cases, when (18) admits two Lie point symmetries, they satisfy the relationship ½G1 ; G2 ¼ λG2 ;
70
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U
4
U
4
4
2
2
2
2
4
Y
4
2
2
2
2
4
4
4
Y
Fig. 17. Behaviour of (34) in the charged fluid, with f 2 a 0 (f 2 ¼ 2, α ¼ 7 2). (a) α 4 0 and (b) α o 0.
U
2
U
2
2
1
1
1
1
2
Y
2
1
1
1
1
2
2
2
Y
Fig. 18. Behaviour of (32) in the charged fluid when f 2 ¼ 0, g2 40 and α a 0. The system predicts node and saddle points (g2 ¼ 2 and α ¼ 7 3). (a) Stable node at ð0; 0Þ and (b) Unstable node at ð0; 0Þ.
U
U U
2
2 1.5
1
1.0
1
0.5 2
1
1
2
Y
2
1
1
1
1
2
2
2
Y 1.5
1.0
0.5
0.5
1.0
1.5
Y
0.5 1.0 1.5
Fig. 19. Behaviour of (32) in the charged fluid when f 2 ¼ 0. The system predicts a node when g2 o 0 and α a 0, and a centre when α ¼ 0 (g2 ¼ 2 and α ¼ 7 3). (a) g2 o 0 and α 4 0. (b) g2 o 0 and α o 0. (c) α ¼ 0.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U 2
U
U
2
2
1
1
71
1
2
1
1
2
Y 1.0
0.5
0.5
1.0
1.5
Y 2.0
1.0
0.5
0.5
1.0
1.5
Y 2.0
1 1
1
2
2
2
Fig. 20. Behaviour of (34) in the neutral perfect fluid when β ¼ 6α2 =25. The system predicts for α ¼ 0 a cusp (f 2 ¼ 1), and for α a 0 a node and a saddle point (f 2 ¼ 6, α ¼ 7 5). (a) α ¼ 0 and (b) α 40. (c) α o 0.
U
2.0
1.5
1.0
U
2
2
1
1
0.5
0.5
1.0
Y
2.0
1.5
1.0
0.5
0.5
1
1
2
2
1.0
Y
Fig. 21. Behaviour of (34) in the neutral perfect fluid when β ¼ 6α2 =25. For α a 0, the system predicts a node and a saddle point (f 2 ¼ 6, α ¼ 7 5).
where λ is a non-zero constant. Since the symmetries are unconnected, the Lie algebra is Lie's Type III. This confirms that the equation cannot be linearized, but can be reduced to quadratures [16].
6. Discussion In view of the foregoing, we observe that the constraints obtained by Kweyama et al. [15] in their quantitative study do not fully integrate with our qualitative study. The special values obtained via the Lie symmetry analysis are not special when it comes to the dynamical systems analysis, and with these values there is no bifurcation. 2 However, in Section 3.3 (where 16f 2 þ 48g 2 M þ 12g 2 α2 4 0) it was difficult, even impossible, to pursue the stability analysis of the system, because the equilibrium points were complicated and there was no way to isolate the real values. In this case, we used the corresponding constraints obtained by Kweyama et al. in the
Lie symmetry analysis to pursue our dynamical systems analysis and obtain the local behaviour of the system. Therefore we can use the dynamical system analysis to test the relevance of symmetries, and of solutions (for example, special solutions or not) obtained from the Lie symmetry analysis. We can also predict the stability of the solutions, even possible bifurcations. On the other hand, in certain robust case of qualitative analysis, we can use Lie symmetry analysis to investigate constraints that will enable us to study the system locally. Thus we conclude and advise that it is quite important to utilize both techniques of analysis when dealing with differential equations.
Acknowledgements P.M.T.D. and K.S.G. thank the University of KwaZulu-Natal and the National Research Foundation of South Africa for ongoing support.
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References [1] R.L. Anderson, N.H. Ibragimov, Lie–Bäcklund Transformations in Applications, SIAM, Philadelphia, 1979. [2] D. Aubin, A.D. Dalmedico, Writing the history of dynamical systems and chaos: longue durée and revolution, disciplines and cultures, Hist. Math. 29 (3) (2002) 273–339. [3] G.W. Bluman, S. Kumei, Symmetries and Differential Equations, SpringerVerlag, New York, 1989. [4] G.W. Bluman, G.J. Reid, S. Kumei, New classes of symmetries for partial differential equations, J. Math. Phys. 29 (4) (1988) 806–811. [5] A. Clark, DynPac (version 11.03): A Dynamical Systems Package for Mathematica 7, New York, 2009. [6] S. Dimas, D. Tsoubelis, SYM: a new symmetry-finding package for mathematica, in: Proceedings of the 10th International Conference in MOdern GRoup ANalysis, 2005, pp. 64–70. [7] M.C. Faulkes, Charged spheres in general relativity, Can. J. Phys. 47 (18) (1969) 1989–1994. [8] G. Gaeta, P. Morando, On the geometry of lambda-symmetries and PDE reduction, J. Phys. A: Math. Gen. 37 (2004) 6955–6975. [9] K.S. Govinder, P.G.L. Leach, The nature and uses of symmetries of ordinary differential equations, S. Afr. J. Sci. 92 (1996) 23–28. [10] K.S. Govinder, P.G.L. Leach, Integrability analysis of the Emden–Fowler equation, J. Nonlinear Math. Phys. 14 (3) (2007) 443–461. [11] A.K. Head, LIE, a PC program for Lie analysis of differential equations, Comput. Phys. 77 (1993) 241–248. [12] R.S. Irving, Integers, Polynomials, and Rings: A Course in Algebra, SpringerVerlag, New York, 2004. [13] M.J. Jacobson Jr, H.C. Williams, Solving the Pell Equation, Springer-Verlag, New York, 2009. [14] C.M. Khalique, F.M. Mahomed, B.P. Ntsime, Group classification of the generalised Emden–Fowler equation, Nonlinear Anal.: Real World Appl. 10 (2009) 3387–3395.
[15] M.C. Kweyama, K.S. Govinder, S.D. Maharaj, Noether and Lie symmetries for charged perfect fluids, Class. Quantum Gravity 28 (2011) 1–13. [16] S. Lie, Differentialgleichungen, AMS Chelsea Publishing, Providence, 1967. [17] D.G. Luenberger, Introduction to Dynamical Systems: Theory, Models, and Applications, John Wiley & Sons, New York, 1979. [18] S.D. Maharaj, P.G. Leach, R. Maartens, Expanding spherically symmetric models without shear, Gen. Relativ. Gravit. 28 (1996) 35–50. [19] C.M. Mellin, F.M. Mahomed, P.G.L. Leach, Solutions of the generalized Emden– Fowler equations with two symmetries, Int. J. Nonliner Mech. 29 (1994) 529–538. [20] C. Muriel, J.C. Romero, First integrals, integrating factors and λ-symmetries of second-order differential equations, J. Phys. A: Math. Theor. 42 (2009) 365207–365223. [21] P.J. Olver, Applications of Lie Groups to Differential Equations, Springer-Verlag, New York, 1993. [22] L. Padulo, M.A. Arbib, System Theory, Saunders, Philadelphia, 1974. [23] C. Procesi, An Approach through Invariants and Representations, Springer Science þ Business Media, New York, 2007. [24] E. Pucci, G. Saccomandi, On the reduction methods for ordinary differential equations, J. Phys. A: Math. Gen. 35 (2002) 6145–6155. [25] C.W. Soh, F.M. Mahomed, Noether symmetries of y″ ¼ f ðxÞyn with applications to non-static spherically symmetric perfect fluid solutions, Class. Quantum Gravity 16 (1999) 3553–3566. [26] S.H. Strogatz, Nonlinear Dynamics and Chaos with Applications to Physics, Biology, Chemistry and Engineering, Perseus Books, New York, 1994. [27] L.M. Surhone, M.T. Timpledon, S.F. Marseken, Saddle-Node Bifurcation, VDM Verlag Dr. Mueller AG & Co. Kg, Beau Bassin, 2010. [28] Wolfram Research, Inc., Mathematica version 7, Wolfram Research, UrbanaChampaign, 2008. [29] J.S.W. Wong, On the generalized Emden–Fowler equation, SIAM Rev. 17 (1975) 339–360. [30] C. Zhang, B. Zheng, L. Wang, Dynamical systems and related algebra with applications, Abstr. Appl. Anal. 2013 (2013) 2.
Contents lists available at ScienceDirect
International Journal of Non-Linear Mechanics journal homepage: www.elsevier.com/locate/nlm
The interplay of group and dynamical systems analysis: The case of spherically symmetric charged fluids in general relativity P.M. Tchepmo Djomegni, K.S. Govinder n School of Mathematics, Statistics and Computer Science, University of KwaZulu–Natal, Private Bag X54001, Durban 4000, South Africa
art ic l e i nf o
a b s t r a c t
Article history: Received 18 February 2014 Received in revised form 27 February 2014 Accepted 28 February 2014 Available online 12 March 2014
We investigate the relationship between the Dynamical Systems analysis and the Lie Symmetry analysis of ordinary differential equations. We undertake this investigation by looking at a relativistic model of self-gravitating charged fluids. Specifically we look at the impact of specific parameters obtained from Lie Symmetries analysis on the qualitative behaviour of the model. Steady states, stability and possible bifurcations are explored. We show that, in some cases, the Lie analysis can help to simply the dynamical systems analysis. & 2014 Elsevier Ltd. All rights reserved.
Keywords: Symmetries Dynamical systems Emden–Fowler equation
1. Introduction Most of the real world problems (in biology, finance, economics, industry, etc.), and many fundamental laws of physics and chemistry are formulated in the form of differential (or difference) equations. Various methods for solving or analysing such equations have been developed. In the late 19th century, Marius Sophus Lie unified many of these methods by introducing the notion of (what has become known as) Lie groups [23]. The Lie theory of differential equations has been phenomenally successful in determining solutions to differential equations [3,21]. It is a useful tool that can be applied to either find solutions explicitly or it can be used to classify equations via equivalence transformations. A major hurdle has been the oftentimes tedious calculations involved in finding the symmetries. However, with the advent of very capable computer packages [11,6], this disadvantage has been overcome. Symmetries can be easily calculated for a variety of equations and then used to obtain solutions, if possible. Another very useful approach to differential equations is that of dynamical systems analysis [26] in which the long-term behaviour of a system is investigated by focusing on linearization around equilibrium points. This approach has its genesis in Newtonian mechanics, and emphasizes on qualitative rather than quantitative questions [2,17,22]. For example, it was eventually realized that equations describing the motion of the three-body problem (sun, earth and moon) were difficult to solve analytically [26]. Instead of focusing
n
Corresponding author. E-mail addresses: [email protected] (P.M. Tchepmo Djomegni), [email protected] (K.S. Govinder). http://dx.doi.org/10.1016/j.ijnonlinmec.2014.02.010 0020-7462/& 2014 Elsevier Ltd. All rights reserved.
only of the exact positions of the planets at all times, people looked at their stability. Thus, the qualitative analysis is helpful, particularly when the exact solution of an equation cannot be found (and also can give more useful information even when exact solutions exist). Though these two approaches look as they belong to different areas of mathematics, they have some structures in common. For instance, in a natural way, the equivariant bifurcation theory can be viewed as an application of Lie groups in symmetric systems [30]. Therefore, it is natural to consider the use of both approaches when analysing any differential equation of interest. In what follows, we present the results of applying both methods to an Emden–Fowler equation of index three. Such equations have a lengthy history [29,19,25,10,14] but, we believe, have not been approached in this two-fold manner. We will show how each method can be used to obtain interesting information about the behaviour of the solutions.
2. The model The metric governing the shear-free motion of a spherically symmetric perfect fluid can be written as 2
2
2
2
ds ¼ e2νðt;rÞ dt þe2λðt;rÞ ½dr þr 2 ðdθ2 þ sin θ dϕ2 Þ
ð1Þ
where ν and λ are the gravitational potentials. If we impose the presence of an electromagnetic field, the usual Einstein field equations must be modified and supplemented with Maxwell's equations. The resulting Einstein–Maxwell system is then given by λ2t 1 4λr E2 4 4λ ρ ¼ 3 2ν 2λ 2λrr þ λ2r þ ð2Þ r e e r e
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
p¼
1 1 2νr 2λr E2 þ þ 4 4λ ð 3λ2t 2λtt þ2νt λt Þ þ 2λ λ2r þ 2νr λr þ 2ν r r e e r e
ð3Þ
p¼
1 1 νr λr E2 2 2 þ þλ ð 3λ 2λ þ2ν λ Þ þ ν þ ν þ tt t t rr rr 4 4λ t r 2ν 2λ r r e e r e
ð4Þ ð5Þ
E ¼ r 2 eλ ν Φr
ð6Þ
Er ¼ sr 2 e3λ
ð7Þ
Here, ρ is the energy density, p is the isotropic pressure, s is the proper charge density of the fluid and we interpret E as the total charge contained within the sphere of radius r centred around the origin of the coordinate system. The electromagnetic field is present via Φr. Integrating (5) and combining (3) and (4) and integrating allow us to reduce the system to 4λr E2 4 4λ ρ ¼ 3e2h e 2λ 2λrr þ λ2r þ ð8Þ r r e " # 1 2λr E2 λ 2 3λ þ 2h e e λr þ 4 λ p¼ r λt e3λ r e
Eq. (18) is the fundamental non-linear ordinary differential equation which determines the behaviour of self-gravitating charged fluids in general relativity [15]. Once y is determined, Kweyama et al. [15] found that G¼a
0 ¼ νr λt λtr
ð9Þ
∂ ∂ þ ðby þ cÞ ∂x ∂y
a″ ¼ 2b
0
ð20Þ
b″ ¼ 2fc
ð21Þ
c″ ¼ 0
ð22Þ
0
af þ ð2a0 þ bÞf ¼ 3cg
ð23Þ
ag 0 þ ð2a0 þ 2bÞg ¼ 0:
ð24Þ
This system was reduced to 2b ¼ a0 þ α
λr 2E2 ¼ F~ 4 λ eλ λrr λ2r r r e
ð11Þ
E ¼ r 2 eλ ν Φr
ð12Þ
Er ¼ sr 2 e3λ ;
ð13Þ
where h ¼ hðtÞ and F~ ¼ F~ ðrÞ are arbitrary functions of integration. This means that we only need to solve (11) in order to obtain all the other unknown functions. Using the transformation [7,15] x ¼ r2
ð14Þ
y ¼ eλ
ð15Þ
f ðxÞ ¼
F~ 4r 2
ð16Þ
gðxÞ ¼
E2 2r 6
ð17Þ
c ¼ c0 þc1 x
ð26Þ
g ¼ g 2 a 3 exp
Z
α dx ; a
ð27Þ
Z Z Z α dx α dx f 2 3g 2 ca 3=2 exp dx ; f ¼ a 5=2 exp 2a a ð28Þ where α, c0, c1, f2 and g2 are arbitrary constants. Setting Z dx X¼ a Z Z Z b dx c b dx exp dx; Y ¼ y exp a a a
ð29Þ ð30Þ
(18) can be transformed into autonomous form. When c a0, a satisfied the non-linear fourth order equation Z 0 5a α α dx caa⁗ þ c c0 a a‴ ¼ 12g 2 c3 a 3 exp ; ð31Þ þ a 2 2 and (18) became α2 Y ¼ g 2 Y 3 þ f 2 Y 2 þ N: Y″ þ αY 0 þ M þ 4
ð32Þ
When c ¼0
we can rewrite (11) as y″ ¼ f ðxÞy2 þgðxÞy3 :
ð25Þ
ð10Þ
ð19Þ
is a point symmetry of (18), provided the functions a(x), b(x), and c (x) satisfied the following system of ordinary differential equations:
t
eν ¼ λt e h
59
ð18Þ
a ¼ a0 þ a1 x þ a2 x2 ;
ð33Þ
and (18) became Remark. (1) In this approach, our interest is only in point symmetries. We acknowledge that a variety of other symmetries exist, including potential [4], Lie–Bäcklund/generalized [1,21], Lambda [20,24,8] and non-local [9]. Expanding our study to those symmetries may indeed yield other useful results but they are outside the scope of our work. (2) In general, (18) does not admit any Lie point symmetries. As a result, it cannot be linearized under a point transformation. In what follows, we investigate under what conditions the equation does admit point symmetries. We first look at conditions under which it admits a single Lie point symmetry (and so is part of the equivalence class of autonomous second order differential equations). Thereafter, we investigate additional conditions under which the equation admits a second Lie point symmetry. This is important as the possession of two Lie point symmetries guarantees that the equation can be reduced to quadratures.
Y″ þ αY 0 þ βY ¼ g 2 Y 3 þ f 2 Y 2 ;
ð34Þ
where 1 1 β ¼ a21 þ a0 a2 þ α2 ; 4 4
ð35Þ
and M, N, a0, a1 and a2 are constants. The quantities M and N are constants of integration and are given by [15] M¼
1 1 aa″ a02 2f 2 I þ 3g 2 I 2 2 4
ð36Þ
and
Z α dx 1 1 ac0 a0 c þ αc N ¼ a 1=2 exp 2a 2 2 1 1 02 α2 2 3 I þ f 2 I 2g 2 I ; aa″ a þ 2 4 4
ð37Þ
60
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
where Z Z α dx dx: I¼ ca 3=2 exp 2a
α2 U0 ¼ M þ Y αU þ N: 4
ð38Þ
ð44Þ
Since N is constant, the system (44) is linear.
Since (19) can be transformed into G1 ¼
∂ ∂X
1. For M a α2 =4, ð4N=ð4M þα2 Þ; 0Þ is the only equilibrium point. At that point, the Jacobian matrix is given by 0 1 0 1 2 B C ð45Þ A ¼ @ M þα α A: 4
ð39Þ
via (29)–(30), Eqs. (32) and (34) admit only one symmetry. Kweyama et al. [15] then presented conditions under which Eqs. (32) and (34) could have an additional symmetry. Our intention is to investigate the impact of all the parameters and of those conditions on the qualitative behaviour of our underlying equations in the phase plane near the equilibrium points.
Thus τ ¼ α, Δ ¼ M þα2 =4. (a) If M o α2 =4, then Δ o 0. Therefore, the equilibrium point is a saddle point. (b) If α2 =4 o M, then Δ 4 0. Thus τ2 4Δ ¼ α2 4M α2 ¼ 4M. (i) When M 4 0, τ2 4Δ o0. (A) For α a 0, we have a spiral around the equilibrium point. The spiral is stable if α 4 0, and unstable if α o 0. The phase portrait of the solutions is shown in Fig. 1(a), (b). (B) For α ¼ 0, the only equilibrium point ðN=M; 0Þ is a centre, and the direction is given by the sign of U (see the beginning of the section). The phase portrait is shown in Fig. 1(c). (ii) When α2 =4 o M o 0, then τ2 4Δ 40, and α a 0. Therefore the equilibrium point is a node. It is a stable node if α 4 0 and an unstable node if α o 0. (iii) When M ¼0, then τ2 4Δ ¼ 0. (A) If α ¼ 0, then Δ ¼ 0. Thus we have a whole plane of equilibrium points. (B) If α a0, we have a double eigenvalue λ ¼ τ=2 ¼ α=2, with one associated eigenvector V ¼ ð 2=α; 1Þ. Thus the equilibrium point is a degenerate node. The node is stable if α 4 0, and unstable if α o 0. 2. For M ¼ α2 =4, (44) is linear and the analysis is trivial.
3. Non-zero c The first part of our analysis is focused on Eq. (32). If we set U ¼ Y 0;
ð40Þ
we can rewrite (32) as the two-dimensional non-linear system Y 0 ¼ U ¼ hðY; UÞ
α2 Y αU þN ¼ kðY; UÞ: U 0 ¼ g2 Y 3 þ f 2 Y 2 M þ 4
ð41Þ
The motion in the phase plane is given by the phase portrait of solutions that comes from system (41). The phase portrait is determined by the linear system analysis associated with (41) at the equilibrium points. The equilibrium points occur at points satisfying U ¼ 0;
ð42Þ
and
α2 g2 Y 3 þ f 2 Y 2 M þ Y þ N ¼ 0: 4
ð43Þ
Before proceeding with the general analysis, we observe that the function h is odd in U (i.e., hðY; UÞ ¼ hðY; UÞ), and when α ¼ 0 the function k is even in U (i.e., kðY; UÞ ¼ kðY; UÞ). Therefore the system is reversible when α ¼ 0 [26]. The direction of the vector fields is given by the sign of U in (41). Thus Y increases when U 4 0, and decreases when U o0.
3.2. Neutral perfect fluid Here we take g 2 ¼ 0 and f 2 a 0 [18]. Thus there is no charge within the sphere, but we do have matter.
3.1. Neutral dust Here we take g 2 ¼ 0 and f 2 ¼ 0 [18]. From (27) g ¼0, therefore there is no charge contained within the sphere (E ¼0 in (17)). The system (41) becomes
1. When α ¼ 0, the system (41) becomes Y0 ¼ U
0
U 0 ¼ f 2 Y 2 MY þ N:
Y ¼U
U
4
U
4
4
2
2
2
2
4
Y
4
2
U 2
1
2
2
2
4
4
ð46Þ
4
Y
1
1
2
3
Y
1
2
Fig. 1. Behaviour of the system in the neutral dust case when M 4 0. The system predicts a spiral (α ¼ 7 1, M ¼ 2, N ¼3), or a centre (α ¼ 0, M ¼ 1, N ¼ 1). (a) Stable spiral. (b) Unstable spiral and (c) Centre.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
There are two possible equilibrium points
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 0 M þ M 2 4f 2 N M M 2 4f 2 N @ A @ ðY 0 ; 0Þ ¼ ; 0 ; ðY 1 ; 0Þ ¼ ; 0A; 2f 2 2f 2
ð47Þ provided M 2 4f 2 N Z0. (a) If M 2 4f 2 N ¼ 0, i.e., N ¼ M 2 =4f 2 , then ðM=2f 2 ; 0Þ is the only equilibrium point, and at that point the Jacobian matrix is given by 0 1 A¼ : ð48Þ 0 0 Thus τ ¼ 0 and Δ ¼ 0. We are in a borderline case. We cannot extrapolate from the linear case to the fully non-linear case. However, system (46) yields Y 0 ðf 2 Y 2 MY þ NÞ ¼ UU 0 ;
ð49Þ
i.e., 0
2
0
0
0
f 2 Y Y MY Y þ NY ¼ UU :
ð50Þ
After integration (50) becomes 1 1 1 f Y 3 MY 2 þ NY þ K 0 ¼ U 2 ; 3 2 2 2
ð51Þ
where K0 is a constant. The phase portrait is shown in Fig. 2. (b) If M 2 4f 2 N 4 0, we have two equilibrium points ðY 0 ; 0Þ and ðY 1 ; 0Þ in (47). (I) At ðY 0 ; 0Þ, the Jacobian matrix is 0 1 0qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 A: ð52Þ A¼@ M 2 4f 2 N 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ 0 and Δ ¼ M 2 4f 2 N o 0. Therefore we have a saddle point. (II) At ðY 1 ; 0Þ, as Jacobian matrix we have 0 1 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 @ A: ð53Þ A¼ M 2 4f 2 N 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ 0 and Δ ¼ M 2 4f 2 N 4 0. Therefore the point is a linear centre. Since the system is reversible, then the equilibrium point is a non-linear centre. From (47) if f 2 4 0, the equilibrium points exist when N o M 2 =4f 2 , and then fuse together when N ¼ M 2 =4f 2 , and
disappear when N 4 M 2 =4f 2 . There has been a saddle-node bifurcation [27]. Similarly, if f 2 o 0, the equilibrium points exist when N 4 M 2 =4f 2 , and then fuse when N ¼ M 2 =4f 2 , and are destroyed when N o M 2 =4f 2 . Therefore we still have a saddlenode bifurcation, specifically a blue sky bifurcation. Here, N is the bifurcation parameter (and not f2 as it cannot be zero). We can visualize the bifurcation diagram as shown in Fig. 3. 2. If α a 0, we have the following equilibrium points: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 4M þ α2 þ 16M 2 64f 2 N þ 8Mα2 þ α4 ð54Þ ðY 0 ; 0Þ ¼ @ ; 0A 8f 2 and 0 ðY 1 ; 0Þ ¼ @
4M þ α2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 16M 2 64f 2 N þ 8Mα2 þ α4 ; 0 A; 8f 2
ð55Þ
provided 16M 2 64f 2 N þ8Mα2 þ α4 Z 0. (a) If we assume that 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, then: (I) At ðY 0 ; 0Þ, the Jacobian matrix is given by 0 1 0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi @ A: 1 ð56Þ A¼ 16M 2 64f 2 N þ 8Mα2 þ α4 α 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Thus τ ¼ α and Δ ¼ 14 16M 2 64f 2 N þ 8Mα2 þ α4 o 0. Therefore the point is a saddle point. (II) At ðY 1 ; 0Þ, the Jacobian matrix given by 0 1 0 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q A; ð57Þ A¼@ 1 16M 2 64f 2 N þ 8Mα2 þα4 α 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi yields τ ¼ α and Δ ¼ 14 16M 2 64f 2 N þ 8Mα2 þ α4 40. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (i) If τ2 4Δ ¼ α2 16M 2 64f 2 N þ8Mα2 þ α4 4 0, then the equilibrium point is a node. The node is stable when α 4 0, and unstable when α o 0. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (ii) If τ2 4Δ ¼ α2 16M 2 64f 2 N þ 8Mα2 þα4 o 0, we have a spiral. It is a stable spiral for α 4 0, and unstable spiral for α o 0. (b) If we assume that 16M 2 64f 2 N þ8Mα2 þ α4 ¼ 0, ðð4M þ α2 Þ=8f 2 ; 0Þ is the only equilibrium point, and at that point the Jacobian matrix is given by 0 1 A¼ : ð58Þ 0 α Thus τ ¼ α and Δ ¼ 0. We are in a borderline case. Since our system is non-linear and difficult to solve analytically, we have only plotted the full system numerically with specific values of parameters just to show the behaviour in the phase space in Fig. 4(a), (b) (we used the package Dynpac [5]).
U 2
1
1
61
1
2
3
Y
1
2 Fig. 2. Behaviour of the system in the neutral perfect fluid case where α ¼ 0 (f 2 ¼ 1, M ¼2, and N ¼1).
From (54) and (55), when α 4 0 we have a bifurcation (because when α o 0 all the equilibrium points are unstable, and there is no change of stability). Thus for f 2 4 0, the two equilibrium points exist when N o ð16M 2 þ 8Mα2 þα4 Þ=64f 2 , then fuse when N ¼ ð16M 2 þ8Mα2 þ α4 Þ=64f 2 , and disappear when N 4 ð16M 2 þ 8Mα2 þ α4 Þ=64f 2 . A saddle-node bifurcation occurred.
Likewise, for f 2 o 0 the opposite happens. We have a blue sky bifurcation, with N as the bifurcation parameter. The bifurcation diagram is shown in Fig. 5.
62
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
Y
Y Stable node
N
Saddle point
N
Stable spiral Saddle point
Fig. 3. Bifurcation diagram Y ¼ f ðNÞ (f 2 ¼ 7 1 and M¼ 2). (a) f 2 4 0: saddle-node bifurcation. (b) f 2 o 0: blue sky bifurcation.
2
U 3
U 3
2
2
1
1
1
1
2
Y
2
1
1
1
1
2
2
3 2
2
Y
3
Fig. 4. Behaviour of the system when 16M 64f 2 N þ 8Mα þ α ¼ 0 (the linear analysis classified as a borderline case). We have plotted using f 2 ¼ 1, M¼ 1, N ¼ 1, α ¼ 7 2. (a) α 4 0 and (b) α o 0. 2
4
We use this condition because δ could be zero. It is clear that any of these points Y0, Y1, Y2 could be zero.
3.3. Charged fluid We now insist on g 2 a0. Thus E a 0 and the matter is charged. Eq. (43) admits as solutions
2
2
Y0 ¼
f2 16f 2 þ 48g 2 M þ 12g 2 α2 δ ffiffiffi ; pffiffiffi þ þ p 3g 2 6 3 4g 2 δ 12 3 2g 2
ð59Þ
Y1 ¼
pffiffiffi pffiffiffi 2 f2 ð1 þ i 3Þð16f 2 þ 48g 2 M þ 12g 2 α2 Þ ð1 i 3Þδ ffiffiffi ffiffiffi p p ; 3 3 3g 2 12 4g 2 δ 24 2g 2
ð60Þ
pffiffiffi pffiffiffi 2 f ð1 i 3Þð16f 2 þ 48g 2 M þ 12g 2 α2 Þ ð1 þi 3Þδ p p ffiffiffi ffiffiffi Y2 ¼ 2 ; 3g 2 12 3 4g 2 δ 24 3 2g 2
ð61Þ
3
3
Y 00 ¼
ðf g 3 27g 52 NÞ1=3 f2 þ 2 2 ; 3g 2 3g 22
ð64Þ
Y 11 ¼
pffiffiffi 3 ð1 i 3Þðf 2 g 32 27g 52 NÞ1=3 f2 ; 3g 2 6g 22
ð65Þ
Y 22 ¼
pffiffiffi 3 ð1 þ i 3Þðf 2 g 32 27g 52 NÞ1=3 f2 3g 2 6g 22
ð66Þ
as solutions. Since we are interested in real values, the only possible equilibrium point is ðY 00 ; 0Þ. At that point, the Jacobian matrix given by
2
provided 16f 2 þ 48g 2 M þ 12g 2 α2 a 0 [12,13,28], with δ ¼ ð 128f 2 576f 2 g 2 M 1728g 22 N 144f 2 g 2 α2 þ δ1 Þ1=3 ;
2
1. If 16f 2 þ 48g 2 M þ 12g 2 α2 ¼ 0, i.e., M ¼ ð 4f 2 3g 2 α2 Þ=12g 2 , Eq. (43) admits
ð62Þ
and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 δ1 ¼ ð 128f 2 576f 2 g 2 M 1728g22 N 144f 2 g2 α2 Þ2 4ð16f 2 þ 48g2 M þ 12g 2 α2 Þ3 :
ð63Þ
0
0 2=3 5 B 3 3 A ¼ @ ðf 2 g 2 27g 2 NÞ 3 3g 2
1
1
C α A; 3
yields τ ¼ α and Δ ¼ ðf 2 g 32 27g 52 NÞ2=3 =3g 32 .
ð67Þ
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
63
Y
Y Stable node
N
Saddle point
N
Stable spiral Saddle point
Fig. 5. Bifurcation diagram Y ¼ f ðNÞ (f 2 ¼ 0:5, M ¼ 1, α ¼ 3 for figure (a), and M ¼ 1, f 2 ¼ 1 and α ¼ 2 for figure (b)). (a) f 2 40: saddle-node bifurcation and (b) f 2 o 0: blue sky bifurcation.
U 3
U 3
2
2
1
1
1
1
2
3
4
Y
1
1
1
1
2
2
3
3
2
3
4
Y
pffiffiffi 2 Fig. 6. Nodes predicted when 16f 2 þ 48g2 M þ 12g2 α2 ¼ 0 in the charged fluid (f 2 ¼ 3, g 2 ¼ 1, M¼ 0, N ¼ 2, α ¼ 7 2 3). (a) Stable node and (b) Unstable node.
3
(a) If f 2 27g 22 N a 0 and g 2 4 0, then Δ o 0. Therefore ðY 00 ; 0Þ is a saddle point. 3 (b) If f 2 27g 22 N a 0 and g 2 o 0, then Δ 40. If we assume that α ¼ 0, ðY 00 ; 0Þ is a non-linear centre (since the system is reversible). However, if α a 0, the point is stable when α 4 0, and unstable when α o 0. Note that 3
τ2 4Δ ¼ α2 þ
4ðf 2 g 32 27g 52 NÞ2=3 : 3g 32
ð68Þ
(i) If τ2 4Δ 4 0, the point is a node. (ii) If τ2 4Δ o 0, we have a spiral. (iii) If τ2 4Δ ¼ 0, the linear equilibrium point is a degenerate node. This is a borderline case and we have numerically solved the full system to see the effect on the behaviour, as shown in Fig. 6(a), (b). 3 (c) If f 2 27g 22 N ¼ 0, then Δ ¼ 0. We are also in a borderline case. The behaviour for different values of parameters is shown in Fig. 7(a), (b). 2 2. If 16f 2 þ48g 2 M þ 12g 2 α2 o 0, then δ in (62) is real. Therefore Y0 in (59) is real, whereas Y1 and Y2 in (60) and (61) are still complex. Thus ðY 0 ; 0Þ is the only equilibrium point. At that point, the Jacobian matrix is given by 0
0
2 B A ¼ @ 3g Y 2 þ2f Y M þ α 2 0 2 0 4
1
1
C α A:
ð69Þ
Therefore τ ¼ α and Δ ¼ 3g 2 Y 20 2f 2 Y 0 þ ðM þ α2 =4Þ (which cannot be zero). (a) If Δ o 0, ðY 0 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, the equilibrium point is a non-linear centre. However, for α a0, τ2 4Δ ¼ 12g 2 Y 20 þ 8f 2 Y 0 4M. The equilibrium point is stable when α 4 0, and unstable when α o 0. If τ2 4Δ 4 0, then ðY 0 ; 0Þ is a node. If τ2 4Δ o 0, then ðY 0 ; 0Þ is a spiral. If τ2 4Δ ¼ 0, the linear point is a degenerate node. As before, we solve the full non-linear system numerically to determine its behaviour (see Fig. 8). 2 3. If 16f 2 þ 48g 2 M þ 12g 2 α2 4 0, Y0, Y1 and Y2 in (59), (60), (61) can be real or complex, but we are only interested in real values. It is difficult to transform them into the form A þ iB (where A and B are real), because δ in (62) can also be complex, and not easily transformable in this form. We will restrict ourselves to the constraint obtained in [15], with g 2 40. It was shown that 2
M¼
f2 α2 ; 3g 2 36
3
N¼
f2 2α2 f 2 27g 22 27g 2
led to the additional symmetry ∂ α αf ∂ eðα=3ÞX Yþ 2 : G2 ¼ eðα=3ÞX ∂X 3 9g 2 ∂Y
ð70Þ
ð71Þ
64
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
and then integrate to obtain
By substituting M and N in (59)–(61) by (70), we have three equilibrium points ! pffiffiffi 2α f2 f2 ðY 0 ; 0Þ ¼ ; 0 ; ðY 1 ; 0Þ ¼ pffiffiffiffiffi ; 0 ; 3g 2 3g 2 3 g 2 ! pffiffiffi f 2α ð72Þ ðY 2 ; 0Þ ¼ 2 þ pffiffiffiffiffi ; 0 : 3g 2 3 g 2
(a) Assume that α a 0. (I) At ðY 0 ; 0Þ, the Jacobian matrix given by 0 1 0 1 2 A A ¼ @ 2α α 9
Δ¼
ð73Þ
2α2 : 9
ð74Þ
Since Δ 4 0, τ2 4Δ ¼ α2 8α2 =9 ¼ α2 =9 40. Therefore the eigenvalues are real and have the same sign, hence the equilibrium point is a node. If α 40, then τ o 0. Thus ðY 0 ; 0Þ is a stable node. If α o0, then τ 4 0. Thus ðY 0 ; 0Þ is an unstable node. (II) At ðY 1 ; 0Þ and ðY 2 ; 0Þ, we find as Jacobian matrix 0
0 A ¼ @ 4α2 9
1 α
1 2 g Y 4 þ f 3 Y 3 MY 2 þ 2NY þ 2K 1 ; 2 2 3
ð79Þ
where K1 is a constant. The behaviour is obtained by sketching the above implicit function (79) as shown in Fig. 9.
4. Zero c
yields τ ¼ α;
U2 ¼
1 A:
ð75Þ
The second part of our analysis concerns the non-linear second order equation (34) Y″ þ αY 0 þ βY ¼ g 2 Y 3 þ f 2 Y 2 : ð80Þ 1 2 Since β ¼ γ þ α , with γ an arbitrary constant, this equation is the 4 same as (32) with M ¼ γ and N ¼0. 4.1. Neutral perfect fluid Here we take g 2 ¼ 0. Whatever the value of f2 the condition N ¼0 (according to the previous section) does not affect the existence and the stability of the equilibrium point. Therefore we have the same behaviour of solutions in the phase space. However, there is no bifurcation (since N is constant). 4.2. Charged fluid Now we take g 2 a 0. Setting U ¼ Y 0;
Thus 2
4α Δ¼ : 9
ð76Þ
Since Δ o0, ðY 1 ; 0Þ and ðY 2 ; 0Þ are saddle points. (b) If α ¼ 0, ð f 2 =3g 2 ; 0Þ is the only equilibrium point, and at that point τ ¼ 0;
Δ ¼ 0:
ð77Þ
This is a borderline case. Here we can show the global behaviour of the full system by reducing (41) to the equation α2 Y αU þ N Y 0 ; ð78Þ UU 0 ¼ g 2 Y 3 þf 2 Y 2 M þ 4
3
2
ð81Þ
Eq. (80) becomes the two-dimensional non-linear system Y0 ¼ U U 0 ¼ g 2 Y 3 þf 2 Y 2 βY αU: The equilibrium points are solutions of the system U ¼ 0;
ð83Þ
g 2 Y 3 þ f 2 Y 2 βY ¼ 0:
ð84Þ
Eq. (84) admits as solutions Y 0 ¼ 0;
ð85Þ
U 3
U 3
2
2
1
1
1
1
2
3
Y
3
2
1
1
1
1
2
2
3 Fig. 7. Behaviour of the system when Δ ¼ 0 and
ð82Þ
2
3
Y
3 2 16f 2 þ 48g 2 M þ 12g 2 α2
¼ 0 in the charged fluid (f 2 ¼ 3, g 2 ¼ 1, M ¼ 4, N ¼ 1, α ¼ 7 2). (a) α 4 0. (b) α o 0.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
2
Fig. 8. Spirals node predicted when
U 3
U 3
2
2
1
1
1
1
2
3
Y
2
2
3
3
2 16f 2 þ 48g2 M þ 12g2 α2 o 0
1
τ ¼ α;
Y
Y2 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ f 2 þ 4g 2 β 2g 2
1
1
C C: α A
ð90Þ
and M and N given by 70
Therefore 2
;
ð86Þ
τ ¼ α;
Δ¼
f 2 þ 4g 2 β þf 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ 4g 2 β
2g 2
:
ð91Þ
In this case Δ a 0. :
ð87Þ 2
2
We have four possible cases: f 2 þ 4g 2 β o 0 and β a0, f 2 þ4g 2 β 4 0 2 2 and β a 0, f 2 þ 4g 2 β 4 0 and β ¼ 0, or f 2 þ4g 2 β ¼ 0. When α ¼ 0, the system is reversible, and the direction of the vector fields is given by the sign of U in (82). Thus Y increases when U 4 0, and decreases when U o 0. 2
ð89Þ
0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B 2 2 A¼B @ f 2 þ4g 2 β þ f 2 f 2 þ 4g 2 β 2g 2
3
2g 2
Δ ¼ β:
0
2
Y1 ¼
Y
(a) If β o0, then Δ o 0. Therefore the origin is a saddle point. (b) If β 4 0, for α ¼ 0 the origin is a non-linear centre (since the system is reversible). However for α a 0, note that τ2 4Δ ¼ α2 4β. The point is stable when α 4 0, and unstable when α o 0. If 0 o β o α2 =4, the origin is a node. If β 4α2 =4, we have a spiral. If β ¼ α2 =4, the origin is a linear degenerate node. As in the other cases, the behaviour of the full system (34) is obtained by a numerical simulation, as shown in Fig. 10 (a), (b). 2 2. If f 2 þ 4g 2 β 4 0 and β a0, we have two additional equilibrium points ðY 1 ; 0Þ and ðY 2 ; 0Þ given by (86) and (87). I. At ðY 1 ; 0Þ, the Jacobian matrix is
1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 f 2 þ 4g 2 β
3
Thus
1
2 16f 2 þ 48g2 M þ 12g2 α2 4 0,
2
by the simulated system (41) (f 2 ¼ 3, g2 ¼ 1, M ¼3, N ¼ 2, α ¼ 7 2). (a) Stable spiral. (b) Unstable spiral.
1
Fig. 9. Phase portrait for (f 2 ¼ 3, g 2 ¼ 1, α ¼ 0).
1 1
2
2
1
1
U 3
3
2
65
1. If f 2 þ4g 2 β o 0 and β a 0, there is only one equilibrium point, the origin. At the origin, the Jacobian matrix is given by ! 0 1 A¼ : ð88Þ β α
(a) If Δ o 0, then ðY 1 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, ðY 1 ; 0Þ is a centre. However, for α a 0, the point is stable when α 40, and unstable when α o 0. Note that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 f 2 þ 4g 2 βÞ 2ðf þ 4g β þ f 2 2 2 τ2 4Δ ¼ α2 þ : ð92Þ g2 (i) If τ2 4Δ 4 0, then ðY 2 ; 0Þ is a node. (ii) If τ2 4Δ o 0, then we have spiral node. (iii) If τ2 4Δ ¼ 0, then ðY 1 ; 0Þ is a linear degenerate node. We have solved the system numerically to get the behaviour of the non-linear analysis (see Fig. 11).
66
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
(II) At ð f 2 =g 2 ; 0Þ, the Jacobian matrix given by
(II) At ðY 2 ; 0Þ, the Jacobian matrix is 0
0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi B 2 2 f þ 4g β f f 2 þ4g 2 β A¼B 2 2 @ 2
1
C C: α A
2g 2
Therefore 2
τ ¼ α;
Δ¼
f 2 þ4g 2 β f 2
0
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 f 2 þ 4g 2 β
2g 2
ð93Þ
1 0 1 B f2 C A¼@ 2 A α g2
ð94Þ
yields τ ¼ α and Δ ¼ f 2 =g 2 (which cannot be zero Þ. (a) If g 2 4 0, then Δ o0. Therefore we have a saddle point. (b) If g 2 o 0, then Δ 40. For α ¼ 0, we have a centre. 2 However, for α a0, note that τ2 4Δ ¼ α2 þ 4f 2 =g 2 . The point is stable when α 4 0, and unstable when α o 0. (i) If α2 4 4f 2 =g , then the point is a node.
1
2
:
In this case Δ a 0. (a) If Δ o 0, then ðY 2 ; 0Þ is a saddle point. (b) If Δ 4 0, for α ¼ 0, ðY 2 ; 0Þ is a centre. However, for α a 0, the point is stable when α 4 0, and unstable when α o 0. Note that
τ2 4Δ ¼ α2 þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 f 2 þ 4g 2 β f 2 f 2 þ 4g 2 β g2
:
ð95Þ
(i) If τ2 4Δ 4 0, then ðY 1 ; 0Þ is a node. (ii) If τ2 4Δ o 0, then we have spiral node. (iii) If τ2 4Δ ¼ 0, then ðY 2 ; 0Þ is a linear degenerate node. We have solved the system numerically to get the behaviour of the non-linear analysis (see Fig. 11). 2 3. If f 2 þ 4g 2 β 4 0 and β ¼ 0, i.e., f 2 a 0, from (85)–(87), we have two equilibrium points, ð0; 0Þ and ð f 2 =g 2 ; 0Þ. (I) At ð0; 0Þ, the Jacobian matrix is given by 0 1 A¼ : ð96Þ 0 α
2
2
Thus τ ¼ α and Δ ¼ f 2 =4g 2 . (a) If g 2 4 0 and β a 0, then Δ o 0. Therefore the origin is a saddle point. (b) If g 2 o 0 and β a0, then Δ 4 0. For α ¼ 0, we have a centre. 2 However, for α a 0, note that τ2 4Δ ¼ α2 þf 2 =g 2 . In this case, the origin is stable when α 40, and unstable when α o0. 2 (i) If α2 4 f 2 =g 2 , then the origin is a node. 2 2 (ii) If α o f 2 =g 2 , we have a spiral.
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
Y
1.0
0.5
0.5
0.5
0.5
1.0
1.0
1.5 Fig. 10. Node at the origin in charged fluid when
2
(ii) If α2 o 4f 2 =g , we have a spiral. 2 2 (iii) If α2 ¼ 4f 2 =g , then the origin is a linear degenerate 2 2 node. As in the other cases, we simulated the full system numerically to get the behaviour as shown in Fig. 11. 2 2 4. If f 2 þ 4g 2 β ¼ 0, i.e., β ¼ f 2 =4g 2 , from (86)–(87) we can say that we have two equilibrium points ð0; 0Þ and ð f 2 =2g 2 ; 0Þ for β a 0, and only one, the origin, for β ¼ 0. (I) At f 2 =2g 2 ; 0 , the Jacobian matrix given by 0 1 A¼ ð98Þ 0 α yields τ ¼ α and Δ ¼ 0. This is a borderline case, so the possible phase portrait is obtained by doing a numerical simulation. (II) At the origin, the Jacobian matrix is given by 0 1 0 1 B f2 C ð99Þ A¼@ 2 A: α 4g 2
Thus τ ¼ α and Δ ¼ 0. We are in a borderline case. As previously done, we have simulated the full system numerically to show the possible behaviour at this point. The results are indicated in Fig. 12.
1.0
ð97Þ
1.0
Y
1.5 2 f 2 þ 4g2 β o 0
and β ¼ α =4 4 0. (a) Stable node. (b) Unstable node. 2
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72 2
(iii) If α2 ¼ f 2 =g 2 , then the origin is a linear degenerate node. Since we cannot extrapolate to the non-linear case, we just simulate the system (see Fig. 13). (c) If β ¼ 0, then f 2 ¼ 0 and Δ ¼ 0. The origin is the only equilibrium point. Since we are in a borderline case, we simulated the system numerically to show the possible behaviour as shown in Fig. 14.
From (85)–(87), a pitchfork bifurcation could occur, provided f 2 ¼ 0 and α 4 0 (all the equilibrium points are unstable when α o 0, and there is no change of stability). Thus for g 2 4 0, when β r0 the origin is the only equilibrium point, and is unstable. Two unstable equilibrium points appear in symmetrical pairs when β 4 0, and the origin becomes stable (from (89)). There is a supercritical pitchfork bifurcation. However, for g 2 o 0, when β o 0 there are three equilibrium points. The origin is unstable whereas the other two are stable (from (89), (91) and (94)). When β ¼ 0, the symmetrical pair points disappear but the origin remains unstable, and when β 4 0 the origin becomes stable. In this case there is a subcritical pitchfork bifurcation, with β as the bifurcation parameter (g2 cannot be a bifurcation parameter since it is different to zero). We can see the bifurcation diagrams in Fig. 15(a), (b).
67
additional symmetry G2 ¼ eðα=5Þ X
∂ 2α αM α3 ∂ þ eðα=5Þ X Y þ ; þ ∂X 5 5f 2 500f 2 ∂Y
ð100Þ
provided N¼
M 2 Mα2 49α4 þ þ : 4f 2 8f 2 40 000f 2
ð101Þ
From Section 3.2 this parameter value for N belongs to the region where 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, but it is not a critical value in terms of the behaviour of the system. The stability analysis shows two steady states: one at ðM=2f 2 þ 49α2 =200f 2 ; 0Þ which is a saddle point, and the other at ðM=2f 2 þ α2 =200f 2 ; 0Þ which is a node. The node is stable if α 4 0, and is unstable if α o0. Here, there is no bifurcation. We can see the behaviour of the system as shown in Fig. 16(a), (b). 2. (a) For Eq. (34), Kweyama et al. [15] found when f 2 a 0 and αa0
G2 ¼ eðα=3Þ X
∂ α 2α3 ∂ eðα=3Þ X Yþ ∂X 3 9f 2 ∂Y
ð102Þ
as an additional symmetry, provided the following conditions apply: β¼
5. Comparison with symmetry analysis
4α2 ; 9
2
g2 ¼
f2 : 2α2
ð103Þ
In the previous section we provided the general qualitative behaviour for (32) (when g 2 ¼ 0 and f 2 a 0) and for (34) (when g 2 a 0, and when g 2 ¼ 0 and f 2 a 0). For these cases Kweyama et al. [15] provided special values of some parameters which allowed for an additional symmetry. We consider the implication of those parameters on our qualitative analysis.
From Section 4.2 the parameters g2 and β in (103) are not special – they do not change the stability of the system, but 2 they correspond to the region where f 2 þ 4g 2 β 4 0 and β a0. In this region we have three equilibrium points: ð0; 0Þ and ð 4α2 =3f 2 ; 0Þ which are saddle points, and 2α2 =3f 2 ; 0 which is a node. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 17(a), (b)). (b) When f 2 ¼ 0 (g 2 a 0), Kweyama et al. [15] found
1. Consider Eq. (32) in the neutral perfect fluid case ðg 2 ¼ 0 and f 2 a 0Þ. Kweyama et al. [15] obtained (when α a 0Þ the
G2 ¼ eðα=3Þ X
U 3
U 3
2
2
1
1
1
1
2
3
Y
1
∂ α ∂ eðα=3Þ X Y ∂X 3 ∂Y
1
1
1
2
2
3
3
2
ð104Þ
3
Y
pffiffiffi Fig. 11. Linear degenerate nodes at the equilibrium points predict by (34) (f 2 ¼ 3, g 2 ¼ 1, β ¼ 2, α ¼ 7 8). (a) Stable node at ð0; 0Þ and (b) Unstable node at ð0; 0Þ.
68
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U
U
1.0
2
2
1
1
0.5
0.5
1.0
1.5
2.0
Y
1.0
0.5
0.5
1
1
2
2
1.0
1.5
2.0
Y
2
Fig. 12. Changes of the system when α2 ¼ 4f 2 =g2 (f 2 ¼ 1, g 2 ¼ 1, α ¼ 7 2). (a) α 40 and (b) α o 0.
1.0
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
Y
1.0
0.5
0.5
0.5
0.5
1.0
1.0
1.5
1.5
1.0
1.5
2.0
Y
Fig. 13. Degenerate node at the origin obtained after simulation of the system (f 2 ¼ 1, g2 ¼ 1, β ¼ 0:25, α ¼ 7 1). (a) Stable node and (b) Unstable node.
as an additional symmetry, provided β¼
2α2 : 9
ð105Þ 2
In this case β is also not special. Note that f 2 þ 4g 2 β ¼ 89g 2 α2 . 2 (i) If g 2 4 0 and α a 0, then f 2 þ 4g 2 β 4 0. Thus from pffiffiffi Section pffiffiffi 4.2 we p have three equilibrium points: ð0; 0Þ, ð 2α=3 g ; 0Þ ffiffiffi pffiffiffi and ð 2α=3 g ; 0Þ. The stability analysis shows that the origin is a node (because 0 o β o α2 =4Þ, whereas the two others points are saddle points. The node is stable when α 4 0, and unstable when α o 0. The behaviour is shown in Fig. 18(a), (b). 2 (ii) If g 2 o 0 and α a 0, then f 2 þ 4g 2 β o 0. Therefore, from Section 4.2 the only equilibrium point, the origin, is a
node (since 0 o β o α2 =4Þ. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 19(a), (b)). (iii) When α ¼ 0, the origin is again the only equilibrium point, and at that point, the determinant Δ ¼ 0. This is a borderline case. The behaviour has been obtained in Section 4.2 by simulating the system numerically, as shown in Fig. 19(c). 3. Consider Eq. (34) in the neutral perfect fluid case ðg 2 ¼ 0 and f 2 a 0Þ. Maharaj et al. [18] obtained an additional symmetry subject to the condition β¼ 7
6α2 : 25
ð106Þ
The analysis in this case corresponds to the one in Section 3.2 (with M ¼ β α2 =4 and N ¼ 0Þ. The value of β is not special.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
1.0
U 1.5
U 1.5
U 1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
0.5
1.0
1.5
Y 2.0
1.0
0.5
0.5
1.0
1.5
Y 2.0
1.0
0.5
0.5
0.5
0.5
0.5
1.0
1.0
1.0
1.5
1.5
1.5
69
1.0
1.5
Y 2.0
Fig. 14. Behaviour of the system (34) in the case where β ¼ 0 and f 2 ¼ 0. (a) α 4 0. (b) α o 0 and (c) α ¼ 0.
Y
Y stable unstable
stableβ
unstable
unstable
β
unstable
stable
β
stable
Fig. 15. Bifurcation diagram in charged fluid (g2 ¼ 7 1). (a) Supercritical bifurcation and (b) Subcritical bifurcation.
U 1.5
U 1.5
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
Y
0.5
0.5
0.5
0.5
1.0
1.0
1.5
1.0
1.5
Y
1.5 2
Fig. 16. Behaviour of (32) in the neutral perfect fluid case when α a 0 and N ¼ M =4f 2 þ Mα2 =8f 2 þ 49α4 =40 000f 2 (f 2 ¼ 18, α ¼ 7 10, M ¼ 2). (a) Stable node, saddle point and (b) Saddle point, unstable node.
(a) When α ¼ 0, then β ¼ 0. Thus from Section 3.2 the origin is the only equilibrium point (since M 2 4f 2 N ¼ 0Þ, and at that point the trace τ ¼ 0 and the determinant Δ ¼ 0. We simulate the system numerically to obtain the behaviour as shown in Fig. 20(a). (b) When α a0, consider β ¼ 6α2 =25. Since 16M 2 64f 2 N þ 8Mα2 þ α4 4 0, from Section 3.2 we have two equilibrium points, ð0; 0Þ and ð6α2 =25f 2 ; 0Þ. The stability analysis showed that the origin is a node, and ð6α2 =25f 2 ; 0Þ is a saddle point. The node is stable when α 4 0, and unstable when α o 0 (see Fig. 21(a), (b)).
However, for β ¼ 6α2 =25, our equilibrium points reduce to ð0; 0Þ which is a saddle point, and ð 6α2 =25f 2 ; 0Þ which is a node (see Fig. 21(a), (b)). We observe that the system behaves differently depending on the sign of β in (106). Note that, in all cases, when (18) admits two Lie point symmetries, they satisfy the relationship ½G1 ; G2 ¼ λG2 ;
70
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U
4
U
4
4
2
2
2
2
4
Y
4
2
2
2
2
4
4
4
Y
Fig. 17. Behaviour of (34) in the charged fluid, with f 2 a 0 (f 2 ¼ 2, α ¼ 7 2). (a) α 4 0 and (b) α o 0.
U
2
U
2
2
1
1
1
1
2
Y
2
1
1
1
1
2
2
2
Y
Fig. 18. Behaviour of (32) in the charged fluid when f 2 ¼ 0, g2 40 and α a 0. The system predicts node and saddle points (g2 ¼ 2 and α ¼ 7 3). (a) Stable node at ð0; 0Þ and (b) Unstable node at ð0; 0Þ.
U
U U
2
2 1.5
1
1.0
1
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1
1
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1
1
1
2
2
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Y 1.5
1.0
0.5
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1.5
Y
0.5 1.0 1.5
Fig. 19. Behaviour of (32) in the charged fluid when f 2 ¼ 0. The system predicts a node when g2 o 0 and α a 0, and a centre when α ¼ 0 (g2 ¼ 2 and α ¼ 7 3). (a) g2 o 0 and α 4 0. (b) g2 o 0 and α o 0. (c) α ¼ 0.
P.M. Tchepmo Djomegni, K.S. Govinder / International Journal of Non-Linear Mechanics 62 (2014) 58–72
U 2
U
U
2
2
1
1
71
1
2
1
1
2
Y 1.0
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Y 2.0
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Y 2.0
1 1
1
2
2
2
Fig. 20. Behaviour of (34) in the neutral perfect fluid when β ¼ 6α2 =25. The system predicts for α ¼ 0 a cusp (f 2 ¼ 1), and for α a 0 a node and a saddle point (f 2 ¼ 6, α ¼ 7 5). (a) α ¼ 0 and (b) α 40. (c) α o 0.
U
2.0
1.5
1.0
U
2
2
1
1
0.5
0.5
1.0
Y
2.0
1.5
1.0
0.5
0.5
1
1
2
2
1.0
Y
Fig. 21. Behaviour of (34) in the neutral perfect fluid when β ¼ 6α2 =25. For α a 0, the system predicts a node and a saddle point (f 2 ¼ 6, α ¼ 7 5).
where λ is a non-zero constant. Since the symmetries are unconnected, the Lie algebra is Lie's Type III. This confirms that the equation cannot be linearized, but can be reduced to quadratures [16].
6. Discussion In view of the foregoing, we observe that the constraints obtained by Kweyama et al. [15] in their quantitative study do not fully integrate with our qualitative study. The special values obtained via the Lie symmetry analysis are not special when it comes to the dynamical systems analysis, and with these values there is no bifurcation. 2 However, in Section 3.3 (where 16f 2 þ 48g 2 M þ 12g 2 α2 4 0) it was difficult, even impossible, to pursue the stability analysis of the system, because the equilibrium points were complicated and there was no way to isolate the real values. In this case, we used the corresponding constraints obtained by Kweyama et al. in the
Lie symmetry analysis to pursue our dynamical systems analysis and obtain the local behaviour of the system. Therefore we can use the dynamical system analysis to test the relevance of symmetries, and of solutions (for example, special solutions or not) obtained from the Lie symmetry analysis. We can also predict the stability of the solutions, even possible bifurcations. On the other hand, in certain robust case of qualitative analysis, we can use Lie symmetry analysis to investigate constraints that will enable us to study the system locally. Thus we conclude and advise that it is quite important to utilize both techniques of analysis when dealing with differential equations.
Acknowledgements P.M.T.D. and K.S.G. thank the University of KwaZulu-Natal and the National Research Foundation of South Africa for ongoing support.
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