Lecture Notes:

Theoretical Methods in Industrial Physics

-

Stability and Bifurcation

Thomas Christen

FS 2014

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Contents

1 Introduction 2

2 General Dynamic Systems 3

3 Bifurcation of Steady States 5

4 Linear Stability Analysis 5

5 Symmetries 8

6 Collective Coordinates 11

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1 Introduction

When a technical system is investigated in the framework of a mathematical model, twoimportant questions concern the existence and number of possible state(s), and the proper-ties of these states. Of particular interest are the stability properties of a state. These arerelevant in two respects. First, in many cases the states of interest must be stable in orderthat the system can perform its intended task. Instabilities must then be avoided. Secondly,certain technologies are intrinsically reliant on instabilities in order to be able to performthe intended task. Examples are devices which switch between two different stable states;the switching process runs then usually through an unstable state that separates the twostable states (see Fig. 1 a)). There exist technical systems which even work completely inunstable (controlled) states. (A prominent example is the unstable basic airframe of combataircrafts, which require complex control systems; by this the aircraft can be lighter, smaller,more maneuverable, and more efficient. Another example is the high-voltage direct currentcircuit breaker based on passive resonant circuit instability. The instability is here used forcreating a zero-crossing of the current, which is needed for current interruption and natu-rally occurs for alternating voltage but not for direct voltage).The minimum procedure of a stability investigation involves at least two steps: (i) the de-termination of the basic states that have to be investigated, and (ii) their stability analysis.In most cases the latter refers to a linear stability analysis, which involves infinitesimal per-turbations and to which we will restrict ourselves in the following. More detailed treatisesthan given in these short lecture notes can be found in Refs. [1, 2, 3]. Prior to a moregeneral discussion, an illustrative example is in order.

feedback loop

rOn

Input In Output On

v >1In vIn

v1/r

Î

O

O

State 1 State 2

State 3

I

II

III

a) b)

c)

λ

n+1

Figure 1: a) A switching processes (I to III) from a stable state 1 to a stable state 2 (potentialminima) passes through an unstable state 3 (a saddle). b) The feedback model of Example 1 fora system that amplifies an input In with amplification factor v and couples back the amount rOn

of the output into the input at time n + 1. c) Plot of the steady state solutions O (solid curve:stable, dashed curve: unstable (and unphysical)) as a function of the control parameter product rvwith divergence ar rv = 1. Dashed-dotted curve: growth rate λ(rv) of a small perturbation of thesteady-state solution.

Example 1 Feedback instabilityConsider the feedback loop sketched in Fig. 1 b). An input In (≥ 0) at (discrete) time nis linearly transformed by a device into an output On = vIn, where v (> 0). For v > 1 theinput is amplified. A fraction r (0 ≤ r ≤ 1) gets back to the input at time n + 1, and is

2

added to the external input I, which gives In+1 = I + rOn. An example for such a systemis audio-feedback, e.g., a microphone connected to a loudspeaker via an amplifier. It is well-known that if the microphone comes too close to the loudspeaker, the acoustic feedback (r)can be so strong that a shrill sound appears (”feedback instability”). Other examples thatcan be described by this feedback model exist in many disciplines, also non-technical oneslike climate science, ecology, or even psychology. The basic model equation for this systemis obviously

In+1 = I + rvIn . (1)

For step (i) we determine the steady-state (= time independent) solution In ≡ I of Eq. (1)and find I = I/(1− rv), hence

O =vI

1− rv. (2)

The divergence for rv → 1 corresponds to the mentioned feedback instability. For rv > 1the output becomes negative which is unphysical; however from a mathematical point ofview a solution is existing.1

One still has to investigate if the solution (2) can be realized at all in practice, or if it isunstable. For this second step (ii), one adds an (infinitesimally) small perturbation ∆I0 atn = 0 to I and determines the behavior of ∆In for n → ∞. If ∆In decays to zero, the stateis stable; if ∆In increases, it is unstable, and if it remains constant, it is called ”marginallystable”. Substitution of In = I +∆In in Eq. (1) leads to

∆In+1 = rv∆In , (3)

which can be solved with the ansatz

∆In = ∆I0 exp(λn) , (4)

with λ = ln(rv). Stability implies λ < 0, i.e. rv < 1. Consequently, the solution I is stableat rv < 1. The situation is sketched in Fig. 1 c).

From this simple example one can learn several rather general facts (which will be partlydiscussed in subsequent sections):

• There are four different types of quantities: the time (n; later also continuous time andspace variables), the state variables (In or On; governed by model equations), controlparameters (v, r, I; they can be externally adjusted), and small fluctuations (∆I0; dueto noise or other disturbances which are not fully under control).

• The states of the system are characterized by relations between the system variablesand the control parameters (I = I(I , r, v) or O = O(I , r, v)). A graph of O as afunction of a control parameter (see Fig. 1 c)) is an example for a bifurcation diagram(the origin of the name ”bifurcation” (Verzweigung) will become clear below).

• Parameter values exist where solutions disappear or appear (here rv = 1).

• The stability behavior of I is characterized by a growth rate λ, which becomes also afunction of the control parameters for parameterized I. Regions in control parameterspace where λ is (positive) negative belong to (un)stable states. These regions are sep-arated by boundaries where λ is zero. (This is a special (but often occurring) case. Ingeneral one has to look at the real part of the growth rate. Furthemore, λ may changeits sign also via crossing a pole). The parameter values where solutions disappear orappear are related to these stability boundaries (critical parameter values).

2 General Dynamic Systems

Although the examples below will be simple and specific, the general framework is firstintroduced. It holds for general system Σ described by a (maybe multi-component) state

1The mathematical divergence at rv = 1 is also unphysical; there will always be a saturation or limitationof O by effects not considered in this model. For example, the amplifier has only limited power, or theloudspeaker will break, etc ...

3

variable2 ϕ ∈ Λ (phase space or state space Λ) and a dynamic equation (the state maydepend on time t)

∂tϕ = Fµ[ϕ] (5)

with initial conditions ϕ(t0) = ϕ0 (and boundary conditions if Eq. (5) is a partial dif-ferential equation in space-time). In Eq. (5), µ ∈ Υ (parameter space Υ) is a (maybemulti-component) control parameter (parameter values that can be controlled or tuned ex-ternally). Practically relevant questions are

• Existence: are there solutions (”states”) ϕµ(t) ?

• Uniqueness/ambiguity: how many solutions exist ?

• Type of states: are they steady states (i.e., time independent solutions), time periodicstates, non-periodic time dependent states (quasi-periodic, chaotic, ...) ?

• Stability: Which states can be realized ? Are they globally stable, metastable, ... ?

• Are there symmetries and conserved quantities ?

• Are there thermodynamic equilibrium states ? Are there steady (flow) states nearthermodynamic equilibrium ?

• Existence of a stationarity principle (Ljapunov functional, potential, action, entropyproduction rate, ...) ?

�

�

�

V

�>0 �<0

�>0

�<0

�

�

�

�(i) (ii) (iii)

Figure 2: Bifurcation diagrams for the cases discussed in Example 2 (stable: solid, unstable:dashed, cf. Sect. 4). (i) Stability exchange at intersection points. (ii) Stability exchange at aturning point (saddle node bifurcation). (iii) Stability exchange at a pitch-fork bifurcation; alongthe solution ϕ = 0 the stability changes at the crossing point; along the other solution branch, thestability changes twice due to the turning point and the crossing point, hence the branch µ = ϕ2

remains stable.

2In analogy to phase transition theory, ϕ is sometimes called order parameter.

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3 Bifurcation of Steady States

All answers to the just listed questions are functions of µ. There are manifolds in Λ definedby µ-parameterized solutions ϕµ. The parameter space Υ is divided in regions associated

with different properties of the system Σ, i.e., number N of solutions ϕ(n)µ (n = 1, ..., N),

different stability properties, etc. It is thus convenient to discuss the solutions in a so-calledbifurcation diagram in the space Υ × Λ (i.e., (µ, ϕ)). Bifurcation theory describes how so-lutions or states of a system appear and/or disappear when control parameters are varied.Typical application examples are related to stability issues of devices, or to switching pro-cesses, as will be discussed in the exercises 3 and 4, respectively. In the following, we willconsider some simple cases that can be graphically illustrated and illuminate the mathemat-ical background.Bifurcation theory is not restricted to steady states, but is much more general, and can beapplied to general mathematical objects (including of course time dependent states).

Example 2a Single-component state with one control parameterConsider Λ = Υ = R. The bifurcation diagram consists then of curves in the plane (µ, ϕ)(Fig. 2). Equation (5) is an ordinary differential equation for a real function ϕµ(t). The

function Fµ = −V ′µ can be expressed as derivative of a ”potential” Vµ(ϕ) with a µ-dependent

shape. Its optima are the steady states, which are functions of µ and are denoted by ϕ(n)(µ)(n labels the different states).

(i) V = µ2ϕ − ϕ3/3 =⇒ Fµ = ϕ2 − µ2 =⇒ ϕ(1)(µ) = −µ, ϕ(2)(µ) = µ. There aretwo solution branches which intersect at µ = 0. Such a situation is called ”transcriticalbifurcation”.

(ii) V = −µϕ + ϕ3/3 =⇒ Fµ = µ − ϕ2 =⇒ ϕ(1)(µ) = −√µ, ϕ(2)(µ) =

√µ. Here µ ≥ 0,

for negative µ there is no steady state solution. This bifurcation is called ”saddle-nodebifurcation”, because at µ = 0 a saddle and a minimum of the potential merge.

(iii) V = −µϕ2/2 + ϕ4/4 =⇒ Fµ = µϕ − ϕ3 =⇒ If µ < 0, only one solution exists:ϕ(2) = 0. For µ ≥ 0 there are three solutions, ϕ(1)(µ) = −√

µ, ϕ(2)(µ) = 0, ϕ(3)(µ) =√µ.

This bifurcation is called ”pitchfork bifurcation”.

Example 3a Two-dimensional state spaceConsider ϕ = (ϕ1, ϕ2) and Fµ = (F1(ϕ1, ϕ2;µ), F2(ϕ1, ϕ2;µ)). The steady states are thengiven by the intersection points of the two implicit curves F1 = 0 and F2 = 0 (see Fig. 4a)). Saddle node bifurcations correspond to the appearance or disappearance of intersectionpoints when the curves move in the x1-x2-plane when µ is varied.

Figure 3: The spectrum of the linear stability analysis depends on the control parameter(s) µ.An instability occurs at a critical parameter µc when the largest real part of the eigenvaluesbecomes positive.

4 Linear Stability Analysis

Once the basic states are found, their stability must be determined. In the framework oflinear stability analysis, one has to investigate wether an infinitesimal disturbance δϕ of the

5

Tr

det

Sa

dd

len

od

eb

ifu

rca

tio

n

Hopf bifurcation

a) c)F2(�1, �2)=0

b)Im(λ)

(1)

(2)

(3)

(4)

(5)

(1) (1) (2) (2)(5) (5)

(3)

(3)

(4)

(4)

(1)

F1(�1, �2)=0

�1

�2

Re(λ)

Figure 4: a) Determination of steady states for a dynamic system with two state variables (x1, x2)via the intersection points of the curves F1 = F2 = 0. By changing control parameters, intersectionpoints can disappear or appear (saddle-node bifurcations). b) For the two stability eigenvalues λ1,2

five different cases (1)-(5) exist, depending on if they are real or complex and on their location inthe complex plane. c) Stability discussion in the plane of det(DF ) = λ1λ2 and Tr(DF ) = λ1 + λ2;at a saddle-node bifurcation det changes sign, while a sign-change of Tr at positive det correspondsto a Hopf bifurcation, where the complex conjugate eingenvalue pair cross the imaginary axis.

state under investigation decays or increases. We again focus on steady states, i.e. time-independent solutions ϕµ of Fµ[ϕ] = 0. Linearization of Eq. (5) leads to ∂tδϕ = DFµ[ϕµ]δϕ.

Here, DF represents the first derivative (the ”Jacobian”), and terms of higher order in δϕare neglected. If λk = λk(µ) are the eigenvalues of DF and δϕk the associated eigenvectors(”eigen-modes”), disturbances behave as δϕk ∝ exp(λkt). The index k labels the eigenvaluespectrum, which can be discrete or continuous. As usual, the spectrum is ordered accordingto the real parts of the eigenvalues, such that Re(λk) > Re(λj) for k < j. Hence, for a realdiscrete spectrum λ0 > λ1 > ... > λk > λk+1 > .... Only if for all real parts Re(λk) < 0holds, all disturbances will decay and the state ϕµ is stable. For the discrete real spectrum,the stability requirement is, for instance, λ0 < 0. On the other hand, if at least one realpart is positive, the state is unstable. The case of Re(λ) = 0 will be discussed later.

The eigenvalues depend on the control parameters µ. If a control parameter is varied, aninstability is said to occur if the largest real part changes its sign from negative to positive.Linear stability analyis can thus be summarized as follows. Determine the spectrum of DF ,i.e. solve the linear eigenvalue problem

λδϕ = DFµ[ϕµ]δϕ . (6)

The state ϕµ is stable if the spectrum λk(µ) lies in the left half of the complex plane. Atinstability the control parameter value is called critical, µc. If the dimension Υ is larger thanone, not critical points but rather critical hyper-surfaces in Υ exist, where the largest realpart of the eigenvalue spectrum vanishes. The states on this surface are the marginally stablestates. The eigenvector δϕ (unstable mode) belonging to the critical eigenvalue character-izes the physical disturbance which grows exponentially when the state becomes unstable.Below we will see that eigenvalues can identically vanish due to symmetries, which must bedistinguished from a zero eigenvalue at an instability.

Example 2b Single-component state with one control parameterWe return to Example 2a and Fig. 2. Linearization yields DFµ = F ′

µ = −V ′′µ . Hence

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�

�

(a) (b)

k

x

kc

(c)

(i) (ii)

�cδφ

� = �c

� > �c

� < �c

� > �c

� < �c

(ii) (i)

Re(λ)

Figure 5: a) Subcritical bifurcation, where the unstable state is curved into the parameter regionbelow µc. At instability, the critical perturbation is not immediately saturated be grows to a newstate far from the initial state. (Thin arrows: vector field, thick arrows, hysteresis loop). b) Inspatially extended systems with translation symmetry, the stability problem of the uniform stateis characterized by λ(k), where k is the wave number of the perturbation. The zero-crossing ofthe maximum of Re(λ(k)) is sketched in (i) for different control parameter values. The kc of theunstable mode determines the spatial period of the growing structure. For kc = 0 the unstablemode is uniform, for finite kc and Im(λ(kc)) it is a wave. It can also happen (see (ii)) that themaximum of Re(λ(k)) is a k = 0 and the dispersion is diffusive, Re(λ) = −Dµk

2, and the effectivediffusion constant becomes negative for µ > µc (diffusion instability). c) In spatially extendedsystems with flow, one distinguishes between (i) absolute instabilities, where the amplitude of asmall perturbation growths at given locations, and convective instabilities, where the perturbationgrowths in a co-moving frame (but decays at every x for sufficiently large times).

minima of Vµ are stable and maxima (and saddle points) are unstable. One finds for theabove examples

(i) F ′µ = 2ϕ =⇒ λ(1)(µ) = −2µ, λ(2)(µ) = 2µ. One concludes that ϕ(1) is unstable and

ϕ(2) is stable for µ < 0, and they exchange their stability properties at the crossing point(µ = 0). In Fig. 2, stable and unstable states are indicated by solid and dashed curves,respectively.

(ii) F ′µ = −2ϕ =⇒ λ(1)(µ) = 2

√µ, λ(2)(µ) = −2

√µ (recall µ ≥ 0). The stability behavior

of ϕ(1,2) is exchanged at the turning point, µ = 0.(iii) F ′

µ = µ − 3ϕ2 =⇒ For µ < 0, the only solution is stable, λ(2)(µ) = µ. For pos-

itive µ, the three solutions have the following eigenvalues: λ(1)(µ) = −2µ, λ(2)(µ) = µ,λ(3)(µ) = −2µ. Obviously, ϕ(1,3) are stable, while ϕ(2) is now unstable at positive µ. Notethat µ = 0 is both an intersection point and a turning point. While the solution ϕ(2) that isonly intersected, changes stability, the solution that is intersected at its turning point keepsits stability behavior.

If the dimension of the state space Λ is larger than one, a potential V (ϕ) with F = −∇ϕVdoes not exist in general. It is then not possible to illustrate the bifurcation behavior interms of maxima, minima, or saddle points of a global potential. However, due to topologicalreasons there are still general ”bifurcation rules” associated with stability change, and whichfollow from the theorem of implicit functions (see also Example 3). We list here the mostimportant ones.

• The stabilities of solutions change at an intersection point in the bifurcation diagram

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(as in Fig. 2 (i)).

• The stability of a solution changes at a turning point in the bifurcation diagram (asin Fig. 2 (ii)).

• If a stability exchange occurs twice at the same point (e.g. crossing at a turning point),the stability of the corresponding solution branch remains (as in Fig. 2 (iii)).

Example 3b Two-dimensional state spaceThe Jacobian needed for stability analysis of a steady state given by F − 1 = F2 = 0 is(Df)kl = ∂Fk/∂ϕl and is characterized by the two invariants trace Tr and determinant det.The stability eigenvalues are then λ1,2 = Tr/2 ±

√Tr2/4− det. Two cases can appear,

λ1,2 are both real, or they are complex conjugates. The cases are shown in Fig. 4 b) andc). Stability requires that the real parts of λ1,2 are negative, i.e. Tr < 0 and det > 0. Ifdet = λ1λ2 crosses zero as a function of a control parameter, a single eigenvalue changes sign,hence a saddle-node bifurcation occurs. If Tr crosses zero but det does not, two complexconjugate eigenvalues cross the imaginary axis, where the sign of Re(λ) changes. This caseis called a Hopf bifurcation and can result in time-periodic states (limit cycle).

There is a large zoo of different bifurcation scenarios involving generic (structurally sta-ble) and non-generic (structurally unstable) cases. An extensive specialist dictionary exists[3]. In the following some practically relevant terms are mentioned.One distinguishes between super- und subcritical bifurcations. Supercritical bifurcations arecharacterized by the continuous growth of the bifurcating steady state solution from zerowhen µc is crossed, as in Fig. 2 (iii). In a sub-critical bifurcation, the new state extendsto the sub-critical parameter region and is thus unstable near bifurcation. At instability, aperturbation increases quickly to another state which is far from the initial state. A typicalexample is the instability which occurs in systems with hysteresis loops, as is shown in Fig.5 a).Often systems spatially extended are to be considered, and the states can be space depen-dent. In linear stability analysis, perturbations of uniform (spatially constant) steady states

can be decomposed into Fourier components, δϕ exp(λt+ ikx), where for simplicity a singlespace dimension is assumed (in general, other sets of eigenfunctions may be used). Thegrowth rate is a function of wave number k, hence λ = λµ(k) (dispersion relation). Instabil-ities can then be classified according to the values of k and λi (or frequency ω = −Im(λ))at instability. The unstable mode behaves as follows:

• kc = 0;ωc = 0: uniform, non-oscillating

• kc = 0;ωc = 0: uniform, oscillating

• kc = 0;ωc = 0: spatially periodic, non-oscillating

• kc = 0;ωc = 0: spatially periodic, oscillating (growing wave)

Localized spatio-temporal instabilities can be classified according to there convectionbehavior. An instability is called absolute if a perturbation growths at given location x, andis convective if perturbations decay at every x for t → ∞ but grow in a co-moving frame.Convective instabilities are common in hydrodynamic systems.

5 Symmetries

The presence of symmetries can (i) lead to simplifications when solving for the ground state,and (ii) can have important consequences for the stability and other properties of a statewhich breaks the symmetry. For instance, case (iii) of example 2 had mirror symmetryϕ → −ϕ. This symmetry group has two elements, namely the inversion ϕ → −ϕ and theidentity ϕ → ϕ. It is obvious, that any solution ϕ which breaks this symmetry (i.e., which isnot invariant if mirrored; to be concrete: ϕ = 0), can be used to create another solution byits mirror image. The number of elements of the group properties of the symmetry groupare related to the number of solutions, which can be generated by the group operations froma symmetry breaking solution. Most important is that if a control parameter dependentstate ϕ(µ) breaks a continuous symmetry of the system Σ, then an identically vanishingeigenvalue exists, i.e., λ(µ) = 0 ∀µ. The associated eigenvector is called Goldstone mode.

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Figure 6: Plane vector field F (with ϕ ∈ R2) with rotation symmetry. g is a rotation with anglebetween 0 and 2π.

Goldstone modes are not only relevant in the context of stability analysis. They are usefulfor describing and modelling the behavior of complex systems with the help of collectivecoordinates, as will be discussed in the next section. Due to its relevance, we will now brieflyexplain in more detail the effect of symmetries and the origins of Goldstone modes.

Definition: According to Eq. (5), one may consider F as a ”velocity field”, which definestrajectories that are the solutions of Eq. (5) (we drop here the index µ). A diffeomorphismg (i.e., bijectiv, g and g−1 differentiable) is called a symmetry of F , if F is invariant underg, i.e., F [gϕ] = gF [ϕ]. The symmetry maps solutions (trajectories) of (5) to solutions(trajectories) of this equation. The symmetry operations of an equation form a group, thesymmetry group.

Let F be symmetric under the (Lie) group {gs | s ∈ S ≡ differentiable manifold; g0 = 1}.Examples are the group of continuous translations in d-dimensional space, or rotations in R2

or R3. If ϕ0, with ϕ0 = ϕs ≡ gsϕ0 for s = 0 (this inequality defines ”symmetry breaking”) isa stationary solution of F [ϕ] = 0, then also ϕs is a stationary solution, F [ϕs] = 0 (∀s ∈ S).In other words, if a symmetry breaking solution is known, one can create more solutions(the symmetry manifold) by the action of the symmetry group. From dF [ϕs]/ds = 0 andthe chain rule one obtains

DF [ϕs]∂sϕs = 0 , (7)

i.e., δϕGM ≡ ∂sϕs is a Goldstone mode. The meaning of this zero eigenvalue is that a ”dis-placement” of an otherwise stable state in direction of the Goldstone mode (i.e., tangentialto the symmetry manifold) does not lead to a restoring force. As will be shown in Sect. 6,a weak symmetry breaking force can lead to a slow motion of the state on the symmetrymanifold.

Example 3 Translational symmetryConsider F [ϕ] = f(ϕ) + ∂2

xϕ, with x, t, and ϕ(x, t) real (and one-dimensional). If ϕ0(x)is a spatially inhomogeneous steady state, then also gsϕ0(x) ≡ ϕ0(x − s) is a steady statefor all s ∈ R. The spectrum of the linear stability problem has an eigenvalue λ ≡ 0 witheigenvector ϕ′

0. It is clear that δϕ = ϕ′0δs is the infinitesimal displacement of ϕ0. The

equation ∂t(ϕ′0δs) = 0 implies that a spatial displacement of a stable solution ϕ0(x) is not

restored. In the presence of weak noise, for instance, this can lead to a Brownian motionof s, i.e., a diffusion of the solution ϕ in x-space. The eigenvalue spectrum has the form0 = λ0 > λ1 > ... > λk > λk+1 > ....

Example 4 Generalized third law of KepplerAppropriate elaboration of symmetry properties can provide relevant information on thebehavior of a system. As an example, consider Newtons law x(= d2x/dt2) = F withhomogeneous force F (kx) = kdF (x). Consider the group gs acting on (x, t) and defined

9

by gs(x, t) = (x exp(βs), t exp(αs)). This ”stretching” transforms the acceleration as x 7→exp((β − 2α)s)x, and the force F 7→ exp(βsd)F . The Newton equation remains invariantif β(1 − d) = 2α, i.e. if this relation between α and β holds, gs is a continuous symmetrygroup ∀s. If a particle needs a time T1 for a trajectory of length L1, then a particle movingon the stretched trajectory with length L2 = L1 exp(β) needs the time T2 = T1 exp(α) withα = β(1− d)/2. Consequently,

T2

T1=

(L2

L1

) 1−d2

.

For d = −2 this gives Keppler’s third law T 2 ∝ L3, which is proven to be a consequence ofthe specific symmetry of the force.

Exercise 3 Stability analysis for thermal runawayConsider the one-dimensional heat conduction problem

ρc∂tT = λ∂2xT + σ

(U

L

)2

with boundary conditions ±λ∂xT = αT at x = ±L/2. Here ρ, c, λ, and U are mass den-sity, specific heat, heat conductivity, and applied voltage. The conductivity is given byσ(T ) = σ0(1 + T/T0). Assume first homogeneous Dirichlet boundary conditions, i.e., aninfinite heat transfer coefficient α → ∞.

a) Transform all quantities and the heat equation to dimensionless units and determinethe control parameter µ which appears as the factor in front of the dimensionless heat sourceterm.

b) Calculate the solution T (x), sketch the bifurcation diagram (µ, Tmax), and determinethe critical parameter value µc.

c) Perform a stability analysis by solving the associated eigenvalue problem, and deter-mine again µc. How changes the critical value µc as a function of α?

(d) Determine the order of magnitude of the critical voltage and the characteristic timescale for a typical polymeric electric insulation material.

Exercise 4 Interruption of an electric currentIf one tries to interrupt an electrical current by separating two contacts, a conductive plasmachannel (electric arc) is created, the ionization of which is sustained by the ohmic heatingUI. Here, U and I are voltage and current of the arc, respectively. A simple model for thearc conductance G = I/U , which reproduces qualitatively the arc behavior, is given by (theMayr model)

dG

dt=

G

τ

(UI

K− 1

)where τ is a typical charge-carrier generation-recombination time, and K is the coolingpower (heat radiation, convection, etc.); both parameters are assumed to be constant.

a) Give a physical interpretation of this equation. Determine the steady states for (i)imposed current I (current control) and (ii) imposed voltage U (voltage control) and discusstheir stability properties.

b) Consider a series connection of the arc, a voltage source U0, and a series resistance R.(i) Show with a sketch in the I-U diagram, that a saddle-node bifurcation occurs. (ii) Whichlimit cases for U0 and R correspond to the cases (i) and (ii) of a)? (iii) Write the dynamicequation for the unitless variable g = RG, determine the relevant control parameter µ, findall steady state solutions, and discuss the bifurcation diagram in the µ − g plane. Whathappens at current interruption when the driving voltage U0 is lowered?

c) Discuss interruption of an alternating current near the zero crossing for large R andlarge U0, with U0 = U0 sin(ωt), by linearizing the voltage in time and comparing the con-

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ductance decay time and the traversal time of the monostable U0-region (near t = 0).3

Figure 7: Symmetry manifold ϕ(0)s with direction of Goldstone mode ∂sϕ

(0)s and perpendicular

direction ∂tϕ(1)s .

6 Collective Coordinates

If the stable state of a system breakes a continuous symmetry, there exists a manifold ofnon-equivalent solutions described by the parameter s. Often, physical systems (Eq. (5))can be decomposed such that F = F (0) + ϵF (1), where F (0) is symmetric with respect to asymmetry group gs, while F (1) is not symmetric, but ϵ is small such that F (1) acts only asa weak disturbance. In that case, one can make the ansatz for the solution

ϕ = ϕ(0)s + ϵϕ(1)

s + ... (8)

where ϕ(0)s is a symmetry breaking steady state of the undisturbed system, i.e., F (0)[ϕ

(0)s ] = 0.

The weak symmetry breaking force has two effects. First, it weakly deforms the shape of

ϕ(0)s . This deformation, which is described by ϵϕ

(1)s in leading order, is only ”weak” if the

force is small compared to the ”restoring” forces, which are related to the finite eigenvalues

of the stability spectrum of the undisturbed system. Secondly, it drives the state ϕ(0)s in

direction of the Goldstone mode, where the restoring force vanishes. In other words, thesymmetry ”parameter” s becomes a slow dynamic variable and is thus time dependent,s = s(t). Slow means, that the velocity s is of order ϵ. Insertion of the expansion in Eq. (5)yields

∂sϕs+ ϵ∂tϕ(1)s = ϵDF [ϕ(0)

s ]ϕ(1)s + ϵF (1)[ϕ(0)

s ] +O(ϵ2) .

In lowest order of ϵ, rearrangement gives

ϵDF [ϕ(0)s ]ϕ(1)

s − ϵ∂tϕ(1)s = ∂sϕs− ϵF (1)[ϕ(0)

s ] . (9)

The previous equation can be understood as a linear inhomogeneous equation for ϕ(1)s of the

form Aϕ(1)s = h. There exists a solvability condition (Fredholm alternative) which requires

that h is perpendicular to the kernel of the adjoint A†. In our context, this means thatscalar product of the right hand side of Eq. (9) and the (adjoint) Goldstone mode vanishes.The geometrical meaning is that the dynamic equation (5) is projected onto the symmetrymanifold of F (0). The result is an ordinary differential equation for the collective coordinates given by

s = ϵ< ∂sϕs | F (1) >

< ∂sϕs | ∂sϕs >. (10)

The following exercise should clarify these issues.

3This problem can also be analytically calculated under the assumption, that τω ≪ 1, and that for t < 0I = −It is imposed (current control), and for t > 0 the voltage is imposed (voltage control).

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Exercise 5 One-dimensional domain wall subject to a weak forceConsider a 1-d system (space coordinate x) with state variable ϕ(x, t), and boundary con-ditions ϕ(x → −∞) = −1 and ϕ(x → ∞) = 1. Assume F [ϕ] = ∂2

xϕ − dW/dϕ + f withW (ϕ) = ϕ2(ϕ2 − 2), and f(ϕ, x) is a weak disturbance which breaks the symmetry of theunperturbed system.

a) What are the symmetries of the unperturbed system. Sketch a steady state ϕ(x, s)of the undisturbed system (f = 0) (Hint: use the analogy of the steady state equation withNewton’s equation for ”coordinate” ϕ and ”time” x). Discuss the stability of the solution,and show that the zero eigenvalue corresponds to the translation symmetry (Hint: for thestability spectrum, use the analogy to the Schrodinger equation).

b) Derive the equation for the domain wall motion in the form s = −dH(s)/ds from Eq.(10).

c) Sketch H(s) for f = v∂xϕ − aϕδ(x), where δ(x) is the Dirac delta function. Discussthe behavior of the domain wall for a > 0 and v > 0. What is the condition for a stablesteady state?

Exercise 6 One-dimensional domain wall between metastable and stable statesIn the previous exercise, the potential was symmetric such that the two stable states hadthe same energy. Determine the domain wall velocity between two bistable states ϕ1 and ϕ2

with energy difference ∆W = W (ϕ1)−W (ϕ2) (but now f ≡ 0).

References

[1] H. Haken, Synergetics, an Introduction, Springer, New York, 1983.

[2] H. Thomas, Nonlinear Dynamics in Solids, Springer, Berlin, 1992.

[3] J. D. Crawford, Introduction to bifurcation theory, Rev. Mod. Phys. 63, 991, 1991.

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