_{1}

^{*}

Inverted pendulum models are commonly used to study the bio-mechanics of biped walkers. In its simplest form, the inverted pendulum consists of a point mass attached to two straight mass-less legs. Most works constrain the motion of the mass to the sagittal plane,
*i.e.* the plane perpendicular to the ground that contains the direction toward the biped is walking. In this article, we remove this constrain to study the oscillations, the mass experiences in the direction perpendicular to the sagittal plane as the biped walks. While small, these lateral oscillations are unavoidable and of importance in the understanding of balance and stability of walkers, as well as walkers induced oscillations in pedestrian bridges.

When a human walks, the sagittal plane refers to the plane perpendicular to the ground that contains the direction toward the person is walking. As the person walks, its center of mass oscillates, both in the vertical direction (perpendicular to the ground) and lateral direction (perpendicular to the sagittal plane). One of our goals is to contribute to the understanding of the lateral oscillations.

The lateral oscillations play an important role in the balance and stability of individuals as they walk. Thus, their understanding is of interest in the field of bio-mechanics. These oscillations are also of interest in the field of robotics, since their understanding and control are likely to help improve the design of stable biped robots. While small, these lateral oscillations are the cause of some observed undesired and unexpected motions of pedestrian bridges when too crowded [

The use of inverted pendulum models to study the bio-mechanics of walking is a common practice. In its simplest form, the inverted pendulum consists of a point mass, which models the center of mass of the individual, attached to two straight mass-less segments, the legs. Works that use inverted pendulum (in its simplest or more sophisticated forms) or similar models to study aspects of the mechanics of biped walkers or related toys or biped robots include [

Different simple models are surveyed in [

Most works using the simplest inverted pendulum model constrain the motion of the center of mass to the sagittal plane. In this article, we remove this constrain. As a consequence, we are able to use this unconstrained inverted pendulum model to study the lateral oscillations the mass experiences as the person walks, i.e. the oscillations in the direction perpendicular to the sagittal plane. We believe and hope that the model and techniques described in this article will be adopted by other researchers and will prove useful in the study of different aspects of the mechanics of biped walkers.

In the next section the model is introduced. In the following section we describe the equations governing the dynamics of the mass while one foot is off the ground. Subsequently, we identify the solutions to the governing equations that correspond to periodic walking. We then report results of numerical simulations. We further explore our model by restricting our attention to the parameter regime of slow walkers. We then also study the short steps parameter regime. We finish the article with a discussion.

We model a human as a point mass m attached to two straight mass-less segments, the legs. Each leg is of length L . The mass m is the common end point of the two legs. The other end point of each leg is its foot. Only the feet can touch the ground. We will call this model of a human the model biped. Next, we describe our modeling assumptions. In the statements below k is any integer.

Assumption 1. At all times, either one or both feet are touching the ground.

The fact that each step takes the same time leads to our next assumption.

Assumption 2. Let T ¯ s > 0 be the time of one step. Both feet are touching the ground only at times t ¯ = t ¯ k = k T ¯ s .

During the time interval ( t ¯ k , t ¯ k + 1 ) , the foot of only one leg is touching the ground. This leg is called the stance leg. The other leg is called the swing leg. For definiteness, we assume the left and right leg alternate being the stance and swing legs as follows.

Assumption 3. The left leg is the stance leg during the time intervals ( t ¯ 2 k , t ¯ 2 k + 1 ) and thus, the right leg is the stance leg during the time intervals ( t ¯ 2 k − 1 , t ¯ 2 k ) .

Assumption 4. During the time interval ( t ¯ k , t ¯ k + 1 ) , the foot of the stance leg remains in the same position.

Observation 1. During the time interval ( t ¯ k , t ¯ k + 1 ) , the only forces acting on the mass m are the force due to gravity, F g , and the force due to the stance leg, F = F ( t ¯ ) , which is parallel to the stance leg and points from the mass in the direction opposite to the foot (see

As it makes contact with the ground at time t ¯ = t ¯ k , the leg that was the swing leg during the time period ( t ¯ k − 1 , t ¯ k ) exerts an impulse on the mass m , changing its momentum, and preventing the mass from falling to the ground. Simultaneously, the leg that was the stance leg during the time period #Math_18#, exerts another impulse on the mass m , changing its momentum further, and giving the mass enough energy to take its next step. This last impulse corresponds to the human pushing off the ground with the foot of the leg that is changing from being the stance leg to being the swing leg. The impulse each leg exerts on the mass must be parallel to that leg and pointing from the mass away from its foot. This discussion leads to the next assumption.

Assumption 5. Let r m ( t ¯ ) , r l ( t ¯ ) and r r ( t ¯ ) denote the position of the mass m and left and right foot at time t ¯ respectively. At times t ¯ = t ¯ k , the mass m feels an impulse of the form J ( t ¯ k ) = α k ( r m ( t ¯ k ) − r l ( t ¯ k ) ) + β k ( r m ( t ¯ k ) − r r ( t ¯ k ) ) for some α k ≥ 0 and β k ≥ 0 , with α k + 2 = α k and β k + 2 = β k for all k . By symmetry, we also have α k + 1 = β k .

Observation 2. Since the legs and feet are mass-less, the motion of the swing leg does not affect the motion of the mass m . It only matters the position where its foot lands.

Assumption 6. When both feet are touching the ground, at t ¯ = t ¯ k , the mass m is in the plane that contains both feet and is perpendicular to the ground.

One could wonder if, when both feet are touching the ground, the mass m could be out of the plane perpendicular to the ground containing the feet. An initial guess could lead us to believe that the mass being slightly ahead of this plane could facilitate the motion and make walking forward more efficient. However, in the context of this simple model, we have proved that there is no periodic walking if the mass is out of this plane. While elementary, this proof is somewhat tedious and lengthy, so we have elected to leave it out of this article.

In

Recall that T ¯ s is the time of one step and r m = r m ( t ¯ ) is the position of the mass m at time t ¯ . The symmetry and periodicity of the walk leads to the next assumption.

Assumption 7. r m ( t ¯ + 2 T ¯ s ) = 2 u e f + r m ( t ¯ ) for all t ¯ . The components of the velocity of m in the walking and vertical directions, e f and e v , are periodic with period T ¯ s and the component of the velocity of m in the lateral direction, e h , is anti-periodic with anti-period T ¯ s .

We now proceed to describe the motion of the mass during the time interval ( 0 , t ¯ 1 ) . Recall that t ¯ 1 = T ¯ s . During this period, the left leg is the stance leg and the right foot does not make contact with the ground. The left leg and the mass form an inverted pendulum. According to Observation 1 and Newton’s third law

m r ″ m = F + F g , (1)

where primes denote derivatives with respect to t ¯ .

Let r ^ be the dimensionless unit vector pointing from the left foot to the mass, r ^ = ( r m − r l ) / L . We define F = ‖ F ‖ / ( m g ) , where g is the acceleration of gravity and we use the notation ‖ a ‖ for the norm of any vector a , i.e. ‖ a ‖ = a 1 2 + a 2 2 + a 3 2 . We introduce the dimensionless time t = g / L t ¯ . Equation (1) becomes

r ^ ¨ = F r ^ − e v , (2)

where dots denote derivatives with respect to t .

Let θ be the polar angle and ϕ be the azimuthal angle,

r ^ = sin θ cos ϕ e h + sin θ sin ϕ e f + cos θ e v (3)

where − π / 2 < ϕ < π / 2 and 0 < θ < π / 2 (see

θ ¨ − ϕ ˙ 2 sin θ cos θ = sin θ (4)

2 θ ˙ ϕ ˙ cos θ + ϕ ¨ sin θ = 0 (5)

− θ ˙ 2 − ϕ ˙ 2 sin 2 θ = F − cos θ . (6)

We will solve these equations during the first step, i.e. in the time interval ( 0 , t 1 ) , where t 1 = T ¯ s g / L . Equation (6) determines F = F ( t ) . The dynamics of the mass m is given by Equations (4) and (5). Multiply Equation (5) by sin θ and integrate once to obtain

ϕ ˙ sin 2 θ = K (7)

for some constant K . Use Equation (7) to eliminate ϕ ˙ from Equation (4) to get θ ¨ − K 2 cos θ sin 3 θ = sin θ . This equation is integrated once after is multiplied by θ ˙ to get

1 2 θ ˙ 2 + K 2 2 sin 2 θ + cos θ = E , (8)

for some E . Note that E is the dimensionless mechanical energy.

The initial azimuthal angle, ϕ 0 = ϕ ( 0 ) , and polar angle, θ 0 = θ ( 0 ) , are related to the parameters displayed in

ϕ w = arctan ( w u ) , (9)

then

ϕ 0 = − π 2 + ϕ w and θ 0 = arcsin ( u 2 + w 2 2 L ) . (10)

Assume the parameters w , u and L are given. For each pair #Math_106# and K , the system of Equations (7) and (8) subjected to the initial conditions (9) and (10) has a unique solution. However, this solution may not correspond to periodic walking as described in this article. In the next section, we list the two necessary and sufficient conditions for periodic walking. These conditions will lead to a relationship between E and K .

In this section we assume that the parameters w , u and L are given and fixed. Thus, θ 0 and ϕ 0 are also given through Equations (9) and (10). We seek to answer the following question: Given that θ ( 0 ) = θ 0 and ϕ ( 0 ) = ϕ 0 , what are the necessary and sufficient conditions for periodic walking?

Condition 1. If θ ( t ) and ϕ ( t ) , solutions of Equations (7) and (8) subjected to the initial conditions (9) and (10), correspond to periodic walking, there exists t 1 > 0 such that θ ( t 1 ) = θ 0 and ϕ ( t 1 ) = π / 2 − ϕ w .

Note that t 1 is the dimensionless time of one step. θ ( t 1 ) = θ 0 simply means that the height of the mass m at the start and the end of a step is the same. The need for ϕ ( t 1 ) = π / 2 − ϕ w is illustrated in

Given any function g = g ( x ) of one variable, we use the standard notations g ( a + ) and g ( a − ) for the limits of g ( x ) as x tends to a from the right (with values of x > a ) and from the left (with values of x < a ). Given two vectors a and b we denote by a ⋅ b = a 1 b 1 + a 2 b 2 + a 3 b 3 their dot product.

Let n ^ l be the unit vector from the left foot to the mass at time t = t 1 and let n ^ r be the unit vector from the right foot to the mass at time t = t 1 . Given Assumption 7, r ^ ˙ ( t 1 + ) = r ^ ˙ ( 0 + ) − 2 ( e h ⋅ r ^ ˙ ( 0 + ) ) e h if r ^ ( t ) correspond to periodic walking. This fact, plus Assumption 5 leads to the second condition for periodic walking.

Condition 2. If θ ( t ) and ϕ ( t ) , solutions of Equations (7) and (8) subjected to the initial conditions (9) and (10), correspond to periodic walking, there exists

α > 0 and β > 0 such that r ^ ˙ ( 0 + ) − 2 ( e h ⋅ r ^ ˙ ( 0 + ) ) e h = r ^ ˙ ( t 1 − ) + α n ^ l + β n ^ r .

In the above condition t 1 is as defined in Condition 1, and r ^ ( t ) is related to θ ( t ) and ϕ ( t ) by Equation (3).

Observation 3. If Conditions 1 and 2 are necessary and sufficient for periodic walking.

Observation 4. If Condition 1 is satisfied, so is Condition 2.

The proof of this observation is simple. One shows that, if Condition 1 is satisfied, r ^ ˙ ( 0 + ) − 2 ( e h ⋅ r ^ ˙ ( 0 + ) ) e h − r ^ ˙ ( t 1 − ) = c 1 e v and n ^ l + n ^ r = c 2 e v for some positive constants c 1 and c 2 , from where we conclude that Condition 2 is satisfied with α = β = c 1 / c 2 . The corresponding calculations are elementary but lengthy so we do not present them here.

The azimuthal angle θ initially decreases and it attains its minimum at the time t = t ⋆ such that θ ˙ ( t ⋆ ) = 0 . The value of this minimum azimuthal angle θ ⋆ = θ ( t ⋆ ) , can be obtained in terms of the constants of motion E and K from Equation (8) by setting θ ˙ = 0

K 2 2 sin 2 θ ⋆ + cos θ ⋆ = E . (11)

Note ϕ ⋆ = ϕ ( t ⋆ ) can be computed as follows

ϕ ⋆ − ϕ 0 = ∫ θ 0 θ ⋆ d ϕ d θ ( θ ) d θ = − ∫ θ ⋆ θ 0 d ϕ d θ ( θ ) d θ . (12)

Due to symmetry, the Condition 1 is equivalent to: The minimum azimuthal angle θ ⋆ is attained at the same time that the polar angle ϕ is zero, i.e.

ϕ ⋆ = 0 . Thus, since d ϕ d θ = ϕ ˙ / θ ˙ , setting ϕ ⋆ = 0 in Equation (12), using the fact

that ϕ 0 = − π / 2 + ϕ w , using Equations (7) and (8) and simple manipulations, we get that Condition 1 is satisfied if and only if

π 2 − ϕ w = ∫ θ ⋆ θ 0 K sin θ 2 ( E − cos θ ) sin 2 θ − K 2 d θ . (13)

On one hand, given θ 0 , Equation (8) implies that E > cos θ 0 . On the other hand, we recall that the force the stance leg exerts on the mass m points from the mass away from its foot. This means that F in Equation (6) is constrained to be positive. Thus, while Equation (6) is not used to solve for θ and ϕ , it does impose a constrain. Using Equations (7) and (8) and simply algebra we get that F = 3 cos θ − 2 E . Thus, the constrain F ≥ 0 reduces to E ≤ ( 3 / 2 ) cos θ . Since cos θ is a decreasing function of θ for 0 ≤ θ < π / 2 , this constrain needs to be verified only for θ = θ 0 . In summary, if the pair ( E , K ) corresponds to periodic walking, we necessarily have cos θ 0 < E ≤ ( 3 / 2 ) cos θ 0 . Note that the constrain E ≤ ( 3 / 2 ) cos θ 0 gives a limit on the speed of the biped. If the biped were to try to walk faster, the foot of its stance leg would lose contact with the ground.

Since ϕ ˙ > 0 , we have that K > 0 (see Equation (7)). On the other hand, given θ 0 and E , Equation (8) implies that the largest value of K possible is the one that makes θ ˙ ( 0 + ) = 0 . Simple algebraic manipulations leads to the following constrain on K once E and θ 0 are given, 0 < K ≤ sin θ 0 2 ( E − cos θ 0 ) .

We summarize the findings in this section, and add to that, in the following observation.

Observation 5. (1) Let θ 0 be given, the pairs ( E , K ) that correspond to periodic walking are the solutions of Equations (11) and (13).

(2) If the pair ( E , K ) corresponds to periodic walking, then cos θ 0 < E ≤ ( 3 / 2 ) cos θ 0 and 0 < K ≤ sin θ 0 2 ( E − cos θ 0 ) .

(3) Let θ 0 be given. For each 1 < E < ( 3 / 2 ) cos θ 0 , there exist a K such that 0 < K ≤ sin θ 0 2 ( E − cos θ 0 ) , and the pair ( E , K ) satisfies Equations (11) and (13) (we prove this point in Appendix 1).

(4) Let θ 0 be given. Numerical simulations strongly suggest that, for each 1 < E < ( 3 / 2 ) cos θ 0 , there exist a unique K such that 0 < K ≤ sin θ 0 2 ( E − cos θ 0 ) , and the pair ( E , K ) satisfies Equations (11) and (13) (i.e. the K from point (3) is unique).

(5) While we were not able to prove it, we have evidence to believe that, if E ≤ 1 and ϕ w < π / 4 , there is no periodic walking. Note that ϕ w ≥ π / 4 means that the feet are wider apart than the length of the steps, not a situation we will be considering. Thus, in the rest of this article, we will restrict our attention to the parameter regime ϕ w < π / 4 and assume that 1 < E < ( 3 / 2 ) cos θ 0

In what follows we will study aspects of the periodic walking of the model biped. The parameters that determine the motion are ϕ w , θ 0 and E . The angles ϕ w and θ 0 determine the geometry of the biped, i.e. the width and length of the steps relative to the length of the legs. The parameter E is a dimensionless mechanical energy.

Let T be the dimensionless time of one step. By symmetry, θ ( T / 2 ) = θ ⋆ , the minimum azimuthal angle (see Equation (11)). Thus, T / 2 = ∫ θ 0 θ ⋆ 1 / θ ˙ d θ . Using Equation (8) and simple manipulations we get

T = 2 ∫ θ ⋆ θ 0 sin θ 2 ( E − cos θ ) sin 2 θ − K 2 d θ . (14)

Let V be the dimensionless average velocity. From

V = 2 sin θ 0 cos ϕ w T . (15)

For the parameter values θ 0 = ϕ w = π / 6 ,

Let A h be the dimensionless amplitude of the lateral oscillations. Note that A h = e h ⋅ ( r ^ ( 0 + ) − r ^ ( T / 2 ) ) . Thus, from Equation (3) and the facts that ϕ 0 = − π / 2 + ϕ w and ϕ = 0 when θ = θ ⋆ , we have

A h = sin ( θ 0 ) sin ( ϕ w ) − sin ( θ ⋆ ) . (16)

Let A v be the dimensionless amplitude of the vertical oscillations. The mass is at its lowest when θ = θ 0 and its highest when θ = θ ⋆ . Thus, A v = e v ⋅ ( r ^ ( T / 2 ) − r ^ ( 0 + ) ) / 2 , which is

A v = cos θ ⋆ − cos θ 0 2 . (17)

With the parameter values θ 0 = ϕ w = π / 6 ,

In

We remind the reader that the dimensionless energy satisfies the constrains 1 < E < ( 3 / 2 ) cos θ 0 . In this section, we will study the dynamics of slow walkers. As our analysis will show, this corresponds to values of the energy of the form

E = 1 + ε 2 , where 0 < ε ≪ 1. (18)

The choice of the square in ε is to simplify future calculations. The fact that ε ≪ 1 will allow us to use asymptotic approximations and as thus, obtain a deeper understanding of the dynamics of the model biped than by numerical simulations alone.

In the Appendix 2 we show that the asymptotic value of the minimum polar angle θ ⋆ (see Equation (11)) valid in the parameter regime of Equation (18) is

θ ⋆ ≈ 2 sin ϕ w cos ( 2 ϕ w ) ε . (19)

In the Appendix 3 we show that the asymptotic value of the dimensionless time of one step, T (see Equation (14)), is

T ≈ − ln ε . (20)

In the above equation, we mean that lim ε → 0 − T / ln ε = 1 . Thus, the dimensionless average velocity V (see Equation (15)) satisfies, in this parameter regime

V ≈ − 2 sin θ 0 cos ϕ w ln ε . (21)

Note that T → ∞ and V → 0 as ε → 0 , consistent with the title of this section: slow walkers. Note also that the asymptotic formula for V is consistent with the plot in

Given Equation (19), we can obtain the asymptotic value of the amplitude of the lateral oscillations from Equation (16):

A h ≈ sin ( θ 0 ) sin ( ϕ w ) − 2 sin ϕ w cos ( 2 ϕ w ) ε . (22)

Compare

In this section, we explore a different parameter regime. Namely, we restrict our attention to small values of the initial azimuthal angle

θ 0 ≪ 1. (23)

This corresponds to the biped taking steps that are much shorter than the lengths of its legs, a realistic parameter regime.

In the Appendix 4, we outline the steps required to get the asymptotic value of the minimum azimuthal angle θ ⋆ and of K in the regime θ 0 ≪ 1 . We obtain

θ ⋆ ≈ θ 0 sin ϕ w − ( 7 − 4 E ) 12 ( E − 1 ) θ 0 3 sin ϕ w cos 2 ϕ w (24)

and

K ≈ θ 0 2 ( E − 1 ) sin ϕ w . (25)

The asymptotic value of the dimensionless time required by the biped to take one step, T , is obtained from Equations (14), (24) and (25) and simple manipulations

T ≈ 2 E − 1 ∫ θ 0 sin ϕ w θ 0 θ d θ θ 2 − θ 0 2 sin 2 ϕ w = 2 E − 1 θ 0 cos ϕ w . (26)

Thus, V , the dimensionless average velocity of the biped satisfies

V ≈ 2 ( E − 1 ) (27)

(see Equation (15)).

Making use of Equations (24) and (16) we get the asymptotic value of the amplitude of the lateral oscillations

A h ≈ ( 3 − 2 E ) 4 ( E − 1 ) θ 0 3 sin ϕ w cos 2 ϕ w . (28)

Note that A h becomes small fast. It is of the order of θ 0 3 for small values of θ 0 .

On the other hand, the amplitude of vertical oscillations goes to zero quadratically as θ 0 → 0 , since simple calculations lead to

A v ≈ cos 2 ϕ w 4 θ 0 2 . (29)

Explicit Time Dependence of the Azimuthal Angle θ and Polar Angle ϕ When θ 0 ≪ 1As previously defined, T is the dimensionless time required by the biped to take one step. Note that θ is a decreasing function of t in the time interval 0 < t < T / 2 . Thus, using Equation (8), the asymptotic value of K (Equation (25)), the asymptotic approximations sin θ ≈ θ and cos θ ≈ 1 when θ ≪ 1 , and simple algebra we get

d θ d t ≈ − 2 ( E − 1 ) θ θ 2 − θ 0 2 sin 2 ϕ w . (30)

Using the initial conditions θ ( 0 + ) = θ 0 , this simple separable first order equation can be integrated analytically to get

#Math_334# (31)

On the other hand, Equation (7) leads to

d ϕ d t ≈ θ 0 2 ( E − 1 ) sin ϕ w θ 2 . (32)

Note that θ is given by Equation (31). Plugging that expression for θ as a function of t into Equation (32), integrating, and using the initial condition ϕ ( 0 + ) = ϕ 0 = − π / 2 + ϕ w , leads to the following formula

ϕ ( t ) ≈ { − arctan ( cos ϕ w sin ϕ w − 2 ( E − 1 ) θ 0 sin ϕ w t ) if 0 ≤ t ≤ T 2 − ϕ ( T − t ) if T 2 ≤ t ≤ T . (33)

In this article, we use a very simple inverted pendulum model to explore aspects of the mechanics of biped walkers. The novelty of this article is that we do not restrict the motion of the mass of the pendulum to the sagittal plane. As a consequence, we were able to study the lateral oscillations of the center of mass as the biped walks. These oscillations were beyond the capability of the simplest inverted pendulum models when the motion of the mass was restricted to the sagittal plane.

We performed numerical simulations and explore different parameter regimes with the use of asymptotic techniques. Our analysis shows that the inverted pendulum model remains simple enough to study, even when the mass is not restricted to move in the sagittal plane. We believe and hope the approach introduced in this paper will prove useful and be adopted by other researchers to study different aspects of the dynamic of biped walkers.

Goldsztein, G.H. (2017) Modeling Walking with an Inverted Pendulum Not Constrained to the Sagittal Plane. Numerical Simulations and Asymptotic Expansions. Applied Mathematics, 8, 57-76. http://dx.doi.org/10.4236/am.2017.81006

1. Proof of Point (3) in Observation 5

Let 0 < θ 0 < π / 2 fixed. For any parameters E and K satisfying 2 ( E − cos θ 0 ) sin 2 θ 0 > K > 0 we define α = α ( E , K ) , to be the unique solution of

K 2 2 sin 2 α + cos α = E (34)

that satisfies 0 < α < θ 0 (that such a unique α exists is an easy calculus exercise). Taking partial derivatives of the above equation with respect to K ,

and solving for ∂ α ∂ K , we obtain

∂ α ∂ K = K sin α K 2 cos α + sin 4 α > 0 , (35)

since 0 < α < π / 2 . Thus, given E fixed, α is a continuous and increasing function of K in the interval 0 < K ≤ sin θ 0 2 ( E − cos θ 0 ) . It can be easily shown that

lim K → 0 + α = { 0 if E ≥ 1 arccos ( E ) if 0 < E < 1 (36)

It also easily follows from the definition of α that

lim K → sin θ 0 2 ( E − cos θ 0 ) − α ( E , K ) = θ 0 . (37)

Our findings regarding α are summarized in the following observation.

Observation 6. Fix E > cos θ 0 . Regard α = α ( E , K ) , defined in Equation (34), as a function of K on the interval ( 0 , sin θ 0 2 ( E − cos θ 0 ) ) . Then, α

is a continuous and increasing function of K that satisfies α ( E , 0 + ) = 0 if E > 1 ; α ( E , 0 + ) = arccos ( E ) if 0 < E ≤ 1 ; and Equation (37).

Consider E a fixed parameter. We define

f ( θ ) = 2 ( E − cos θ ) sin 2 θ . (38)

Assume θ 0 is fixed. Assume 0 < α < θ 0 < π / 2 , where α was defined in Equation (34). As always, also assume E > cos θ 0 . We define

I = I ( K , E ) = ∫ α θ 0 K sin θ f ( θ ) − K 2 d θ . (39)

Since α depends on E and K , and the function f depends of E , the integral I is a function of the two parameters E and K . Note that, given Equation (34),

K = f ( α ) . (40)

Let K 0 = f ( θ 0 ) . Let ε = K 0 2 − K 2 . Assume 0 < ε ≪ 1 . This implies that K 0 − K ≪ 1 , and thus, we also have that 0 < θ 0 − α ≪ 1 . Then, we have K 0 2 − ε = K 2 = f ( α ) ≈ f ( θ 0 ) + f ′ ( θ 0 ) ( α − θ 0 ) = K 0 2 + f ′ ( θ 0 ) ( α − θ 0 ) . As a con- sequence, we have the approximation

α ≈ θ 0 − ε f ′ ( θ 0 ) . (41)

Since K 2 = K 0 2 − ε and K 0 2 = f ( θ 0 ) , we have that K 2 = f ( θ 0 ) − ε , and thus, the argument in the square root in the integral of Equation (39) is f ( θ ) − K 2 = f ( θ ) − f ( θ 0 ) + ε . For θ ∈ ( α , θ 0 ) we have 0 < θ 0 − θ ≪ 1 since we are assuming 0 < θ 0 − α ≪ 1 . Thus, f ( θ ) − f ( θ 0 ) + ε ≈ f ′ ( θ 0 ) ( θ − θ 0 ) + ε , or equivalently

f ( θ ) − K 2 ≈ f ′ ( θ 0 ) ( θ − θ 0 ) + ε . (42)

Using Equations (41) and (42), plus K ≈ K 0 , and θ ≈ θ 0 , and simple algebra, we have the following asymptotic approximation (in the regime ε ≪ 1 )

I ( K , E ) ≈ ∫ θ 0 − ε f ′ ( θ 0 ) θ 0 K 0 sin θ 0 f ′ ( θ 0 ) ( θ − θ 0 ) + ε d θ (43)

Make now the change of variable z = − f ′ ( θ 0 ) ( θ − θ 0 ) / ε (note that f ′ ( θ 0 ) > 0 ) and some simple algebra to get

I ( K , E ) ≈ K 0 ε sin θ 0 f ′ ( θ 0 ) ∫ 0 1 d z 1 − z d z = 2 K 0 ε sin θ 0 f ′ ( θ 0 ) . (44)

The limit K → sin θ 0 2 ( E − cos θ 0 ) from below corresponds to ε → 0 + . We summarize our findings in the following observation.

Observation 7.

lim K → sin θ 0 2 ( E − cos θ 0 ) − I ( K , E ) = 0 , (45)

Assume now that cos θ 0 > 2 / 3 and E is such that 1 < E < ( 3 / 2 ) cos θ 0 . Both E and θ 0 are fixed. Assume K ≪ 1 . Given Observation 6, we have that α ≪ 1 .

Let θ 1 be such that α ≪ θ 1 ≪ 1 and θ 1 ≤ θ 0 . We split the integral I (Equation (39)) in two, I = I 1 + I 2 , where

#Math_425# (46)

Since θ ≪ 1 for all θ ∈ ( α , θ 1 ) , we have f ( θ ) ≈ 2 ( E − 1 ) θ 2 and sin θ ≈ θ in the integrand of I 1 (see Equation (38) for the definition of f ). Since α ≪ 1 and K = f ( α ) (see Equation (40)), it can also be very easily seen that α ≈ K / 2 ( E − 1 ) . Thus, we have, after simple manipulation, the following asymptotic approximation for I 1 ,

I 1 ≈ ∫ K 2 ( E − 1 ) θ 1 d θ θ 2 ( E − 1 ) θ 2 K 2 − 1

Making the change of variables x 2 = 2 ( E − 1 ) θ 2 / K 2 − 1 , and using that α ≪ θ 1 , we get

I 1 ≈ ∫ 0 ∞ d x x 2 + 1 = π 2 .

Note that I 2 can be bounded as follows

I 2 ≤ ∫ θ 1 θ 0 K sin θ 1 f ( θ 1 ) − K 2 d θ = K θ 0 sin θ 1 f ( θ 1 ) − K 2

which tends to zero as K → 0 + , since θ 1 can be selected independent of K . Thus, we have proved the following observation

Observation 8.

lim K → 0 + I ( K , E ) = π 2 if 1 < E < 3 2 cos θ 0 . (47)

Observations 7 and 8 show that, if cos θ 0 > 2 / 3 , for each E such that 1 < E < ( 3 / 2 ) cos θ 0 , there exists at least one K such that 0 < K < 2 ( E − cos θ 0 ) , for which I ( K , E ) = π / 2 − ϕ w (this is the same as Equation (13)). Those parameters ( E , K ) give us a periodic walking and the corresponding α equals θ ⋆ . This proves Point (3) in Observation 5.

2. Calculations leading to Equation (19)

Consider E a fixed parameter. Let f ( θ ) be as defined in Equation (38). Note that Equation (11) implies that K = f ( θ ⋆ ) . Thus, Equation (13) implies that θ ⋆ is the solution of

π 2 − ϕ w = I , where I = ∫ θ ⋆ θ 0 f ( θ ⋆ ) sin θ f ( θ ) − f ( θ ⋆ ) d θ . (48)

Recall that we are in the parameter regime of Equation (18), i.e. E = 1 + ε 2 where ε ≪ 1 . Assume θ ⋆ ≈ C ε for some constant C . This assumption will be verified later. Let θ 1 be such that ε ≪ θ 1 ≪ 1 and θ 1 ≤ θ 0 . We split the integral I (Equation (48)) in two, I = I 1 + I 2 , where

I 1 = ∫ θ ⋆ θ 1 f ( θ ⋆ ) sin θ f ( θ ) − f ( θ ⋆ ) d θ and I 2 = ∫ θ 1 θ 0 f ( θ ⋆ ) sin θ f ( θ ) − f ( θ ⋆ ) d θ . (49)

We first note that I 2 can be bounded as follows

| I 2 | ≤ θ 0 f ( θ ⋆ ) sin θ 1 f ( θ 1 ) − f ( θ ⋆ ) → 0 as ε → 0 (50)

because θ 1 can be chosen to be independent of ε and f ( θ ⋆ ) → 0 as ε → 0 .

Next, we compute the asymptotic value of I 1 . Since θ 1 ≪ 1 , for θ ⋆ ≤ θ ≤ θ 1 we have f ( θ ) ≈ ( 2 ε 2 + θ 2 ) θ 2 and sin θ ≈ θ . As a consequence,

f ( θ ) − f ( θ ⋆ ) ≈ ( 2 ε 2 + θ 2 ) θ 2 − ( 2 ε 2 + ( θ ⋆ ) 2 ) ( θ ⋆ ) 2 = ( θ 2 − ( θ ⋆ ) 2 ) ( θ 2 + 2 ε 2 + ( θ ⋆ ) 2 )

Thus, we have the following approximation for I 1

I 1 ≈ θ ⋆ 2 ε 2 + ( θ ⋆ ) 2 ∫ θ ⋆ θ 1 d θ θ θ 2 − ( θ ⋆ ) 2 θ 2 + 2 ε 2 + ( θ ⋆ ) 2 . (51)

Making the change of variable 2 ( ε 2 + ( θ ⋆ ) 2 ) z 2 = θ 2 − ( θ ⋆ ) 2 , some algebraic manipulation, and using that θ ⋆ ≈ C ε for some constant C and ε ≪ θ 1 , we get

I 1 ≈ a 2 2 ε 2 + ( θ ⋆ ) 2 θ ⋆ ∫ 0 ∞ d z ( z 2 + a 2 ) z 2 + 1 , where a 2 = ( θ ⋆ ) 2 2 ( ε 2 + ( θ ⋆ ) 2 ) . (52)

This last integral can be computed analytically to get

I 1 ≈ arctan 2 ε 2 ( θ ⋆ ) 2 + 1 . (53)

Given Equation (50), we have I ≈ I 1 . Thus, Equations (48) and (53) lead to

π 2 − ϕ w ≈ arctan 2 ε 2 ( θ ⋆ ) 2 + 1 . (54)

This equation can be easily solved to give Equation (19).

3. Derivation of Equation (20)

We now proceed to compute the asymptotic value of the dimensionless time of one step, i.e. T , given by Equation (14), in the parameter regime E = 1 + ε 2 , with ε ≪ 1 .

Let θ 1 be such that ε ≪ θ 1 ≪ 1 and θ 1 ≤ θ 0 . We split the integral in Equation (14) in two, to get T = 2 J 1 + 2 J 2 , where

J 1 = ∫ θ ⋆ θ 1 sin θ f ( θ ) − f ( θ ⋆ ) d θ and J 2 = ∫ θ 1 θ 0 sin θ f ( θ ) − f ( θ ⋆ ) d θ , (55)

where f ( θ ) is a defined in Equation (38), and we remind the reader that K 2 = f ( θ ⋆ ) .

We first note that J 2 remains bounded as ε → 0 . Note that J 2 does not go to 0, but it does remain bounded independently of ε . This results from very simple facts so we skip the details.

Next, we compute the asymptotic value of J 1 . Since θ 1 ≪ 1 , for θ ⋆ ≤ θ ≤ θ 1 we have f ( θ ) ≈ ( 2 ε 2 + θ 2 ) θ 2 and sin θ ≈ θ . As a consequence,

f ( θ ) − f ( θ ⋆ ) ≈ ( 2 ε 2 + θ 2 ) θ 2 − ( 2 ε 2 + ( θ ⋆ ) 2 ) ( θ ⋆ ) 2 = ( θ 2 − ( θ ⋆ ) 2 ) ( θ 2 + 2 ε 2 + ( θ ⋆ ) 2 )

Thus, we have the following approximation for J 1

J 1 = ∫ θ ⋆ θ 1 θ θ 2 − ( θ ⋆ ) 2 θ 2 + 2 ε 2 + ( θ ⋆ ) 2 d θ . (56)

Making the change of variable z 2 = θ 2 − ( θ ⋆ ) 2 , some algebraic manipulation, and using that θ ⋆ ≈ C ε for some constant C and ε ≪ θ 1 , we get

J 1 ≈ ∫ 0 θ 1 2 − ( θ ⋆ ) 2 d z z 2 + 2 ( ε 2 + ( θ ⋆ ) 2 ) ≈ − ln 2 ( ε 2 + ( θ ⋆ ) 2 ) ≈ − ln ε . (57)

In the above equation we have used the facts that ε ≪ θ 1 ≪ 1 and Equation (19). This shows the validity of Equation (20).

4. Derivation of Equations (24) and (25)

Let f ( θ ) = 2 ( E − cos θ ) sin 2 θ . Note that K 2 = f ( θ ⋆ ) , and after some manipulation Equation (13) becomes

π 2 − ϕ w = ∫ θ ⋆ θ 0 K sin θ f ( θ ) − K 2 d θ . (58)

Let ε = θ 0 . Let a = K / ε and b = f ( ε ) / ε . Make the change of variables ε 2 ( x 2 + a 2 ) = f ( θ ) . Some manipulations show

K sin θ f ( θ ) − K 2 d θ = 2 ε 2 a sin θ f ′ ( θ ) d x . (59)

Note that

sin θ f ′ ( θ ) = 2 sin 4 θ + 4 ( E − cos θ ) sin 2 θ cos θ = 2 sin 4 θ + 2 cos θ f ( θ )

from where we get

sin θ f ′ ( θ ) = 2 sin 4 θ + 2 ε 2 ( x 2 + a 2 ) cos θ . (60)

Next, we observe that f ( θ ) ≈ 2 ( E − 1 ) θ 2 + O ( ε 4 ) , and thus, since ε 2 ( x 2 + a 2 ) = f ( θ ) , we have

θ ≈ ε x 2 + a 2 2 ( E − 1 ) + O ( ε 3 ) . (61)

From the last two equations we conclude that

sin θ f ′ ( θ ) ≈ 2 ε 4 ( x 2 + a 2 2 ( E − 1 ) ) 2 + 2 ( 1 − ε 2 ( x 2 + a 2 ) 4 ( E − 1 ) ) ε 2 ( x 2 + a 2 ) + O ( ε 6 ) .

Further calculation and expanding in powers of ε lead to

2 ε 2 a sin θ f ′ ( θ ) = a x 2 + a 2 − ( 2 − E ) 4 ( E − 1 ) 2 a ε 2 + O ( ε 4 ) . (62)

Note that x = 0 when θ = θ ⋆ , and that x = b 2 − a 2 when θ = θ 0 = ε . Thus, after the change of variables, Equation (58) becomes

#Math_540# (63)

Further elementary operations and expansions in powers of ε leads to Equations (24) and (25).