6.5 Method of Lagrange Multiplier
7.1.1 Motivation
We have seen in the last that a conservative mechanical system:
m¨x = −∇V (x)
has the first integral H = 12m ˙x2+V (x), which is invariant for all t. It can be derived by multiplying the Newton law of motion by ˙x, we get
0 = m¨x ˙x + ∇V (x) ˙x = d dt
1
2m ˙x2+ V (x)
. From this, we get
1
2m| ˙x|2+ V (x) = E for some constant E. When x is a scalar, we can obtain
˙ x = ±
r2
m(E − V (x)) Then x(t) can be obtained by method of separation of variable:
Z dx
q2
m(E − V (x))
= ± Z
dt.
155
In the derivation above, the quantity H = 12m| ˙x|2+ V (x) plays a key role. We can express H in a more symmetric way. Define p = m ˙x, called momentum. Express
H(x, p) = p2
2m+ V (x).
Then Newton’s mechanics is equivalent to
x˙ = Hp(x, p)
˙
p = −Hx(x, p). (7.1)
An advantage to express the Newton mechanics in this form is that it is easier to find invariants of the flow.
Definition 7.1. A quantity f (x, p) is called an invariant of the Hamiltonian flow (7.1) if d
dtf (x(t), p(t)) = 0.
From chain rule, we see that f is invariant under the Hamiltonian flow (7.1) if and only if d
dtf (x(t), p(t)) = fxHp− fpHx = 0.
Theorem 7.1. If a Hamiltonian H is independent of t, then H is invariant under the corresponding Hamiltonian flows (7.1).
Remarks
• General Lagrangian mechanics:
δx Z t1
t0
L(x1(t), ..., xn(t), ˙x1(t), ..., ˙xn(t)) dt = 0 can also be written in the form of Hamiltonian system
˙
xi = Hpi(x1, ..., xn, p1, ..., pn),
˙
pi = −Hxi(x1, ..., xn, p1, ..., pn).
This will be derived in later section.
• If H is invariant under certain group action, then there are corresponding invariants of the Hamiltonian flow. For instance, if the flow is in two dimensions, say H(x1, x2, p1, p2). Sup-pose H is invariant under x17→ x1+ c for any c, that is,
H(x1+ c, x2, p1, p2) = H(x1, x2, p1, p2), for any c ∈ R.
An immediate consequence is that
˙
p1 = −Hx1 = 0.
Thus, p1is an invariant of this Hamiltonian flow. We can eliminate it right away. The system becomes smaller! We shall come back to this point at the end of this chapter.
7.1. HAMILTONIAN SYSTEMS 157 7.1.2 Trajectories on Phase Plane
A conservative quantity of a time-independent Hamiltonian flow
x = H˙ y(x, y)
˙
y = −Hx(x, y) (7.2)
is the Hamiltonian H itself. That is, along any trajectory (x(t), y(t)) of (7.2), we have d
dtH(x(t), y(t)) = Hxx + H˙ yy = H˙ xHy + Hy(−Hx) = 0.
In two dimensions, the trajectories of a Hamiltonian system in the phase plane are the level sets of its Hamiltonian.
Example Some linear and nonlinear oscillators are governed by a restoration potential V . The equation of motion in Newton’s mechanics is
m¨x = −V0(x)
where V is a restoration potential. Define the momentum y = mv and the total energy H(x, y) = y2
2m+ V (x), 1. Harmonic oscillator: H(x, y) = 12y2+k2x2.
2. Duffing oscillator: H(x, y) = 12y2−δ2x2+x44. 3. Cubic potential: H(x, y) = 12 y2− x2+ x3.
4. Simple pendulum: H(x, y) = 12y2−glcos x.
You can plot the level sets of H to see the trajectories. In particular, you should pay more attentions on critical points, homoclinic and heteroclinic orbits.
Example Consider fluid flows in a two dimensional domain Ω. The flow is represented as a vector field V : Ω → R2, or in component form: V(x, y) = (u(x, y), v(x, y)). The flow is called a potential flow if it is incompressible and irrotational . That is
∇ · V = 0, ∇ × V = 0.
In component form, they are
ux+ vy = 0, incompressible vx− uy = 0, irrotational.
From divergence theorem, the first equation yields that there exists a function called stream function ψ(x, y) such that
u(x, y) = ψy(x, y), v(x, y) = −ψx(x, y).
Indeed, from this divergence free condition, we can define the stream function ψ(x, y) by the line integral:
ψ(x, y) = Z (x,y)
(−v(x, y)dx + u(x, y)dy).
1. The starting point of the line integral is not important. What is relevant is the derivatives of ψ. We can choose any point as our starting point. The corresponding ψ is defined up to a constant, which disappears after taking differentiation.
2. By the divergence theorem, the integral is independent of path in a simply connected domain.
Hence, ψ is well-defined on simply connected domain. You can check that ψy = u and ψx = −v. If the domain is not simply connected, the steam function may be a multiple valued function. We shall not study this case now.
The second equation for the velocity field yields that
∂2ψ
∂x2 +∂2ψ
∂y2 = 0.
This equation is called a potential equation.
The particle trajectory (which flows with fluid flow) is governed by
˙
x = u(x, y) = ψy(x, y)
˙
y = v(x, y) = −ψx(x, y).
This is a Hamiltonian flow with Hamiltonian ψ(x, y). The theory of potential flow can be analyzed by complex analysis. You can learn this from text books of complex variable or elementary fluid mechanics.
Here are two examples for the potential flow: let z = x + iy 1. ψ(z) = Im(z2) = 2xy,
2. ψ(z) = Im(z + 1/z) = y −x2+yy 2.
The first one represent a jet. The second is a flow passing a circle .
Example The magnetic field B satisfies divB = 0. For two-dimensional steady magnetic field B = (u, v), this reads
ux+ vy = 0.
The magnetic field lines are the curves which are tangent to B at every points on this line. That is, it satisfies
˙
x = u(x, y) = ψy(x, y)
˙
y = v(x, y) = −ψx(x, y)
where ψ is the stream function corresponding to the divergent free field B.
7.1. HAMILTONIAN SYSTEMS 159 Example Linear hamiltonian flow. If we consider
H(x, y) = ax2
2 + bxy + cy2 2 the corresponding Hamiltonian system is
x˙
˙ y
=
b c
−a −b
x y
(7.3)
7.1.3 Equilibria of a Hamiltonian system
In this subsection, we want to investigate the property of the equilibria of the Hamiltonian flows.
These equilibria are the critical points of the Hamiltonian H.
Definition 7.2. If ∇H(¯x, ¯y) = 0, then (¯x, ¯y) is called a critical point of H. Such a critical point is said to be non-degenerate if the hessian ofH at (¯x, ¯y) (i.e. the matrix d2H(¯x, ¯y)) is non-singular.
Since H is usually convex in y variable in mechanical problems, we may assume that Hyy > 0 at the equilibrium. Notice that this assumption eliminates the possibility of any local maximum of H.
To study the stability of an equilibrium (¯x, ¯y) of the Hamiltonian system (7.2), we linearize it around (¯x, ¯y) to get
˙
u = Au,
where A is the Jacobian of the linearized system of (7.2) at an equilibrium (¯x, ¯y) A =
Hyx Hyy
−Hxx −Hxy
(¯x,¯y)
.
Since the trace part T of A is zero, its eigenvalues are λi= ±1
2 q
Hyx2 − HxxHyy|(¯x,¯y), i = 1, 2.
We have the following possibilities.
• H has minimum at (¯x, ¯y). This is equivalent to HxxHyy− Hxy2 > 0 at (¯x, ¯y) because we already have Hyy > 0 from assumption. This is also equivalent to λi i = 1, 2 are pure imaginary. Thus, (¯x, ¯y) is a center.
• H has a saddle at (¯x, ¯y). This is equivalent to HxxHyy−Hxy2 < 0 at (¯x, ¯y). The corresponding two eigenvalues are real and with opposite signs. Hence the equilibrium is a saddle.
• H cannot have a local maximum at (¯x, ¯y) because the assumption Hyy> 0.
We summarize it by the following theorem.
Theorem 7.2. Assuming that (¯x, ¯y) is a non-degenerate critical point of a Hamiltonian H and assumingHyy(¯x, ¯y) > 0. Then
1. (¯x, ¯y) is a local minimum of H iff (¯x, ¯y) is a center of the corresponding Hamiltonian flow.
2. (¯x, ¯y) is a saddle of H iff (¯x, ¯y) is a saddle of the corresponding Hamiltonian flow..
The examples we have seen are
1. Simple pendulum: H(x, p) = 12p2−glcos x.
2. Duffing oscillator: H(x, p) = 12p2−δ2x2+x44.
3. Cubic potential: H(x, p) = 12 p2− x2+ x3.
In the case of simple pendulum, (2nπ, 0) are the centers, whereas (2(n + 1)π, 0) are the saddles.
In the case of Duffing oscillator, (±√
δ, 0) are the centers, while (0, 0) is the saddle. In the last example, the Hamiltonian system reads
x˙ = p
˙
p = x −32x2. (7.4)
The state (0, 0) is a saddle, whereas (3/2, 0) is a center.
Below, we use Maple to plot the contour curves the Hamiltonian. These contour curves are the orbits.
> with(DEtools):
> with(plots):
> E := yˆ2/2+xˆ3/3-delta*xˆ2/2;
E := 1 2y2+1
3x3−1 2δ x2
Plot the level set for the energy. Due to conservation of energy, these level sets are the orbits.
> contourplot(subs(delta=1,E),x=-2..2,y=-2..2,grid=[80,80],contours
> =[-0.3,-0.2,-0.1,0,0.1,0.2,0.3],scaling=CONSTRAINED,labels=[‘s‘,‘s’‘],
> title=‘delta=1‘);
7.2. GRADIENT FLOWS 161
delta=1
–2 –1 1 2
s’
–1.5 –1 –0.5 0.5 1 1.5
s
7.2 Gradient Flows
In many applications, we look for a strategy to find a minimum of some energy function or entropy function. This minimal energy state is called the ground state. One efficient way is to start from any state then follow the negative gradient direction of the energy function. Such a method is called the steepest descent method. The corresponding flow is called a (negative ) gradient flow. To be precise, let us consider an energy function ψ(x, y). We consider the ODE system:
x˙ = −ψx(x, y)
˙
y = −ψy(x, y). (7.5)
Along any of such a flow (x(t), y(t)), we have dψ
dt(x(t), y(t)) = ψxx + ψ˙ yy = −(ψ˙ x2+ ψy2) < 0, unless the flow reaches a minimum of ψ.
The gradient flow of ψ is always orthogonal to the Hamiltonian flow of ψ. For if
x˙ = ψy(x, y)
˙
y = −ψx(x, y)
ξ˙ = −ψx(ξ, η)
˙
η = −ψy(ξ, η) then
˙
x(t) · ˙ξ(t) + ˙y(t) · ˙η(t) = 0.
Thus, the two flows are orthogonal to each other. We have seen that ψ is an integral of the Hamilto-nian flow. Suppose φ is an integral of the gradient flow (7.5) (that is, the gradient flows are the level sets of φ), then the level sets of ψ and φ are orthogonal to each other.
Example 1. Let ψ = (x2− y2)/2. Then the gradient flow satisfies hand, the Hamiltonian flow is given by
x˙ = ψy = y
˙
y = −ψx= −x
Its solutions are given by x = A sin(t + t0), y = A cos(t + t0). The integral is ψ = (x2+ y2)/2.
In fact, 12ln(x2 + y2) is also an integral of the Hamiltonian flow. The complex valued function ψ + iφ = ln z.
Example 3. In general, the hamiltonian
ψ(x, y) = ax2
2 + bxy + cy2 2 the corresponding Hamiltonian system is
x˙
Find the corresponding integral φ of the gradient flow by yourself.
Example 4. Let
7.2. GRADIENT FLOWS 163 By the separation of variable
dy
y = dx
−x + x3, we get
ln y =
Z dx
−x + x3 = − ln |x| +1
2ln |1 − x| +1
2ln |1 + x| + C.
Hence, the solutions are given by
φ(x, y) := x2y2
1 − x2 = C1. Remarks.
• We notice that if ψ is an integral of an ODE system, so is the composition function h(ψ(x, y)) for any function h. This is because
d
dth(ψ(x(t), y(t)) = h0(ψ)d
dtψ(x(t), y(t)) = 0.
• If (0, 0) is the center of ψ, then (0, 0) is a sink of the corresponding gradient flow.
• If (0, 0) is a saddle of ψ, it is also a saddle of φ.
The properties of a gradient system are shown in the next theorem.
Theorem 7.3. Consider the gradient system
x˙ = −ψx(x, y)
˙
y = −ψy(x, y)
Assume that the critical points of ψ are isolated and non-degenerate. Then the system has the following properties.
• The equilibrium is either a souce, a sink, or a saddle. It is impossible to have spiral structure.
• If (¯x, ¯y) is an isolated minimum of ψ, then (¯x, ¯y) is a sink.
• If (¯x, ¯y) is an isolated maximum of ψ, then (¯x, ¯y) is a source.
• If (¯x, ¯y) is an isolated saddle of ψ, then (¯x, ¯y) is a saddle.
To show these, we see that the Jacobian of the linearized equation at (¯x, ¯y) is the Hessian of the function ψ at (¯x, ¯y): is
−
ψxx ψxy
ψxy ψyy
Its eigenvalues λi, i = 1, 2 are
−1 2
T ±p
T2− 4D ,
where T = ψxx+ ψyy, D = ψxxψyy− ψxy2 . From
T2− 4D = (ψxx− ψyy)2+ 4ψxy2 ≥ 0
we have that the imaginary part of the eigenvalues λi are 0. Hence the equilibrium can only be a sink, a source or a saddle.
Recall from Calculus that whether the critical point (¯x, ¯y) of ψ is a local maximum, a local min-imum, or a saddle, is completed determined by λ1, λ2 < 0, λ1, λ2 > 0, or λ1λ2 < 0, respectively.
On the other hand, whether the equilibrium (¯x, ¯y) of (7.5) is a source, a sink, or a saddle, is also completed determined by the same conditions.
Homeworks.
1. Consider a linear ODE
x˙
(b) Show that the system is a gradient system if and only if b = c, i,e. the matrix is sym-metric.
7.3 Simple pendulum
Motion on a given curve in a plane A curve (x(s), y(s)) in a plane can be parametrized by its arc length s. If the curve is prescribed as we have in the case of simple pendulum, then the motion is described by just a function s(t). By Newton’s law, the motion is governed by
m¨s = f (s),
where f (s) is the force in the tangential direction of the curve. For instance, suppose the curve is given by y = y(s), and suppose the force is the uniform garvitational force −mg(0, 1), then the force in the tangential direction is
f (s) = (dx ds,dy
ds) · [−mg(0, 1)] = −mgdy ds. Thus, the equation of motion is
¨ Hence, the equation of motion is
ml ¨θ = −mg sin θ,
7.3. SIMPLE PENDULUM 165 7.3.1 global structure of phase plane
We are interested in all possible solutions as a function of its parameters E and t0. The constant t0 is unimportant. For the system is autonomous, that is its right-hand side F (y) is independent of t. This implies that if y(t) is a solution, so is y(t − t0) for any t0. The trajectories (y(t), ˙y(t)) and (y(t − t0), ˙y(t − t0)) are the same curve in the phase plane (i.e. y- ˙y plane). So, to study the trajectory on the phase plane, the relevant parameter is E. We shall take the simple pendulum as a concrete example for explanation. In this case, V (θ) = − cos(θ)g/l.
As we have seen that
θ˙2
2 + V (θ) = E, (7.7)
the total conserved energy. We can plot the equal-energy curve on the phase plane.
CE := {(θ, ˙θ) | θ˙2 2 −g
l cos θ = E} (7.8)
This is the trajectory with energy E. These trajectories can be classified into the follow categories.
1. No trajectory: For E < −g/l, the set {(θ, ˙θ)|θ˙22 −gl cos θ = E} is empty. Thus, there is no trajectory with such E.
2. Equilibria: For E = −g/l, the trajectories are isolated points (2nπ, 0), n ∈ Z. These correspond to equibria, namely they are constant state solutions
θ(t) = 2nπ, for all t.
3. Bounded solutions. For −g/l < E < g/l, the trajectories are bounded closed orbits. Due to periodicity of the cosine function, we see from (7.8) that (θ, ˙θ) is on CE if and only if (θ + 2nπ, ˙θ) is on CE. We may concentrate on the branch of the trajectory lying between (−π, π), since others are simply duplications of the one in (−π, π) through the mapping (θ, ˙θ) 7→ (θ + 2nπ, ˙θ).
For θ ∈ (−π, π), we see that the condition θ˙2
2 −g
l cos θ = E implies
E +g
l cos θ ≥ 0, or
cos θ ≥ −El g . This forces θ can only stay in [−θ1, θ1], where
θ1= cos−1(−El/g).
The condition −g/l < E < g/l is equivalent to 0 < θ1 < π. The branch of the trajectory CE velocity ˙θ = 0 and the corresponding angle θ = θ1, the largest absolute value. The value θ1 is called the amplitude of the pendulum.
We integrate the upper branch of this closed orbit by using the method of separation of vari-able:
We may normalize t0 = 0 because the system is autonomous (that is, the right-hand side of the differential equation is independent of t). Let us denote
t1 :=
The first integral is t1, whereas the second integral is −ψ(θ). Thus, ψ(θ) = 2t1− t.
As θ varies from θ1to −θ1, 2t1− t varies from t1to −t1, or equivalently, t varies from t1to 3t1. Hence, the solution for t ∈ [t1, 3t1] is
θ(t) := φ(2t1− t).
7.3. SIMPLE PENDULUM 167 We notice that
θ(t) = φ(2t1− t) = θ(2t1− t) for t ∈ [2t1, 3t1]
At t = 3t1, θ(3t1) = −θ1and ˙θ(3t1) = 0. We can continue the time by integrating the upper branch of CE again. This would give the same orbit. Therefore, we can extend θ periodically with period T = 2t1 by:
θ(t) = θ(t − 2nT ) for 2nT ≤ t ≤ 2(n + 1)T.
4. Another equilibria: For E = g/l, the set CE contains isolated equilibria:
{((2n + 1)π, 0)|n ∈ Z} ⊂ CE = {(θ, ˙θ) | θ˙2 2 −g
l cos θ = E}
These equilibria are saddle points, which can be seen by linearizing the system at these equi-libria. The nonlinear system is ¨θ +glsin θ = 0. Near the equilibrium ((2n + 1)π, 0), we write the solution (θ, ˙θ) = ((2n + 1)π + u, ˙u), where (u, ˙u) is the perturbation, which is small.
Plug into the equation, we get
¨ u + g
l sin((2n + 1)π + u) = 0.
For small u, we get the linearized equation
¨ u −g
lu ≈ 0.
The characteristic roots of this linearized system are ±pg/l. Thus, 0 is a saddle point of the linearized system.
5. Heteroclinic orbits: We can connect two neighboring saddle points (−π, 0) and (π, 0). This can be thought as a limit of the above case with E → g/l from below. For E = g/l, Using the polar stereographic projection:
u = tan θ
Integrate this, we get
Here, we normalize a constant t00 = 0, which is just a shift of time. It is nothing to do with the orbit. Solve u, we obtain
u = 1 − et0 θ ∈ R. By using the method of separation of variable, we get
Z θ
Let us call the left-hand side of the above equation by ψ(θ). Notice that ψ(θ) is a monotonic increasing function defined for θ ∈ (−∞, ∞), because ψ0(θ) > 2(E−g/l)1 > 0. The range of
From the periodicity of the cosine function, we have for 2nπ ≤ θ ≤ 2(n + 1)π,
t = ψ(θ) =
7.4. CYCLOIDAL PENDULUM – TAUTOCHRONE PROBLEM 169 7.3.2 Period
Let us compute the period for case 3 in the previous subsection. Recall that
T = where 0 < θ1 = arccos(−El/g) < π is the amptitude of the pendulum. By the substitution
u = sin(θ/2)
where k = sin(θ1/2). This integral is called an elliptic integral. This integral cannot be expressed as an elementary function. But we can estimate the period by using
1 ≥ 1 − k2u2 ≥ 1 − k2 for −1 ≤ u ≤ 1 and usingR1
−11/√
1 − u2du = π, the above elliptic integral becomes
2π
Using Taylor expansion for (1 − k2u2)−1/2, expand the elliptic integral f (k) =
Z 1
−1
du
p(1 − u2)(1 − k2u2)
in Taylor series in k for k near 0. You may use Maple to do the integration.
7.4 Cycloidal Pendulum – Tautochrone Problem
7.4.1 The Tautochrone problem
The period of a simple pendulum depends on its amptitude y11. A question is that can we design a pendulum such that its period is independent of its amptitude. An ancient Greek problem called
1Indeed, k = sin(y1/2)
tautochrone problem answers this question. The tautochrone problem is to find a curve down which a bead placed anywhere will fall to the bottom in the same amount of time. Thus, such a curve can provide a pendulum with period independent of its amptitude. The answer is the cycloid. The cycloidal pendulum oscillates on a cycloid. The equation of a cycloid is
x = l(θ + π + sin θ). The equation of motion on cycloidal pendulum is
¨ s = −g
4ls,
a linear equation! Its period is T = 2πpl/g, which is independent of the amplitude of the oscilla-tion.
Which planar curves produce linear oscillators?
The equation of motion on a planar curve is
¨
s = −gdy ds.
The question is: what kind of curve produce linear oscillator. In other word, which curve gives dy/ds = ks. This is an ODE for y(s). Its solution is
y(s) = k 2s2. Since s is the arc length of the curve, we have
x0(s)2+ y0(s)2 = 1.
7.4. CYCLOIDAL PENDULUM – TAUTOCHRONE PROBLEM 171 Thus, the planar curve that produces linear restoration tangential force is a cycloid.
Ref. http://mathworld.wolfram.com
7.4.2 Construction of a cycloidal pendulum
To construct a cycloidal pendulum2, we take l = 1 for explanation. We consider the evolute of the cycloid
x = π + θ + sin θ, y = −1 − cos θ. (7.11)
In geometry, the evolute E of a curve C is the set of all centers of curvature of that curve. On the other hand, if E is the evolute of C, then C is the involute of E. An involute of a curve E can be constructed by the following process. We first wrape E by a thread with finite length. One end of the thread is fixed on E. We then unwrape the thread. The trajectory of the other end as you unwrape the thread forms the involute of E. We shall show below that the evolute E of a cycloid C is again a cycloid. With this, we can construct a cycloidal pendulum as follows. We let the mass P is attached by a thread of length 4 to one of the cusps of the evolute E. Under the tension, the thread is partly coincide with the evolute and lies along a tangent to E. The mass P then moves on the cycloid C.
Next, we show that the motion of the mass P lies on the cycloid C. The proof consists of three parts.
1. The evolute of a cycloid is again a cycloid. Suppose C is expressed by (x(θ), y(θ)). We recall that the curvature of C at a particular point P = (x(θ), y(θ)) is defined by dα/ds, where α = arctan( ˙y(θ)/ ˙x(θ)) is the inclined angle of the tangent of C and ds = p
˙
x2+ ˙y2dθ is the infinitesimal arc length. Thus, the curvature, as expressed by parameter θ, is given by
κ = dα
The center of curvature of C at P = (x, y) is the center of the osculating circle that is tangent to C at P . Suppose P0 = (ξ, η) is its coordinate. Then P P0 is normal to C (the normal (nx, ny) is (− ˙y, ˙x)/p
˙
x2+ ˙y2) and the radius of the osculating circle is 1/κ. Thus, the coordinate of the center
2Courant and John’s book, Vol. I, pp. 428.
of curvature is
ξ = x + 1
κnx = x − ˙yx˙2+ ˙y2
˙
x¨y − ˙y ¨x, η = y + 1
κny = y + ˙xx˙2+ ˙y2
˙
x¨y − ˙y ¨x. When (x(θ), y(θ)) is given by the cycloid equation (7.11),
x = π + θ + sin θ, y = −1 − cos θ, −π ≤ θ ≤ π, we find that its evolute
ξ = π + θ − sin θ, η = 1 + cos θ, (7.12)
is also a cycloid.
2. The evolute of C is the envelope of its normals. We want to find the tangent of the evolute E and show it is identical to the normal of C. To see this, we use arc length s as a parameter on C.
With this, the normal (nx, ny) = (−y0, x0) and the curvature κ = x0y00− y0x00, where0is d/ds. The evolute is
ξ = x − ρy0, η = y + ρx0, (7.13)
where ρ = 1/κ. Thus, the evolute E is also parametrized by s. Since x02+ y02= 1, we differentiate it in s to get x0x00+ y0y00= 0. This together with κ = x0y00− y0x00yield
x00= −y0/ρ, = y00= x0/ρ.
Differentiating (7.13) in s, we can get the tangent of the evolute E:
ξ0 = x0− ρy00− ρ0y0= −ρ0y0, η0 = y0+ ρx00+ ρ0x0 = ρ0x0, (7.14) Therefore,
ξ0x0+ η0y0 = 0.
This means that the tangent (ξ0, η0) of the evolute at the center of curvature is parallel to the normal direction (−y0, x0) of the curve C. Since both of them pass through (ξ, η), they are coincide. In other words, the normal to the curve C is tangent to the evolute E at the center of curvature.
3. The end point of the thread P lies on the cycloid C. We show that the radius of curvature plus the length of portion on E where the thread is attched to is 4. To see this, we denote the acr length on the evolute E by σ. The evolute E, as parametrized by the arc length s of C is given by (7.13). Its arc length σ satisfies
dσ ds
2
= ξ02+ η02= (−ρ0y0)2+ (ρ0x0)2 = ρ02
7.5. THE ORBITS OF PLANETS AND STARS 173 Here, we have used (7.14). Hence, σ02 = ρ02. We take s = 0 at θ = π ((x, y) = (π, −2)). We choose s > 0 when θ > π. We take σ(0) = 0 which corresponds to (ξ, η) = (π, 2). We call this point A (the cusp of the cycloid E). We also choose σ(s) > 0 for s > 0. Notice that ρ0(s) < 0.
From these normalization, we have
σ0(s) = −ρ0(s).
Now, as the mass moves along C to a point P on C, the center of curvature of C at P is Q which is on the evolute E. We claim that
length of the arc AQ on E + the length of the straight line P Q = 4.
To see that, the first part above is Z s
0
σ0ds = − Z s
0
ρ0ds = ρ(0) − ρ(s).
The second part is simply the radius of curvature ρ(s). Hence the above sum is ρ(0) = 4.
Homework.
1. Given a family of curves Γλ : {(x(t, λ), y(t, λ))|t ∈ R}, a curve E is said to be the envelop of Γλif
(a) For each λ, Γλ is tangent to E. Let us denote the tangent point by Pλ¿ (b) The envelop E is made of Pλwith λ ∈ R.
Now consider the family of curves to be the normal of a cycliod C, namely Γθ = (x(θ) + tnx(θ), y(θ) + tny(θ)),
where (x(θ), y(θ)) is given by (7.11) and (nx, ny) is its normal. Using this definition of envelop, show that the envelop of Γθis the cycloid given by (7.12).
7.5 The orbits of planets and stars
7.5.1 Centrally directed force and conservation of angular momentum
The motion of planets or stars can be viewed as a particle moving under a centrally directed field of force:
F = F (r)ˆer,
where r is the distance from the star to the center, r is the position vector from the center to the star and
ˆ er= r
r,
is the unit director. The equation of motion of the star is
¨
r = F (r)ˆer. Define the angular momentum L = r × ˙r. We find
dL
dt = ˙r × ˙r + r × ¨r = F (r)r × ˆer= 0.
Hence , L is a constant. A function in the state space (r, ˙r) is called an integral if it is unchanged along any orbits. The integrals can be used to reduce number of unknowns of the system. The
Hence , L is a constant. A function in the state space (r, ˙r) is called an integral if it is unchanged along any orbits. The integrals can be used to reduce number of unknowns of the system. The