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2009-01-15

1.5 オイラー・ラグランジュ方程式 /The Euler-Lagrange Equations

| 03:10 | 1.5 オイラー・ラグランジュ方程式 /The Euler-Lagrange Equations - naoya_t@SICM(べ、べつにあなたのためじゃないんだからね) を含むブックマーク はてなブックマーク - 1.5 オイラー・ラグランジュ方程式 /The Euler-Lagrange Equations - naoya_t@SICM(べ、べつにあなたのためじゃないんだからね) 1.5 オイラー・ラグランジュ方程式 /The Euler-Lagrange Equations - naoya_t@SICM(べ、べつにあなたのためじゃないんだからね) のブックマークコメント

停留作用の法則は配置空間における系の実現可能な経路を、その作用が停留値(stationary value)をもつものとして性格づける。初等微積分学において我々は、関数の臨界点(critical point)とはその微分が消滅する点であると習う。同様に、その経路に沿った作用は停留であるような経路群は連立微分方程式の解である。オイラーラグランジュ方程式(Euler-Lagrange equations) あるいは単にラグランジュ方程式(Lagrange equations)と呼ばれるこの体系は、そのおかげで力学系の運動の計算に停留作用の法則が使えるようになる上、力学の変分的定式化とニュートン的定式化を関連づけてくれるlinkである*1

ラグランジュ方程式 /Lagrange equations

Lが時刻・座標・速度に依存した系のラグランジュ関数で、qがその上で作用S\[q\](t_1,t_2)が(経路の両端点を固定した変種の経路に対して)停留になるような座標経路であるなら、

D(\delta_2L\circ\Gamma\[q\])-\delta_1L\circ\Gamma\[q\]=0

であることが分かるだろう。ここで L は局所タプルをとる実数値関数である;\delta_1L\delta_2Lは、一般化座標位置と一般化速度成分に関するLの偏微分を意味する*2

関数\delta_2Lは局所タプルを、一般化速度の各成分に関するLの微分を成分とする構造にマッピングする。関数Γ[q]は時刻を局所タプルにマッピングする:\Gamma\[q\](t)=(t,q(t),Dq(t),...)。従って合成関数\delta_1L\circ\Gamma\[q\]および\delta_2L\circ\Gamma\[q\]は共に、時刻を1引数とする関数である。ラグランジュ方程式は、\delta_2L\circ\Gamma\[q\]微分が常に\delta_1L\circ\Gamma\[q\]に等しくなると表明する。ラグランジュ関数が与えられればラグランジュ方程式は、実現可能な経路によって必ず満たされる通常の連立微分方程式を形成する*3

1.5.1 ラグランジュ方程式微分 /Derivation of the Lagrange Equations

停留作用の法則が、実現可能な経路は一組の常微分方程式を満たすことを示唆していることを示そう。まず、経路の変動に伴い経路依存関数がどのように変動するかを調べるツールを開発しよう。それから、これらのツールを作用に適用し、ラグランジュ方程式を導きたい。

経路を変動させる

変分とか変化とか変動とか何となく書いてるけど、定訳があったり用語の違いがはっきり決まってるならそれに従うかも(その方が他の資料をあたる時に便利)。自分にとってわかりやすいかどうかを優先させるなら、カタカナないし原語ママで書いたほうが良い場合がある。pathとかvaryとかそのままでいいよもう。

  • vary / variation
  • change

経路qに依存する関数 f[q] があるとしよう。経路の変動につれこの関数はどのように変動するだろうか?

qを座標経路、q+ε^η をvaryされた経路、ここで関数 ηは 経路qに足すことのできるような経路的関数、要素εはスケールファクターである。経路q上の関数fの変分(variation) δηf[q] を*4

\delta_\eta f\[q\]=\lim{\epsilon →0}(\frac{f\[q+\epsilon\eta\]-f\[q\]}\epsilon)

と定義する。fの変分とは、経路内の小さな変分に対する関数fの変化への一次近似である。fの変分はηに依存する。

簡単な例として、同一経路関数 I[q] = q の変分を挙げる。この定義を適用すれば

 δηI[q] = limε→0(...) = η

伝統的にδηI[q]は簡単にδqと書く。他の例として、経路の微分を返す経路関数の変分を挙げよう:

 g[q]=Dq,... δηg[q] = limε→0(...) = Dη

伝統的にδηg[q]はδDqと書く。

変分は微分を用いて表現することができる。g(ε)=f[q+ε^η]とするなら

~1.22

変分には以下のような微分的な性質がある。パス依存関数f,g,定数cについて:

~1.23

~1.24

~1.25

Fを経路非依存関数、gを経路依存関数とする、と

~1.26

演算子D(微分)とδ(変分)は以下の意味でcommuteする:

~1.27

似たような意味で、変分は積分ともcommuteする。

経路依存関数fが特定の経路qについて、その経路内での小さな変化に対し停留であるなら、それは、

特定の関数ηの小さな倍数をqに足した結果のそれらの変分sのサブセットについても停留にちがいない。

なので、任意のηについての式δηf[q]=0 は、qの近傍の経路の小さな変分について関数fが停留であることを示唆する。

問題1.7 δの性質

δが式1.23〜1.27の性質をもつことを示せ。

問題1.8 δの実装

a. Suppose we have a procedure f that implements a path-dependent function: for path q and time t it has the value ((f q) t). The procedure delta computes the variation ( f)[q](t) as the value of the expression ((((delta eta) f) q) t). Complete the definition of delta:

(define (((delta eta) f) q)
  ...  )

b. Use your delta procedure to verify the properties of listed in exercise 1.7 for simple functions such as implemented by the procedure f:

(define (f q)
  (compose 
    (literal-function 'F (-> (UP Real Real Real) Real))
    (Gamma q)))

This implements a one-degree-of-freedom path-dependent function that depends on the local tuple of the path at each moment. You should compute both sides of the equalities and compare the results.

Varying the action

The action is the integral of the Lagrangian along a path:

For a realizable path q the variation of the action with respect to any variation that preserves the endpoints, (t1) = (t2) = 0, is zero:

Variation commutes with integration, so the variation of the action is

Using the fact that

which follows from equations (1.20) and (1.21), and using the chain rule for variations (1.26), we get*5

Integrating the last term of equation (1.32) by parts gives

For our variation we have (t1) = (t2) = 0, so the first term vanishes.

Thus the variation of the action is zero if and only if

The variation of the action is zero because, by assumption, q is a realizable path. Thus (1.34) must be true for any function that is zero at the endpoints.

We retain enough freedom in the choice of the variation that the factor in the integrand multiplying is forced to be zero at each point along the path. We argue by contradiction: Suppose this factor were nonzero at some particular time. Then it would have to be nonzero in at least one of its components. But if we choose our to be a bump that is nonzero only in that component in a neighborhood of that time, and zero everywhere else, then the integral will be nonzero. So we may conclude that the factor in curly brackets is identically zero:53

This is just what we set out to obtain, the Lagrange equations.

A path satisfying Lagrange's equations is one for which the action is stationary, and the fact that the action is stationary depends only on the values of L at each point of the path (and at each point on nearby paths), not on the coordinate system we use to compute these values. So if the system's path satisfies Lagrange's equations in some particular coordinate system, it must satisfy Lagrange's equations in any coordinate system. Thus the equations of variational mechanics are derived the same way in any configuration space and any coordinate system.

Harmonic oscillator

For an example, consider the harmonic oscillator. A Lagrangian is

Then

The Lagrangian is applied to a tuple of the time, a coordinate, and a velocity. The symbols t, x, and v are arbitrary; they are used to specify formal parameters of the Lagrangian.

Now suppose we have a configuration path y, which gives the coordinate of the oscillator y(t) for each time t. The initial segment of the corresponding local tuple at time t is

So

and

so the Lagrange equation is

which is the equation of motion of the harmonic oscillator.

Orbital motion

As another example, consider the two-dimensional motion of a particle of mass m with gravitational potential energy - µ/r, where r is the distance to the center of attraction. A Lagrangian is54

where and are formal parameters for rectangular coordinates of the particle, and v and v are formal parameters for corresponding rectangular velocity components. Then55

Similarly,

Now suppose we have a configuration path q = ( x, y ), so that the coordinate tuple at time t is q(t) = ( x(t), y(t) ). The initial segment of the local tuple at time t is

So

and

The component Lagrange equations at time t are

問題1.9 Lagrange's equations

Derive the Lagrange equations for the following systems, showing all of the intermediate steps as in the harmonic oscillator and orbital motion examples.

a. A particle of mass m moves in a two-dimensional potential V(x, y) = (x2 + y2)/2 + x2 y - y3/3, where x and y are rectangular coordinates of the particle. A Lagrangian is L(t; x, y; vx, vy) = (1/2) m (vx2 + vy2) - V(x, y).

b. An ideal planar pendulum consists of a bob of mass m connected to a pivot by a massless rod of length l subject to uniform gravitational acceleration g. A Lagrangian is L(t, , ) = (1/2) m l2 2 + m g l cos . The formal parameters of L are t, , and ; measures the angle of the pendulum rod to a plumb line and is the angular velocity of the rod.56

c. A Lagrangian for a particle of mass m constrained to move on a sphere of radius R is L(t; , ; , ß) = (1/2) m R2 (2 + (ß sin )2). The angle is the colatitude of the particle and is the longitude; the rate of change of the colatitude is and the rate of change of the longitude is ß.

問題1.10 Higher-derivative Lagrangians

Derive Lagrange's equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form L(t, q, , ) with terms are57

In general, these equations, first derived by Poisson, will involve the fourth derivative of q. Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?

1.5.2 Computing Lagrange's Equations

The procedure for computing Lagrange's equations mirrors the functional expression (1.18), where the procedure Gamma implements :58

(define ((Lagrange-equations Lagrangian) q)
  (- (D (compose ((partial 2) Lagrangian) (Gamma q)))
     (compose ((partial 1) Lagrangian) (Gamma q))))

The argument of Lagrange-equations is a procedure that computes a Lagrangian. It returns a procedure that when applied to a path q returns a procedure of one argument (time) that computes the left-hand side of the Lagrange equations (1.18). These residual values are zero if q is a path for which the Lagrangian action is stationary.

Observe that the Lagrange-equations procedure, like the Lagrange equations themselves, is valid for any generalized coordinate system. When we write programs to investigate particular systems, the procedures that implement the Lagrangian function and the path q will reflect the actual coordinates chosen to represent the system, but we use the same Lagrange-equations procedure in each case. This abstraction reflects the important fact that the method of derivation of Lagrange's equations from a Lagrangian is always the same; it is independent of the number of degrees of freedom, the topology of the configuration space, and the coordinate system used to describe points in the configuration space.

自由粒子 /The free particle

Consider again the case of a free particle. The Lagrangian is implemented by the procedure L-free-particle. Rather than numerically integrating and minimizing the action, as we did in section 1.4, we can check Lagrange's equations for an arbitrary straight-line path t ( at + a0, bt + b0, ct + c0 ):

(define (test-path t)
  (up (+ (* 'a t) 'a0)
      (+ (* 'b t) 'b0)
      (+ (* 'c t) 'c0)))
(print-expression
 (((Lagrange-equations (L-free-particle 'm))
   test-path)
  't))
→ (down 0 0 0)

That the residuals are zero indicates that the test path satisfies the Lagrange equations.59

We can also apply the Lagrange-equations procedure to an arbitrary function:60

(show-expression
 (((Lagrange-equations (L-free-particle 'm))
   (literal-function 'x))
  't))
→ (* (((expt D 2) x) t) m)

The result is an expression containing the arbitrary time t and mass m, so it is zero precisely when D2 x = 0, which is the expected equation for a free particle.

調和振動子(振り子) /The harmonic oscillator

Consider the harmonic oscillator again, with Lagrangian (1.16). We know that the motion of a harmonic oscillator is a sinusoid with a given amplitude, frequency, and phase:

Suppose we have forgotten how the constants in the solution relate to the physical parameters of the oscillator. Let's plug in the proposed solution and look at the residual:

(define (proposed-solution t)
  (* 'a (cos (+ (* 'omega t) 'phi))))

(show-expression
 (((Lagrange-equations (L-harmonic 'm 'k))
   proposed-solution)
  't))

The residual here shows that for nonzero amplitude, the only solutions allowed are ones where ( k - m 2 ) = 0 or = (k/m)1/2.

問題1.11

Compute Lagrange's equations for the Lagrangians in exercise 1.9 using the Lagrange-equations procedure. Additionally, use the computer to perform each of the steps in the Lagrange-equations procedure and show the intermediate results. Relate these steps to the ones you showed in the hand derivation of exercise 1.9.

問題1.12

a. Write a procedure to compute the Lagrange equations for Lagrangians that depend upon acceleration, as in exercise 1.10. Note that Gamma can take an optional argument giving the length of the initial segment of the local tuple needed. The default length is 3, giving components of the local tuple up to and including the velocities.

b. Use your procedure to compute the Lagrange equations for the Lagrangian

Do you recognize the resulting equation of motion?

c. For more fun, write the general Lagrange equation procedure that takes a Lagrangian of any order, and the order, to produce the required equations of motion.

*1:48

*2:49

*3:50

*4:51

*5:52

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