機遇創生論 1.6.1

(define (((delta eta) f) q)
  (define (g epsilon)
    (f (+ q (* epsilon eta))))
  ((D g) 0))

(define eta (literal-function 'eta))

(define q (literal-function 'q))

— Patrick Eli Catach

Code

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

and

(define (f q) (compose (literal-function 'f) q))

have the same effect. If you execute

(show-expression ((((delta eta) f) q) 't))

you will get

$\displaystyle{\delta_\eta f[q] = \eta(t) Df(q(t))}$

It seems not to make sense, because $\displaystyle{f}$ is functional. Its input is a path function, not a real number. We cannot differentiate it with respect to its argument, which is a path.

It actually make sense in this special case because now

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

In other words, $\displaystyle{f}$ is a function of one argument only. And variation itself can be expressed in terms of differentiation:

$\displaystyle{ \begin{aligned} g( \epsilon ) &= f[q + \epsilon \eta] \\ \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{g(\epsilon) - g(0)}{\epsilon} \right) = D g(0) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\delta_\eta f[q] \\ &= \left. \frac{d}{d \epsilon} f[q + \epsilon \eta] \right|_{x=a} \\ \end{aligned}}$

No. This explanation is not correct. Even if $\displaystyle{f}$ is a function of one argument only, it is still a path function. Its input argument is a path, not a number. Different path inputs will give different outputs.

The correct explanation should be the following.

(define (f q) (compose (literal-function 'f) q)) 

(define eta (literal-function 'eta))

(define q (literal-function 'q))

These lines actually imply that $\displaystyle{f}$ is just a path-independent function.

$\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{f[q+\epsilon \eta]-f[q]}{\epsilon} \right) \\ \end{aligned}}$

If the function $\displaystyle{f}$ is just a path-independent,

$\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{f(q+\epsilon \eta)-f(q)}{\epsilon} \right) \\ \end{aligned}}$

For the time $\displaystyle{t}$ that $\displaystyle{\eta(t) \ne 0}$ ,

$\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \eta \lim_{\epsilon \to 0} \left( \frac{f(q+\epsilon \eta)-f(q)}{\epsilon \eta} \right) \\ &= \eta(t) D f(q(t)) \\ \end{aligned}}$

So the above lines of code are not correct, because they do not represent a path-dependent function. The correct code (for one-dimensional path $\displaystyle{q(t)}$ ) should be

(define (((delta eta) f) q)
  (define (g epsilon)
    (f (+ q (* epsilon eta))))
    ((D g) 0))

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

(define q (literal-function 'q (-> Real (UP Real))))

(define eta (literal-function 'eta (-> Real (UP Real))))

(print-expression ((((delta eta) f) q) 't))

(show-expression ((((delta eta) f) q) 't))

The result is

$\displaystyle{ \begin{aligned} &\delta_\eta f[q] \\ &= D \eta(t) \partial_2 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) + \eta(t) \partial_1 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) \\ \end{aligned}}$

But this has a problem. It seems that $\displaystyle{f}$ is still not a function with a path as input. It is just a function with 3 real numbers as inputs.

Note that the path-dependent function $\displaystyle{f}$ is not really a general functional. A general functional can depend on an input function in any ways. For example,

$\displaystyle{F[y] = y(3)}$

$\displaystyle{F[y] = \int_0^1 y(x) dx}$

A functional that defines the arc length of a curve:

$\displaystyle{L[y] = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx}$

– The Calculus of Variations

– Brian Keng

The function $\displaystyle{f}$ is not an ordinary function because its input is not a real number, but a path-function $\displaystyle{q(t)}$ . In other words, it does not depend on individual values of $\displaystyle{q(t)}$ at different individual times. Instead, its output depends on the whole path $\displaystyle{q(t)}$ . However, the function $\displaystyle{f}$ is not a general functional either.

It is not a general functional in the sense that its input is not a general mathematical function, but a (function that represents a) path $\displaystyle{q(t)}$ . Also, $\displaystyle{f}$ depends on $\displaystyle{q}$ in a specific way:

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

$\displaystyle{ \begin{aligned} &f[q] \\ &= f(\Gamma[q]) \\ &= f \left( (t, q(t), Dq(t)) \right) \\ &= f(t, q(t), Dq(t)) \\ \end{aligned}}$

So, besides being a (non-general) functional, $\displaystyle{f[q]}$ , with a path, $\displaystyle{q}$ , as input, $\displaystyle{f}$ can also be regarded as an ordinary function that has 3 real number inputs: $\displaystyle{t}$ , $\displaystyle{q(t)}$ , and $\displaystyle{Dq(t)}$ .

We can vary with respect to a path (the $\displaystyle{q}$ itself), but we cannot differentiate with respect to a path. We can only differentiate with respect to the values of the path $\displaystyle{q(t)}$ .

For simplicity, we consider the case that the path $\displaystyle{q(t)}$ is one-dimensional. Only in the sense that $\displaystyle{f}$ can also be regarded as an ordinary function that has 3 real number inputs, it can use the chain rule of ordinary differentiation:

— Me@2020-06-20 01:17:39 PM

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Monthly Archives: June 2020

機遇創生論 1.6.1

SICM, 4

Ex 1.8.2.2 Implementation of $\delta$