This article is the final installment of the "Natural Extremum" series and likely the last article of 2010. In this wonderful year of 2010, everyone must have gained a lot, and BoJone has also grown significantly. As 2010 comes to a close, BoJone and Scientific Space wish everyone more happiness and joy in the new year, and faster progress on the path of science.
In this article, BoJone will discuss the most fundamental principles of finding extrema. This line of thought was inspired by the "Feynman Lectures on Physics" by the genius Richard Feynman. We will perform a brief analysis of finding extrema for functions (differentiation) and for functionals (calculus of variations).
I. Finding Extrema of Functions
For a function y=f(x), suppose it reaches a maximum value at x=x_0. Then, for a very small increment \Delta x, it is obvious that: f(x_0+\Delta x) \leq f(x_0) \tag{3} According to the Taylor series, we have: f(x_0+\Delta x) = f(x_0) + f'(x_0)\Delta x \tag{4} Here, we have omitted quadratic and higher-order terms because the Mean Value Theorem tells us that the sum of the remaining terms is still just a quadratic term (second-order infinitesimal). That is to say, it cannot "shake the status of f'(x_0)\Delta x." Substituting (4) into (3), we get: f'(x_0)\Delta x \leq 0 It is important to note that f'(x_0) is a fixed value, while \Delta x is a variable that can be either positive or negative. Thus, we obtain: f'(x_0) \leq 0 \quad \text{and} \quad f'(x_0) \geq 0 Consequently, f'(x_0) = 0.
We can also replace the term "maximum" with "minimum" and swap \leq and \geq; the discussion follows similarly, and the result is the same. Thus, we conclude: f'(x)=0 is a necessary condition for a function f(x) to have a local maximum or minimum.
II. Finding Extrema of Functionals
Discussions regarding the brachistochrone and catenary problems ultimately boil down to the following problem:
Find a function y=f(x) passing through (x_1, y_1) and (x_2, y_2) that satisfies the condition that the integral \int_{x_1}^{x_2} F(x, y, \dot{y}) dx is a maximum (or minimum).
Suppose the function y=y(x) is the desired function. For a very small increment function \varepsilon = \varepsilon(x), where \varepsilon(x_1) = \varepsilon(x_2) = 0, then y = y(x) + \varepsilon(x) is also a function passing through (x_1, y_1) and (x_2, y_2). Then: \int_{x_1}^{x_2} F(x, y+\varepsilon, \dot{y}+\dot{\varepsilon}) dx \leq \int_{x_1}^{x_2} F(x, y, \dot{y}) dx \tag{5} Using the multivariate Taylor series to expand F(x, y+\varepsilon, \dot{y}+\dot{\varepsilon}), we get: F(x, y+\varepsilon, \dot{y}+\dot{\varepsilon}) = F(x, y, \dot{y}) + \frac{\partial F}{\partial y}\varepsilon + \frac{\partial F}{\partial \dot{y}}\dot{\varepsilon} Here, we similarly omit quadratic and higher-order terms. Substituting this into equation (5), we obtain: \int_{x_1}^{x_2} \left( \frac{\partial F}{\partial y}\varepsilon + \frac{\partial F}{\partial \dot{y}} \dot{\varepsilon} \right) dx \leq 0 \tag{6} There is a technique for handling \int \left( \frac{\partial F}{\partial \dot{y}} \dot{\varepsilon} \right) dx using "integration by parts" from Mathematical Analysis: \int \left( \frac{\partial F}{\partial \dot{y}} \dot{\varepsilon} \right) dx = \int \left( \frac{\partial F}{\partial \dot{y}} d\varepsilon \right) = \frac{\partial F}{\partial \dot{y}}\varepsilon - \int \left[ \frac{d\left(\frac{\partial F}{\partial \dot{y}}\right)}{dx}\varepsilon \right] dx Substituting this into equation (6), we get: \left( \frac{\partial F}{\partial \dot{y}}\varepsilon \right) \bigg|_{x_1}^{x_2} + \int_{x_1}^{x_2} \left[ \frac{\partial F}{\partial y} - \frac{d\left(\frac{\partial F}{\partial \dot{y}}\right)}{dx} \right] \varepsilon dx \leq 0 Since \varepsilon(x_1) = \varepsilon(x_2) = 0, the term \left( \frac{\partial F}{\partial \dot{y}}\varepsilon \right) \bigg|_{x_1}^{x_2} = 0. Similarly, since \varepsilon can be positive or negative, we must have: \int_{x_1}^{x_2} \left[ \frac{\partial F}{\partial y} - \frac{d\left(\frac{\partial F}{\partial \dot{y}}\right)}{dx} \right] \varepsilon dx = 0 This equation must hold for all \varepsilon = \varepsilon(x), so the value inside the brackets must be zero: \frac{\partial F}{\partial y} - \frac{d\left(\frac{\partial F}{\partial \dot{y}}\right)}{dx} = 0 \tag{7} By swapping maximum and minimum and swapping \leq and \geq, the discussion remains the same. Thus, we conclude: equation (7) is a necessary condition for the integral \int_{x_1}^{x_2} F(x, y, \dot{y}) dx to be an extremum.
Equation (7) is the famous (two-dimensional form of the) Euler-Lagrange equation.
Using a similar approach, the equation can be extended to more dimensions and higher orders (e.g., if F contains \ddot{y} terms). It is not hard to see that the underlying logic is consistent: Assume extremum \to set increment \to first-order expansion \to compare with original value \to analysis and simplification \to derive equation \to solve equation. Although the processing details vary, the principle remains unchanged. Therefore, this can be considered the most fundamental approach to handling extremum problems.
Since this article is intended as a conceptual guide rather than a professional tutorial, the discussion ends here. For specific details, you can consult relevant content on Wikipedia.
Calculus of Variations: http://zh.wikipedia.org/zh/Calculus_of_variations
Euler-Lagrange Equation: http://zh.wikipedia.org/zh-sg/Euler-Lagrange_equation
The "Natural Extremum" series has come to an end, and 2010 is also coming to a close. Despite many lingering attachments and regrets, we have still gained much and grown much in 2010. May we all carry the most beautiful hopes to welcome the upcoming 2011, growing slowly and moving forward steadily amidst the sunshine and the rain, experiencing science and appreciating truth! On the path of science, I wish to continue moving forward with many science enthusiasts!
End of "Natural Extremum" series.
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