Introduction to Variational Methods

This article can be used as a simple entry to the variational method, including the following four parts:

1. Basic concepts of Function
2. Prefix Theorem
   
3. Derivation of the Euler-Laplace Equation
4. Specific Application

I. Basic concepts of functional

The birth of the variational method is traced back to the "shortest line problem" proposed by Johann Bernoulli (1667-1748) In 1696. This problem is an extreme value problem, however, unlike ordinary functions for extreme value calculation, the independent variable of its target function is not a number, but a function. Due to the novelty of the problem, it soon attracted some interest. Johann's brother Jacob Bernoulli (1654-1705) gave a more general solution. Later, Euler (1707 ~ 1783) and Lago (1736-1813) on this basis, the Euler-Lago equation is obtained, and a general solution to this type of problem is given.

What the variational method needs to deal with isMap functions to real numbersSuch a ing is calledFunctionalBecause we do not want to write a textbook, we will not provide a precise definition here. We will use the following example to briefly describe it.

Set the function $ Y = f (x) \ geq 0 $ to be continuous on $ [0, 1] $, then \ begin {Align *} Q [F (x)] = \ int_0 ^ 1 F (x) \ mbox {d} X \ end {Align *} is a functional, its input is a non-negative continuous function on $ [] $, and the output is the area enclosed between the function and the $ x $ axis in the range of $ [] $.

Obviously, functional is the promotion of the concept of function. The only difference is thatDifferent Independent VariablesIn order to obtain the extreme values of a functional, the most direct idea is to apply the function to obtain the extreme values. However, before going deeper, we need to briefly introduce some basic concepts of functional, includingContinuous functional,Linear Function,Extreme functional valuesAndFunctional VariationThey correspond to continuous functions, linear functions, function extreme values, and function differentiation respectively.

• Continuous Function: For function $ Q [Y (x)] $, if $ Y (x) $ changes $ \ Delta y (x) $, the change amount of $ q $ can be any small, so the function $ Q [Y (x)] $ is continuous. For example, for function \ begin {Align *} Q [Y (x)] = \ int_a ^ B y (x) \ mbox {d} X \ end {Align, $ Q [Y (x) \ in C [a, B] $ is defined. For any $ \ Epsilon> 0 $, As long as \ begin {Align *} \ Max _ {A \ Leq x \ Leq B} | y_1 (x)-y (X) | <\ frac {\ Epsilon} {B-a} \ end {Align *} has \ begin {Align *} \ left | Q [Y_1 (x)] -Q [Y (x)] \ right | = \ left | \ int_a ^ B \ left (y_1 (x)-y (x) \ right) \ mbox {d} X \ right | \ Leq \ int_a ^ B | y_1 (x)-y (X) | \ mbox {d} x <\ int_a ^ B \ frac {\ Epsilon} {B-a} \ mbox {d} X = \ Epsilon \ end {Align *} That is $ Q [Y (x)] $ is a continuous functional.
• Extreme functional values: For curves $ Y (x) $, \ begin {Align *} \ max_x \ | y_1 (x)-y (X) | all consecutive curves of \ Leq \ varepsilon \ end {Align *} $ Y_1 (x) $ constitute a set of $ Y (x) $ \ varepsilon $-neighborhood, if any curve $ Y_1 (x) $ in the neighborhood has \ begin {Align *} Q [Y_1 (x)] \ Leq Q [Y (x)] \ end {Align *} is called the function $ Q [Y (x)] $. Obtain the extreme value in a $ \ varepsilon $-neighborhood of $ Y (x) $.
• Linear Function: if any constant $ C_1 $ and $ C_2 $, $ Q [F (x)] $ meets \ begin {Align *} Q [C_1 Y_1 (X) + C_2 Y_2 (x)] = C_1 Q [Y_1 (x)] + C_2 Q [Y_2 (x)] \ end {Align *} indicates that $ Q [F (x)] $ is a continuous functional.
• Functional variation: Let's look at an example. Set the function $ Q [Y (x)] = \ int_a ^ B y ^ 2 (x) \ mbox {d} x $, function $ Y_1 (x) = y (x) + \ Delta y (x) $ is a disturbance to $ Y (x) $, then $ Q [Y (x)] $ increment: \ begin {Align *} \ delta Q & = Q [Y_1 (x)]-Q [Y (x)] = Q [Y (X) + \ Delta y (x)]-Q [Y (x)] \\&=\ int_a ^ B (Y (x) + \ Delta y (x )) ^ 2 \ mbox {d} X-\ int_a ^ B y ^ 2 (x) \ mbox {d} X \ & = \ int_a ^ B 2 Y (X) \ Delta y (x) \ mbox {d} X + \ int_a ^ B (\ Delta y (x )) ^ 2 \ mbox {d} X \ end {Align *} visible $ \ delta q $ consists of two parts. The first item is recorded as \ begin {Ali GN *} \ int_a ^ B 2 Y (x) \ Delta y (x) \ mbox {d} X = T [Y (x), \ Delta y (x)] \ end {Align *} When $ Y (x) $ is fixed, $ t [Y (x), \ Delta y (x)] $ is a linear functional about $ \ Delta y (x) $. \ Begin {Align *} \ left | \ int_a ^ B (\ Delta y (x )) ^ 2 \ mbox {d} X \ right | \ Leq \ left (\ Max _ {A \ Leq x \ Leq B} | \ Delta y (x) | \ right) ^ 2 (B-) \ end {Align *} For the second item, \ begin {Align *} \ lim _ {\ Max _ {A \ Leq x \ Leq B} | \ Delta y (X) | \ rightarrow 0} \ frac {\ int_a ^ B (\ Delta y (x )) ^ 2 \ mbox {d} x} {\ Max _ {A \ Leq x \ Leq B} | \ Delta y (X) |}& \ Leq \ lim _ {\ Max _ {A \ Leq x \ Leq B} | \ Delta y (X) | \ rightarrow 0} \ frac {\ left (\ Max _ {A \ Leq x \ Leq B} | \ Delta y (x) | \ right) ^ 2 (B-)} {\ Max _ {A \ Leq x \ Leq B} | \ Delta y (X) |}\\<\lim _ {\ Max _ {A \ Leq x \ Leq B} | \ Delta y (X) | \ rightarrow 0} \ Max _ {A \ Leq x \ Leq B} | \ Delta y (x) | (B-) \ & = 0 \ end {Align *} is $ \ int_a ^ B (\ Delta y (x )) ^ 2 \ mbox {d} x $ is an infinitely small high order about $ \ Delta y (x) $, which is recorded as $ O (\ Delta y (x) $, so \ begin {Align *} \ delta q = T [Y (x), \ Delta y (x)] + O (\ Delta y (x )) \ end {Align *} compares the concepts of function differentiation and finds out the similarities between them. Similarly, we can say that if we perform a small increment for the independent variable $ Y (x) $ \ Delta y (x) $, the corresponding functional value increment $ \ delta q $ can be divided into two parts: the first part $ t [Y (x), \ Delta y (x)] $ is the linear function about $ \ Delta y (x) $. The second part is the high-order infinity of $ \ Delta y (x) $, then, $ t [Y (x), \ Delta y (x)] $ is called the variational function $ Q [Y (x)] $.

Ii. Preliminary Theorem

We need the following preliminary theorem before deriving the Euler-Laplace equation.

If the function $ Y = f (x) $ is continuous on $ [a, B] $ and \ begin {Align *} \ int_a ^ B f (x) \ ETA (X) \ mbox {d} x = 0 \ end {Align *} where $ \ ETA (x) $ has a continuous derivative on $ [a, B] $, $ \ ETA (A) = \ ETA (B) = 0 $ and $ | \ ETA (x) | <\ Epsilon $ ($ \ Epsilon $ is an arbitrary positive number ), then the function $ f (x) $ is constant at $ [a, B] $ equal to $0 $.

Proof: using the reverse verification method, assume that $ x_0 \ In (a, B) $ causes $ F (x_0)> 0 $, from $ f (x) $ continuity knowledge has a positive value $ \ Delta $ so that $ f (x)> 0 $ is present when $ | x-x_0 | <\ Delta $. Current function \ begin {Align *} \ PSI (x) =\begin {cases} 0 & X \ in [, x_0-\ Delta] \ e ^ {\ frac {1} {(X-x_0) ^ 2-\ Delta ^ 2} & X \ In (x_0-\ delta, x_0 + \ delta) \ 0 & X \ in [x_0 + \ delta, B] \ end {cases} \ end {Align *} Apparently $ \ PSI () = \ PSI (B) = 0 $. In addition, $ \ PSI (x) $ has a continuous derivative on $ [a, B] $ (let's make a hypothesis here and give a proof later ), select the appropriate $ A $ to make $ \ ETA (x) = A \ PSI (x) $ meet $ | \ ETA (x) | <\ Epsilon $, therefore, $ \ ETA (x) $ meets all the conditions, SO \ begin {Align *} \ int_a ^ B f (x) \ ETA (X) \ mbox {d} X = \ Int _ {x_0-\ Delta} ^ {x_0 + \ Delta} f (x) \ ETA (X) \ mbox {d} x> 0 \ end {Align *} conflicts, so $ x_0 $ does not exist, that is, $ f (x) $ in $ (A, B) $ is not greater than $0 $. Likewise, it can be proved that $ f (x) $ is not less than $0 $ on $ (a, B) $, so it can only be $ f (x) $ constant on $ (a, B) $ is equal to $0 $, and is known by $ f (x) $ continuity $ F (a) = F (B) = 0 $, therefore, the constant value of $ f (x) $ in $ [a, B] $ is $0 $.

Finally, it is proved that $ \ PSI (x) $ has a continuous derivative on $ [a, B] $, you only need to consider the two points $ x_0 + \ Delta $ and $ x_0-\ Delta $. Specifically, we can take two steps to prove that the Left and Right derivatives are equal, that is, the derivatives exist, it is proved that the Left and Right limits of the derivative are equal, that is, the continuity is true.

First consider the right derivative, yi zhi \ begin {Align *} \ lim _ {x \ rightarrow (x_0 + \ delta) ^ +} \ frac {\ PSI (X) -\ PSI (x_0 + \ delta)} {X-(x_0 + \ delta) }=\ LiM _ {x \ rightarrow (x_0 + \ delta) ^ +} \ frac {0-0} {X-(x_0 + \ delta)} = 0 \ end {Align *} Then consider the left derivative, yi Zhi \ begin {Align *} \ lim _ {x \ rightarrow (x_0 + \ delta) ^-} \ frac {\ PSI (X) -\ PSI (x_0 + \ delta)} {X-(x_0 + \ delta)} & = \ lim _ {x \ rightarrow (x_0 + \ delta) ^-} \ frac {e ^ {\ frac {1} {(X-x_0) ^ 2-\ D ELTA ^ 2 }}- 0} {X-(x_0 + \ delta )} \\&=\ LiM _ {T \ rightarrow-\ infty} \ frac {e ^ t} {\ left (\ frac {1} {t} + \ Delta ^ 2 \ right) ^ {\ frac {1} {2}-\ Delta} \ left (t = \ frac {1} {(X-x_0) ^ 2-\ Delta ^ 2} \ right) \\& =\ LiM _ {T \ rightarrow-\ infty} \ frac {e ^ t }{\ frac {1} {2} \ left (\ frac {1 }{ t} + \ Delta ^ 2 \ right) ^ {-\ frac {1} {2 }}\ left (-\ frac {1} {t ^ 2} \ right )} \\<=\ LiM _ {T \ rightarrow-\ infty} (-2) E ^ t ^ 2 \ left (\ frac {1} {t} + \ Delta ^ 2 \ right) ^ {\ frac {1} {2 }}\ end {Align *} among them, the second and last equal signs are introduced by the l' Hospital rule. Note \ begin {Align *} \ lim _ {T \ rightarrow-\ infty} \ left (\ frac {1} {t} + \ Delta ^ 2 \ right) ^ {\ frac {1} {2 }}=\ Delta \ end {Align *} And \ begin {Align *} \ lim _ {T \ rightarrow-\ infty} e ^ t ^ 2 = \ lim _ {T \ rightarrow \ infty} \ frac {t ^ 2} {e ^ t} = \ lim _ {T \ rightarrow \ infty} \ frac {2 t} {e ^ t} = \ lim _ {T \ rightarrow \ infty} \ frac {2} {e ^ t} = 0 \ end {Align *} left Derivative it is also $0 $, because the Left and Right derivatives are equal, the derivative of $ \ PSI (x) $ in $ x_0 + \ Delta $ is $0 $.

Next we will prove the continuity of the derivative. We will first consider the right limit, because when $ x> x_0 + \ Delta $ \ PSI (x) \ equiv 0 $, therefore, when $ x> x_0 + \ Delta $ \ PSI '(x) \ equiv 0 $, then $ \ lim _ {x \ rightarrow (x_0 + \ delta) ^ +} \ PSI '(x) = 0 $.

Consider the left limit. When $ x <x_0 + \ Delta $ has \ begin {Align *} \ PSI '(X) = e ^ {\ frac {1} {(X-x_0) ^ 2-\ Delta ^ 2 }}\ frac {-2 (x-x_0 )} {(X-x_0) ^ 2-\ Delta ^ 2) ^ 2} \ end {Align *} So \ begin {Align *} \ lim _ {x \ rightarrow (x_0 + \ delta) ^-} \ PSI '(X) & =\ LiM _ {x \ rightarrow (x_0 + \ delta) ^-} e ^ {\ frac {1} {(X-x_0) ^ 2-\ Delta ^ 2 }}\ frac {-2 (x-x_0)} {(X-x_0) ^ 2-\ Delta ^ 2) ^ 2 }\\=\ LiM _ {T \ rightarrow-\ infty} (-2) E ^ t ^ 2 \ left (\ frac { 1} {t} + \ Delta ^ 2 \ right) ^ {\ frac {1} {2 }}\\ left (t = \ frac {1} {(X-x_0) ^ 2-\ Delta ^ 2} \ right) \ & = 0 \ end {Align *}. Therefore, the left and right limits are equal. Therefore, $ \ PSI '(x) $ is continuous at $ x_0 + \ Delta $, in summary, $ \ PSI (x) $ has a continuous derivative at $ x_0 + \ Delta $. Likewise, it can prove $ \ PSI (X) $ has a continuous derivative at $ x_0-\ Delta $.

Iii. Derivation of the Euler-Laplace Equation

With the preliminary theorem, let's proceed to the subject and consider the following form of functional \ begin {Align *} Q [Y (x)] = \ int_a ^ B f (x, Y (x), y' (x) \ mbox {d} X \ end {Align *} where $ f (x, y (x), y' (x )) $ is a continuous function of three variables. When the vertex $ X, Y $ is in a bounded domain on the plane, $ f (x, y (x), y' (x )) $ and until the second-order partial derivatives are continuous.

The basic idea of extreme functional values is to applyFermat TheoremAssume that $ Q [F (x)] $ gets the extreme value at $ Y (x) $, and any function $ \ ETA (x) $ meets $ \ ETA () = \ ETA (B) = 0 $ with continuous derivatives, consider $ Y (x) $ functions in a certain field $ Y_1 (x) = y (X) + \ Alpha \ ETA (x) $, when $ \ Alpha $ is small enough, there should be $ Q [Y_1 (x)] \ Leq Q [Y (x)] $. Because the function $ Q [Y (x) + \ Alpha \ ETA (x)] = \ PSI (\ alpha) $ is also a function of $ \ Alpha $, therefore, the Fermat Theorem suggests that $ \ PSI '(0) = 0 $.

And \ begin {Align} \ PSI '(0) = \ PSI' (\ alpha) | _ {\ alpha = 0} & = \ left. \ frac {\ mbox {d }}{\ mbox {d} \ Alpha} \ left (\ int_a ^ B f (x, y (X) + \ Alpha \ ETA (x), y' (x) + \ Alpha \ ETA '(x) \ mbox {d} X \ right) \ right | _ {\ alpha = 0} \ nonumber \\\ label {Euler-Lago} & =\ left. \ left (\ int_a ^ B \ frac {\ mbox {d} f (x, y (x) + \ Alpha \ ETA (x), y' (X) + \ Alpha \ ETA '(x) }{\ mbox {d} \ Alpha} \ mbox {d} X \ right) \ right | _ {\ alpha = 0 }\\& = \ Left. \ left (\ int_a ^ B f_y \ frac {\ mbox {d} (Y (x) + \ Alpha \ ETA (x ))} {\ mbox {d} \ Alpha} + F _ {y'} \ frac {\ mbox {d} (y' (X) + \ Alpha \ ETA '(x) }{\ mbox {d} \ Alpha} \ mbox {d} X \ right) \ right | _ {\ alpha = 0} \ nonumber \\&=\ left. \ left (\ int_a ^ B f_y \ ETA (x) + F _ {y'} \ ETA '(x) \ mbox {d} X \ right) \ right | _ {\ alpha = 0} \ nonumber \\\&=\ int_a ^ B f_y \ ETA (X) \ mbox {d} X + \ int_a ^ B F _ {y'} \ ETA '(x) \ mbox {d} X \ nonumber \\& = \ Int_a ^ B f_y \ ETA (x) \ mbox {d} X + \ left. F _ {y'} \ ETA (x) \ right | _ A ^ B-\ int_a ^ B \ ETA (X) \ mbox {d} f _ {y'} \ (\ ETA (A) = \ ETA (B) = 0) \ nonumber \\\=\ int_a ^ B \ left (f_y-\ frac {\ mbox {d} f _ {y' }}{\ mbox {d} x} \ Right) \ ETA (x) \ mbox {d} X \ nonumber \ end {Align} among them (\ ref {Euler-Laplace }) the order of points and derivation is exchanged (the correctness will be proved later ). So \ begin {Align *} \ int_a ^ B \ left (f_y-\ frac {\ mbox {d} f _ {y' }}{\ mbox {d} x }\ right) \ ETA (X) \ mbox {d} x = 0 \ end {Align *} By the probe theorem \ begin {Align *} f_y = \ frac {\ mbox {d} f _ {y '}} {\ mbox {d} x} \ end {Align.

(\ Ref {Euler-Laplace }) in formula, the third equal sign can exchange points and the order of derivation is because \ begin {Align *} \ frac {\ PSI (\ Alpha + \ Delta \ alpha) -\ PSI (\ alpha) }{\ Delta \ Alpha} <=\ frac {q [Y (x) + (\ Alpha + \ Delta \ alpha) \ ETA (x)]-Q [Y (x) + \ Alpha \ ETA (x)]} {\ Delta \ Alpha }\\\&=\ int_a ^ B \ frac {f (x, y (x) + (\ Alpha + \ Delta \ alpha) \ ETA (x), y' (x) + (\ Alpha + \ Delta \ alpha) \ ETA '(x)-f (x, y (X) + \ Alpha \ ETA (x), y' (x) + \ Alpha \ ETA '(x)} {\ Delta \ alpha} \ Mbox {d} X \\\&=\ int_a ^ B F _ {\ Alpha} (X, Y (x) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA (x), y' (x) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA '(x )) \ mbox {d} X \ (\ Theta \ In (0, 1 )) the last equal sign in \ end {Align *} is because of the mean value theorem (the continuity and testability of $ F $ are required; otherwise, the mean value theorem cannot be used ). Therefore, we can see from the continuity of each partial derivative of $ F $ for $ \ forall \ Epsilon> 0 $, you can always find a sufficiently small $ \ Delta \ Alpha $ to make \ begin {Align *} | f _ {\ Alpha} (X, Y (X) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA (x), y' (x) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA '(x)-f _ {\ Alpha} (X, Y (x) + \ Alpha \ ETA (x), y' (X) + \ Alpha \ ETA '(x )) | <\ Epsilon \ end {Align *} So \ begin {Align *} & \\\\\ left | \ frac {\ PSI (\ Alpha + \ Delta \ alpha) -\ PSI (\ alpha) }{\ Delta \ Alpha}-\ int_a ^ B F _ {\ Alpha} (X, Y (X) + \ Alpha \ ETA (x), y' (x) + \ Alpha \ ETA '(x )) \ mbox {d} X \ right | \ & \ Leq \ int_a ^ B | f _ {\ Alpha} (X, Y (X) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA (x), y' (x) + (\ Alpha + \ Theta \ Delta \ alpha) \ ETA '(x)-f _ {\ Alpha} (X, Y (x) + \ Alpha \ ETA (x), y' (X) + \ Alpha \ ETA '(x) | \ mbox {d} X \\& <\ Epsilon (B-) \ end {Align *} When $ \ Epsilon \ rightarrow 0 $ \ Delta \ Alpha \ rightarrow 0 $, so \ begin {Align *} \ frac {\ mbox {d }}{\ mbox {d} \ Alpha} \ left (\ int_a ^ B f (x, y (X) + \ Alpha \ ETA (x), y' (x) + \ Alpha \ ETA '(x) \ mbox {d} X \ right) = \ int_a ^ B \ frac {\ mbox {d} f (x, y (x) + \ Alpha \ ETA (x), y' (X) + \ Alpha \ ETA '(x) }{\ mbox {d} \ Alpha} \ mbox {d} X \ end {Align *}

Iv. Specific applications

Next we will look at three specific applications.

• Find the curves with the shortest length among all the curves connected by $ () $ and $ () $.The shortest length of a straight lineIn the following example, we use the variational method to solve the problem, that is, to obtain the extreme values of the functional \ begin {Align *} \ int_0 ^ 1 \ SQRT {1 + y' ^ 2} \ mbox {d} X \ end {Align, note $ f (x, y (x), y' (x) = \ SQRT {1 + y' ^ 2} $, so \ begin {Align *} f_y & = 0 \ f _ {y'} & =\ frac {y'} {\ SQRT {1 + y' ^ 2 }}\ end {Align *} is known by the Euler-Laplace equation \ begin {Align *} \ frac {\ mbox {d }}{\ mbox {d} x} \ left (\ frac {y'} {\ SQRT {1 + y' ^ 2} \ right) = 0 \ end {Align *} So \ begin {Align *} \ frac {y'} {\ SQRT {1 + y' ^ 2 }}= C \ end {Align *} $ C $ is a constant, sorted \ begin { Align *} y' = \ PM \ frac {c} {\ SQRT {1-C ^ 2 }}\ end {Align *} the slope of the optimal curve is a constant, that is, a straight line, the boundary condition is $ Y = x $.
• The target function of the regression problem can be written as \ begin {Align *} \ min_y \ mathbb {e} [l] = \ int _ {\ boldsymbol {x} \ int_t L (t, Y (\ boldsymbol {x}) P (\ boldsymbol {x}, T) \ mbox {d} t \ mbox {d} \ boldsymbol {x} \ end {Align *} where $ y $ is the function to fit the observed data, $ T $ is the observed value of the sample $ \ boldsymbol {x} $, $ L $ is the loss function, and the target is to find the optimal $ y $ to minimize the expected loss. In particular, if the loss function is a square loss, that is, $ L = (Y (\ boldsymbol {x})-T) ^ 2 $, then remember \ begin {Align *} f (x, y (x), y '(x) = \ int_t (Y (\ boldsymbol {x})-T) ^ 2 P (\ boldsymbol {x}, T) \ mbox {d} t \ end {Align *} So \ begin {Align *} f_y & =\ int_t 2 (Y (\ boldsymbol {x})-T) P (\ boldsymbol {x}, T) \ mbox {d} t \ f _ {y'} & = 0 \ end {Align *} known by the Euler-lagiron equation \ begin {Align *} \ int_t t p (\ boldsymbol {x }, t) \ mbox {d} t = \ int_t y (\ boldsymbol {x}) P (\ boldsymbol {x}, T) \ mbox {d} t = Y (\ boldsymbol {x}) P (\ boldsymbol {x }) \ end {Align *} So \ begin {Align *} y (\ boldsymbol {x}) = \ frac {\ int_t t p (\ boldsymbol {x}, T) \ mbox {d} t} {P (\ boldsymbol {x}) }=\ int_t t p (t | \ boldsymbol {x }) \ mbox {d} t = \ mathbb {e} [T | \ boldsymbol {x}] \ end {Align *} is the least squares regression.The optimal fitting function is the expected condition for a given input..
• \ Begin {Align} \ label {Gauss 1} \ int _ {-\ infty} ^ \ infty p (x) \ mbox {d} X & = 1 \ label {Gauss 2} \ int _ {-\ infty} ^ \ infty x P (x) \ mbox {d} X & =\ MU \ label {Gauss 3} \ int _ {-\ infty} ^ \ infty (X-\ mu) ^ 2 p (x) \ mbox {d} X & = \ Sigma ^ 2 \ end {Align} probability distribution, whereThe largest entropy is normal distribution..
  Due to constraints optimization problems, the introduction of the resource multiplication operator $ \ lambda_1 $, $ \ lambda_2 $, and $ \ lambda_3 $, the target function can be written as \ begin {Align *} \ min _ {p (x)} \ int _ {-\ infty} ^ \ infty p (x) \ ln p (x) \ mbox {d} X-\ lambda_1 \ left (\ int _ {-\ infty} ^ \ infty p (x) \ mbox {d} X-1 \ right)-\ lambda_2 \ left (\ int _ {-\ infty} ^ \ infty x P (x) \ mbox {d} X-\ MU \ right)-\ lambda_3 \ left (\ int _ {-\ infty} ^ \ infty (X-\ mu) ^ 2 p (x) \ mbox {d} X-\ Sigma ^ 2 \ right) \ end {Align *} about $ p (x) $ is proposed separately as $ f (x, P (x), p '(x) $, then \ begin {Align *} f (x, p (x), p' (x )) = p (x) \ ln p (x)-\ lambda_1 p (x)-\ lambda_2 x P (x)-\ lambda_3 (X-\ mu) ^ 2 p (x) \ end {Align *} So \ begin {Align *} F_p & = \ ln p (x) + 1-\ lambda_1-\ lambda_2 X-\ lambda_3 (X-\ mu) ^ 2 \ f _ {P'} & = 0 \ end {Align *} known by the Euler-Laplace equation \ begin {Align} \ label {Gauss 4} p (x) = \ mbox {exp} (-1 + \ lambda_1 + \ lambda_2 x + \ lambda_3 (X-\ mu) ^ 2) \ end {Align} Will (\ ref {Gauss 4}) back to (\ ref {Gauss 1}), (\ ref {Gauss 2}), (\ ref {Gauss 3 }) make $ Y = x-\ Mu $ get \ begin {Align} \ label {Gauss 5} \ int _ {-\ infty} ^ \ infty \ mbox {exp }(- 1 + \ lambda_1 + \ lambda_2 (Y + \ mu) + \ lambda_3 y ^ 2) \ mbox {d} y & = 1 \ label {Gauss 6} \ int _ {-\ infty} ^ \ infty (Y + \ mu) \ mbox {exp} (-1 + \ lambda_1 + \ lambda_2 (Y + \ mu) + \ lambda_3 y ^ 2) \ mbox {d} y & =\ MU \ label {Gauss 7} \ int _ {-\ infty} ^ \ infty y ^ 2 \ mbox {ex P} (-1 + \ lambda_1 + \ lambda_2 (Y + \ mu) + \ lambda_3 y ^ 2) \ mbox {d} y & =\ Sigma ^ 2 \ end {Align} (\ ref {Gauss 6}) $-\ Mu $ (\ ref {Gauss 5 }) \ begin {Align} \ label {Gauss 8} \ int _ {-\ infty} ^ \ infty Y \ mbox {exp} (-1 + \ lambda_1 + \ lambda_2 \ mu + \ lambda_2 y + \ lambda_3 y ^ 2) \ mbox {d} y = 0 \ end {Align} dropped the non-zero constant item $ \ mbox {exp} (-1 + \ lambda_1 + \ lambda_2 \ mu) $, then $2 \ lambda_3 $ (\ ref {Gauss 8}) $ + \ lambda_2 $ (\ ref {Gauss 5}) \ begin {Align *} \ lambda_2 =\int _ {-\ infty} ^ \ infty (\ lambda_2 + 2 \ lambda_3 y) \ mbox {exp} (\ lambda_2 y + \ lambda_3 y ^ 2) \ mbox {d} y = \ lim _ {Y \ rightarrow \ infty} \ mbox {exp} (\ lambda_2 y + \ lambda_3 y ^ 2) -\ lim _ {Y \ rightarrow-\ infty} \ mbox {exp} (\ lambda_2 y + \ lambda_3 y ^ 2) \ end {Align *} note that $ \ lambda_2 $ and $ \ lambda_3 $ are finite constants, so they tend to be infinite with $ y $, $ \ lambda_2 y + \ lambda_3 y ^ 2 $ either trend $ \ infty $, or trend $-\ infty $, and \ begin {alig N *} \ lim _ {x \ rightarrow \ infty} \ mbox {exp} (x) & =\ infty, \ lim _ {x \ rightarrow-\ infty} \ mbox {exp} (X) = 0 \ end {Align *} So it may only be $ \ lambda_2 = 0 $ and \ begin {Align *} \ lim _ {Y \ rightarrow \ infty} \ lambda_2 y + \ lambda_3 y ^ 2 & =-\ infty, \ lim _ {Y \ rightarrow-\ infty} \ lambda_2 y + \ lambda_3 y ^ 2 =-\ infty \ end {Align *} So $ \ lambda_3 <0 $. At this time, by (\ ref {Gauss 5 }) \ begin {Align *} 1 & =\int _ {-\ infty} ^ \ infty \ mbox {exp} (-1 + \ lambda_1 + \ lambda_3 y ^ 2) \ mbox {d} y \\\&=\ mbox {exp} (-1 + \ lambda_1) \ int _ {-\ infty} ^ \ infty \ mbox {exp} \ left (-\ frac {(\ SQRT {-2 \ lambda_3} y) ^ 2} {2} \ right) \ mbox {d} Y \ & =\ mbox {exp} (-1 + \ lambda_1) \ frac {1} {\ SQRT {-2 \ lambda_3 }\int _ {-\ infty} ^ \ infty \ mbox {exp} \ left (-\ frac {z ^ 2} {2} \ right) \ mbox {d} Z \ left (Z = \ SQRT {-2 \ lambda_3} Y \ right) \ & =\ mbox {exp} (-1 + \ lambda_1) \ frac {\ SQRT {2 \ PI }}{\ SQRT {-2 \ lambda_3 }}\ end {Align *} Then \ begin {Align} \ label {Gauss 9 }\ mbox {exp} (-1 + \ lambda_1) =\ SQRT {-\ frac {\ lambda_3 }{\ PI }}\ end {Align} by (\ ref {Gauss 7 }) \ begin {Align *} \ Sigma ^ 2 & =\ SQRT {-\ frac {\ lambda_3 }{\ PI }}\ int _ {-\ infty} ^ \ infty y ^ 2 \ mbox {exp} (\ lambda_3 y ^ 2) \ mbox {d} Y \ & = 2 \ SQRT {-\ frac {\ lambda_3 }{\ PI }}\ int_0 ^ \ infty y ^ 2 \ mbox {exp }( \ lambda_3 y ^ 2) \ mbox {d} y \\\&=\ SQRT {-\ frac {\ lambda_3 }{\ PI }}\ frac {1 }{\ lambda_3 }\ int_0 ^ \ infty y (2 \ lambda_3 Y \ mbox {exp} (\ lambda_3 y ^ 2 )) \ mbox {d} y \\&=\ left. \ SQRT {-\ frac {\ lambda_3} {\ PI }}\ frac {Y \ mbox {exp} (\ lambda_3 y ^ 2 )} {\ lambda_3} \ right | _ 0 ^ \ infty-\ SQRT {-\ frac {\ lambda_3} {\ PI }}\ frac {1} {\ lambda_3} \ int_0 ^ \ infty \ mbox {exp} (\ lambda_3 y ^ 2) \ mbox {d} Y \ & =- \ SQRT {-\ frac {\ lambda_3 }{\ PI }}\ frac {1 }{\ lambda_3} \ frac {1} {2} \ SQRT {-\ frac {\ PI} {\ lambda_3 }\\\& =-\ frac {1} {2 \ lambda_3} \ end {Align *} Then \ begin {Align} \ label {Gauss 10} \ lambda_3 =-\ frac {1} {2 \ Sigma ^ 2} \ end {Align} Will (\ ref {Gauss 9}) and (\ ref {Gauss 10}) (\ ref {Gauss 4}) can get \ begin {Align *} p (x) = \ mbox {exp} (-1 + \ lambda_1 + \ lambda_3 (X-\ mu) ^ 2) = \ SQRT {-\ frac {\ lambda_3} {\ PI }}\ mbox {exp} (\ lambda_3 (X-\ mu) ^ 2) = \ frac {1} {\ SQRT {2 \ pi \ Sigma ^ 2 }}\ mbox {exp} \ left (-\ frac {(X-\ mu) ^ 2} {2 \ Sigma ^ 2} \ right) \ end {Align *}

Introduction to Variational Methods