数学专业外文翻译--多元函数的极值

外文文献EXTREME VALUES OF FUNCTIONS OF SEVERALREAL VARIABLES1. Stationary PointsDefinition 1.1 Let n R D ⊆ and R D f →:. The point a D a ∈ is said to be:(1) a local maximum if )()(a f x f ≤for all points x sufficiently close to a ;(2) a local minimum if )()(a f x f ≥for all points x sufficiently close to a ;(3) a global (or absolute) maximum if )()(a f x f ≤for all points D x ∈;(4) a global (or absolute) minimum if )()(a f x f ≥for all points D x ∈;;(5) a local or global extremum if it is a local or global maximum or minimum. Definition 1.2 Let n R D ⊆ and R D f →:. The point a D a ∈ is said to be critical or stationary point if 0)(=∇a f and a singular point if f ∇ does not exist at a .Fact 1.3 Let n R D ⊆ and R D f →:.If f has a local or global extremum at the point D a ∈, then a must be either:(1) a critical point of f , or(2) a singular point of f , or(3) a boundary point of D .Fact 1.4 If f is a continuous function on a closed bounded set then f is bounded and attains its bounds.Definition 1.5 A critical point a which is neither a local maximum nor minimum is called a saddle point.Fact 1.6 A critical point a is a saddle point if and only if there are arbitrarily small values of h for which )()(a f h a f -+ takes both positive and negative values.Definition 1.7 If R R f →2: is a function of two variables such that all second order partial derivatives exist at the point ),(b a , then the Hessian matrix of f at ),(b a is the matrix⎪⎪⎭⎫ ⎝⎛=yy yxxy xx f f f f H where the derivatives are evaluated at ),(b a . If R R f →3: is a function of three variables such that all second order partial derivatives exist at the point ),,(c b a , then the Hessian of f at ),,(c b a is the matrix⎪⎪⎪⎭⎫ ⎝⎛=zz zy zx yz yy yx xz xy xx f f f f f f f f f H where the derivatives are evaluated at ),,(c b a .Definition 1.8 Let A be an n n ⨯ matrix and, for each n r ≤≤1,let r A be the r r ⨯ matrix formed from the first r rows and r columns of A .The determinants det(r A ),n r ≤≤1,are called the leading minors of ATheorem 1.9(The Leading Minor Test). Suppose that R R f →2:is a sufficiently smooth function of two variables with a critical point at ),(b a and H the Hessian of f at ),(b a .If 0)det(≠H , then ),(b a is:(1) a local maximum if 0>det(H 1) = f xx and 0<det(H )=2xy yy xx f f f -;(2) a local minimum if 0<det(H 1) = f xx and 0<det(H )=2xy yy xx f f f -;(3) a saddle point if neither of the above hold.where the partial derivatives are evaluated at ),(b a .Suppose that R R f →3: is a sufficiently smooth function of three variables with a critical point at ),,(c b a and Hessian H at ),,(c b a .If 0)det(≠H , then ),,(c b a is:(1) a local maximum if 0>det(H 1), 0<det(H 2) and 0>det(H 3);(2) a local minimum if 0<det(H 1), 0<det(H 2) and 0>det(H 3);(3) a saddle point if neither of the above hold.where the partial derivatives are evaluated at ),,(c b a .In each case, if det(H )= 0, then ),(b a can be either a local extremum or a saddleExample. Find and classify the stationary points of the following functions:(1) ;1),,(2224+++++=xz z y y x x z y x f(2) ;)1()1(),(422++++=x y x y y x fSolution. (1) 1),,(2224+++++=xz z y y x x z y x f ,so)24),(3z xy x y x f ++=∇（i )2(2y x ++j )2(x z ++kCritical points occur when 0=∇f ,i.e. when(1) z xy x ++=2403(2) y x 202+=(3) x z +=20Using equations (2) and (3) to eliminate y and z from (1), we see that 021433=--x x x or 0)16(2=-x x ,giving 0=x ,66=x and 66-=x .Hence we have three stationary points: ）（0,0,0,）（126,121,66-- and ）（126,121,66--. Since y x f xx 2122+=,x f xy 2=,1=xz f ,2=yy f ,0=yz f and 2=zz f ,the Hessian matrix is⎪⎪⎪⎭⎫ ⎝⎛+=201022122122x x y x H At ）（126,121,66--, ⎪⎪⎪⎪⎭⎫ ⎝⎛=201023/613/66/11H which has leading minors 611>0, 039631123/63/66/11det >=-=⎪⎪⎭⎫ ⎝⎛ And det 042912322>=--=H .By the Leading Minor Test, then, ）（126,121,66--is a local minimum. At ）（126,121,66--, ⎪⎪⎪⎪⎭⎫ ⎝⎛--=201023/613/66/11H which has leading minors 611>0,039631123/63/66/11det >=-=⎪⎪⎭⎫ ⎝⎛ And det 042912322>=--=H .By the Leading Minor Test, then, ）（126,121,66--is also a local minimum. At ）（0,0,0, the Hessian is⎪⎪⎪⎭⎫ ⎝⎛=201020100HSince det 2)(-=H , we can apply the leading minor test which tells us that this is a saddle point since the first leading minor is 0. An alternative method is as follows. In this case we consider the value of the expressionhl l k k h h l k h f f D ++++=+++-=22240,0,00,0,0）（）（，for arbitrarily small values of h, k and l. But for very small h, k and l , cubic terms and above are negligible in comparison to quadratic and linear terms, sothat hl l k D ++≈22.If h, k and l are all positive, 0>D . However, if 0=k and0<h and h l <<0,then 0<D .Hence close to ）（0,0,0,f both increases and decreases, so ）（0,0,0 is a saddle point.(2) 422)1()1(),(++++=x y x y y x f so))1(4)1(2(),(3+++=∇x y x y x f i ))1(2(2+++x y j .Stationary points occur when 0=∇f ,i.e. at )0,1(-.Let us classify this stationary point without considering the Leading Minor Test (in this case the Hessian has determinant 0 at )0,1(- so the test is not applicable). Let.0,10,1422h k h k k h f f D ++=++---=）（）（Completing the square we see that .43)2(222h h k D ++=So for any arbitrarily small values of h and k , that are not both 0, 0>D and we see that f has a local maximum at )0,1(-.2. Constrained Extrema and Lagrange MultipliersDefinition 2.1 Let f and g be functions of n variables. An extreme value of f (x )subject to the condition g (x) = 0, is called a constrained extreme value and g (x ) = 0 is called the constraint.Definition 2.2 If R R f n →: is a function of n variables, the Lagrangian function of f subject to the constraint 0),,,(21=n x x x g is the function of n+1 variables),,,,(),,,(),,,,(212121n n n x x x g x x x f x x x L λλ+=where is known as the Lagrange multiplier.The Lagrangian function of f subject to the k constraints0),,,(21=n i x x x g ,k i ≤≤1, is the function with k Lagrange multipliers,i λk i ≤≤1,∑=+=ki n n n x x x g x x x f x x x L 1212121),,,(),,,(),,,,( λλTheorem 2.3 Let R R f →2: and ),(00y x P = be a point on the curve C, withequation g(x,y) = 0, at which f restricted to C has a local extremum.Suppose that both f and g have continuous partial derivatives near to P and that P is not an end point of C and that 0),(00≠∇y x g . Then there is some λ such that ),,(000z y x is a critical point of the Lagrangian Function),(),(),,(y x g y x f y x L λλ+=.Proof. Sketch only. Since P is not an end point and 0≠∇g ,C has a tangent at P with normal g ∇.If f ∇ is not parallel to g ∇at P , then it has non-zero projection along this tangent at P .But then f increases and decreases away from P along C ,so P is not an extremum. Hence f ∇and g ∇are parallel and there is some¸such that g f ∇-=∇λ and the result follows.Example. Find the rectangular box with the largest volume that fits inside the ellipsoid 1222222=++cz b y a x ,given that it sides are parallel to the axes. Solution. Clearly the box will have the greatest volume if each of its corners touch the ellipse. Let one corner of the box be corner (x, y, z) in the positive octant, then the box has corners (±x,±y,±z) and its volume is V= 8xyz .We want to maximize V given that 01222222=-++cz b y a x . (Note that since the constraint surface is bounded a max/min does exist). The Lagrangian is⎪⎪⎭⎫ ⎝⎛-+++=18),,,(222222c z b y a x xyz z y x L λλ and this has critical points when 0=∇L , i.e. when,.280,28022b y zx y L a x yz x L λλ+=∂∂=+=∂∂=⎪⎪⎭⎫ ⎝⎛-++=∂∂=+=∂∂=10,2802222222c z b y a x z L c z xy z L λ (Note that λL will always be the constraint equation.) As we want to maximize V we can assume that 0≠xyz so that 0,,≠z y x .)Hence, eliminating λ, we get,444222zxy c y zx b x yz a -=-=-=λ so that 2222b x a y = and .2222c y b z =But then 222222c z b y a x ==so 2222222231ax c z b y a x =++= or 3a x =,which implies that 3b y = and 3c z = (they are all positive by assumption). So L has only one stationary point ),3,3,3(λc b a (for some value of λ, which we could work out if we wanted to). Since it is the only stationary point it must the required max and the max volume is3383338abc c b a =.中文译文 多元函数的极值1. 稳定点定义1.1 使n R D ⊆并且R D f →:. 对于任意一点D a ∈有以下定义:(1)如果)()(a f x f ≤对于所有x 充分地接近a 时，则)(a f 是一个局部极大值;(2)如果)()(a f x f ≥对于所有x 充分地接近a 时，则)(a f 是一个局部极小值;(3)如果)()(a f x f ≤对于所有点D x ∈成立，则)(a f 是一个全局极大值（或绝对极大值）;(4) 如果)()(a f x f ≥对于所有点D x ∈成立，则)(a f 是一个全局极小值（或绝对极小值）; (5) 局部极大（小）值统称为局部极值；全局极大（小）值统称为全局极值.定义 1.2 使n R D ⊆并且R D f →:.对于任意一点D a ∈，如果0)(=∇a f ，并且对于任意奇异点a 都不存在f ∇，则称a 是一个关键点或稳定点.结论 1.3 使n R D ⊆并且R D f →:.如果f 有局部极值或全局极值对于一点D a ∈, 则a 一定是:(1)函数f 的一个关键点, 或者(2)函数f 的一个奇异点, 或者(3)定义域D 的一个边界点.结论 1.4 如果函数f 是一个在闭区间上的连续函数，则f 在区间上有边界并且可以取到边界值.定义 1.5 对于任一个关键点a ，当a 既不是局部极大值也不是局部极小值时，a 叫做函数的鞍点.结论 1.6 对于一个关键点a 是鞍点当且仅当h 任意小时，对于函数)()(a f h a f -+取正值和负值.定义 1.7 如果R R f →2: 是二元函数，并且在点),(b a 处所有二阶偏导数都存在，则则根据函数f 在点),(b a 处导数，有f 在点),(b a 处的Hessian 矩阵为：⎪⎪⎭⎫ ⎝⎛=yy yx xy xxf ff f H . 推广：如果R R f →3: 是三元函数，并且在点),,(c b a 处所有二阶偏导数都存在，则根据函数f 在点),,(c b a 处导数，有f 在点),,(c b a 处的Hessian 矩阵为：⎪⎪⎪⎭⎫ ⎝⎛=zz zyzxyz yy yxxz xy xxf f f f f f f f f H . 定义 1.8 矩阵A 是n n ⨯ 阶矩阵，并且对于每一个都有n r ≤≤1,从矩阵A 中选取左上端的r 行和r 列，令其为r r ⨯阶的矩阵r A .则行列式det(r A ),n r ≤≤1,叫做矩阵A 的顺序主子式.定理 1.9 假如R R f →2:是一个充分光滑的二元函数，且在点),(b a 处稳定，其Hessian 矩阵为H .如果0)det(≠H ，则根据偏导数判定),(b a 点是：(1) 一个局部极大值点， 如果0>det(H 1) = f xx 并且0<det(H )=2xy yy xx f f f -; (2) 一个局部极小值点， 如果0<det(H 1) = f xx 并且0<det(H )=2xy yy xx f f f -;(3) 一个鞍点，如果点),(b a 既不是局部极大值点也不是局部极小值点. 假如R R f →3:是一个充分光滑的三元函数，且在点),,(c b a 处稳定，其Hessian 矩阵为H .如果0)det(≠H ，则根据偏导数判定),,(c b a 点是： (1) 一个局部极大值点， 如果当0>det(H 1), 0<det(H 2) 并且 0>det(H 3)时; (2) 一个局部极小值点， 如果当0<det(H 1), 0<det(H 2) 并且 0>det(H 3)时; (3) 一个鞍点，如果点),,(c b a 既不是局部极大值点也不是局部极小值点. 在不同的情况下 ,当det(H )= 0时, 点),(b a 是一个局部极值点，或者是一个鞍点.例. 确定下列函数的稳定点并说明是哪一类点: (1) ;1),,(2224+++++=xz z y y x x z y x f (2) ;)1()1(),(422++++=x y x y y x f 解. (1) 1),,(2224+++++=xz z y y x x z y x f ,so)24),(3z xy x y x f ++=∇（i )2(2y x ++j )2(x z ++k当0=∇f 时有稳定点,也就是说, 当(1) z xy x ++=2403 (2) y x 202+= (3) x z +=20时,将方程(2)和方程(3)带入到方程(1)可以消去变量y 和z, 由此可以得到021433=--x x x 即0)16(2=-x x ,得0=x ,66=x 和66-=x .因此我们可以得到函数的三个稳定点:）（0,0,0,）（126,121,66--和）（126,121,66--. 又因为y x f xx 2122+=,x f xy 2=,1=xz f ,2=yy f ,0=yz f 和2=zz f ,则Hessian 矩阵为⎪⎪⎪⎭⎫⎝⎛+=201022122122x x y x H在点）（126,121,66--处, ⎪⎪⎪⎪⎭⎫⎝⎛=201023/613/66/11H则顺序主子式611>0, 039631123/63/66/11>=-=并且行列式042912322>=--=H .根据主子式判定方法，则点）（126,121,66--是一个局部极小值点.在点）（126,121,66--处, ⎪⎪⎪⎪⎭⎫⎝⎛--=201023/613/66/11H则顺序主子式 611>0,039631123/63/66/11>=-=-- 并且行列式042912322>=--=H .根据主子式判定方法，则点）（126,121,66--也是一个极小值点.在点）（0,0,0处，Hessian 矩阵为⎪⎪⎪⎭⎫⎝⎛=201020100H因此det 2)(-=H ,根据主子式判定方法，第一主子式为0，由此我们可以知道该点是一个鞍点. 下面是另一种计算方法，在这种情况下，我们考虑现在下面函数表达式hl l k k h h l k h f f D ++++=+++-=22240,0,00,0,0）（）（，的值，对于任意h, k 和l 无限小时. 担当h, k 和l 非常小时, 三次及三次以上方程相对线性二次方程时可忽略不计,则原方程可为hl l k D ++≈22.当h, k 和l 都为正时,0>D .然而， 当0=k 、0<h 和h l <<0,则0<D .因此当接近）（0,0,0时,f 同时增加或者同时减少, 所以 ）（0,0,0是一个鞍点. (2) 422)1()1(),(++++=x y x y y x f so))1(4)1(2(),(3+++=∇x y x y x f i ))1(2(2+++x y j .当0=∇f 时有稳定点,也就是说, 当在)0,1(-时.现在我们在不考虑主子式判定方法的情况下为该稳定点进行分类（因为在)0,1(-时Hessian 矩阵的行列式为0，所以该判定方法在此刻无法应用）.令.0,10,1422h k h k k h f f D ++=++---=）（）（配成完全平方的形式为.43)2(222h h k D ++=所以对h 和k 为任意小时（h 和k 都不为0），有0>D ，因此我们可以确定函数f 在点)0,1(-处有局部极大值.2. 条件极值和Lagrange 乘数法定义 2.1 函数f 和函数g 都是n 元函数.对于限制在条件g (x) = 0下的函数f (x )的极值叫做函数的条件极值，函数g (x ) = 0叫做限制条件.定义 2.2 如果函数R R f n →: 是一个n 元函数, 则对应于函数f 的Lagrange 函数在限制条件0),,,(21=n x x x g 下的函数是一个n +1元函数),,,,(),,,(),,,,(212121n n n x x x g x x x f x x x L λλ+=这就是著名的Lagrange 乘数法.对应于函数f 的Lagrange 函数在k 个限制条件0),,,(21=n i x x x g ,k i ≤≤1时, 带有k 个i λk i ≤≤1,的Lagrange 函数为：∑=+=ki n n n x x x g x x x f x x x L 1212121),,,(),,,(),,,,( λλ定理 2.3 使R R f →2:并且),(00y x P =是曲线C 上的一个点, 有方程 g(x,y) = 0成立，则在限制条件C 上函数f 有局部极值.假设函数f 和函数g 在点P 都有连续的偏导数，点P 不是曲线C 的端点，且0),(00≠∇y x g . 因此存在λ的值使得点),,(000z y x 是Lagrange 函数的关键点),(),(),,(y x g y x f y x L λλ+=.证明.仅仅描述. 因为点P 不是曲线C 的端点，且0≠∇g ,则曲线C 在点P 处的切线与g ∇有关.如果f ∇在点P 处与g ∇平行，则函数在点P 处的切线有非零值.但另一方面函数 f 的值随着P 在C 的运动增加减小,所以点P 不是极值点. 因为f ∇和g ∇平行，所以存在λ使得g f ∇-=∇λ成立.例. 求内接于椭球1222222=++cz b y a x 的体积最大的长方体的体积，长方体的各个面平行于坐标面解：明显地，当长方体的体积最大时，长方体的各个顶点一定在椭球上. 设长方体的一个顶点坐标为(x, y, z) (x>0, y>0, z>0), 则长方体的其他顶点坐标分别为(±x,±y,±z)，并且长方体的体积为V= 8xyz.我们要求V 在条件01222222=-++cz b y a x 下的最大值. (注意：因为约束条件是有边界的，故其一定存在极大或者极小值). 其Lagrange 函数为⎪⎪⎭⎫⎝⎛-+++=18),,,(222222c z b y a x xyz z y x L λλ并且存在稳定点当0=∇L 时，也就是说，当,.280,280,280222cz xy z L b y zx y L a x yz x L λλλ+=∂∂=+=∂∂=+=∂∂=⎪⎪⎭⎫ ⎝⎛-++=∂∂=10222222c z b y a x z L 时.(注意：λL 是约束方程.要想求得体积V 的最大值，假设0≠xyz ，则可得0,,≠z y x .)因此, 用其他式子表示λ, 我们可以得到,444222zxyc y zx b x yz a -=-=-=λ 消去λ，有2222b x a y =和.2222c y b z =进而得出 222222cz b y a x ==，因此有2222222231ax c z b y a x =++=或者得出3a x =,同理可得出3by =和3c z = (根据假设可得x, y, z 都是正值).所以函数 L 有且仅有一个稳定点),3,3,3(λc b a (λ为某一计算可得到的常数). 又因为该点是函数L 的唯一稳定点，则该稳定点一定是所要求的最大值点，故其体积的最大值为3383338abcc b a =.。