数学及应用数学专业外文翻译--多元函数的极值

1、外文原文EXTREME VALUES OF FUNCTIONS OF SEVERALREAL VARIABLES1. Stationary PointsDefinition 1.1 Let D 二 Rn and f:D'R. The point a a D is said to be:(1) a local maximum if f (x) < f (a) for all points x sufficiently close to a;(2) a local minimum if f (x) _ f (a) for all points x sufficiently close

2、 to a;(3) a global (or absolute) maximumif f (x)三 f (a) for all points x D;(4) a global (or absolute) minimum if f (x) _ f (a) for all points x D;(5) a local or global extremum if it is a local or global maximum or minimum.Definition 1.2 Let D 二 Rn and f:DrR. The point a a D is said to be critical o

3、r stationary point if f (a) = 0 and a singular point if f does not exist at a.Fact 1.3 Let D 三 Rn and f : D R .If f has a local or global extremum at the point a D , 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 contin

4、uous function on a closed bounded set thenf 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 f (a h)

5、 - f (a) takes both positive and negative values.2Definition 1.7 If f : R R is a function of two variables such that all second order partial derivatives exist at the point (a,b), then theHessian matrix of f at(a,b) is the matrix where the derivatives are evaluated ata, b).fxxf xyf yxf yy3If f : R R

6、 is a function of three variables such that all second order partial derivatives exist at the point (a,b, c), then the Hessian off at (a,b,c) is the matrixfxxfxyH =f yxfyyfzxfzyfxzfzzfyzwhere the derivatives are evaluated ata, b, c).Definition 1.8 Let A be an n n matrix and, for each 1_r_n,let Ar be

7、 the r r matrix formed from the first r rows and r columns of A .The determinants det(Ar ),1 - r - n,are called theleading minors of A2Theorem 1.9(The Leading Minor Test). Suppose that f: R > Ris a sufficiently smooth function of two variables with a critical point at (a,b) and H the Hessian off

8、at(a,b) .If det(H ) : 0 , then (a, b) is:2(1) a local maximum if 0>det(Hi) = fxx and 0<det(H)= fxxfyy - fxy；2(2) a local minimum if 0< det(Hi) = fxx and 0<det(H)= fxx fyy - fxy ；(3) a saddle point if neither of the above hold.where the partial derivatives are evaluated at(a,b).3Suppose t

9、hat f : R R is a sufficiently smooth function of three variables with a critical point at (a,b,c)and Hessian H at(a,b,c) .If det(H ) = 0 , then (a, b, c) is:(1) a local maximum if 0>det(Hi), 0<det(H2)and 0>det(H3)；(2) a local minimum if 0< det(Hi), 0<det(H2)and 0>det(H3);(3) a sadd

10、le point if neither of the above hold.where the partial derivatives are evaluated at(a,b,c).In each case, ifdet(H)= 0, then (a,b) can be either a local extremum or a saddleoKe y Points. A continuous function on a closed bounded set is bounded and achieves its bounds. To find the extreme values of a

11、function on a closed bounded set it is necessary to consider the value of the function at stationary points( "f = 0), singular points ，Vf ,、( does not exist) and boundary points(points on the edge of the set). Stationary points can be classified as local maxima , local minim a or saddle points.

12、 If The Leading Minor Test 1.9 is not applicable, the stationary point must be classified by directly applying Definition 1.1 and Fact 1.6. For example in the two variable case, if f has a stationary point at (a,b),we consider the sign off (a +h,b + h) - f (a,b)for arbitrarily small, positive and ne

13、gative values of h and k (that are not both zero).Example. Find and classify the stationary points of the following functions:一 一4222(1) f (x, y, z) = x x y y z xz 1;224(2) f (x, y) =y (x 1) y (x 1);Solution. (1) f (x,y, z) =x4 x2y y2 z2 xz 1,so'、f (x, y) = (4x3 2xy z) i (x2 2y) j (2z x) kCritic

14、al points occur when f = 0 ,i.e. when3(1) 0 = 4x2xy z_2(2) 0 = x2y0 = 2z xUsing equations (2) and (3) to eliminate y and z from (1), we see3312that4x -x -x = 0or x(6x -1)=0,giving x = 0,x =火 and x .-丑.Hence 61,6、we have three stationary Points:(0，0，0)，(T，-，-)and/616、( ，一, ).612 122Since fxx =12x 2y

15、, fxy = 2x, f* =1, fyy = 2, fyz = 0 andfzz =2,the Hessianmatrix is二 2二12x2+2y2x1H = 2x20102<J-/,616、At (， ，),6121211/6 <6/3 1H =恋/320102<Jwhich has leading minors 11 >0,det11/6e'6/346/3、 11 6 =2 J 39=3 0A , ,22And det H =312-2=4 0 .By the Leading Minor Test, then, 9/、61、. 6、(,)61212i

16、s a local minimum.At (- 61. 6、，一,),612 1211/6-6/31-6/320102>which has leading minors 11 >0,76/3、 11det11/6.6/36=3 022 12And det H =-2=4 0.By the Leading Minor Test, then, 39(- - ,-,) is also a local minimum.612 12At (0,0,0) , the Hessian is001、H = 0 2 0J 0 2,Since det( H ) 二 一2 , we can apply

17、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 expression4222D=f (0,0,0) - f (0 + h,0 + k,0 + l) =h +hk + k +l + hl ,for arbitrarily small values of h, k and l. But f

18、or very small h, k and l, cubic terms and above are negligible in comparison to quadratic and linear terms, so22thatD ： k l hl .If h, k and l are all positive, D 0 . However, if k=0 and h <0 and 0 <l < h ,then D <0.Hence close to (0,0,0) , f both increases and decreases, so(0,0,0) is a s

19、addle point.(2) f (x, y) = y2 (x 1)2y (x 1)4sof(x, y) =(2(x 1)y 4(x 1)3) i (2y (x2 1) j.Stationary points occur when f =0 ,i.e. at (7,0).Let us classify this stationary point without considering the Leading Minor Test (in this case the Hessian has determinant 0 at - 1,0) so the test is not applicabl

20、e). Let224D=f( 1,0) f( 1+h,0+k)=k +hk+h.3h2j- .So for any arbitrarily smallh2 2Completing the square we see thatD = (k ) 2values of h and k, that are not both 0, D 0 and we see that f has a local maximum at (-1,0).2. Constrained Extrema and Lagrange MultipliersDefinition 2.1 Let f and g be functions

21、 of n variables. An extreme value of f(x) subject to the condition g(x) = 0, is called a constrained extreme valueand g(x) = 0 is called the constraint.Definition 2.2 If f : Rn R is a function of n variables, the Lagrangian function of f subject to the constraint g(x1,x2,xn) =0 is the function of n+

22、1variablesL ( x1, x2 , , x n 1 ) = f ( x1 , x2 , xn ) . .3g (x1, x2 , xn，where is known as theLagrange multiplier.The Lagrangian function of f subject to thek constraintsgi(x1,x2, , ,xn) =0 ,1 三i 三k , is the function with kLagrange multipliers,i ,1 < i < kkL(Xi,X2, ,xn, 1)= f (Xi,X2； ,xn) 二-.g

23、(x1,x2； ,xn) i=1Key Points. To find the extreme values of f subject to the constraint g(x) = 0:Xd,Xo： ,Xn(1) calculate L, remembering that it is a function of the n+1 variables 1, 2, , nand(2) find values 0fx1,x2,Xnsuch thatL(x1,x2,Xn,)一0 (you do not have toexplicitly find the corresponding values o

24、f ):(3) evaluate f at these points to find the required extrema., Note that the equation ' L = 0 is equivalent to the equations-I0 = fx " 'gx风 1 ,1 Mi Mn and g(x1,X2，Xn) = 0So, in the two variable case, we have Lagranianfunction L(x, y, ) - f (x, y)g(x,y)and are solving the equations:0

25、= fx+®,0=fy+，g, and g(x,y) = 0. With more than one constraint we solve the equation2Theorem 2.3 Let f : R ' R and P = (x0, y0) 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 ne

26、ar to P and that P is not an endpoint of C and that ig(x0,y0)=0. Then there is some such that (x0,y0,z0)isa critical point of the Lagrangian FunctionL(x, y, ) f (x, y) g(x,y).Proof. Sketch only. SinceP is not an end point and ' g : 0 ,C has a tangent at Pwith normal g .If f is not parallel to g

27、at p , then it has non-zero projection along this tangent at P .But then f increases and decreases away fromP along C ,so P is not an extremum. Hence' f and g are parallel and there is some?bch that ' f - -，' g and the result follows.Example. Find the rectangular box with the largest vol

28、ume that fits inside the222ellipsoid 鼻y2 z2 = 1 ,given that it sides are parallel to the axes.a b cSolution. 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, ytz)

29、 and its volume is V= 8xyz.b224 7 = 0 . (Note that since the c2 xWe want to maximize V given that aconstraint surface is bounded a max/min does exist). The Lagrangian isL(x, y, z, ) =8xyzj x2 aand this has critical points when ' L = 0, i.e. when；x0 =fyCFL0 =:zx= 8yz 2, a= 8zx 2' -y2,.b=8xy 2

30、 1 -zr, cL0 =；z22x j.1b2 c2(Note that L , will always be the constraint equation.) As we want to maximizeV we can assume thaXyz=0 so that x, y,z = 0 .)Hence, eliminating , we get2 yz .2 zx 2 xy - -4a =-4b = -4c 一 2 22 2.2 2so that y a = x b and z b2 2 .y c . But thenb22z=soc1=当ab2or x=-a,which impli

31、es that ./3by=、3and,3(they are all positive byassumption). SoL has only one stationary point/ a b c(,1) (for some value of 3 - 3 . 3,which we could work out if we wanted to). Since it is the only stationary point itmust the required max and the max volume isa b c 8abc8.33 .33.3 .中文译文多元函数的极值1.稳定点定义1.

32、1使D£ Rn并且f : D t R.对于任意一点aw D有以下定义：(1)如果f(x)Mf(a)对于所有x充分地接近a时，则f (a)是一个局部极大值;(2)如果f(x) > f(a)对于所有x充分地接近a时，则f (a)是一个局部极小值;(3)如果f (x) < f (a)对于所有点xw D成立，则f(a)是一个全局极大值(或绝对 极大值)；(4)如果f (x) > f (a)对于所有点xw D成立，则f (a)是一个全局极小值(或绝 对极小值)；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DRn并且f :Dt R.对于任意一点

33、aw D ,如果中(a)=0,并且对于任意奇异点a都不存在Vf ,则称a是一个关键点或稳定点.结论1.3使D J Rn并且f : Dt R.如果f有局部极值或全局极值对于一点a w D ,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f(a +h) -f (a)取正值和负值.定义1.7如果f：R2T R

34、是二元函数，并且在点(a,b)处所有二阶偏导数都存在，则则根据函数f在点(a,b)处导数，有f在点(a,b)处的Hessian矩阵为：xyyyxx yx推广：如果f：R3T R是三元函数，并且在点(a,b,c)处所有二阶偏导数都存在,则根据函数f在点(a,b, c)处导数,有f在点(a,b,c)处的Hessian矩阵为:xxxyxzyxyyyzzxzyzz/定义1.8矩阵A是n m n阶矩阵，并且对于每一个都有1 M r M n,从矩阵A中选取左上端的r行和r歹1，令其为r”阶的矩阵Ar.则行列式det( A ),1 W r W n ,叫做矩阵A的顺序主子式.定理1.9假如f ：R2t R是一

35、个充分光滑的二元函数，且在点(a,b)处稳定，其Hessian矩阵为H .如果det(H),0 ,则根据偏导数判定(a,b)点是：(1) 一个局部极大值点,如果0>det(H1) = fxx并且0<det(H尸fjyy - fxy;(2) 一个局部极小值点,如果0<det(H1) = fxx并且0<det(H尸f/ fyy - f。;(3) 一个鞍点，如果点(a,b)既不是局部极大值点也不是局部极小值点.假如f ：R3t R是一个充分光滑的三元函数，且在点(a,b,c)处稳定，其Hessian矩阵为H .如果det(H) #0 ,则根据偏导数判定(a,b,c)点是：(1

36、) 一个局部极大值点， 如果当0>det(H1), 0<det(H2)并且0>det(H3)时；(2) 一个局部极小值点， 如果当0<det(H1), 0<det(H2)并且0>det(H3)时；(3) 一个鞍点，如果点(a,b, c)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尸0时，点(a,b)是一个局部极值点，或者是一个鞍百 八、.关键点.在有界闭集上的连续函数有边界，并且可以取到其边界值.当确定函数在有界闭集上的极值时，必须考虑函数在稳定点(即7f =0时)，奇例.确定下列函数的稳定点并说明是哪一类点：Z/、- ,、4222,(

37、1) f (x, y,z) = x x y y z xz 1;224(2) f (x, y) = y (x 1) y (x 1);解.(1) f (x, y, z) = x4 x2y y2 z2 xz 1 ,so七 f (x, y) = (4x3 2xy z) i (x2 2y) j (2z x) k当* = 0时有稳定点也就是说，当- -3(1) 0= 4x2xyz2(2) 0=x22y(3) 0= 2zx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4x3 -x3 - - X 0 02即x(6x2 -1) =0，得x = 0, x =三6和x = _运.因此我

38、们可以得到函数的三个稳定 66点：(0。0),616,一12,612）和（fzz = 2,则 Hessian又因为 fxx =12x2 +2y, fxy =2x, fxz=1, £¥丫 = 2,£丫7=0和矩阵为彳2x2+2y 2x 厂H = 2x 20< 1 0 2.，在点（112各处,V6/3 1、200 2>则顺序主子式11八>0, 611/66/376/3 11 62 一 3 一9二3 0并且行列式H喘一（2。.根据主子式判定方法,则点（*今 是一个局部极小值点.在点（-11211/6H = -<6/31-5/6/3 1'2

39、002,则顺序主子式11八>0, 611/6- 6 / 3-V6/3二3 022 12.616并且行列式系号一,0.根据主子式判定方法，则点（-丁-石板也是一个极小值点在点(0,0,0)处,Hessian矩阵为001、H = 0 2 0U 0 2,因此det(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4222D=f (0,0,0) - f (0 + h,0 + k,0 + l) =h +hk + k +l + hl ,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

40、上方程 相对线性二次方程时可忽略不计，则原方程可为D全k2+12+hl .当h, k和l都为正 时,D >0.然而，当k=0、h M0和0<l <|h ,则D <0.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) f (x, y) = y2 (x 1)2 y (x 1)4so_3 _2_"(x, y) =(2(x 1) y 4(x 1) ) i (2y (x 1) j.当7f =0时有稳定点，也就是说，当在(-1,0)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(-1,0)时Hessian矩阵

41、的行列式为0,所以该判定方法在此刻无法应用) 令224D=f( 1,0) f( 1+h,0+k)=k +hk+h.22配成完全平方的形式为D =(k +h)2 +”.所以对h和k为任意小时(h和k都不24为0),有D >0,因此我们可以确定函数f在点(-1,0)处有局部极大值.2.条件极值和Lagrang噪数法定义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f ：RnT R是一个n元函数，则对应于函数f的Lagrange函数在限制条件g(Xi,X2，Xn)=0下的函数是一个n+1元函数L(Xi,X2, ,Xn,)= f (Xi,X2, ,Xn)g(Xi,X2, , Xn),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gi (x1,x?，Xn) = 0,1 w i w k时，带有k个九,1 Wi Wk的Lagrange函数为:kL(X1,X2, ,Xn, ' ) = f(X1,X2, ,Xn) 一2：/，g(X1, X2, ,Xn)关键点.确定在限制条件g(x) = 0