泰勒公式外文翻译教学内容

泰勒公式外文翻译
Taylor's Formula and the Study of Extrema
1. Taylor's Formula for Mappings
Theorem 1. If a mapping Y U f →: from a neighborhood ()x U U = of a point x in a normed space X into a normed space Y has derivatives up to order n -1 inclusive in U and has an n-th order derivative ()()x f n at the point x, then
()()()()()⎪⎭⎫ ⎝⎛++++=+n n n h o h x f n h x f x f h x f !
1,Λ (1) as 0→h .
Equality (1) is one of the varieties of Taylor's formula, written here for rather general
classes of mappings.
Proof. We prove Taylor's formula by induction.
For 1=n it is true by definition of ()x f ,.
Assume formula (1) is true for some N n ∈-1.
Then by the mean-value theorem, formula (12) of Sect. 10.5, and the induction hypothesis, we obtain.
()()()()()()()()()()()()()⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭
⎫ ⎝⎛-+++-+≤⎪⎭
⎫ ⎝⎛+++-+--<<n n n n n n h o h h o h h x f n h x f x f h x f h x f n h f x f h x f 11,,,,10!11sup !1x θθθθθΛΛ，
as 0→h .
We shall not take the time here to discuss other versions of Taylor's formula, which are sometimes quite useful. They were discussed earlier in detail for numerical functions. At this point we leave it to the reader to derive them (see, for example, Problem 1 below).
2. Methods of Studying Interior Extrema
Using Taylor's formula, we shall exhibit necessary conditions and also sufficient conditions for an interior local extremum of real-valued functions defined on an open subset of a normed space. As we shall see, these conditions are analogous to the differential conditions already known to us for an extremum of a real-valued function of a real variable.
Theorem 2. Let R U f →: be a real-valued function defined on an open set U in a normed space X and having continuous derivatives up to order 11≥-k inclusive in a neighborhood of a point U x ∈ and a derivative ()()x f k of order k at the point x itself.
If ()()()0,,01,==-x f x f k Λ and ()()0≠x f k , then for x to be an extremum of the function f it is: necessary that k be even and that the form
()()k k h x f be semidefinite,
and
sufficient that the values of the form ()()k k h x f on the unit sphere 1=h be bounded away from zero; moreover, x is a local minimum if the inequalities
()()0>≥δk k h x f ,
hold on that sphere, and a local maximum if
()()0<≤δk k h x f ,
Proof. For the proof we consider the Taylor expansion (1) of f in a neighborhood of x. The assumptions enable us to write
()()()()()k k k h h h x f k x f h x f α+=-+!1
where ()h α is a real-valued function, and ()0→h α as 0→h .
We first prove the necessary conditions.
Since ()()0≠x f k , there exists a vector 00≠h on which
()()00≠k k h x f . Then for values of the real parameter t sufficiently close to zero,
()()()()()()k k k th th th x f k x f th x f 0000!1α+=-+
()()()k
k k k t h th h x f k ⎪⎭⎫ ⎝⎛+=000!1α
and the expression in the outer parentheses has the same sign as
()()k k h x f 0. For x to be an extremum it is necessary for the left-hand side (and hence also the right-hand side) of this last equality to be of constant sign when t changes sign. But this is possible only if k is even.
This reasoning shows that if x is an extremum, then the sign of the difference ()()x f th x f -+0 is the same as that of ()()k k h x f 0 for sufficiently small t; hence in that case there cannot be two vectors 0h , 1h at which the form ()()x f k assumes values with opposite signs.
We now turn to the proof of the sufficiency conditions. For definiteness we consider the case when ()()0>≥δk k h x f for 1=h . Then
()()()()()k k k h h h x f k x f h x f α+=-+!1
()()()k k k h h h h x f k ⎪⎪⎪⎭
⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛=α!1 ()k h h k ⎪⎭⎫ ⎝⎛+≥αδ!1
and, since ()0→h α as 0→h , the last term in this inequality is positive for all vectors
0≠h sufficiently close to zero. Thus, for all such vectors h,
()()0>-+x f h x f ,
that is, x is a strict local minimum.
The sufficient condition for a strict local maximum is verified similiarly.
Remark 1. If the space X is finite-dimensional, the unit sphere ()1;x S with center at X x ∈, being a closed bounded subset of X, is compact. Then the continuous function
()()()()k k i i i i k k h h x f h x f ⋅⋅∂=ΛΛ11 (a k-form) has both a maximal and a minimal value on ()1;x S . If
these values are of opposite sign, then f does not have an extremum at x. If they are both of the same sign, then, as was shown in Theorem 2, there is an extremum. In the latter case, a sufficient condition for an extremum can obviously be stated as the equivalent requirement that the form ()()k k h x f be either positive- or negative-definite.
It was this form of the condition that we encountered in studying realvalued functions on n R . Remark 2. As we have seen in the example of functions R R f
n →:, the semi-definiteness of the form ()()k k h x f exhibited in the necessary conditions for an extremum is not a sufficient criterion for an extremum.
Remark 3. In practice, when studying extrema of differentiable functions one normally uses only the first or second differentials. If the uniqueness and type of extremum are obvious from the meaning of the problem being studied, one can restrict attention to the first
differential when seeking an extremum, simply finding the point x where ()0,=x f
3. Some Examples
Example 1. Let ()()R R C L ;31∈ and ()[]()R b a C f ;,1∈. In other words, ()()321321,,,,u u u L u u u α is a continuously differentiable real-valued function defined in 3R and ()x f x α a smooth real-valued function defined on the closed interval []R b a ⊂,.
Consider the function
()[]()R R b a C F →;,:1 (2)
defined by the relation
()[]()()f F R b a C f α;,1∈
()()()
R dx x f x f x L b
a ∈=⎰,,, (3) Thus, (2) is a real-valued functional defined on the set of functions ()[]()R
b a C ;,1.
The basic variational principles connected with motion are known in physics and
mechanics. According to these principles, the actual motions are distinguished among all the conceivable motions in that they proceed along trajectories along which certain functionals have an extremum. Questions connected with the extrema of functionals are central in optimal control theory. Thus, finding and studying the extrema of functionals is a problem
of intrinsic importance, and the theory associated with it is the subject of a large area of
analysis - the calculus of variations. We have already done a few things to make the transition from the analysis of the extrema of numerical functions to the problem of finding and
studying extrema of functionals seem natural to the reader. However, we shall not go deeply into the special problems of variational calculus, but rather use the example of the functional
(3) to illustrate only the general ideas of differentiation and study of local extrema considered above.
We shall show that the functional (3) is a differentiate mapping and find its differential. We remark that the function (3) can be regarded as the composition of the mappings
()()()()()x f x f x L x f F ,1,,= (4)
defined by the formula
()[]()[]()R b a C R b a C F ;,;,:11→ (5)
followed by the mapping
[]()()()R dx x g g F R b a C g b
a ∈=∈⎰2;,α (6) By properties of the integral, the mapping 2F is obviously linear and continuous, so that its differentiability is clear.
We shall show that the mapping 1F is also differentiable, and that
()()()()()()()()()
()x h x f x f x L x h x f x f x L x h f F ,,3,2,1.,,,∂+∂= (7) for ()[]()R b a C h ;,1∈.
Indeed, by the corollary to the mean-value theorem, we can write in the present case
()()()i i i u u u L u u u L u u u L ∆∂--∆+∆+∆+∑=3
2131321332211,,,,,,
()()()()()()∆
⋅∂-∆+∂∂-∆+∂∂-∆+∂≤<<u L u L u L u L u L u L 331221111
0sup θθθθ ()()i
i i i u L u u L i ∆⋅∂-+∂≤=≤≤=3,2,110max max 33,2,1θθ (8)
where ()321,,u u u u = and ()321,,∆∆∆=∆. If we now recall that the norm
()1c f of the function f in ()[]()R b a C ;,1 is ⎭⎬⎫⎩⎨⎧c c f f ,,max (where c f is the maximum absolute value of the function on the closed interval []b a ,), then,
setting x u =1, ()x f u =2, ()x f u ,3=, 01=∆, ()x h =∆2, and ()x h ,3=∆, we obtain from inequality (8),
taking account of the uniform continuity of the functions ()3,2,1,,,321=∂i u u u L i , on bounded subsets of 3R , that
()()()()()()()()()()()()()()()
()x h x f x f x L x h x f x f x L x f x f x L x h x f x h x f x L b x ,,3,2,,,0,,,,,,,,max ∂-∂--++≤≤ ()()1c h o = as ()01
→c h But this means that Eq. (7) holds.
By the chain rule for differentiating a composite function, we now conclude that the
functional (3) is indeed differentiable, and
()()()()()()()()()()⎰∂+∂=b
a
dx x h x f x f x L x h x f x f x L h f F ,,3,2,,,,, (9) We often consider the restriction of the functional (3) to the affine space consisting of the functions ()[]()R b a C f ;,1∈ that assume fixed values ()A a f =, ()B b f = at the endpoints of the closed interval []b a ,. In this case, the functions h in the tangent space ()1f TC , must have the value zero at the endpoints of the closed interval []b a ,. Taking this fact into account, we may
integrate by parts in (9) and bring it into the form
()()()()()()()
()⎰⎪⎭⎫ ⎝⎛∂-∂=b a dx x h x f x f x L dx d x f x f x L h f F ,3,2,,,,, (10) of course under the assumption that L and f belong to the corresponding class ()2C .
In particular, if f is an extremum (extremal) of such a functional, then by Theorem 2 we have ()0,=h f F for every function ()[]()R b a C h ;,1∈ such that ()()0==b h a h . From this and relation (10) one can easily conclude (see Problem 3 below) that the function f must satisfy the equation
()()()()()()0,,,,,3,2=∂-∂x f x f x L dx d x f x f x L (11)
This is a frequently-encountered form of the equation known in the calculus of variations as the Euler-Lagrange equation.
Let us now consider some specific examples.
Example 2. The shortest-path problem
Among all the curves in a plane joining two fixed points, find the curve that has minimal length.
The answer in this case is obvious, and it rather serves as a check on the formal
computations we will be doing later.
We shall assume that a fixed Cartesian coordinate system has been chosen in the plane, in which the two points are, for example, ()0,0 and ()0,1 . We confine ourselves to just the curves that are the graphs of functions ()[]()R C f ;1,01∈ assuming the value zero at both ends of the closed interval []1,0 . The length of such a curve
()()()⎰+=1
02
,1dx x f f F (12)
depends on the function f and is a functional of the type considered in Example 1. In this case the function L has the form
()()233211,,u u u u L +=
and therefore the necessary condition (11) for an extremal here reduces to the equation ()()
()012,,=⎪⎪⎪⎭⎫ ⎝⎛+x f x f dx d from which it follows that ()
()()常数≡+x f x f 2,,1 (13)
on the closed interval []1,0
Since the function 21u u
+ is not constant on any interval, Eq. (13) is possible only if
()≡x f ,const on []b a ,. Thus a smooth extremal of this problem must be a linear function whose graph passes through the points ()0,0 and ()0,1. It follows that ()0≡x f , and we arrive at the closed interval of the line joining the two given points.
Example 3. The brachistochrone problem
The classical brachistochrone problem, posed by Johann Bernoulli I in 1696, was to find the shape of a track along which a point mass would pass from a prescribed point 0P to another fixed point 1P at a lower level under the action of gravity in the shortest time.
We neglect friction, of course. In addition, we shall assume that the trivial case in which both points lie on the same vertical line is excluded.
In the vertical plane passing through the points 0P and 1P we introduce a rectangular
coordinate system such that 0P is at the origin, the x-axis is directed vertically downward, and the point 1P has positive coordinates ()11,y x .We shall find the shape of the track among the graphs of smooth functions defined on the closed interval []1,0x and satisfying the condition ()00=f ,()11y x f =. At the moment we shall not take time to discuss this by no means
uncontroversial assumption (see Problem 4 below).
If the particle began its descent from the point 0P with zero velocity, the law of variation of its velocity in these coordinates can be written as
gx v 2= (14)
Recalling that the differential of the arc length is computed by the formula
()()()()dx x f dy dx ds 2
,221+=+= (15) we find the time of descent
()()()⎰+=102
,121
x dx x x f g f F (16) along the trajectory defined by the graph of the function ()x f y = on the closed interval []1,0x . For the functional (16) ()()12
33211,,u u u u u L +=,
and therefore the condition (11) for an extremum reduces in this case to the equation ()
()()012,,=⎪⎪⎪⎪
⎪⎭
⎫
⎝⎛⎪⎭⎫ ⎝⎛+x f x x f dx d ,
from which it follows that ()()()
x c x f x f =+2,,1 (17)
where c is a nonzero constant, since the points are not both on the same vertical line. Taking account of (15), we can rewrite (17) in the form
x c ds dy = (18)
However, from the geometric point of view
ϕcos =ds dx
,ϕsin =ds dy
(19)
where ϕ is the angle between the tangent to the trajectory and the positive x-axis.
By comparing Eq. (18) with the second equation in (19), we find ϕ22sin 1
c x = (20) But it follows from (19) an
d (20) that dx dy d dy =ϕ,2222sin 2
sin c c d d tg d dx tg d dx ϕ
ϕϕϕϕϕϕ=⎪⎪⎭⎫
⎝⎛
==,
from which we find
()b c y +-=ϕϕ2sin 221
2 (21)
Setting a c =221
and t =ϕ2, we write relations (20) and (21) as
()
()b t t a y t a x +-=-=sin cos 1 (22)
Since 0≠a , it follows that 0=x only for πk t 2=,Z k ∈. It follows from the form of the function
(22) that we may assume without loss of generality that the parameter value 0=t corresponds to the point ()0,00=P . In this case Eq. (21) implies 0=b , and we arrive at the simpler form
()
()t t a y t a x sin cos 1-=-= (23)
for the parametric definition of this curve.
Thus the brachistochrone is a cycloid having a cusp at the initial point 0P where the tangent is vertical. The constant a, which is a scaling coefficient, must be chosen so that the curve (23) also passes through the point 1P . Such a choice, as one can see by sketching the curve (23), is by no means always unique, and this shows that the necessary condition (11) for an extremum is in general not sufficient. However, from physical considerations it is clear which of the
possible values of the parameter a should be preferred (and this, of course, can be confirmed by direct computation).
泰勒公式和极值的研究
1.映射的泰勒公式
定理1 如果从赋范空间X 的点x 的邻域()x U U =到赋范空间Y 的映射Y U f →:在U 中有直到n-1阶(包括n-1在内)的导数，而在点x 处有n 阶导数。

()()x f n ，那么当0→h 时有
()()()()()⎪⎭⎫ ⎝⎛++++=+n n n h o h x f n h x f x f h x f !
1,Λ (1) 等式(1)是各种形式的泰勒公式中的一种，这一次它确实是对非常一般的函数 类写出来的公式了。

我们用归纳法证明泰勒公式(1)。

当1=n 时，由()x f ,的定义，(1)式成立。

假设(1)对N n ∈-1成立。

于是根据有限增量定理，5章中公式(12)和所作的归纳假设，我们得到，当0→h 时成立。

()()()()()()()()()()()()()⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭
⎫ ⎝⎛-+++-+≤⎪⎭
⎫ ⎝⎛+++-+--<<n n n n n n h o h h o h h x f n h x f x f h x f h x f n h f x f h x f 11,,,,10!11sup !1x θθθθθΛΛ，
这里我们不再继续讨论其他的，有时甚至是十分有用的泰勒公式形式。

当时，在研究数值函数时，曾详细地讨论过它们。

现在我们把它们的结论提供给读者(例如，可参看练习1)。

2.内部极值的研究
我们将利用泰勒公式指出定义在赋范空间的开集上的实值函数在定义域内部取得局部极值的必要微分条件和充分微分条件。

我们将看到，这些条件类似于我们熟知的实变量的实值函数的极值的微分条件。

定理2 设R U f →:是定义在赋范空间X 的开集U 上的实值函数，且f 在某个点U x ∈的邻域有直到11≥-k 阶.(包括
k-1阶在内的)导映射，在点x 本身有k 阶导映射
()()x f k 。

如果()()()0,,01,==-x f x f k Λ且()()0≠x f k ，那么为使x 是函数f 的极值点
必要条件是: k 是偶数，()()k k h x f 是半定的。

充分条件是: ()()k k h x f 在单位球面1=h 上的值不为零；这时，如果在这个球面上
()()0>≥δk k h x f ，
那么x 是严格局部极小点；如果
()()0<≤δk k h x f ，
那么x 是严格局部极大点
为了证明定理，我们考察函数f 在点x 邻域内的泰勒展开式。

由所作的假设可得
()()()()()k k k h h h x f k x f h x f α+=-+!1，
其中()h α是实值函数，而且当0→h 时。

()0→h α
我们先证必要条件。

因为()()0≠x f k ，所以有向量00≠h ，使()()00≠k k h x f 。

于是，对于充分接近于零的实参
量t ，
()()()()()()k k k th th th x f k x f th x f 0000!1α+=-+ ()()()k k k k t h th h x f k ⎪⎭
⎫ ⎝⎛+=000!1α
括号内的表达式与()()k k h x f 0同号。

为使x 是极值点，当t 变号时最后一个等式的左边(从而右边)必须不改变符号。

这只有当k 为偶数时才可能。

上述讨论表明，如果x 是极值点，那么对于充分小的t ，差()()x f th x f -+0的符号与
()()k k h x f 0相同，因而在这种情况下不可能有两个向量0h ，1h ，使()()x f k 在它们上的取值有不同的符号。

我们转到极值充分条件的证明。

为了确定起见，我们研究
()()0>≥δk k h x f ，当 1=h
的情况。

这时
()()()()()k k k h h h x f k x f h x f α+=-+!1 ()()()k k k h h h h x f k ⎪⎪⎪⎭
⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛=α!1 ()k h h k ⎪⎭
⎫ ⎝⎛+≥αδ!1， 又因0→h 时()0→h α，所以不等式的右端对于所有充分接近于零的向量0≠h 均为正。

因而对所有这些向量h
()()0>-+x f h x f ，
即x 是严格局部极小点。

严格局部极大点的充分条件可类似地验证。

注1 如果空间X 是有限维的，那么以点X x ∈为中心的单位球面()1;x S 是X 中的有界闭集，因而是紧集。

这时，连续函数()()()()k k i i i i k k h h x f h x f ⋅⋅∂=ΛΛ11（k-形式）在()1;x S 上
有最大值和最小值。

如果最大值和最小值异号，那么函数f 在点x 没有极值。

如果它们同号，那么像定理2所指出的，f 在点x 有极值。

在后一种情况下，显然极值的充分条件可叙述为与它等价的形式:形式()()k k h x f 是定的(正定的或负定的)。

我们在研究n R 中的实值函数时所遇到的正是这种形式的极值条件。

注2 像我们在函数R R f n →:的例子所看到的那样，在极值的必要条件中所说的形式()()k k h x f 的半定性还不是极值的充分条件。

注3 实际上，在研究可微函数的极值时，通常只利用一阶微分或一阶和二阶微分。

如果根据所研究问题的意义，极值点的唯一性及极值的特性是显然的，那么在求极值点时就可只用一阶微分:求满足()0,=x f 的点x 。

3.一些例子
例1 设()()R R C L ;31∈，而()[]()R b a C f ;,1∈，换句话说，
()()3
21321,,,,u u u L u u u α 是定义在3R 中的连续可微的实值函数，而()x f x α是定义在区间[]R b a ⊂,上的光滑实值函数。

我们研究函数
()[]()R R b a C F →;,:1
（2）
它由以下关系式给出
()[]()()f F R b a C f α;,1∈
()()()R dx x f x f x L b a
∈=⎰,
,, （3）
因此，(2)是定义在函数集()[]()R b a C ;,1上的实泛函。

在物理学和力学中，与运动密切相关的基本变分原理是众所周知的。

根据这些原理，在所有可能的运动中真实运动的特点是，它们总是沿着使某些泛函有极值的轨道进行。

与泛函的极值有关的问题是最优控制理论中的中心问题。

因此，寻求和研究泛
函的极值是重要的独立课题，分析中以大量篇幅讨论这个课题的理论，这就是变分学，为使读者对从数值函数的极值分析到寻求和研究泛函的极值的转变不感到突然
的，我们己做了某些工作。

但是我们不准备深入讨论变分法的专门问题，仅以泛函(3)为例说明上面讲过的微分法和局部极值研究的一般思想。

我们要证明泛函(3)是可微映射并求出它的微分。

首先指出，函数(3)可以看作由公式
()()()()()x f x f x L x f F ,1,,= (4) 给出的映射
()[]()[]()R b a C R b a C F ;,;,:11→ (5) 和映射
[]()()()R dx x g g F R b a C g b a ∈=∈⎰2;,α
(6) 的复合。

由积分的性质，映射2F 显然是线性连续映射，因而它的可微性问题是明显的。

我们来证明1F 也是可微的，而且
()()()()()()()()()()x h x f x f x L x h x f x f x L x h f F ,,3,2,1.,,,∂+∂= (7) 其中()[]()R b a C h ;,1∈。

事实上，由有限增量定理的推论，在我们的情况下可得
()()()i i i u u u L u u u L u u u L ∆∂--∆+∆+∆+∑=3
2131321332211,,,,,,
()()()()()()∆
⋅∂-∆+∂∂-∆+∂∂-∆+∂≤<<u L u L u L u L u L u L 331221111
0sup θθθθ ()()i
i i i u L u u L i ∆⋅∂-+∂≤=≤≤=3,2,110max max 33,2,1θθ (8)
其中()321,,u u u u =，()321,,∆∆∆=∆。

如果记起()[]()R b a C ;,1中函数f 的范数()1c f 是⎭⎬⎫⎩⎨⎧c c f f ,,max (其中c f 是函数在区间[]b a ,上的最大模)，那么设x u =1,()x f u =2,()x f u ,3=,01=∆,()x h =∆2和()x h ,3=∆，考虑到函数()
3,2,1,,,321=∂i u u u L i 在3R 的有界子集上的一致连续性，从不等式(8)得到 ()()()()()()()()()()()()()()()
()x h x f x f x L x h x f x f x L x f x f x L x h x f x h x f x L b x ,,3,2,,,0,,,,,,,,max ∂-∂--++≤≤ ()()1c h o = 当()01
→c h 时 而这意味着等式(7)成立。

现在，根据复合映射的微分定理断定，泛函(3)确实可微，并且
()()()()()()()()()()⎰∂+∂=b
a dx x h x f x f x L x h x f x f x L h f F ,
,3,2,,,,, (9) 经常把泛函(3)限制在那样一些函数()[]()R b a C f ;,1∈的仿射空间上，它们在区间[]b a ,的端点取固定的值()A a f =,()B b f =。

在这种情况下，切空间()1f TC 中的函数h 在区间[]b a ,的端点应该有零值。

考虑到这一点，在这种情况下，利用分部积分，显然可把等式(9)化为
()()()()()()()
()⎰⎪⎭⎫ ⎝⎛∂-∂=b a dx x h x f x f x L dx d x f x f x L h f F ,3,2,,,,, (10) 当然要预先假设L 和f 属于相应的函数类()2C 。

特别地，如果f 是这个泛函的极值点(极值曲线)，那么根据定理2，对于任意使得()()0==b h a h 的函数()[]()R b a C h ;,1∈均有()0,=h f F 。

由此，由(10)不难推出(见练习3)，函数f 应该满足欧拉-拉格朗日方程。

()()()()()()0,,,,,3,2=∂-∂x f x f x L dx d x f x f x L (11)
这是在变分学中被称为欧拉-拉格朗日方程的特殊的形式，现在研究具体的例子。

例2 短程线问题
在平面内连接两个固定点的曲线中，求长度最小的那条曲线。

在这种情况下，答案是显然的，宁愿把它作为对以下推理的一个检验。

我们将认为，在平面上给出了笛卡儿坐标系，在该坐标系中不妨认为点()0,0和()0,1是给定的点。

我们只限于研究那些曲线，它们是在区间[]1,0的端点取零值的函数()[]()R C f ;1,01∈的图像，这种曲线的长度
()()()⎰+=1
02
,1dx x f f F (12) 依赖于函数f 且是例1中所研究的那种类型的泛函。

在所给的情况下，函数L 有形式 ()()233211,,u u u u L +=，
因此，在这里极值的必要条件(11)归结为方程 ()()
()012
,,=⎪⎪⎪⎭⎫ ⎝⎛+x f x f dx d ， 由它推出，在区间[]1,0上 ()
()()常数≡+x f x f 2,,1 (13)
因为函数21u u
+是严格单调增的增函数，所以(13)只有在[]b a ,上()≡x f ,常数时才能成
立。

这样一来，要求的光滑极值函数应是线性函数，其图形通过点()0,0,()0,1。

由此推出()0≡x f ，于是我们得到，连接两个已知点的直线段为所求的曲线。

例3 最速降线(或捷线)问题
这个于1696年由约翰·伯努利首先提出的关于捷线的经典问题乃是寻求那样的沟槽形式，质点沿着该沟槽在重力作用下在最短的时间内从已知点0P 滑落到另一更低的固定点1P 。

当然，我们忽略摩擦力。

此外，在以后的研究中我们不考虑两个点位于同一铅垂 线上的平凡情况。

在通过点0P ，1P 的铅垂平面内，引进直角坐标系，使得点0P 是它的原点，横轴垂直向下，而点1P 有正的坐标()11,y x 。

我们将只在定义在区间[]1,0x 上满足条件()00=f ，
()11y x f =的光滑函数的图形中寻求沟槽的形式。

我们暂且不讨论这个绝非无可争议的假定(见习题4)。

如果质点从点0P 以零速度开始自己的运动，那么在所选择的坐标系中它的速度的变化规律为 gx v 2= (14)
回忆弧长微分的计算公式， ()()()()dx x f dy dx ds 2,221+=+= (15)
求出沿着由定义在区间[]1,0x 上的函数()x f y =的图形确定的轨道运动的时间
()()()
⎰+=102,121x dx x x f g f F (16)
对于泛函(16), ()()12
33211,,u u u u u L +=,
因此，极值的必要条件(11)在所给情况下归结为方程 ()()()012
,,=⎪⎪
⎪⎪
⎪
⎭
⎫ ⎝⎛⎪⎭⎫ ⎝⎛+x f x x f dx d ,
从它推出 ()()()
x
c x f x f =+2,,1 (17)
其中c 是不为零的常数。

(因为0P ，1P 不在同一铅垂线上!)
考虑到(15)，方程(17)可以改写为
x c ds dy
= (18)
但从几何的观点有
ϕcos =ds dx ,ϕsin =ds dy ,
（19）
其中ϕ是曲线的切线与横轴正方向间的夹角。

比较方程(18)和(19)的第二个方程，我们求出
ϕ22sin 1
c x = (20)
但从(19)和(20)得到
dx dy d dy =ϕ,2222sin 2sin c c d d tg d dx tg d dx ϕϕϕϕϕϕϕ=⎪⎪⎭⎫ ⎝⎛==,
由此求出
()b c y +-=ϕϕ2sin 221
2 (21) 设a c =221
和t =ϕ2，可把关系式(20)和(21)改写为
()
()b t t a y t a x +-=-=sin cos 1 (22)
因为0≠a ，所以只在πk t 2=,Z k ∈时0=x 。

从函数(22)的形式推出，不失一般性，可以认为点()0,00=P 对应于参量0=t 。

这时0=b ，我们得到所求曲线的更简单的参数形式
()
()t t a y t a x sin cos 1-=-= (23)
这样，捷线是摆线，它在起始点0P 是具有铅垂切线的尖点。

常数a ，即同位相似系数，应当这样选择，它使曲线(23)也通过点1P 。

画出曲线(23)后，可以发现，这样的选择不总是唯一的，这也证实了极值的必要条件(11)一般来说不是充分的。

然而从物理意义考虑，在参数a 的可能的值中应该选取怎样的值是显然的(其实也可以用直接计算来证实)。