1610912322-b551b095a53deaf3d3fbd1ed05ae9b84 (824701), страница 86
Текст из файла (страница 86)
8.1 show,g (y) o(h) = o(h) as h → 0,f (x + h) − f (x) = f (x)h + o(h) = O(h) + o(h) = O(h)ando f (x + h) − f (x) = o O(h) = o(h)as h → 0,as h → 0.Consequently,α(x; h) = o(h) + o(h) = o(h)as h → 0,and the theorem is proved.When rewritten in coordinate form, Theorem 3 means that if x is an interior pointof the set X and⎞⎛∂1 f 1 (x) · · · ∂m f 1 (x)......⎟ j⎜f (x) = ⎝.⎠ = ∂i f (x),..∂1 f n (x)···∂m f n (x)and y = f (x) is an interior point of the set Y and⎛⎞∂1 g 1 (y) · · · ∂n g 1 (y)......⎜⎟ kg (y) = ⎝.⎠ = ∂j g (y),..∂1 g k (y)···∂n g k (y)then⎛∂1 (g 1 ◦ f )(x)..⎜(g ◦ f ) (x) = ⎝.···...∂1 (g k ◦ f )(x)···⎞∂m (g 1 ◦ f )(x)..⎟ l⎠ = ∂i g ◦ f (x) =.∂1 g 1 (y)..⎜=⎝.···...∂m (g k ◦ f )(x)⎞⎛∂n g 1 (y)∂1 f 1 (x)....⎟⎜⎠⎝..∂1 g k (y)···∂1 f n (x)⎛∂n g k (y)= ∂j g l (y) · ∂i f j (x) .···...···⎞∂m f 1 (x)..⎟⎠=.∂m f n (x)In the equality l∂i g ◦ f (x) = ∂j g l f (x) · ∂i f j (x)(8.34)summation is understood on the right-hand side with respect to the index j over itsinterval of variation, that is, from 1 to n.8.3 The Basic Laws of Differentiation443In contrast to Eqs.
(8.31 ), (8.32 ), and (8.33 ), relation (8.34) is nontrivial evenin the sense of elementwise equality of the matrices occurring in it.Let us now consider some important cases of the theorem just proved.b. The Differential and Partial Derivatives of a Composite Real-ValuedFunctionLet z = g(y 1 , . . . , y n ) be a real-valued function of the real variables y 1 , . . .
, y n , eachof which in turn is a function y j = f j (x 1 , . . . , x m ) (j = 1, . . . , n) of the variablesx 1 , . . . , x m . Assuming that the functions g and f j are differentiable (j = 1, . . . , n),)let us find the partial derivative ∂(g◦f(x) of the composition of the mappings f :∂x iX → Y and g : Y → R.According to formula (8.34), in which l = 1 under the present conditions, wefind(8.35)∂i (g ◦ f )(x) = ∂j g f (x) · ∂i f j (x),or, in notation that shows more detail,∂g ∂y 1∂(g ◦ f ) 1∂g ∂y n∂zmx=(x)=,...,x·+···+·=∂x i∂x i∂y n ∂x i∂y 1 ∂x i= ∂1 g f (x) · ∂i f 1 (x) + · · · + ∂n g f (x) · ∂i f n (x).c.
The Derivative with Respect to a Vector and the Gradient of a Functionat a PointConsider the stationary flow of a liquid or gas in some domain G of R3 . The term“stationary” means that the velocity of the flow at each point of G does not varywith time, although of course it may vary from one point of G to another. Suppose, for example, f (x) = f (x 1 , x 2 , x 3 ) is the pressure in the flow at the pointx = (x 1 , x 2 , x 3 ) ∈ G.
If we move about in the flow according to the law x = x(t),where t is time, we shall record a pressure (f ◦ x)(t) = f (x(t)) at time t. The rateof variation of pressure over time along our trajectory is obviously the derivatived(f ◦x)dt (t) of the function (f ◦ x)(t) with respect to time. Let us find this derivative,assuming that f (x 1 , x 2 , x 3 ) is a differentiable function in the domain G. By the rulefor differentiating composite functions, we find∂f d(f ◦ x)∂f ∂f (t) = 1 x(t) ẋ 1 (t) + 2 x(t) ẋ 2 (t) + 3 x(t) ẋ 3 (t),dt∂x∂x∂xi(8.36)where ẋ i (t) = dxdt (t) (i = 1, 2, 3).Since the vector (ẋ 1 , ẋ 2 , ẋ 3 ) = v(t) is the velocity of our displacement at time tand (∂1 f, ∂2 f, ∂3 f )(x) is the coordinate notation for the differential df (x) of the4448 Differential Calculus in Several Variablesfunction f at the point x, Eq.
(8.36) can also be rewritten asd(f ◦ x)(t) = df x(t) v(t),dt(8.37)that is, the required quantity is the value of the differential df (x(t)) of the functionf (x) at the point x(t) evaluated at the velocity vector v(t) of the motion.In particular, if we were at the point x0 = x(0) at time t = 0, thend(f ◦ x)(0) = df (x0 ) v,dt(8.38)where v = v(0) is the velocity vector at time t = 0.The right-hand side of (8.38) depends only on the point x0 ∈ G and the velocityvector v that we have at that point; it is independent of the specific form of thetrajectory x = x(t), provided the condition ẋ(0) = v holds.
That means that thevalue of the left-hand side of Eq. (8.38) is the same on any trajectory of the formx(t) = x0 + vt + α(t),(8.39)where α(t) = o(t) as t → 0, since this value is completely determined by giving thepoint x0 and the vector v ∈ T R3x0 attached at that point. In particular, if we wishedto compute the value of the left-hand side of Eq. (8.38) directly (and hence also theright-hand side), we could choose the law of motion to bex(t) = x0 + vt,(8.40)corresponding to a uniform motion at velocity v under which we are at the pointx(0) = x0 at time t = 0.We now give the followingDefinition 1 If the function f (x) is defined in a neighborhood of the point x0 ∈ Rmand the vector v ∈ T Rmx0 is attached at the point x0 , then the quantityDv f (x0 ) := limt→0f (x0 + vt) − f (x0 )t(8.41)(if the indicated limit exists) is called the derivative of f at the point x0 with respectto the vector v or the derivative along the vector v at the point x0 .It follows from these considerations that if the function f is differentiable at thepoint x0 , then the following equality holds for any function x(t) of the form (8.39),and in particular, for any function of the form (8.40):Dv f (x0 ) =d(f ◦ x)(0) = df (x0 ) v.dt(8.42)8.3 The Basic Laws of Differentiation445In coordinate notation, this equality saysDv f (x0 ) =∂f∂f(x0 )v 1 + · · · + m (x0 )v m .∂x∂x 1(8.43)In particular, for the basis vectors e1 = (1, 0, .
. . , 0), . . . , em = (0, . . . , 0, 1) thisformula implies∂f(x0 ) (i = 1, . . . , m).∂x iBy virtue of the linearity of the differential df (x0 ), we deduce from Eq. (8.42)that if f is differentiable at the point x0 , then for any vectors v1 , v2 ∈ T Rmx0 and anyλ1 , λ2 ∈ R the function has a derivative at the point x0 with respect to the vector(λ1 v1 + λ2 v2 ) ∈ T Rmx0 , and thatDei f (x0 ) =Dλ1 v1 +λ2 v2 f (x0 ) = λ1 Dv1 f (x0 ) + λ2 Dv2 f (x0 ).(8.44)If Rm is regarded as a Euclidean space, that is, as a vector space with an innerproduct, then (see Sect. 8.1) it is possible to write any linear functional L(v) as theinner product &ξ, v' of a fixed vector ξ = ξ(L) and the variable vector v.In particular, there exists a vector ξ such thatdf (x0 )v = &ξ, v'.(8.45)Definition 2 The vector ξ ∈ T Rmx0 corresponding to the differential df (x0 ) of thefunction f at the point x0 in the sense of Eq. (8.45) is called the gradient of thefunction at that point and is denoted grad f (x0 ).Thus, by definition+,(8.46)df (x0 )v = grad f (x0 ), v .If a Cartesian coordinate system has been chosen in Rm , then, by comparingrelations (8.42), (8.43), and (8.46), we conclude that the gradient has the followingrepresentation in such a coordinate system:∂f∂f, .
. . , m (x0 ).(8.47)grad f (x0 ) =∂x∂x 1We shall now explain the geometric meaning of the vector grad f (x0 ).Let e ∈ T Rmx0 be a unit vector. Then by (8.46)De f (x0 ) = grad f (x0 ) cos ϕ,(8.48)where ϕ is the angle between the vectors e and grad f (x0 ).Thus if grad f (x0 ) = 0 and e = , grad f (x0 ),−1 grad f (x0 ), the derivativeDe f (x0 ) assumes a maximum value. That is, the rate of increase of the function f(expressed in the units of f relative to a unit length in Rm ) is maximal and equalto , grad f (x0 ), for motion from the point x0 precisely when the displacement is4468 Differential Calculus in Several Variablesin the direction of the vector grad f (x0 ). The value of the function decreases mostsharply under displacement in the opposite direction, and the rate of variation of thefunction is zero in a direction perpendicular to the vector grad f (x0 ).The derivative with respect to a unit vector in a given direction is usually calledthe directional derivative in that direction.Since a unit vector in Euclidean space is determined by its direction cosinese = (cos α1 , .
. . , cos αm ),where αi is the angle between the vector e and the basis vector ei in a Cartesiancoordinate system, it follows that+,∂f∂fDe f (x0 ) = grad f (x0 ), e = 1 (x0 ) cos α1 + · · · + m (x0 ) cos αm .∂x∂xThe vector grad f (x0 ) is encountered very frequently and has numerous applications. For example the so-called gradient methods for finding extrema of functionsof several variables numerically (using a computer) are based on the geometric property of the gradient just noted. (In this connection, see Problem 2 at the end of thissection.)Many important vector fields, such as, for example, a Newtonian gravitationalfield or the electric field due to charge, are the gradients of certain scalar-valuedfunctions, known as the potentials of the fields (see Problem 3).Many physical laws use the vector grad f in their very statement. For example,in the mechanics of continuous media the equivalent of Newton’s basic law of dynamics ma = F is the relationρa = − grad p,which connects the acceleration a = a(x, t) in the flow of an ideal liquid or gasfree of external forces at the point x and time t with the density of the mediumρ = ρ(x, t) and the gradient of the pressure p = p(x, t) at the same point and time(see Problem 4).We shall discuss the vector grad f again later, when we study vector analysis andthe elements of field theory.8.3.3 Differentiation of an Inverse MappingTheorem 4 Let f : U (x) → V (y) be a mapping of a neighborhood U (x) ⊂ Rm ofthe point x onto a neighborhood V (y) ⊂ Rm of the point y = f (x).
Assume that fis continuous at the point x and has an inverse mapping f −1 : V (y) → U (x) thatis continuous at the point y.Given these assumptions, if the mapping f is differentiable at x and the tanm−1 :gent mapping f (x) : T Rmx → T Ry to f at the point x has an inverse [f (x)]8.3 The Basic Laws of Differentiation447m−1 : V (y) → U (x) is differentiable at the pointT Rmy → T Rx , then the mapping fy = f (x), and the following equality holds: −1 )*−1f(y) = f (x) .Thus, mutually inverse differentiable mappings have mutually inverse tangentmappings at corresponding points.Proof We use the following notation:f (x) = y,f (x + h) = y + t,t = f (x + h) − f (x),so thatf −1 (y) = x,f −1 (y + t) = x + h,h = f −1 (y + t) − f −1 (y).We shall assume that h is so small that x + h ∈ U (x), and hence y + t ∈ V (y).It follows from the continuity of f at x and f −1 at y thatt = f (x + h) − f (x) → 0 as h → 0(8.49)h = f −1 (y + t) − f −1 (y) → 0 as t → 0.(8.50)andIt follows from the differentiability of f at x thatt = f (x)h + o(h)as h → 0,(8.51)that is, we can even assert that t = O(h) as h → 0 (see relations (8.17) and (8.18)of Sect.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
