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Derivatives of Vector Fields. Derivative Theory for Vector Fields Is a Straightfor- Ward Extension of That for Scalar Fields. Gi 18.024 SPRING OF 2008 DER. DERIVATIVES OF VECTOR FIELDS Derivatives of vector fields. Derivative theory for vector fields is a straightfor- ward extension of that for scalar fields. Given f : D ⊂ Rn ! Rm a vector field, in components, f consists of m scalar fields of n variables. That is, f(x) = (f1(x); : : : ; fm(x)); n and each fi : D ⊂ R ! R is a scalar field, where i = 1; : : : ; m. It seems natural to define the differentiability of f, thus, componentwise. We recall that if f : D ⊂ Rn ! R is a scalar field and a 2 int(D) then the derivative of f at a (if exists) is given by the gradient rf(a) = (D1f(a;D2f(a);:::;Dnf(a)): For our purposes, it is convenient to understand the derivative of f as a row matrix rather than as a vector. We will write as Df(a) = [D1f(a;D2f(a);:::;Dnf(a)]: Then, f is differentiable at a (with the total derivative Df(a) if (*) f(a + h) − f(a) = Df(a)h + khkE(h; a); and jE(h; a)j ! 0 as khk ! 0. Here the first term on the right side is matrix multiplication of Df(a) and the column matrix 2 3 h1 6 . 7 h = 4 . 5 : hn Definition 1. Let f : D 2 Rm and D ⊂ Rn be open. In scalar form, f(x) = (f1(x1; : : : ; xn); : : : ; fm(x1; : : : ; xn)): We say that f is differentiable at a 2 D if each of the component functions f1; : : : ; fm is differentiable at a (in the sense discussed in [Apo, Section 8.11] or above). Furthermore, by the derivative of f at a we mean the matrix 2 3 D1f1(a) D2f1(a) ··· Dnf1(a) 6 D1f2(a) D2f2(a) ··· Dnf2a) 7 6 7 Df(a) = 6 . 7 : 4 . 5 D1fm(a) D2fm(a) ··· Dnfm In other words, Df(a) is the matrix whose ith row is given as the derivative Dfi(a) of the ith coordinate function of f. It is often called the jacobian matrix∗ of f. An- other notation for this matrix is @(f ; ··· ; f ) 1 m : @(x1; ··· ; xn) ∗It is named after the Prussian mathematician Carl Gustav Jacob Jacobi (1804-1851). 1 2 DERIVATIVES OF VECTOR FIELDS In [Apo, Section 8.18], the differentiability of a vector field f is given, alterna- tively, by directly extending the Taylor’s formula (*) in the vector-field setting. Our definition then is equivalent to that given in [Apo], which is stated below. Theorem 2. The vector field f is differentiable at a if and only if f(a + h) − f(a) = Df(a)h + khkE(h; a) and kE(h; a)k ! 0 as khk ! 0. Here, f; h and E are written as column matrices. Proof. We observe that both sides of the equation represent column matrices. Con- sidering the ith entries of these matrices, we obtain the following equation fi(a + h) − fi(a) = Dfi)(a) · h + khkEi(h; a): Now if f is differentiable at a if and only if each fi is, and fi is differentiable at a if and only if jEi(h; a)j ! 0 as khk ! 0. But, jEi(h)j ! 0 as khk ! 0 for each i if and only if kE(h)k ! 0 as khk ! 0, and we are done. The following result is immediate from our definition of differenentiability. Theorem 3. If f is differentiable at a then f is continuous at a. The chain rule for derivatives of vector fields. Before considering the chain rule for vector fields, let us take the chain rule for scalar fields we have already proved [Apo, Section 8.15] and reformulate it in terms of matrices. Let f : D ⊂ Rn ! R be a scalar field defined in an open ball about a, and let n x : I ⊂ R ! R be a vector-valued function defined in an open interval about t0. Let x(t0) = a and x(I) ⊂ D. If f is differentiable at a and if x is differentiable at t0, then we have shown in Theorem 8.8 of [Apo] that f ◦ x is differentiable at t0 and its derivative is given by d f(x(t )) = rf(x(t )) · x0(t ): dt 0 0 0 We now write this formula in scalar form as d @f dx @f dx (f ◦ x) = 1 + ··· + n ; dt @x1 dt @xn dt or, we write it in matrix form as 2 dx1 3 d @f @f dt 6 . 7 (f ◦ x) = ··· . : dt @x1 @xn 4 5 dxn dt Recalling the definition of the Jacobian matrix Df the latter formula is recognized as d (f(x(t)) = Df(x(t))Dx(t): dt (Note that Df is a row matrix while Dx is by definition a column matrix.) This is the form of the chain rule which is useful to extend to vector fields. Let f : Rn ! Rp and g : Rp ! Rm be vector fields such that the composition field h = g ◦ f is defined in an open ball about a 2 Rn. Let f(a) = b 2 Rp. We write these fields as f(x) = f(x1; : : : ; xn); g(y) = g(y1; : : : ; yp): DERIVATIVES OF VECTOR FIELDS 3 If f and g are differentiable at a and b, respectively, then partial derivatives of h exist at a by applying the chain rule for each hi(x) = gi(f(x)) as @h @g @f @g @f i = i 1 + ··· + i p : @xj @y1 @xj @yp @xj This is recognized as matrix multiplication 2 3 Djf1 6 . 7 [D1gi D2gi ··· Dpgi] 4 . 5 : Djfp In other words, its multiplication of the ith row of Dg and the jth column of Df. Thus, the Jacobian matrix of h is expected to satisfy the matrix equation Dh(a) = Dg(b)Df(a): Not exactly. If f and g are differentiable then we know that the partial deriva- tives of the composite function h exist. But, the mere existence of the partial deriva- tives of the function hi does not imply the differentiability of hi, and thus, one needs to provide a separate argument that h is differentiable. One may avoid giving a separate proof of the differentiability of h by assum- ing a stronger hypothesis, namely, that both f and g are continuously differentiable. Then, the formula, p X Djhi(x) = Dlgi(f(x))Djfl(x); l=1 which we have proved, shows that Djhi is continuous, that is, h is continuously differentiable. Therefore, h is differentiable. We summarize these. Theorem 4. Let f : Rn ! Rp and g : Rp ! Rm be vector fields defined in open balls about a 2 Rn and b 2 Rp, respectively. Let f(a) = b. If f and g are continuously differ- entiable on their respective domains, then h(x) = g ◦ f(x) is continuously differentiable on its domain and Dh(x) = Dg(f(x) · Df(x): Most applications of the chain rule we will use in future are in the C1-class setting, and thus this theorem will suffice. But, one may remove the C1 condition. Please read the statement and its proof of [Apo, Theorem 8.11]. Example 5 (Polar coordinates). Given a function f(x; y) defined on the (x; y) plane, we introduce polar coordinates x = r cos θ and y = r sin θ and f becomes a function of r and θ as φ(r; θ) = f(r cos θ; r sin θ). Then, @φ @f @x @f @y @f @f = + = cos θ + sin θ @r @x @r @y @r @x @y and @φ @f @f = −r sin θ + r cos θ: @θ @x @y 4 DERIVATIVES OF VECTOR FIELDS The second derivative is computed as @2φ @ @φ @ @f @f = = cos θ + sin θ @r2 @r @r @r @x @y @ @f @f @(cos θ) @ @f @f @(sin θ) = cos θ + + sin θ + @r @x @x @r @r @y @y @r @ @f @ @f @ @f @ @f = cos θ + sin θ cos θ + cos θ + sin θ sin θ @x @x @y @x @x @y @y @y @2f @2f @2f = + + sin θ cos θ: @x2 @y2 @x@y An analogous formula for @2φ/∂θ2 is worked in [Apo, Example 3] and @2φ/∂r@θ is in [Apo, Section 8.22, Exercise 5]. Derivatives of inverses. Recall that if f(x) is a differentiable real-valued function of a single variable x, and if f 0(x) > 0 for x 2 [a; b] (or f 0(x) < 0), then f is strictly increasing (or decreasing), and it has an inverse, say g. Furthermore, g is differentiable and its derivative satisfies 1 g0(f(x)) = : f 0(x) Part of this theory extends to vector fields. Theorem 6. Let f : Rn ! Rn be defined in an open ball of a and f(a) = b. Suppose that f has an inverse, say g. If f is differentiable at a and if g is differentiable at b then Dg(b) = (Df(a))−1: In interpretation, the jacobian matrix of the inverse function is just the inverse of the jacobian matrix. Proof. We observe that g ◦ f(x) = x. Since f and g are differentiable and so is the composite function g ◦ f, by the chain rule, it follows that Dg(b)Df(a) = In; which implies the assertion. The theorem shows that in order for the differentiable function f to have a dif- ferentiable inverse, it is necessary that the Jacobian matrix Df have rank n. Roughly speaking, it is also sufficient for f to have an inverse. More precisely, one has the fol- lowing Inverse Function Theorem, which is one of the fundamental results in math- ematical analysis.
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