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Chapter 2 Differentiation in Higher Dimensions

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Chapter 2 Differentiation in Higher Dimensions Chapter 2 Di®erentiation in higher dimensions 2.1 The Total Derivative Recall that if f : R ! R is a 1-variable function, and a 2 R, we say that f is di®erentiable f(a+h)¡f(a) 0 at x = a if and only if the ratio h tends to a ¯nite limit, denoted f (a), as h tends to 0. There are two possible ways to generalize this for vector ¯elds f : D! Rm; D ⊆ Rn; for points a in the interior D0 of D. (The interior of a set X is de¯ned to be the subset X0 obtained by removing all the boundary points. Since every point of X0 is an interior point, it is open.) The reader seeing this material for the ¯rst time will be well advised to stick to vector ¯elds f with domain all of Rn in the beginning. Even in the one dimensional case, if a function is de¯ned on a closed interval [a; b], say, then one can properly speak of di®erentiability only at points in the open interval (a; b). The ¯rst thing one might do is to ¯x a vector v in Rn and saythat f is di®erentiable along v i® the following limit makes sense: 1 lim (f(a + hv) ¡ f(a)) : h!0 h When it does, we write f 0(a; v) for the limit. Note that this de¯nition makes sense because a is an interior point. Indeed, under this hypothesis, D contains a basic open set U containing a, and so a + hv will, for small enough h, fall into U, allowing us to speak of f(a + hv). This 1 derivative behaves exactly like the one variable derivative and has analogous properties. For example, we have the following Theorem 1 (Mean Value Theorem for scalar ¯elds) Suppose f is a scalar ¯eld. Assume 0 f (a + tv; v) exists for all 0 · t · 1. Then there is a to with 0 · to · 1 for which 0 f(a + v) ¡ f(a) = f (a + t0v; v). Proof. Put Á(t) = f(a + tv). By hypothesis, Á is di®erentiable at every t in [0; 1], and 0 0 Á (t) = f (a + tv; v). By the one variable mean value theorem, there exists a t0 such that 0 Á (t0) is Á(1) ¡ Á(0), which equals f(a + v) ¡ f(a). Done. When v is a unit vector, f 0(a; v) is called the directional derivative of f at a in the direction of v. The disadvantage of this construction is that it forces us to study the change of f in one direction at a time. So we revisit the one-dimensional de¯nition and note that the condition 0 for di®erentiabilityµ there is equivalent¶ to requiring that there exists a constant c (= f (a)), f(a + h) ¡ f(a) ¡ ch such that lim = 0. If we put L(h) = f 0(a)h, then L : R ! R is h!0 h clearly a linear map. We generalize this idea in higher dimensions as follows: De¯nition. Let f : D! Rm (D ⊆ Rn) be a vector ¯eld and a an interior point of D. Then f is di®erentiable at x = a if and only if there exists a linear map L : Rn ! Rm such that jjf(a + u) ¡ f(a) ¡ L(u)jj (¤) lim = 0: u!0 jjujj Note that the norm jj ¢ jj denotes the length of vectors in Rm in the numerator and in Rn in the denominator. This should not lead to any confusion, however. Lemma 1 Such an L, if it exists, is unique. Proof. Suppose we have L; M : Rn ! Rm satisfying (*) at x = a. Then jjL(u) ¡ M(u)jj jjL(u) + f(a) ¡ f(a + u) + (f(a + u) ¡ f(a) ¡ M(u))jj lim = lim u!0 jjujj u!0 jjujj jjL(u) + f(a) ¡ f(a + u)jj · lim u!0 jjujj jjf(a + u) ¡ f(a) ¡ M(u)jj + lim = 0: u!0 jjujj 2 Pick any non-zero v 2 Rn, and set u = tv, with t 2 R. Then, the linearity of L; M implies that L(tv) = tL(v) and M(tv) = tM(v). Consequently, we have jjL(tv) ¡ M(tv)jj lim = 0 t!0 jjtvjj jtj jjL(v) ¡ M(v)jj = lim t!0 jtj jjvjj 1 = jjL(v) ¡ M(v)jj: jjvjj Then L(v) ¡ M(v) must be zero. De¯nition. If the limit condition (¤) holds for a linear map L, we call L the total deriva- tive of f at a, and denote it by Taf. It is mind boggling at ¯rst to think of the derivative as a linear map. A natural question n which arises immediately is to know what the value of Taf is at any vector v in R . We will show in section 2.3 that this value is precisely f 0(a; v), thus linking the two generalizations of the one-dimensional derivative. Sometimes one can guess what the answer should be, and if (*) holds for this choice, then it must be the derivative by uniqueness. Here are two examples which illustrate this. (1) Let f be a constant vector ¯eld, i.e., there exists a vector w 2 Rm such that f(x) = w, for all x in the domain D. Then we claim that f is di®erentiable at any a 2 D0 with derivative zero. Indeed, if we put L(u) = 0, for any u 2 Rn, then (*) is satis¯ed, because f(a + u) ¡ f(a) = w ¡ w = 0. (2) Let f be a linear map. Then we claim that f is di®erentiable everywhere with Taf = f. Indeed, if we put L(u) = f(u), then by the linearity of f, f(a + u) ¡ f(a) = f(u), and so f(a + u) ¡ f(a) ¡ L(u) is zero for any u 2 Rn. Hence (*) holds trivially for this choice of L. Before we leave this section, it will be useful to take note of the following: Lemma 2 Let f1; : : : ; fm be the component (scalar) ¯elds of f. Then f is di®erentiable at a i® each fi is di®erentiable at a. Moreover, T f(v) = (T f1(v); T f2(v); : : : ; T fn(g)). An easy consequence of this lemma is that, when n = 1, f is di®erentiable at a i® the following familiar looking limit exists in Rm: f(a + h) ¡ f(a) lim ; h!0 h 3 0 0 allowing us to suggestively write f (a) instead of Taf. Clearly, f (a) is given by the vector 0 0 0 (f1(a); : : : ; fm(a)), so that (Taf)(h) = f (a)h, for any h 2 R. n Proof. Let f be di®erentiable at a. For each v 2 R , write Li(v) for the i-th component of (Taf)(v). Then Li is clearly linear. Since fi(a + u) ¡ fi(u) ¡ Li(u) is the i-th component of f(a + u) ¡ f(a) ¡ L(u), the norm of the former is less than or equal to that of the latter. This shows that (*) holds with f replaced by fi and L replaced by Li. So fi is di®erentiable for any i. Conversely, suppose each fi di®erentiable. Put L(v) = ((Taf1)(v);:::; (Tafm)(v)). Then L is a linear map, and by the triangle inequality, Xm jjf(a + u) ¡ f(a) ¡ L(u)jj · jfi(a + u) ¡ fi(a) ¡ (Tafi)(u)j: i=1 It follows easily that (*) exists and so f is di®erentiable at a. 2.2 Partial Derivatives n Let fe1; : : : ; eng denote the standard basis of R . The directional derivatives along the unit vectors ej are of special importance. 0 De¯nition. Let j · n. The jth partial derivative of f at x = a is f (a; ej), denoted by @f (a) or Djf(a). @xj @f @f Just as in the case of the total derivative, it can be shown that (a) exists i® i (a) @xj @xj exists for each coordinate ¯eld fi. Example: De¯ne f : R3 ! R2 by f(x; y; z) = (exsin(y); zcos(y)): @f All the partial derivatives exist at any a = (x ; y ; z ). We will show this for and leave 0 0 0 @y it to the reader to check the remaining cases. Note that 1 ex0sin(y0+h) ¡ ex0sin(y0) cos(y + h) ¡ cos(y ) (f(a + he ) ¡ f(a)) = ( ; z 0 0 ): h 2 h 0 h We have to understand the limit as h goes to 0. Then the methods of one variable calculus x0sin(y0) show that the right hand side tends to the ¯nite limit (x0cos(y0)e ; ¡z0sin(y0)), which 4 @f is (a). In e®ect, the partial derivative with respect to y is calculated like a one variable @y @f derivative, keeping x and z ¯xed. Let us note without proof that (a) is (sin(y )ex0sin(y0); 0) @x 0 @f and (a) is (0; cos(y )). @z 0 It is easy to see from the de¯nition that f 0(a; tv) equals tf 0(a; v), for any t 2 R. This 1 1 follows as h (f(a + h(tv)) ¡ f(a)) = t th (f(a + (ht)v) ¡ f(a)). In particular the Mean Value 0 Theorem for scalar ¯elds gives fi(a + hv) ¡ f(a) = hfi (a + t0hv) = hfi(a + ¿v) for some 0 · ¿ · h. We also have the following Lemma 3 Suppose the derivatives of f along any v 2 Rn exist near a and are continuous at a.
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