NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS

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1 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS SAMEER CHAVAN Abstract. This is the first part of Notes on Multivariable Calculus based on the classical texts [6] and [5]. We present here the geometric theory of functions of several real variables. Contents 1. Introduction and Motivation 2 2. Preliminaries The Usual Topology in the Euclidean Spaces Linear Transformations The Plus Topology in the Euclidean Spaces* 7 3. Vector-valued Functions in Several Real Variables Limits Continuity Plus-Continuity* Partial Derivatives and Continuity Partial Derivatives and Plus-Continuity* Directional Derivatives Partial Derivatives of Higher Order Differentiation Properties of Derivatives The Matrix Representation of Derivatives A Criterion for Differentiability The Chain Rule Applications A First Ordered Partial Differential Equation The Multi-dimensional Mean Value Theorem The Inverse Function Theorem Statement and Examples A Contraction Principle Proof of the Inverse Function Theorem The Implicit Function Theorem Statement and Examples Proof of the Implicit Function Theorem 29 Acknowledgements 31 References 31 1

2 2 SAMEER CHAVAN 1. Introduction and Motivation This is the Differential Calculus part of Notes on Multivariable Calculus prepared for the IISER course MTH-201. In these notes, we present the geometric theory of Calculus in Several Dimensions. Our treatment here assumes the knowledge of the preceding IISER courses: Linear Algebra MTH-101 and Calculus MTH In this part, we study vector-valued functions f : U V, where U and V are (open) subsets of R n and R m respectively. Notice that such an f(x) can be rewritten as (f 1 (x),, f m (x)) for x U, where each f i is a scalar-valued function of n variables. In this study, the following basic problems regarding the system of (non-linear) equations may be addressed: (1) (Implicit Function Problem) For f : R n+m R n and for y R n, find sufficient conditions under which a unique x R m for the system f(x, y) = 0 of n equations can be obtained. (2) (Inverse Function Problem) For f : R n R n and for y R n, find sufficient conditions under which the system y = f(x) of n equations can be solved for x in terms of y. Let us look at the linear versions of the aforementioned problems: (1) For an n (n + m) matrix A : R n+m R n and for y R n, find sufficient conditions under which a unique solution x R m for the system A(x, y) = 0 of n linear equations can be obtained. (2) For an n n matrix A : R n R n and for y R n, find sufficient conditions under which the system y = Ax of n linear equations can be solved for x in terms of y. We have already encountered with these problems in Linear Algebra MTH-101 ([3]). Theorem 1.1. (The Linear Implicit Function Theorem, Theorem 9.27 of [6]) For an n (n + m) matrix A : R n+m R n and x R n and y R m, let A 1 x = A(x, 0) and A 2 y = A(0, y). If A 1 is invertible then there corresponds to every y R m a unique x R n such that A(x, y) = 0. Proof. Note that A(x, y) = 0 A 1 x + A 2 y = 0 x = (A 1 ) 1 A 2 y. This completes the proof of the theorem. Problem 1.2 : Show that the system a 11 x a 1n x n = y 1 (1.1).. a n1 x a nn x n = y n can be rewritten in the form B(x, 1) = 0 for x R n for some n (n + 1) matrix B. Conclude the following: (1) If {a ij } 1 i,j n is invertible then there exists a unique x R n such that the non-homogeneous system (1.1) admits a solution. (2) (1.1) is the linear version of the Inverse Function Problem.

3 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 3 Before we turn to the discussion of multivariable non-linear considerations, it is advisable to look at the one-variable Inverse Function Problem. Theorem 1.3. (One-dimensional Inverse Function Theorem, Theorem 13 of Chapter 3 of [5]) If f : (a, b) f(a, b) is a differentiable map and f (x) 0 for every x (a, b) then for any y f(a, b) there exists a unique x (a, b) such that y = f(x). Proof. Clearly, for any y f(a, b) there exists an x (a, b) such that y = f(x). We need only to take care of the Uniqueness Part. Since f is never zero, by the Intermediate Value Property of Derivatives (Darboux s Theorem) MTH-101 one of the following holds true: (1) f (x) > 0 for every x (a, b). (2) f (x) < 0 for every x (a, b). Without loss of generality, we may assume that (1) holds. If a < s < t < b then by the Mean Value Theorem there exists at least one θ (s, t) such that f(s) f(t) = f (θ)(s t). It follows that f(s) f(t) < 0 if s < t. Thus f is strictly monotone. Hence f is one-one and the proof of the Uniqueness Part is over. The following example shows that the assumption that f (x) 0 for every x (a, b) is needed in Theorem 1.3. Problem 1.4 : Define f : ( 1, 1) f( 1, 1) by f(0) = 0 and ( ) 1 f(t) = t + 2t 2 sin if t 0. t Verify: (1) f is differentiable, f (0) 0, but f is not injective. (2) There exists a t ( 1, 1) such that f (t) = 0. Observe that in the linear as well as the one-variable case, we have obtained global solutions to IFP s. The multivariable picture is quite different. For example, consider the function f : R 2 R 2 given by f(x, y) = (exp(x) cos(y), exp(x) sin(y)) (x, y R). It is easy to see that f(0, 0) = (1, 0) = (0, 2π) and that the Jacobian of f is non-zero at any point of R 2! We conclude the section with a calculus problem which involves implicit differentiation. Problem 1.5 : Suppose the equation f(x, y) = 0 determines y as a differentiable function of x. Suppose further that the partial derivatives of f with respect to x and y exist. If y (x, y(x)) 0 then dy dx = x (x, y(x)) (x, y(x)). y Surprisingly, the non-vanishing of y is also sufficient to justify the assumption that y is a differentiable function of x. This is a special case of the so-called the Implicit Function Theorem.

4 4 SAMEER CHAVAN 2. Preliminaries In this section, we record a few prerequisites pertaining to the Euclidean spaces and the linear transformations The Usual Topology in the Euclidean Spaces. Consider the vector space R n {x = (x 1,, x n ) : x i R for i = 1,, n} over the real field R with respect to the following operations: (x 1,, x n ) + (y 1,, y n ) = (x 1 + y 1,, x n + y n ), α(x 1,, x n ) = (αx 1,, αx n ). Endow R n with the usual inner-product x, y R n : n x, y R n x i y i (x = (x 1,, x n ), y = (y 1,, y n ) R n ). i=1 This inner-product gives rise to the usual norm on R n : x R n + x, x R n (x = (x 1,, x n ) R n ). Whenever there is no ambiguity, we will suppress the suffix and simply write x, y and x in place of x, y R n and x R n respectively. Problem 2.1 : For x, y, z R n and α R, verify the following: N1 (Non-negativity) x 0. N2 (Positive Definiteness) x = 0 if and only if x = 0. N3 (Dilation) αx = α x. I1 (Symmetry) x, y = y, x. I2 (Linearity in the 1st slot) αx + y, z = α x, z + y, z. I3 (Linearity in the 2nd slot) x, αy + z = x, z + α y, z (Use I1 and I2). IN1 (The Cauchy-Schwarz Inequality) x, y x y ( Note that the discriminant x, y 2 x 2 y 2 of the equation n i=1 (x it y i ) 2 = 0, quadratic in t, is non-positive. N4 (Triangle Inequality-I) x + y x + y (Solve x + y, x + y using I2, I3, and then use IN1). N5 (Triangle Inequality-II) x y x + y (Use N4). IN2 (Pythagorean Identity) If x, y = 0 then x + y 2 = x 2 + y 2. We denote by B r (x) the open ball {y R n : y x < r} in R n centered at x R n of radius r > 0. Definition 2.2 : Let U be any subset of R n. We say that U is open in R n if there exists a ball B r (x) U whenever x U. Problem 2.3 : Show that any open interval is open in R, that any open disc is open in R 2, and in general that B r (x) is open in R n. Problem 2.4 : Prove: (1) If {U α } α is an arbitrary collection of open sets in R n then α U α is also open in R n (First verify this for two open subsets of R n ). In particular, R n is open in R n.

5 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 5 (2) If {U m } k m=1 is a finite collection of open sets in R n then k m=1 U m is also open in R n. In particular, the empty set is open in R n. (3) If {U m } m 1 is a countable collection of open sets in R n then m 1 U m need not be open in R n (First try to get a counterexample in the case n = 1). (4) If U is open in R n+m then {y R m : (0, y) U} is open in R m. Definition 2.5 : Let U be any subset of R n. We say that U is closed in R n if the complement R n \ U of U is open in R n. We denote by B r (x) the closed ball {y R n : y x r} in R n centered at x R n of radius r > 0. Problem 2.6 : Show that any closed interval is closed in R, that any closed disc is closed in R 2, and in general that B r (x) is closed in R n. Problem 2.7 : Formulate statements similar to those in Problem 2.4 for closed sets in R n (Use demorgan s Law: R n \ α U α = α {Rn \ U α }). By a sequence in R n we mean an ordered subset {x k (x k 1,, x k n)} k 0 of R n. Note that each {x k i } k 0 is a sequence of real numbers (i = 1,, n). We say that {x k } k 0 is (1) convergent and converges to x R n if for every ɛ > 0 there exists a integer n ɛ > 0 such that x k x < ɛ for all k n ɛ. (2) Cauchy if for every ɛ > 0 there exists a integer n ɛ > 0 such that x k x l < ɛ for all l, k n ɛ. (3) bounded if there exists a real M > 0 such that x k M for every k 1. Remark 2.8 : Notice that every convergent sequence is Cauchy and that every Cauchy sequence is bounded. However, a bounded sequence in R n need not be Cauchy (Example: {(( 1) k, 1)} k 0 ). Problem 2.9 : Let {x k (x k 1,, x k n)} k 0 denote a sequence in R n. Then (1) {x k i } k 0 is Cauchy for all i = 1,, n if and only if so is {x k } k 0. (2) {x k i } k 0 is convergent for all i = 1,, n if and only if so is {x k } k 0. Theorem In R n, every Cauchy sequence is convergent. Proof. Let {x k (x k 1,, x k n)} k 0 denote a sequence. Recall that every Cauchy sequence in R is convergent MTH-102. In view of this fact and Problem 2.9, one has {x k } k 0 is Cauchy if and only if {x k i } k 0 is Cauchy (i = 1,, n) if and only if {x k i } k 0 is convergent (i = 1,, n) if and only if {x k } k 0 is convergent. Problem 2.11 : Let U be a closed subset of R n. If {x k } k 0 U is convergent and converges to x then x U Linear Transformations. Let U and V be vector spaces over R. Recall that T : U V is a linear transformation if T (αu 1 + u 2 ) = αt (u 1 ) + T (u 2 ) for all u 1, u 2 U and for all α R. In case U = V then we will refer to T as a linear operator on U. In these notes, we restrict ourselves to the linear transformations T : R n R m. Recall that the matrix representation of such a T is an m n matrix with real entries.

6 6 SAMEER CHAVAN Definition 2.12 : Let T : R n R m be a linear transformation. We define the (operator) norm of T as follows: T sup T x R m. x R n 1 Remark 2.13 : We make the following useful observations: (1) T x R m T x R n for every x R n. (2) If λ R is such that T x R m λ x R n for every x R n then T λ. Problem 2.14 : If T : R n R m is a linear transformation then show that T is a finite non-negative real number. Solution. Let {e 1,, e n } be the standard basis for R n. Write x = n i=1 x je j for scalars x j R. Verify that T x R m n j=1 x j T e j R m. An easy application of the Cauchy-Schwarz Inequality yields that 1 2 n T x R m x T e j 2 R. m j=1 Hence by Remark 2.13(2) T n i=1 T e i 2 R m. Theorem (Theorem 9.7, [6]) Let T 1 : R n R m, T 2 : R n R m and S : R m R l be linear transformations and let c be a real scalar. Then the following are true: (1) T 1 + T 2 T 1 + T 2 and ct 1 = c T 1. (2) The linear transformation S T 1 : R n R l given by S T 1 (x) S(T 1 (x)) (x R n ) satisfies S T 1 S T 1. Proof. Recall that if T : R n R m is a linear transformation then T x R m T x R n for every x R n and T is finite (Remark 2.13(1) and Problem 2.14). By the Triangle Inequality and Remark 2.13(1) (T 1 + T 2 )x = T 1 x + T 2 x T 1 x + T 2 x T 1 x + T 2 x = ( T 1 + T 2 ) x for any x R n. It follows from Remark 2.13(2) that T 1 + T 2 T 1 + T 2. To see the second part, note that S T 1 x = S(T 1 x) S T 1 (x) S T 1 x (x R n ). Again, we appeal to Remark 2.13(2). In practice, it is extremely hard to compute the norms of arbitrary matrices. However, the following can be easily seen (refer to Section 6 of Chapter II of [2]). Problem 2.16 : If A = (a ij ) is an m n matrix with real entries then the following are true: (1) A i,j a2 ij. (2) A C 1 2 s R 1 2 s, where C s sup 1 j n i=1 m a ij and R s sup 1 i m j=1 n a ij.

7 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 7 n Solution. In view of the solution of Problem 2.14, one has A j=1 Ae j 2 R m, where {e 1,, e n } is the standard basis for R n. Note that Ae i = (a 1j,, a nj ). By the Pythagorean Identity (Problem 2.1) Ae j 2 R = n m i=1 a2 ij (1 j n). The first estimate is now obvious. A straight-forward application of the Cauchy-Schwarz Inequality 2.1(3) yields the second estimate as well. The desired verification is left to the reader. Recall that a linear operator T on R n is invertible if there exists another linear operator, say, S on R n such that T S = I = ST. We will use T 1 to denote the inverse of invertible T. Problem 2.17 : If T is a linear operator on R n then T is invertible if and only if there exists a positive real number c such that T x c x (x R n ). In this case T 0. Hint. Use the Rank-Nullity Theorem. If T is a linear operator on R n then we denote by B r (T ) the set {S : S is a linear operator on R n such that S T < r}, where r is a positive real. The set of invertible linear operators on R n turns out to be open. Theorem (Theorem 9.8, [6]) Let I(R n ) denote the set of invertible linear operators on R n. If T I(R n ) then B T 1 1(T ) I(Rn ). Proof. Let S be in B T 1 1(T ). Thus S T < T 1 1. Hence T 1 1 x = T 1 1 T 1 T x T 1 1 T 1 T x = T x = T x Sx + Sx T x Sx + Sx S T x + Sx. It follows that ( T 1 1 S T ) x Sx (x R n ). Now the desired conclusion follows from Problem The Plus Topology in the Euclidean Spaces*. The present subsection, based on the article [7], provides some basics of the so-called the plus topology in R n. As we will see soon that this topology is appropriate for the study of partial derivatives (see Subsection 3.5). Throughout this subsection, we assume that n 2. For x R n, we use P r (x) to denote n {y R n : y i x i < r and y j = x j for 1 j i n}. i=1 If n = 2 then the set P r (x) looks like a plus sign centered at x. Definition 2.19 : Let U be any subset of R n. We say that U is plus-open in R n if for every x U there exists a real number r > 0 such that P r (x) is contained in U.

8 8 SAMEER CHAVAN Remark 2.20 : If U is open in R n then it is plus-open in R n. A set plus-open in R n need not be open in R n : Problem 2.21 : Verify that the set given below is plus-open in R 2 but not open in R 2 : {(x 1, x 2 ) R 2 : x 2 > 3 x 1 or x 2 < 3 x 1 } {(0, 0)}. Problem 2.22 : Formulate statements similar to those in Problem 2.4 for plusopen sets in R n. 3. Vector-valued Functions in Several Real Variables Recall that a real-valued function f : R R is differentiable at a point x R f(x+h) f(x) if lim h 0 h exists. If the above limit exists then we say that f is differentiable at x with the derivative f f(x+h) f(x) (x) = lim h 0 h. In order to formulate an appropriate notion of differentiability of vector-valued functions in several real variables f : R n R m, we need to take care of two things: (1) Since h R n, we need to make h 0 more explicit. (2) Since division by h R n does not make any sense for n 2, we need to find a suitable replacement for division by h Limits. Let f : R R be a real-valued function and let a be a real number. Recall that lim x a f(x) exists if there exists a real number L such that lim x a f(x) = L = lim x a + f(x). Thus x approaches a either from the left or from the right. The situation is very different when one deals with the limit of a vector-valued function in several real variables. Definition 3.1 : Suppose U and V are open subsets of R n and R m respectively. Let f : U V be a vector-valued function and let a U be given. We say that lim x a f(x) = L R if for every ɛ > 0 there exists a δ > 0 such that f(x) L < ɛ whenever x U and 0 < x a < δ. As in the one-variable case, the following basic assertions hold true. Theorem 3.2. If limit exists, it is unique. Limits preserve real linear combinations as well as products. Proof. Mimic the one-variable arguments. Problem 3.3 : Show that lim (x,y) (0,0) xy x 2 +y 2 xy x 2 +y 2 does not exist. Solution. Note that f(x, y) = is defined for any (x, y) R 2 \ {(0, 0)}, so that f : U V where U = R 2 \ {(0, 0)} and V = R. Suppose there exists a real L such that lim (x,y) (0,0) f(x) = L. Then for ɛ = 1 4 there exists some δ > 0 such that xy x 2 + y 2 L < 1 4 whenever 0 < x 2 + y 2 < δ.

9 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 9 If 0 < x < δ and y = 0 then we must have L < 1 4. If 0 < x = y < must have 1 2 L < 1 4. Thus 1 2 = 1 2 L + L 1 2 L + L < = 1 2, we arrive at a contradiction! Problem 3.4 : Show that lim (x,y) (0,0) x 4 y 4 (x 2 +y 4 ) 3 does not exist. δ 2 then we It may happen that lim x a lim y b f(x, y) lim y b lim x a f(x, y) : Let a = 0 = b and consider the function f(x, y) = x y x+y for all (x, y) R2 such that x + y 0. Then lim x 0 lim y 0 f(x, y) = 1 1 = lim y 0 lim x 0 f(x, y). Theorem 3.5. Let f : R n \ {a} R be a real-valued function not defined possibly at a R n. Let {i 1,, i n } denote a permutation of {1,, n}. If lim f(x), x a lim f(x), lim x in a in x in 1 a in 1 lim lim f(x) exists. Indeed, x i1 a i1 x in a in lim f(x),, x in a in lim lim f(x) = lim f(x). x i1 a i1 x in a in x a lim lim f(x) exist then x i2 a i2 x in a in Proof. Set L = lim x a f(x). Then for ɛ > 0 there exists δ > 0 such that f(x) L < ɛ 2 whenever a x B δ(a). Let {e 1,, e n } denote the standard basis for R n and let h i be such that 0 < h i < δ n. Let R denote the interior of the rectangular box with 2n vertices of the form a ( ( ) n 0 terms), a + hi e i ( ( ) n 1 terms), a + hi e i + h j e j ( ( n 2) terms),, a + h1 e h n e n ( ( n n) terms). Since R Bδ (a) \ {a}, f(x) L < ɛ whenever x R. 2 If we fix h ik (1 k n 1) in the last inequality and let h in 0 then we get lim f(x) L x in a in ɛ 2 whenever x R and x i n = a in. If we fix h ik (1 k n 2) in the last inequality and let h in 1 0 then we get lim lim f(x) L x in a in ɛ 2 whenever x R, x i n 1 = a in 1 and x in = a in. x in 1 a in 1 If we continue this, at the (n 1) th step, we get lim lim f(x) L x i2 a in 1 x in a in < ɛ whenever 0 < x i 1 a i1 < It follows that lim lim f(x) = L. x i1 a i1 x in a in The hypotheses in Theorem 3.5 are all needed: δ n. Example 3.6 : Consider f(x, y) = x sin( 1 y ) ((x, y) R2 \ {(0, 0)}). Then it is easy to see that lim (x,y) (0,0) f(x, y) exists. However, lim lim f(x, y) lim lim f(x, y). x 0 y 0 y 0 x 0

10 10 SAMEER CHAVAN Converse to Theorem 3.5 is not true: Example 3.7 : Consider f(x, y) = easy to see that x 2 y 2 x 2 y 2 +(x y) 2 lim lim f(x, y) = 0 = lim lim f(x, y). x 0 y b y 0 x a However, lim (x,y) (0,0) f(x, y) does not exist Continuity. ((x, y) R 2 \ {0, 0}). Then it is Definition 3.8 : Let B r (x) be as in Subsection 2.1 and let U and V be open subsets of R n and R m respectively. We say that f : U V is continuous at a U if for every ɛ > 0 there exists a δ > 0 such that f(x) f(a) < ɛ whenever x U and x B δ (a). f is continuous in U if it is continuous at every point in U. Remark 3.9 : Continuous functions are in abundance: (1) If T : R n R m is a linear transformation then T is continuous: If T = 0 then any δ works; if T 0 then δ = ɛ T works (see Problem 2.14). In particular, the projection maps π i : R n R given by π i (x 1,, x m ) = x i (1 i n) are continuous. (2) It follows from Theorem 3.2 that real linear combinations as well as products of continuous functions are continuous. In particular, any 2-variable polynomial p given by k p(x 1, x 2 ) = c mn x m 1 x n 2 (x 1, x 2 R) m,n=0 is continuous in R 2, where c mn R for integers m, n 0. (3) If f : U V and g : V W are continuous then g f : U W given by g f(x) g(f(x)) (x U) is also continuous. (4) f = (f 1,, f m ) : R n R m is continuous if and only if f i is continuous for i = 1,, m : Since f i = π i f, continuity of f implies that of f i (1 i m). Also, since f(x) = m i=1 f i(x)e i, continuity of f i (1 i m) imply that of f, where e i is the i th vector in the standard basis for R n. Problem 3.10 : The rational function R given by R(x, y) = x3 x 2 + y 2 is defined everywhere in R 2 except at (0, 0). What value should be assigned to R(0, 0) to make the function continuous at the point (0, 0)? Definition 3.11 : Let f : R n R be a real-valued function. Fix a R n and i {1,, n}. Define f ai : R R by f ai (t) = f(a 1,, a i 1, t, a i+1,, a n ) (t R). We say that f is separately continuous if for every a j R (1 j i n) the function f ai is continuous for i = 1,, n.

11 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 11 Remark 3.12 : Consider the injection maps I ai : R R n given by I ai (t) = (a 1,, a i 1, t, a i+1,, a n ) (1 i n). Since f ai = f I ai and since I ai is continuous, by Remark 3.9(3) the continuity of f : R n R implies the continuity of f ai (i = 1,, n). The following example shows that the continuity of f ai at a i (i = 1,, n) need not ensure the continuity of f at a. Example 3.13 : Consider the function f : R 2 R given by f(0, 0) = 0 and f(x, y) = xy x 2 +y for any (x, y) R 2 \ {(0, 0)}. Let a = (0, 0). Then f 2 a1 = 0 = f a2 is continuous everywhere in R. However f is not continuous at a (see Problem 3.3) Plus-Continuity*. Definition 3.14 : Let U denote a plus-open set in R n. We say that f : U R m is plus-continuous at a U if for every ɛ > 0 there exists a subset V of U containing a and plus-open in R n such that f(x) f(a) < ɛ whenever x V. f is plus-continuous in U if it is plus-continuous at every point in U. Remark 3.15 : Since every open set is plus-open, a continuous function is always plus-continuous. Theorem (Theorem 1, [7]) A real-valued function f : R n R is pluscontinuous if and only if it is separately continuous. Proof. We notice that (1) x P δ (a) if and only if x = (a 1,, a i 1, x i, a i+1,, a n ) with x i a i < δ for some 1 i n. (2) If x P δ (a) then f(x) = f ai (x i ). Suppose f is plus-continuous at a. Thus for every ɛ > 0 there exists a plus-open set V in R n containing a such that f(x) f(a) < ɛ whenever x V. Since V is plus-continuous, there exists δ > 0 such that P δ (a) V. Thus for given ɛ > 0 there exists a δ > 0 such that By (1) and (2) we have f(x) f(a) < ɛ whenever x P δ (a). f ai (x i ) f ai (a i ) < ɛ whenever x i a i < δ. If we let x i = t then it follows that f ai (t) f ai (a i ) < ɛ whenever t a i < δ. Hence f ai is continuous at a i for each i. Conversely, assume that f is separately continuous. Let U be any open subset of R. Let a f 1 (U) {x R n : f(x) U}. Since f(a) U and since U is open in R there exists ɛ > 0 such that B ɛ (f(a)) U. We claim that there exists δ > 0 such that P δ (a) f 1 (B ɛ (f(a))) {x R n : f(x) B ɛ (f(a))} f 1 (U).

12 12 SAMEER CHAVAN Since f is separately continuous, there exist δ i (1 i n) such that f ai (t) f ai (a i ) < ɛ whenever t a i < δ. Set δ = min 1 i n δ i. If we let t = x i then again it follows from (1) and (2) that f(x) f(a) < ɛ whenever x P δ (a). Thus the claim stands verified. In particular, if we let U = B ɛ (a) then there exists a plus-open set V = f 1 (B ɛ (a)) containing a such that f(x) f(a) < ɛ whenever x V. Hence f is plus-continuous at a. A plus-continuous function need not be continuous: If f is as in Example 3.13 then f is plus-continuous everywhere (Theorem 3.16) and discontinuous at the origin (Example 3.13) Partial Derivatives and Continuity. Consider a function f : U R, where U is an open subset of R n. Let {e 1,, e n } denote the standard basis for R n. For x U and 1 i n, define the partial derivative of f with respect to x i to be ( ) f(x + te i ) f(x) (x) lim, t 0 t provided the limit exists. Remark 3.17 : Note the following: (1) The partial derivative (x) is the slope of the hypersurface y = f(x) at the point x in the direction of the x i axis. (2) If U = R n then f ai as given in Definition 3.11 is differentiable at a i if and only if (3) If and g (αf + g) exists at a (i = 1,, n). Indeed, (f ai ) (a i ) = (a). exist then (αf+g) and (fg) exist. Moreover, = α + g (α R), (fg) = f g + g. (This follows from (2) and elementary theory of Calculus MTH-102). Problem 3.18 : Suppose f : R n R is of the form f(x 1,, x n ) = g(x i )h(x 1,, ˇx i,, x n ) where g is a real-valued function of x i and h is a real-valued function of x j s (1 j i n). Show that (x) = g (x i )h(x 1,, ˇx i,, x n ). Problem 3.19 : Let f : R n R be given and let x R n. Define g : R R by g(t) = f(x + te i ). Show that g is differentiable at t 0 if and only if exists at x + t 0 e i. In this case, (x + t 0 e i ) = g (t 0 ). Problem 3.20 : Let f : B r (x) R 2 R be such that x 1 (x) = 0 for every x B r (x). Show that f(x) depends only on x 2.

13 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 13 Hint. Use the Mean Value Theorem of Calculus MTH-102. We have seen in Calculus MTH-102 that differentiability implies continuity. The following example shows that the existence of partial derivatives need not ensure continuity. Problem 3.21 : If f(0, 0) = 0 and f(x, y) = xy x 2 if (x, y) (0, 0), + y2 prove that the origin. x 1 and x 2 exist at every point of R 2, although f is not continuous at Theorem Suppose that f : U R is such that all its partial derivatives exist everywhere, where U is an open subset of R n. Assume further that for some real number M > 0 (x) M (x U, i = 1,, n). Prove that f is continuous in U. Proof. Let x U and choose r > 0 such that B r (x) U. Let {e 1,, e n } denote the standard basis of R n and rewrite h B r (0) as n i=1 h ie i for some h i R. Consider the set {v k } n k=0 of points of Rn : v 0 = 0 v 1 = h 1 e 1, v 2 = h 1 e 1 + h 2 e 2,.. v n = h 1 e h n e n = h. Check that the segments with end points x + v i 1 and x + v i lie in B r (x) for i = 1,, n (Exercise!). Then n (3.2) f(x + h) f(x) = [f(x + v i ) f(x + v i 1 )]. i=1 Consider the function g i : [0, h i ] R by g i (t) = f(x + v i 1 + te i ). Then g i is continuous in [0, h i ], differentiable in (0, h i ), and (x + te i ) = g (t) (t (0, h i )) (Problem 3.19). Hence by the Mean Value Theorem of Calculus there exists θ i (0, h i ) such that g i (h i ) g(0) = h i g (θ i ). In other words, f(x + v i ) f(x + v i 1 ) = h i (x + v i 1 + θ i e i ). Combining this with (3.2) yields n (3.3) f(x + h) f(x) = h i (x + v i 1 + θ i e i ). It follows that i=1 f(x + h) f(x) = n h i (x + v i 1 + θ i e i ) x i=1 i n M h i. Since n i=1 h i converges to 0 as h 0, the proof is over. i=1

14 14 SAMEER CHAVAN If f is as in Problem 3.21 then it follows from Theorem 3.22 that one of the partial derivatives of f is not bounded Partial Derivatives and Plus-Continuity*. Mere existence of the partial derivatives guarantees the plus-continuity. Theorem (Corollary 2, [7]) Suppose f : R n R. If the partial derivatives (i = 1,, n) exist at every point of R n then f is plus-continuous in R n. Proof. This follows at once from Remark 3.17(2) and Theorem Example 3.24 : Let U denote the plus-open subset {(x 1, x 2 ) R 2 : x 2 > 3 x 1 or x 2 < 3 x 1 } {(0, 0)} of R 2. Consider the function f : R 2 R given by f(x, y) = x 2 + y 2 if (x, y) U, = 1 if (x, y) / U. Since x 1 and x 2 exist at every point of U, it follows from Theorem 3.23 that f is plus-continuous in U. However, f is not continuous at the origin. Surprisingly, there exist everywhere discontinuous functions with partial derivatives defined everywhere except for countably many points of R 2 (refer to [7]) Directional Derivatives. Let f : U R be a real-valued function defined in an open subset U of R n and let u = (u 1,, u n ) R n be a unit vector (that is u = 1). The directional derivative of f at x in the direction u is the limit, if it exists, f(x + tu) f(x) ( u f)(x) lim. t 0 t Remark 3.25 : Notice the following: (1) The directional derivative ( u f)(x) is the slope of the hypersurface y = f(x) at the point x in the direction of the unit vector u. (2) If e i denotes the i th vector in the standard basis for R n then ei f =. (3) It may happen that the function f is continuous and that all partial derivatives exist, although u f does not exist for some unit vector u at some x : Define f(0, 0) = 0, and f(x, y) = x 3 2 y x 2 +y 2 if (x, y) (0, 0). Problem 3.26 : Let f : R 2 R be defined by f(x 1, x 2 ) = x 1 x 2. Find the slope of the surface y = f(x 1, x 2 ) at the point (1, 2) in the direction of (3, 4). It can happen that the function f is not continuous at some point, although all directional derivatives exist: Problem 3.27 : Define f(0, 0) = 0, and f(x, y) = x2 y x 4 if (x, y) (0, 0). + y2 Show that all directional derivatives of f exist at 0, but the function is not continuous at 0.

15 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS Partial Derivatives of Higher Order. Suppose f is a real-valued function defined in an open subset U of R n, with the partial derivatives (i = 1,, n). Suppose the partial derivative of with respect to x j exists. Then we define the second-order partial derivatives of f by 2 f x j x i ( ) x j (1 i, j n). Inductively, one may define the k th -order partial derivative k f 1 x ik of f, where 1 k n, and {i 1,, i k } {1,, n}. It can happen that f x jx i 2 f x j at some point, although both second ordered partial derivatives exist: Problem 3.28 : Define f(0, 0) = 0, and Verify the following: 2 f f(x, y) = xy(x2 y 2 ) x 2 + y 2 if (x, y) (0, 0). (1) x 1x 2 and 2 f x 2x 1 exist at every point of R 2. (2) 2 f x 1x 2 (0, 0) = 1 and 2 f x 2x 1 (0, 0) = 1. Lemma (Mean Value Theorem for Partial Derivatives, Theorem 9.40 of [6]) Let f : U R be a real-valued function defined in an open subset U of R 2. Suppose: (1) x 1 and 2 f x 2x 1 exist at every point of U, (2) R is a rectangle in U with vertices (a, b), (a, b+k), (a+h, b), and (a+h, b+k), where h 0 and k 0. Then there exists a point (x, y) in the interior of R such that f(a + h, b + k) f(a + h, b) f(a, b + k) + f(a, b) = Area(R) 2 f x 2 x 1 (x, y). Proof. Very Simple: Apply the Mean Value Theorem of Calculus to g, where g(t) f(t, b + k) f(t, b) to get an x (a, a + h) such that g(a + h) g(a) = hg (x). In view of Problem 3.19, one has [ f(a + h, b + k) f(a + h, b) + f(a, b) f(a, b + k) = h (x, b + k) ] (x, b). x 1 x 1 One more application of MVT to h(t) x 1 (x, t) yields a y (b, b + k) such that [ ] f(a + h, b + k) f(a + h, b) + f(a, b) f(a, b + k) = h k 2 f (x, y). x 2 x 1 This completes the proof of the lemma. Theorem (Equality of Mixed Partial Derivatives, Theorem 9.41 of [6]) Let f : U R be a real-valued function defined in an open subset U of R 2. Assume: Then (1) (2) x 1, 2 f x 2x 1 exist at every point of U, x 2, and 2 f x 2x 1 is continuous at some point (a, b) U. 2 f x 1x 2 exists at (a, b) and 2 f x 1x 2 (a, b) = 2 f x 2x 1 (a, b).

16 16 SAMEER CHAVAN Proof. Since is continuous at (a, b), given ɛ > 0 there exists a δ > 0 such that 2 f (x, y) 2 f (a, b) x 2 x 1 x 2 x 1 < ɛ whenever (x, y) (a, b) < δ. 2 2 f x 2x 1 Let h, k be such that 0 < h, k < δ 2. Consider the rectangle R with vertices (a, b), (a, b + k), (a + h, b), and (a + h, b + k). Notice that R B δ (a, b). Hence, in view of Lemma 3.29, f(a + h, b + k) f(a + h, b) f(a, b + k) + f(a, b) 2 f (a, b) hk x 2 x 1 < ɛ 2 for some (x, y) R. Fix h, and let k 0. Then { 1 (a + h, b) } (3.4) (a, b) 2 f (a, b) h x 2 x 2 x 2 x 1 < ɛ. Since ɛ is arbitrary and (3.4) holds for any h such that 0 < h < δ 2, it follows that 2 f x 1x 2 exists at (a, b) and that 2 f x 1x 2 (a, b) = 2 f x 2x 1 (a, b). This completes the proof of the theorem. 4. Differentiation To motivate the definition of f (x) for the vector-valued functions f, we need to revisit the scalar case: Problem 4.1 : Let f : R R be a real-valued function and let x R. Then the following are equivalent: (4.5) (1) f f(x+h) f(x) (x) lim h 0 h exists. (2) There exist real-valued functions r (remainder function) and D f (x) (linear derivative functional) in R such that f(x + h) f(x) = D f (x)(h) + r(h), lim h 0 r(h) h = 0. If one of the preceding statements holds true then f (x) = D f (x)(1). Solution. (1) = (2) : Define D f (x) : R R by D f (x)(y) f (x) y. Notice that D f (x) is a linear functional. Define r : R R by r(y) f(x + y) f(x) D f (x)(y). Since D f (x) is linear, it follows that r(y) y = = f(x + y) f(x) D f (x)(y) y y f(x + y) f(x) f (x) 0 as y 0. y (2) = (1) : Assume (2). Since D f (x) is linear, D f (x)(h) = hd f (x)(1). If one divides the first identity in (4.5) by h and let h 0 then it follows that f(x+h) f(x) lim h 0 h exists and is equal to D f (x)(1). Thus the derivative of f : R R can be thought of as a linear transformation D f (x) : R R. This viewpoint motivates the following definition.

17 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 17 Definition 4.2 : Suppose U is an open subset of R n, f : U R m, and x U. If there exists a linear transformation A : R n R m such that (4.6) f(x + h) f(x) Ah R m lim h 0 R n = 0, then we say that f is differentiable at x, we write f (x) = A. If f is differentiable at every point x U then we say that f is differentiable in U. Remark 4.3 : The following points are worth-notable: (1) The derivative of a vector-valued function is not a real number, it is a linear transformation! (2) The relation (4.6) can be rewritten as (4.7) f(x + h) f(x) = Ah + r(h), r(h) R m R n 0 as h 0. (3) The remainder r in (4.7) satisfies r(h) R m 0 as h 0. (4) Geometrically, if m = 1 and if f is differentiable at a then the hypersurface Graph(f) {(x, f(x)) : x U} R n+1 has the tangent hyperplane y = f(a) + f (a)(x a) at (a, f(a)). If m = n = 1, then Graph(f) R 2 has the tangent line y = f(a) + f (a)(x a). We will see in the next section that if m = 1, n = 2 then Graph(f) R 3 has the tangent plane y = f(a) + 1 x 1 (a)(x 1 a 1 ) + 1 x 2 (a)(x 2 a 2 ). Problem 4.4 : Consider the 2-variable circular paraboloid P given by y = x x 2 2 (x 1, x 2 R). Show that P has a tangent plane at the point (1, 2, 5) given by y = 2x 1 + 4x 2 5. Hint. Let f(x 1, x 2 ) = x x 2 2, a = (1, 2), and use Remark 4.3(4). Problem 4.5 : Compute the derivative of (1) T : R n R m given by T x = Ax + b where A : R n R m is a linear transformation and b R m. (2) the projection maps π i : R n R given by π i (x 1,, x m ) = x i (1 i n). We conclude the subsection with the following straight-forward generalizations of one-variable assertions. Theorem 4.6. (Theorems 5 and 6 of Chapter 5 of [5]) The following hold true: (1) Derivative is unique if it exists. (2) Differentiability implies continuity.

18 18 SAMEER CHAVAN Proof. (1) : Suppose (4.6) holds for two linear transformations A 1, A 2 : R n R m. Then for 0 h R n (A 1 A 2 )h = (f(x + h) f(x) A 2h) (f(x + h) f(x) A 1 h) f(x + h) f(x) A 2h + f(x + h) f(x) A 1h, the right hand side converges to 0 as h 0. Note that the last inequality is valid even if we replace h by th for some real t and let t 0. Since A 1, A 2 are linear, (A 1 A 2)h = (A1 A2)th th 0 as t 0. Hence A 1 h = A 2 h for any h R n. (2) : Since A is a linear transformation, it follows from the relation (4.7), Problem 2.14, and Remark 4.3(3) that f(x + h) f(x) Ah + r(h) A + r(h) 0 as h Properties of Derivatives. We begin with the following easy exercise. Problem 4.7 : (Linearity of Derivatives) Suppose f, g : R n R m are two differentiable functions. Show that αf +g is also differentiable and that (αf +g) = αf +g for any real α. Proposition 4.8. (Product Rule for Derivatives) Suppose f, g : R n R are two real-valued differentiable functions and let x R n. Then fg is differentiable at x. Indeed, (fg) (x) = f (x)g(x) + f(x)g (x). Proof. Let x be in R n and suppose f (x) = A 1 and g (x) = A 2 with the remainder functions r 1 and r 2 respectively. We first check that fg is differentiable at x. To see that, let r(h) = f(x + h)g(x + h) f(x)g(x) A 1 (h)g(x) A 2 (h)f(x). In view of (4.7), it is easy to see that r(h) = A 1 (h)a 2 (h) + A 1 (h)r 2 (h) + r 1 (h)a 2 (h) Now it suffices to check that r(h) r(h) +f(x)r 2 (h) + r 1 (h)g(x) + r 1 (h)r 2 (h). A 1(h)A 2 (h) + f(x)r 2(h) 0 as h 0. This is easy since + A 1(h)r 2 (h) + r 1(h)g(x) + r 1(h)A 2 (h) + r 1(h)r 2 (h) A 1 A 2 + A 1 r 2 (h) + r 1 (h) A 2 + r 1(h) g(x) + r 1(h) r 2 (h), + f(x) r 2(h) and since the last entity converges to 0 as h 0. Proposition 4.9. (Quotient Rule for Derivatives) Suppose f, g : R n R are two real-valued differentiable functions and let x ( R ) n. If g 0 in some open ball around x then f g is differentiable at x. Indeed, f g (x) = f (x)g(x) f(x)g (x) g(x). 2

19 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 19 Proof. Since g 0 in some open ball B r (x), one can choose 0 h R n so that g(x + h) 0. In view of the Product Rule, we may assume that f 1, the constant ( ) 1 function 1. Thus it suffices to check that g (x) = g (x) g(x). Put g (x) = A 2 2 and consider 1 g(x + h) 1 g(x) + A 2(h) g(x) 2 = g(x) g(x + h) + A 2(h) g(x + h)g(x) g(x) 2 = 1 ( A 2 (h) + r 2 (h) g(x) g(x) + A 2 (h) + r 2 (h) A ) 2(h) g(x) = 1 g(x) ( r2 (h)g(x) A 2 (h)(a 2 (h) + r 2 (h)) (g(x) + A 2 (h) + r 2 (h))g(x) in view of (4.7). As in the last proposition, it can be seen that r(h) 1 1 g(x) + A2(h) g(x) 2 satisfies r(h) 4.2. The Matrix Representation of Derivatives. ) g(x+h) 0 as h 0. Theorem (Theorem 5 of Chapter 5 of [5]) Suppose f : R n R m is differentiable at some x. Then all the directional derivatives of f at x exist. Indeed, Proof. Follows from (4.7). (f (x))(h) = lim t 0 f(x + th) f(x) t (h R n ). Corollary (The Matrix Representation of Derivatives) Let E denote the standard basis {e 1,, e n } of R n. If f = (f 1,, f m ) : R n R m is differentiable at some x then all its partial derivatives exist and ( (f 1 (x))(e i ) =,, ) m (1 i n). In particular, [ ] the matrix representation of f (x) with respect to E is the m n matrix i x j. Proof. Let h = e i in Theorem Problem 4.12 : Show that f : R n R is differentiable and find its matrix representation, where f(x) Ax, x for an n n matrix A with real entries A Criterion for Differentiability. We have seen that if f is a differentiable function then all its partial derivatives exist (Corollary 4.11). We have also seen that the converse to Corollary 4.11 is not true (Problem 3.27). The following theorem provides a partial converse to Corollary Theorem (Theorem 6.2 of Chapter 2 of [4]) Suppose that f = (f 1,, f m ) : U R is a real-valued function in an open subset U of R n. If the partial derivatives i x j (1 i m, 1 j n) exist at each point of U and are continuous in U then f is differentiable. Proof. Let x U, so that B ɛ (x) U for some ɛ > 0. We will show that f is differentiable at x. To see that let h = (h 1,, h n ) R n such that < ɛ. It

20 20 SAMEER CHAVAN follows from (3.3) of the proof of Theorem 3.22 that for some c i. We claim that f(x + h) f(x) f(x + h) f(x) = n i=1 h i i x n (c i ). [ ] x 1 (x),, x n (x) (h) 0 as h 0. To see the claim, consider [ ] f(x + h) f(x) x 1 (x),, x n (x) (h) n i=1 = h i (c i ) n i=1 h i (x) n ( h i = (c i ) ) (x). i=1 Since hi 1, the partial derivatives are continuous, and c i x as h 0, the [ ] claim follows. Thus f is differentiable at x with f (x) = x 1 (x),, x n (x). Example 4.14 : One can use the previous theorem to conclude that the functions given below are differentiable. (1) f(x, y) = sin(xy) (x, y R) (2) g(x, y, z) = xy 2 + z exp(xy) (x, y, z R). 5. The Chain Rule Theorem 5.1. (The Chain Rule, Theorem 9.15 of [6]) Let U and V be two open subsets of R n and R m respectively. Suppose f : U R m and g : V R k are such that (1) f is differentiable at x 0 U, (2) f(u) V, and (3) g is differentiable at f(x 0 ) V. Then the map F : U R k defined by F (x) = g(f(x)) is differentiable at x 0 and F (x 0 ) = g (f(x 0 ))f (x 0 ). Proof. Let y 0 = f(x 0 ), and let A 1 = f (x 0 ) and A 2 = g (f(x 0 )) with the remainder functions r 1 and r 2 respectively. Consider F (x 0 + h) F (x 0 ) A 2 A 1 (h) = g(f(x 0 + h)) g(f(x 0 )) A 2 A 1 (h) = g(f(x 0 ) + A 1 (h) + r 1 (h)) g(f(x 0 )) A 2 A 1 (h) = g (y 0 + h ) g(y 0 ) A 2 A 1 (h)

21 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 21 where h = A 1 (h) + r 1 (h). Since g (y 0 + h ) g(y 0 ) = A 2 (h ) + r 2 (h ), one has F (x 0 + h) F (x 0 ) A 2 A 1 (h) = A 2 (h ) + r 2 (h ) A 2 A 1 (h) It now follows from Problem 2.14 that F (x 0 + h) F (x 0 ) A 2 A 1 (h) = A 2 (h A 1 (h)) + r 2 (h ) = A 2 (r 1 (h)) + r 2 (h ) = A 2(r 1 (h)) + r 2 (h ) A 2 r 1(h) + r 2(h ) h h Since h A 1 + r 1 (h) the right hand in the previous inequality tends to 0 as h 0. Problem 5.2 : Derive the following partial differential equations: r = cos(θ) x + sin(θ) y θ = r sin(θ) x + r cos(θ) y, where (x, y) and (r, θ) denote the rectangular and the polar coordinates respectively. Hint. Apply the Chain Rule to F (r, θ) = f(g(r, θ)) where g(r, θ) = (r cos(θ), r sin(θ)). Problem 5.3 : Let f : R n R be a differentiable function. Prove: (1) If f denotes the gradient vector field ( x 1,, x n ) for f then n ( u f)(x) = ( f)(x), u = (x)u i (Apply the CR to the function F (t) = f(g(t)), where g(t) = x + tu ). (2) If ( f)(x) 0 then ( u f)(x) attains its maximum when u is a positive scalar multiple of ( f)(x) (Use (1) above and 2.1) Applications. In the remaining part of this section, we present several applications of the Chain Rule. Corollary 5.4. (Theorem 10 of Chapter 5 of [5]) A function f = (f 1,, f m ) : U R m is differentiable at p U if and only each f i : U R is differentiable at p, where U is an open subset of R n. Proof. The implication = follows from the Chain Rule and the relation f i = π i f, where π i : R m R is given by π i (x 1,, x m ) = x i (1 i m) (see Problem 4.5(2)). The implication = follows from Problem 4.7 and the relation f(x) = n i=1 f i(x)e i. In order to have a differentiable inverse for a differentiable function f, it is necessary that f be non-singular. Corollary 5.5. (Theorem 7.4 of Chapter 2 of [4]) Suppose f : U R n is a function defined in an open subset U of R n. Let a U and put b = f(a). Suppose further that g : B r (b) R n is such that g(b) = a and g(f(x)) = x (x B t (a)) for some positive reals r and t. If f is differentiable at a and g is differentiable at b then f (a) is an invertible linear transformation such that g (b) = [f (a)] 1. i=1

22 22 SAMEER CHAVAN Proof. If i : R n R n denotes the identity function then the derivative of i is the n n identity matrix (Problem 4.5(1)). Since g(f(x)) = i(x) (x B t (a)), an easy application of the Chain Rule yields that I = g (f(a))f (a). It follows from the Rank-Nullity Theorem ([3]) that f (a) is invertible. The relation g (b) = [f (a)] 1 is now obvious. We conclude this subsection with yet another application of the Chain Rule. Corollary 5.6. (The Mean Value Theorem for real-valued functions, Theorem 7.3 of Chapter 2 of [4]) If f : U R is differentiable in an open subset U of R n and if [x 0, y 0 ] {(1 t)x 0 + ty 0 : t [0, 1]} U, then for some x [x 0, y 0 ]. f(y 0 ) f(x 0 ) = f (x)(y 0 x 0 ) Proof. Apply the Chain Rule to the real-valued function F : [0, 1] R given by F (t) = f(g(t)) where g(t) = (1 t)y 0 + tx A First Ordered Partial Differential Equation. We discuss here an application of differential calculus to partial differential equations (refer to Chapter 9 of [1]). We begin with the following simple example, which illustrates an important difference between ordinary differential equations and partial differential equations. Example 5.7 : The solution space {f : R R : f is differentiable and f (x) = 0 for all x R} of the ordinary differential equation f (x) = 0 is one-dimensional. Indeed, it is equal to the vector space of constant real-valued functions. On the other hand, the solution space {g : R 2 R : g is differentiable and g x (x, y) = 0 for all (x, y) R2 } of the partial differential equation g x (x, y) = 0 is infinite-dimensional. In fact, it contains the vector space of polynomials in y. One can rephrase Problem 3.20 in the following way: Problem 5.8 : A differentiable function f : R 2 R satisfies y (x, y) = 0 ((x, y) R2 ) if and only if there exists a differentiable function g : R R such that f(x, y) = g(x) (y R). The following theorem describes the general solution of a class of homogeneous first ordered partial differential equations. Theorem 5.9. (Theorem 9.1, [1]) For (a, b) R 2 \ {(0, 0)}, consider the first ordered partial differential equation (5.8) a (x, y) + b (x, y) = 0. x y Let f : R 2 R be a differentiable function. equivalent. Then the following assertions are

23 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 23 (1) f satisfies (5.8) for every (x, y) R 2. (2) There exists a differentiable real-valued function g : R R such that f(x, y) = g(bx ay) (x, y R). Proof. (2) = (1) : Suppose there exists a differentiable function g : R R such that f = g h, where h : R 2 R is given by h(x, y) = bx ay ((x, y) R 2. By Theorem 5.1 f (x, y) = g (h(x.y))h (x, y). Since f (x, y) = and since h (x, y) = [b, a], we must have Hence a x x (x, y) = bg (bx ay), (x, y) + b y (x, y) = 0 as desired. y (x, y) = ag (bx ay). [ ] x (x, y), y (x, y) (1) = (2) : Suppose that (1) holds true. Introduce the new variables u and v as follows: u = bx ay, v = y (x, y R). Another application of the Chain Rule yields that v (x, y) = 0 in view of (5.8) (Exercise!). Hence by Problem 5.8 there exists a differentiable function g : R R such that f(x, y) = g(bx ay) (y R). Example 5.10 : Consider the first ordered partial differential equation (5.9) subject to the condition 4 (x, y) + 3 (x, y) = 0 x y f(x, 0) = sin(x) for all x R. Suppose that f : R 2 R is a (differentiable) solution of (5.9). It follows from Theorem 5.9 that there exists a differentiable function g : R R such that f(x, y) = g(3x 4y) (x, y R) with g(3x) = sin(x) (x R). Hence we must have ( ) 3x 4y f(x, y) = sin for any (x, y) R 2. 3 Problem 5.11 : Discuss the solution space of the first ordered partial differential equation subject to the conditions f(0, 0) = 0, 5 (x, y) 2 (x, y) = 0 x y (x, 0) = exp(x) for all x R. x We leave the proof of the following to the interested reader: Theorem For 0 a R n, consider the first ordered partial differential equation (5.10) a 1 x 1 (x) + + a n x n (x) = 0.

24 24 SAMEER CHAVAN Let f : R n R be a differentiable function. equivalent. Then the following assertions are (1) f satisfies (5.10) for every x R n. (2) There exists a differentiable real-valued function g : R R such that ( n ( ) n 1 f(x) = g ( 1) )x k+1 k Π i k a i (x R n ), k 1 k=1 where Π i k a i stands for the product of a 1,, a k 1, a k+1,, a n The Multi-dimensional Mean Value Theorem. Invoke the Mean Value Theorem from Calculus: If f : [a, b] R is a real-valued function continuous in [a, b] and differentiable in (a, b) then there exists a point x (a, b) such that f(b) f(a) = Area([a, b]) f (x). Again, the multivariable picture is quite different: Problem 5.13 : Define f : [π, 2π] R 2 by f(t) = (cos(t), sin(t)). Then f is continuous in [π, 2π] and differentiable in (π, 2π). Show that there is no single θ (π, 2π) such that f(2π) f(π) = Area([π, 2π]) f (θ). The Multi-dimensional Mean Value Theorem is an inequality: Theorem (The Multi-dimensional Mean Value Theorem, Theorem 11 of Chapter 5 of [5]) If f : U R m is differentiable in an open subset U of R n and if [x 0, y 0 ] {(1 t)x 0 + ty 0 : t [0, 1]} U, then f(y 0 ) f(x 0 ) sup f (x) y 0 x 0. x [x 0,y 0] Proof. We may assume that x 0 y 0. Fix any unit vector u R n. By Theorem 5.1 the function g(t) = u, f((1 t)y 0 + tx 0 ) is differentiable and g (t) = u, (f ((1 t)y 0 + tx 0 ))(x 0 y 0 ). By the Mean Value Theorem of Calculus, there exists some θ (0, 1) such that g(1) g(0) = g (θ). That is, u, f(x 0 ) u, f(y 0 ) = u, (f ((1 θ)y 0 + θx 0 ))(x 0 y 0 ). It follows from the Cauchy-Schwarz Inequality that u, f(x 0 ) f(y 0 ) u (f ((1 θ)y 0 + θx 0 ))(x 0 y 0 ) 1 f ((1 θ)y 0 + θx 0 ) x 0 y 0 sup f (x) x 0 y 0. x [x 0,y 0] Since this is true for any unit vector u, one can let u required conclusion. f(x0) f(y0) f(x 0) f(y 0) to get the Corollary Suppose f : U R m is differentiable in an open subset U of R n. Suppose further that U is convex (that is, [x 0, y 0 ] {(1 t)x 0 + ty 0 : t [0, 1]} U for every x 0, y 0 U). If f (x) = 0 for every x U then f is constant. Proof. If f (x) = 0 for every x U then sup x U f (x) = 0. Hence by the last theorem one has f(x 0 ) f(y 0 ) = 0 for every x 0, y 0 U. Hence f(x 0 ) f(y 0 ) = 0 for every x 0, y 0 U.

25 NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS The Inverse Function Theorem Definition 6.1 : Let f : U R m be a vector-valued function differentiable in an open subset U of R n. We say that f is continuous at a U if for every ɛ > 0 there exists a δ > 0 such that f (x) f (a) < ɛ whenever x U and x a < δ. Remark 6.2 : Suppose that f is as in Definition 6.1 and that f (a) is invertible. If we take ɛ < (f (a)) 1 1 then by Theorem 2.18 there exists δ > 0 such that f (x) is invertible for every x U such that x a < δ Statement and Examples. Theorem 6.3. (The Inverse Function Theorem, Theorem 9.24 of [6]) Suppose f : U R n is a vector-valued function differentiable in an open subset U of R n. Suppose, for some a R n, that f (a) : R n R n is an invertible linear transformation and that f is continuous at a. Then (1) there exists an open subset V of R n such that a V U, f(v ) is open in R n, and f V : V f(v ) is injective. (2) f V admits a differentiable inverse. Remark 6.4 : Suppose g : f(v ) V is the differentiable inverse of f V guaranteed by the Inverse Function Theorem. Then (6.11) Thus the system of n equations x = g(f(x)) (x V ). y i = f i (x 1,, x n ) (1 i n) can be solved locally for x 1,, x n in terms of y 1,, y n (that is, if x V and y f(v )). Moreover, the solutions given by (6.11) are unique and differentiable. Example 6.5 : Let f : R 2 R 2 be given by the equation f(x 1, x 2 ) = (2x 2 exp(2x 1 ), x 1 exp(x 2 )) (x 1, x 2 R). Show that there exists an open ball containing (0, 1) that f carries in a one-to-one fashion onto an open ball containing (2, 0). Thus the system of 2 equations y 1 = 2x 2 exp(2x 1 ), y 2 = x 1 exp(x 2 ) can be solved locally for x 1 and x 2 in terms of y 1 and y 2. Solution. In view of Theorem 6.3, it suffices to check that f (0, 1) is an invertible linear transformation. Since f (0, 1) is invertible if and only if its matrix representation is invertible, it suffices to check that ( ) 1 x C = 1 (0, 1) 1 x 2 (0, 1) 2 x 1 (0, 1) 2 x 2 (0, 1) is invertible. The latter one is an easy exercise. Example 6.6 : Let f : R 2 R 2 be given by the equation f(x 1, x 2 ) = (exp(x 1 ) cos(x 2 ), exp(x 1 ) sin(x 2 )) (x 1, x 2 R).