NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS

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 4 2.1. The Usual Topology in the Euclidean Spaces 4 2.2. Linear Transformations 5 2.3. The Plus Topology in the Euclidean Spaces* 7 3. Vector-valued Functions in Several Real Variables 8 3.1. Limits 8 3.2. Continuity 10 3.3. Plus-Continuity* 11 3.4. Partial Derivatives and Continuity 12 3.5. Partial Derivatives and Plus-Continuity* 14 3.6. Directional Derivatives 14 3.7. Partial Derivatives of Higher Order 15 4. Differentiation 16 4.1. Properties of Derivatives 18 4.2. The Matrix Representation of Derivatives 19 4.3. A Criterion for Differentiability 19 5. The Chain Rule 20 5.1. Applications 21 5.2. A First Ordered Partial Differential Equation 22 5.3. The Multi-dimensional Mean Value Theorem 24 6. The Inverse Function Theorem 25 6.1. Statement and Examples 25 6.2. A Contraction Principle 26 6.3. Proof of the Inverse Function Theorem 28 7. The Implicit Function Theorem 28 7.1. Statement and Examples 28 7.2. Proof of the Implicit Function Theorem 29 Acknowledgements 31 References 31 1

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- 102. 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 1 + + a 1n x n = y 1 (1.1).. a n1 x 1 + + 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.

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 SAMEER CHAVAN 2. Preliminaries In this section, we record a few prerequisites pertaining to the Euclidean spaces and the linear transformations. 2.1. 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.

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 2.10. 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. 2.2. 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 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 2.15. (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.

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 2.18. (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 2.17. 2.3. 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 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. 3.1. 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.

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 4 + 1 4 = 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 1 + + 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 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. 3.2. 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.

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). 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 3.16. (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 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). 3.4. 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.

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 3.22. 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 1 + + 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 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. 3.5. Partial Derivatives and Plus-Continuity*. Mere existence of the partial derivatives guarantees the plus-continuity. Theorem 3.23. (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 3.16. 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]). 3.6. 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.

NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 15 3.7. 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 3.29. (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 3.30. (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 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.

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 2 1 + 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 2 1 + 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 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 0. 4.1. 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

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 4.10. (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 4.11. (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 4.10. 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. 4.3. 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 4.11. Theorem 4.13. (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 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)

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). 5.1. 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 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 0. 5.2. 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

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 5.12. 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 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. 5.3. 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 5.14. (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 5.15. 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.

NOTES ON MULTIVARIABLE CALCULUS: DIFFERENTIAL CALCULUS 25 6. 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. 6.1. 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).