1 Derivable Functions

In what follows, we shall work only with real Banach spaces.

Definition 1.1 Let E and F be Banach spaces, ∅ 6= U ⊆ E, U open, a ∈ U . Also, let f : U → F .The function f is said to be derivable or Frechet differentiable at the point a if there exists f

′(a) ∈

L(E, F ) such that

limx → ax 6= a

‖f(x)− f(a)− f′(a) ◦ (x− a)‖

‖x− a‖ = 0.(1.1)

f′(a) is called the derivative of the function f at the point a.

If we put

ωf,a(x) =

f(x)− f(a)− f′(a) ◦ (x− a)

‖x− a‖ , ∀x ∈ U \ {a},

0F , x = a,

then the condition (1.1) becomeslimx→a

ωf,a(x) = 0F .

So, f is derivable at the point a if there exist f′(a) ∈ L(E, F ) and ωf,a : U → F such that

f(x) = f(a) + f′(a) ◦ (x− a) + ‖x− a‖ωf,a(x), ∀x ∈ U(1.2)

andlimx→a

ωf,a(x) = 0F .(1.3)

It is not difficult to prove the following result:

Theorem 1.2 (Uniqueness of the Derivative) If f : U ⊆ E → F is derivable at the point a ∈ U ,then there exists only one linear mapping f

′(a) ∈ L(E, F ) such that (1.1) holds true.

Notice that due to the uniqueness of the derivative f′(a) ∈ L(E,F ), we can define a map

f′: U → L(E, F ).

Also, notice that the domain of the linear mapping f′(a) is the entire space E, despite the fact that

the function f is defined only locally about the point a.

Proposition 1.3 If the function f : U ⊆ E → F is derivable at the point a ∈ U , then f is continuousat the point a.

Proof. Let us suppose that f is derivable at the point a. Therefore, from (1.2) and the continuityof f

′(a), we obtain

‖f(x)− f(a)‖ ≤ (‖f ′(a)‖+ ‖ωf,a(x)‖)‖x− a‖, ∀x ∈ U.

From (1.3), it follows that there exists a neighbourhood U1 ⊆ U and there exists M > 0 such that

‖ωf,a(x)‖ ≤ M, ∀x ∈ U1.

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Hence,‖f(x)− f(a)‖ ≤ (‖f ′

(a)‖+ M)‖x− a‖, ∀x ∈ U1,

which means thatlimx→a

f(x) = f(a),

i.e. f is continuous at the point a.

Remark 1.4 The function f is said to be derivable on U if f is derivable at any point a ∈ U .

Proposition 1.5 If f : U ⊆ E → F is constant on U , then f is derivable on U and f′(a) = 0, ∀a ∈

U .

Proof. Obviously, we have f(x) = f(a) + 0 ◦ (x− a), ∀x ∈ U . Therefore, f is derivable at the pointa and f

′(a) = 0.

Proposition 1.6 If f ∈ L(E, F ), then f is derivable at any point a ∈ E and f′(a) = f, ∀a ∈ E.

Proof. Since f is linear, then

f(x)− f(a)− f(x− a)‖x− a‖ =

0‖x− a‖ = 0, ∀x ∈ E \ {a}.

So, f is derivable at any point a ∈ E and f′(a) = f, ∀a ∈ E.

Exactly like in the one-variable calculus, it is not difficult to prove the following result:

Proposition 1.7 (Linearity of the Derivative) Let E and F be Banach spaces, U ⊆ E, U open,a ∈ U . Also, let f, g : U → F . If f and g are differentiable at the point a, then the sum f + g isdifferentiable at the point a and

(f + g)′(a) = f

′(a) + g

′(a).

Moreover, if α ∈ R, then αf is differentiable at the point a and

(αf)′(a) = αf

′(a).

Very often we need to deal with elaborate functions, obtained by composing simpler ones. Thenext theorem will give us, exactly like in the one-variable calculus, the rule for differentiating com-posite functions.

Theorem 1.8 (Chain Rule) Let E, F, G be Banach spaces, ∅ 6= U ⊆ E, ∅ 6= V ⊆ F , U, V open. Iff : U → V is derivable at the point a ∈ U and g : V → G is derivable at the point f(a) , then g ◦ fis derivable at the point a and

(g ◦ f)′(a) = g

′(f(a)) ◦ f

′(a).

Proof. The proof is left as an exercise.

Remark 1.9 If f′is continuous on U , we shall say that f is continuously differentiable on U .

We shall use the following notation:

C1(U,F ) = {f : U → F | f is continuously differentiable on U}.

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Definition 1.10 Let E,F be Banach spaces, ∅ 6= U ⊆ E, U open, f : U → F, a ∈ U . The functionf is said to be twice (two-times) differentiable at the point a if:

1) ∃V ∈ V(a), V open, such that f is derivable on V ;2) f

′: V → L(E,F ) is derivable at the point a.

The derivative of f′at the point a is denoted by f

′′(a) and it is called the second derivative of f at

the point a.

Remark 1.11 If the function f is two-times differentiable on U and the map

f′′

: U → L(E, L(E, F ))

is continuous on U , then we shall say that f is twice continuously differentiable on Uand we shalluse the notation:

C2(U,F ) = {f : U → F | f is twice continuously differentiable on U}.

Remark 1.12 f′′(a) ∈ L(E, L(E, F )) ' L2(E, E; F ).

By usual induction, we can define

Cn(U,F ) = {f : U → F | f is n-times continuously differentiable on U}.

PuttingC0(U,F ) = {f : U → F | f is continuous on U},

we can defineC∞(U,F ) =

⋂

n∈NCn(U,F ).

A function f ∈ C∞(U,F ) is said to be infinitely (indefinitely) derivable on U .

Particular Case.

Let us briefly consider now the important particular case of one-variable calculus, i.e. the casein which E = F = R. We shall start by recalling the well-known definition of derivability in theone-dimensional case.

Definition 1.13 Let ∅ 6= U ⊆ R be an open set and a ∈ U . The function f : U → R is calledderivable at the point a if

∃ limx→a

f(x)− f(a)x− a

= f′(a) ∈ R.

f′(a) is called the derivative of the function f at the point a.

Equivalently, we can work with the following definition:

Definition 1.14 Let ∅ 6= U ⊆ R be an open set and a ∈ U . The function f : U → R is calledderivable at the point a if there exists wf,a : U → R such that:

1) wf,a is continuous at the point a;2) f(x) = f(a) + wf,a(x) · (x− a), ∀x ∈ U .

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Indeed, the above definitions are equivalent. It is enough to notice that if f satisfies Definition 1.13,then the limit

limx→a

f(x)− f(a)x− a

exists. Letα = lim

x→a

f(x)− f(a)x− a

∈ Rand let

wf,a(x) =

f(x)− f(a)x− a

, ∀x ∈ U \ {a},

α, x = a.

Then, wf,a is continuous at the point a and condition 2) in Definition 1.14 is satisfied.Conversely, if f satisfies Definition 1.14, then the limit

limx→a

f(x)− f(a)x− a

exists and belongs to R. Hence, the above definitions are equivalent. Notice that

wf,a(a) = f′(a).

Let us see now why in the one-variable calculus the derivative of a given function at a given pointwas a real number. From Definition 1.1, the map f

′(a) should be in L(R,R). But L(R,R) ' R, and,

hence, f′(a) ∈ R.

Proposition 1.15 Let ∅ 6= U ⊆ R be an open set and a ∈ U . If the function f : U → R is derivableat the point a, then f is continuous at the point a.

Proof. Since f is derivable at the point a, then there exists wf,a : U → R such that wf,a iscontinuous at the point a and

f(x) = f(a) + wf,a(x) · (x− a), ∀x ∈ U.

Hence, f being composed only by continuous functions, it is continuous.

We shall recall now, without proofs, some well-known properties of derivable functions.

Properties.

1) Let ∅ 6= U ⊆ Rn be an open set and a ∈ U . If the functions f, g : U → R are derivable at thepoint a, then f ± g, λf, λ ∈ R, fg, f/g, g 6= 0, are derivable at the point a.

2) Moreover, if f is injective and continuous on U and f′(a) 6= 0, then the inverse function f−1

is derivable at the point b = f(a) and

(f−1)′(f(a)) =

1f ′(a)

.

Proposition 1.16 (Chain Rule) Let U, V ⊆ R be open nonempty sets and let f : U → R andg : V → R such that f(U) ⊆ V . If f is derivable at the point a ∈ U and g is derivable at the pointb = f(a) ∈ V , then g ◦ f is derivable at the point a and

(g ◦ f)′(a) = g

′(f(a))f

′(a).

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Also, let us recall the following well-known results from the one-variable calculus:

Definition 1.17 Let f : U ⊆ R→ R, U open and nonempty, a ∈ U .a) The function f attains a local minimum at the point a if there exists V ∈ V(a) such that

f(x) ≥ f(a), for any x ∈ V ∩ U .b) The function f attains a local maximum at the point a if there exists V ∈ V(a) such that

f(x) ≤ f(a), for any x ∈ V ∩ U .

The point a is called a local extremum (a minimum or a maximum) for the function f . We shalluse the terms global or absolute maximum and global or absolute minimum to refer to the overallmaximum and minimum values of the function on the range under consideration.

Theorem 1.18 (Fermat) Let f : U ⊆ R→ R, U open and nonempty, a ∈ U . If f is derivable atthe point a and a is a local extremum for f , then f

′(a) = 0.

The points a at which f′(a) = 0 are called stationary points. If a function f has a local extremum

at a point a, then either f′(a) = 0 or f

′(a) does not exist. Such points are called critical points.

Theorem 1.19 (Rolle) Let f : [a, b] → R. If f is continuous on [a, b], derivable on (a, b) andf(a) = f(b), then there exists c ∈ (a, b) such that f

′(c) = 0.

Theorem 1.20 (Lagrange’s Mean-Value Theorem) Let f : [a, b] → R. If f is continuous on [a, b],derivable on (a, b), then there exists c ∈ (a, b) such that

f(b)− f(a)b− a

= f′(c).

The formulaf(b)− f(a) = (b− a)f

′(c)(1.4)

is called Lagrange’s formula of finite increments.

Theorem 1.21 (Cauchy’s Mean-Value Theorem) Let f, g : [a, b] → R. If f, g are continuous on[a, b], derivable on (a, b) and g

′(x) 6= 0, ∀x ∈ (a, b), then g(a) 6= g(b) and ∃ c ∈ (a, b) such that

f(b)− f(a)g(b)− g(a)

=f′(c)

g ′(c).

Exercises.

1) Let f : R→ R,

f(x) =

x2 sin1x

, x 6= 0,

0, x = 0.

Prove that f is derivable on R, but f is not a C1 function on R.

2) Prove that

2 arctan x + arcsin2x

1 + x2= π sgn x, ∀ | x |≥ 1.

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Let us consider now briefly another important case, i.e. the case in which E = Rn and F = Rm.So, let ∅ 6= U ⊆ Rn be an open set, f : U → Rm and a ∈ U . If

f = (f1, . . . , fm),

then f is derivable at the point a if and only if each component fi, with i = 1,m, is derivable at thepoint a ( the componentwise nature of differentiability).

Example. The function f : R2 → R3 defined by

f(x, y) = (x2 + y2, xy, x3 + y3)

is obviously differentiable on R2, since its three components are composed of elementary functions.

2 Partial Derivatives

Definition 2.1 Let E1, E2, . . . , En be Banach spaces and

E =n∏

i=1

Ei.

Let us definepi : E → Ei, pi(x) = xi, ∀x = (x1, . . . , xn) ∈ E

andui : Ei → E, ui(xi) = (0E1 , . . . , 0Ei−1 , xi, 0Ei+1 , . . . , 0En), ∀x = (x1, . . . , xn) ∈ E.

pi are called the canonical projections and ui the canonical injections.

Properties.

1) It is easy to see that pi and ui are linear.

2) pi and ui are continuous. Indeed,

‖pi(x)‖ = ‖xi‖ ≤n∑

k=1

‖xk‖ = ‖x‖1.

Hence, ‖pi‖ ≤ 1, which implies that‖pi(x)‖ ≤ ‖x‖1,

i.e. pi is continuous.On the other hand,

‖ui(xi)‖ = ‖(0, . . . , xi, . . . , 0)‖ =n∑

k=1

‖xk‖ = ‖xi‖.

So, ‖ui‖ = 1, which means that‖ui(xi)‖ = ‖xi‖

and, hence, ui is continuous.

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Also, we can prove thatpi ◦ ui = 1Ei

andn∑

i=1

ui ◦ pi = 1E .

Indeed,(pi ◦ ui)(xi) = pi(0, . . . , xi, . . . , 0) = xi = 1Ei(xi).

andn∑

i=1

(ui ◦ pi)(x) =n∑

i=1

ui(xi) = (x1, 0, . . . , 0) + · · ·+ (0, . . . , 0, xn) = x = 1E(x).

Remark 2.2 Since pi and ui are linear and continuous, it follows immediately that pi and ui arederivable.

Definition 2.3 Let ∅ 6= U ⊆ E, U open and let a ∈ U . For any 1 ≤ i ≤ n, let us defineµa,i : Ei → E as being:

µa,i(xi) = a + ui(xi − ai), ∀xi ∈ Ei,

i.e.µa,i(xi) = (a1, . . . , ai−1, xi, ai+1, . . . , an).

The role played by this map is to fix the variable xi in a.

Let us notice that µa,i is continuous and µ−1a,i (U) is open in Ei.

Definition 2.4 Let E1, E2, . . . , En, F be Banach spaces and E =n∏

i=1Ei. Also, let ∅ 6= U ⊆ E, U

open and f : U → F . The function f is called derivable with respect to xi at the point a ∈ U (orpartial derivable with respect to xi at the point a) if the map f ◦ µa,i : µ−1

a,i (U) → F is derivableat ai ∈ µ−1

a,i (U). In this case, the derivative of the function f ◦ µa,i at the point ai will be called the

partial derivative of f with respect to xi at the point a and we shall denote this derivative by∂f

∂xi(a).

So,∂f

∂xi(a) = (f ◦ µa,i)

′(ai).

Theorem 2.5 Let f : U ⊆ E → F , U open and nonempty. If f is derivable at the point a ∈ U ,then f is (partial) derivable with respect to each variable xi, 1 ≤ i ≤ n, at the point a and

∂f

∂xi(a) = f

′(a) ◦ ui,

f′(a) =

n∑i=1

∂f

∂xi(a) ◦ pi.

Proof. Sinceµa,i(xi) = a + ui(xi − ai), ∀xi ∈ Ei,

it follows that µa,i is derivable at any point t ∈ Ei and (µa,i)′(t) = ui. Then, f ◦µa,i is derivable and

∂f

∂xi(a) = (f ◦ µa,i)

′(ai) = f

′(µa,i(ai)) ◦ (µa,i)

′(ai) = f

′(a) ◦ ui.

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On the other hand,

f′(a) = f

′(a) ◦ 1E = f

′(a) ◦ (

n∑

i=1

ui ◦ pi) =n∑

i=1

f′(a) ◦ ui ◦ pi =

n∑

i=1

∂f

∂xi(a) ◦ pi.

Remark 2.6 Let us notice that∂f

∂xi: U → L(Ei, F )

and∂f

∂xi(a) ∈ L(Ei, F ).

Remark 2.7 Let E1, E2, . . . , En, F1, . . . , Fm be Banach spaces and E =n∏

i=1Ei, F =

m∏j=1

Fj. Also,

let U ⊆ E, U open and nonempty and f : U → F be derivable at the point a. Then, the derivativef′(a) is determined by the matrix

∂f1

∂x1(a) · · · ∂f1

∂xn(a)

.............................

∂fm

∂x1(a) · · · ∂fm

∂xn(a)

In particular, for E = Rn and F = Rm, it is not difficult to see that we have the following result:

Proposition 2.8 Let ∅ 6= U ⊆ Rn, U open, f : U → Rm derivable at the point a ∈ U . The matrixassociated to the linear map f

′(a) : Rn → Rm in the canonical bases of Rn and Rm, called the Jacobi

matrix, is

Mf (a) =

∂f1

∂x1(a) · · · ∂f1

∂xn(a)

..............................

∂fm

∂x1(a) · · · ∂fm

∂xn(a)

If m = n, the determinantJf (a) = det Mf (a)

is called the Jacobian of the function f at the point a.

If f : U ⊆ Rn → Rn, U open and nonempty and if f = (f1, . . . , fn), we shall denote the Jacobianof the vector valued function f at the point a by

Jf (a) =∂(f1, ..., fn)∂(x1, . . . , xn)

(a).

So, if a function is differentiable at a point, its derivative is given by the Jacobian matrix.

Example. If we consider the function f : R2 → R2 defined by

f(x, y) = (x2 + y2, 2xy),

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then its Jacobi matrix at the point (1, 1) is

Mf (1, 1) =

2 2

2 2

and its Jacobian is Jf (1, 1) = 0.

Theorem 2.9 Let E1, E2, . . . , En, F be Banach spaces and E =n∏

i=1Ei. Also, let ∅ 6= U ⊆ E, U

open and f : U → F . The following statements are equivalent:(i) f ∈ C1(U,F );

(ii) f is derivable with respect to each variable xi and∂f

∂xi: U → L(Ei, F ), 1 ≤ i ≤ n, are

continuous.

Let us consider now some important particular cases.

We shall take Ei = F = R. Therefore, E = Rn. Let

f : U ⊆ Rn → R.

For x = (x1, . . . , xn) ∈ Rn, we have:

pi : Rn → R, pi(x) = xi

andui : R→ Rn, ui(xi) = (0, . . . , 0, xi, 0, . . . , 0).

Usually, we denote the canonical projections pi by

pi = d xi

and we call dxi the differential of xi.In this particular case,

f′(a) ∈ L(Rn,R)

and∂f

∂xi(a) ∈ L(R,R) ' R.

Therefore,∂f

∂xi(a) is a real number and, hence, in Theorem 2.5, ”◦ ” becomes the usual product. So,

we get

f′(a) =

n∑

i=1

∂f

∂xi(a) dxi,(2.1)

where∂f

∂xi(a) = lim

xi→ai

f(a1, . . . , ai−1, xi, ai+1, . . . , an)− f(a1, . . . , ai−1, ai, ai+1, . . . , an)xi − ai

.

In fact, {p1, . . . , pn} is the canonical basis for L(Rn,R) and, so, (2.1) is just the unique representation

of the linear map f′(a) in this basis. The partial derivatives

∂f

∂xi(a) are exactly the coefficients of

f′(a) in the canonical basis.

Since f′(a) ∈ L(Rn,R) ' Rn, the derivative f

′(a) can be also regarded as being the vector

f′(a) ' (

∂f

∂x1(a), . . . ,

∂f

∂xn(a)).

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Remark 2.10 Let ∅ 6= U ⊆ Rn, U open, f : U → R derivable at the point a ∈ U . Then, thereexists a unique vector y ∈ Rn such that

f′(a)(x) = 〈x, y〉.

This vector is denoted by gradaf and it is called the gradient of the function f at the point a.

Let us consider now the case in which n = 2. Let (a, b) ∈ U ⊆ R2. Also, let f : U → R, f = f(x, y).In this case,

f′(a, b) ∈ L(R2,R)

andf′(a, b) =

∂f

∂x(a, b)dx +

∂f

∂y(a, b)dy,

where the partial derivatives are given by:

∂f

∂x(a, b) = lim

x→a

f(x, b)− f(a, b)x− a

and∂f

∂y(a, b) = lim

y→b

f(a, y)− f(a, b)y − b

.

Sometimes, the partial derivatives can be denoted by f′x(a, b) and, respectively, f

′y (a, b). In fact,

the partial derivative of the function f with respect to x at the point (a, b) is the usual derivativeat the point a of the one-dimensional function t → f(t, b). Of course, the partial derivative ofthe function f with respect to y at the point (a, b) is the usual derivative at the point b of theone-dimensional function t → f(a, t).

Let us see now which are the interpretations for partial derivatives.On one hand, it is not difficult to see that f

′x represents the rate of change of the function f as

we change x, holding y fixed, while f′y represents the rate of change of f as we change y, holding x

fixed.On the other hand, we know from one-variable calculus that f

′(a) represents the slope of the

tangent line to y = f(x) at x = a. The partial derivatives f′x(a, b) and fy(a, b) also represent the

slopes of some tangent lines. More precisely, the partial derivative fx(a, b) is the slope of the traceof f(x, y) for the plane y = b at the point (a, b). Also, the partial derivative f

′y (a, b) is the slope of

the trace of f for the plane x = a at the point (a, b).

Partial derivatives are very easy to compute: first, we have to fix all but one of the variables andthen we have to take the one-variable derivative with respect to the variable that remains.

Examples.

1) For the function f : R3 → R defined by

f(x, y, z) = x2ey + z,

we get∂f

∂x(a, b, c) =

d

dx(x2eb + c) |x=a= 2aeb,

∂f

∂y(a, b, c) =

d

dy(a2ey + c) |y=b= a2eb

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and∂f

∂z(a, b, c) =

d

dz(a2eb + z) |z=c= 1.

2) Let f : R2 → R be defined by

f(x, y) =√

x2 + y2.

Prove that f is not differentiable at the point (0, 0).

Proof. If f would be differentiable at the point (0, 0), then its partial derivatives at the point(0, 0) would exist. In this case,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)x− 0

= limx→0

| x |x

and∂f

∂y(0, 0) = lim

y→0

f(0, y)− f(0, 0)y − 0

= limy→0

| y |y

.

But these limits don’t exist and, hence, f is not differentiable at the point (0, 0).

3) Define f : R2 → R as follows:

f(x, y) =

xy

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

Test it for differentiability at the point (0, 0).

Proof. If f would be differentiable at the point (0, 0), then f would be continuous at the point(0, 0), i.e. lim

(x,y)→(0,0)f(x, y) = f(0, 0). But lim

(x,y)→(0,0)f(x, y) doesn’t even exist and, hence, f is not

differentiable at the point (0, 0). Still, the partial derivatives∂f

∂x(0, 0) and

∂f

∂x(0, 0) exist and are

equal to zero.

4) Test the following function for differentiability:

f(x, y) =

x2y

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

Obviously, f is differentiable on R2 \ {(0, 0)} and

f′(a, b) = (

∂f

∂x(a, b),

∂f

∂y(a, b)) = (

2ab3

(a2 + b2)2,a2(a2 − b2)(a2 + b2)2

),

for any (a, b) 6= (0, 0).Let us see now what happens at the origin. It is easy to see that f is continuous at the point

(0, 0) and its first partial derivatives exist. Indeed,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)x− 0

= limx→0

0x

= 0

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and, in a similar manner,∂f

∂y(0, 0) = 0. Therefore, if the function f would be differentiable at the

point (0, 0), then the derivative of f at this point should be the zero map and

lim(x, y) → (0, 0)

f(x, y)− f(0, 0)− f′(0, 0) ◦ (x− 0, y − 0)

‖(x− 0, y − 0)‖ = 0,

i.e.

lim(x, y) → (0, 0)

x2y

x2 + y2√

x2 + y2= 0.

But this limit doesn’t exist and, hence, f is not differentiable at the origin.

5) Define f : R2 → R as follows:

f(x, y) =

1 x = y 6= 0,

0 otherwise.

Show that f is not continuous at the point (0, 0), but both partial derivatives∂f

∂x(0, 0) and

∂f

∂x(0, 0)

exist.

It is easy to see that if we let (x, y) to approach the origin along the line x = y, then lim(x,y)→(0,0)

f(x, y) =

1 6= f(0, 0) and this implies the fact that f is not continuous at the origin.However, both partial derivatives exist at the origin. Indeed,

∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)x− 0

= limx→0

0x

= 0

and, in a similar manner,∂f

∂y(0, 0) = 0.

Exercises.

1) Let f : R2 → R be defined by

f(x, y) = ln(1 + 2x2 + 3y2).

Compute f′(1, 2).

2) Test the following function for differentiability:

f(x, y) =

xy√x2 + y2

, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

3) Test the following function, defined on R2, for differentiability:

f(x, y) =

(x2 + y2) sin1

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

12

4) Test the following function for differentiability:

f(x, y) =

x2y3

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

3 Higher Order Partial Derivatives

Definition 3.1 Let E1, E2, . . . , En and F be Banach spaces and E =n∏

i=1Ei. Also, let ∅ 6= U ⊆ E,

U open and f : U → F . The function f is said to be two-times partial derivable at the point a ∈ Uwith respect to the argument xi and then with respect to the argument xj if there exists U0 opensuch that:

1) a ∈ U0 ⊆ U ;2) f is derivable with respect to xi on U0;

3) the map∂f

∂xi: U0 → L(Ei, F ) is derivable with respect to xj at the point a.

The partial derivative of the map∂f

∂xiwith respect to xj at the point a is denoted by

∂2f

∂xj∂xi(a).

Let us notice that∂2f

∂xj∂xi(a) ∈ L(Ej , L(Ei, F )) ' L2(Ej , Ei; F ).

Remark 3.2 Let E1, E2, . . . , En and F be Banach spaces and E =n∏

i=1Ei. Also, let ∅ 6= U ⊆ E, U

open and f : U → F . If f is two-times differentiable at the point a ∈ U , then f is two-times partialderivable at a ∈ U with respect to xi and xj, for any 1 ≤ i, j ≤ n.

Theorem 3.3 Let E and F be Banach spaces, ∅ 6= U ⊆ E, U open, a ∈ U and f : U → F . If f istwo-times differentiable at the point a, then

(f′′(a)x)y = (f

′′(a)y)x, ∀x, y ∈ E.

Let us remember that the map

f′′(a) ∈ L(E, L(E, F ))

was called the second derivative of f at the point a.The function f is two-times differentiable at the point a if there exists an open neighbourhood

V of a such that f is derivable on V and the map f′: V → L(E,F ) is derivable at the point a. In

this case,

∃ limx → ax 6= a

‖f ′(x)− f

′(a)− f

′′(a) ◦ (x− a)‖

‖x− a‖ = 0.

Particular Case.

Let Ei = F = R. Therefore, E = Rn. Let f : U ⊆ Rn → R. In this particular case,

f′′(a) ∈ L2(Rn,Rn;R).

13

Theorem 3.4 (Schwarz) Let f : U ⊆ Rn → R, ∅ 6= U open. If f is two-times derivable at the pointa ∈ U , then

∂2f

∂xi∂xj(a) =

∂2f

∂xj∂xi(a), ∀i, j = 1, n,

f′′(a) =

n∑

i,j=1

∂2f

∂xi∂xj(a) pi ⊗ pj .

(3.1)

In (3.1),

∂2f

∂xi∂xj(a) = lim

xi→ai

∂f

∂xj(a1, . . . , xi, . . . , an)− ∂f

∂xj(a1, . . . , ai, . . . , an)

xi − ai

and pi ⊗ pj is the tensor product of the linear maps pi and pj , defined by

pi ⊗ pj : Rn × Rn → R, (pi ⊗ pj)(x, y) = pi(x)pj(y) = xiyj .

In fact, {pi⊗pj}1≤i,j≤n is the canonical basis of L2(Rn,Rn;R). So, (3.1) is the unique representation

of the bilinear map f′′(a) in this basis. The mixed partial derivatives

∂2f

∂xi∂xj(a) are the coefficients

of f′′(a) in the canonical basis.

Just as the Jacobi matrix was defined for representing (in the canonical bases) the first derivativef′(a) of a function f : U ⊆ Rn → Rm which is derivable at the point a ∈ U , in the case of the second

derivative, for a function f taking values in R, we have the following analogue representation of thesecond derivative:

Theorem 3.5 Let ∅ 6= U ⊆ Rn be an open set and f : U → R be a twice differentiable functionon U . Then, the second derivative of f at the point a is determined, in the canonical basis, by theHessian matrix:

Hf (a) =

∂2f

∂x21

(a) · · · ∂2f

∂x1∂xn(a)

.................................

∂2f

∂xn∂x1(a) · · · ∂2f

∂x2n

(a)

Example. Let f : R3 → R be given by

f(x, y, z) = xy2z3.

Then,

Hf (x, y, x) =

0 2yz3 3y2z2

2yz3 2xz3 6xyz2

3y2z2 6xyz2 6xy2z

.

Remark 3.6 If f ∈ C2(U), then

∂2f

∂xi∂xj(a) =

∂2f

∂xj∂xi(a), ∀i, j = 1, n.

14

Remark 3.7 The existence of the mixed partial derivatives of the second order for a given functiondoesn’t imply their symmetry.

As a counterexample, it is not difficult to see that for the function f : R2 → R, defined by:

f(x, y) =

xyx2 − y2

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0),

one has∂2f

∂x∂y(0, 0) 6= ∂2f

∂y∂x(0, 0).

Indeed, let us compute∂2f

∂x∂y(0, 0). We get:

∂2f

∂x∂y(0, 0) = lim

x→0

∂f

∂y(x, 0)− ∂f

∂y(0, 0)

x− 0.

But∂f

∂y(x, 0) = lim

y→0

f(x, y)− f(x, 0)y − 0

= x

and∂f

∂y(0, 0) = lim

y→0

f(0, y)− f(0, 0)y − 0

= 0.

So,∂2f

∂x∂y(0, 0) = lim

x→0

x− 0x− 0

= 1.

On the other hand, computing∂2f

∂y∂x(0, 0), we obtain:

∂2f

∂y∂x(0, 0) = lim

y→0

∂f

∂x(0, y)− ∂f

∂x(0, 0)

y − 0.

But∂f

∂x(0, y) = lim

x→0

f(x, y)− f(0, y)x− 0

= −y

and∂f

∂x(0, 0) = lim

x→0

f(x, 0)− f(0, 0)x− 0

= 0.

So,∂2f

∂y∂x(0, 0) = lim

y→0

−y − 0y − 0

= −1.

Hence,∂2f

∂x∂y(0, 0) 6= ∂2f

∂y∂x(0, 0).

As already mentioned, for simplicity, we shall use the notation:

pi = dxi.

15

Also, we shall usually omit to write explicitly the symbol ”⊗ ”. So, we have:

f′′(a) =

n∑

i,j=1

∂2f

∂xi∂xj(a) dxidxj .

For the particular case n = 2, we get

f′′(a, b) =

∂2f

∂x2(a, b) dx2 +

∂2f

∂x∂y(a, b) dx dy +

∂2f

∂y∂x(a, b) dy dx +

∂2f

∂y2(a, b) dy2.

Remark 3.8 For m ≥ 2, by induction, we get

∂mf

∂x1 · · · ∂xm(a) =

∂mf

∂xσ(1) · · · ∂xσ(m)(a),

for any permutation σ : {1, . . . , m} → {1, . . . , m}, and

f (m)(a) =n∑

i1,...,im=1

∂mf

∂xi1 · · · ∂xim

(a) pi1 ⊗ · · · ⊗ pim .

Examples.

1) Verify that the function f : R2 → R, defined by f(x, y) = ex cos y is a solution of Laplace’sequation, i.e.

∂2f

∂x2+

∂2f

∂y2= 0.

Indeed, it is easy to see that f ∈ C2(R2). It follows immediately that

∂2f

∂x2= ex cos y

and∂2f

∂y2= −ex cos y.

Therefore, their sum is zero.

2) Compute the second derivative at the point (1, 1) of the function f : R2 → R, defined by:

f(x, y) = ex2+y2.

Obviously, f ∈ C2(R2) and it is not difficult to compute its second order partial derivatives at thepoint (1, 1). As a result, we get

f′′(1, 1) = 6 e2dx2 + 4 e2dx dy + 4 e2dy dx + 6 e2 dy2.

Exercises.

1) Compute the first and the second differential for the function f : R2 → R, f(x, y) =x2 + y2 + ln (1 + x2 + y2) at the point (1, 1).

16

2) Compute the second derivative at the point (2, 2) of the function f : R2 → R2, defined by

f(x, y) = (x2 + y, x + y2).

3) For the function f : R2 → R, defined by:

f(x, y) =

xy sinx2 − y2

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0),

prove that∂2f

∂x∂y(0, 0) 6= ∂2f

∂y∂x(0, 0).

4) Compute the first and the second differential at the point (1, 1) for the function f : R2 → R,defined by

f(x, y) = sin (x2 + y2).

4 Directional Derivatives

We have already seen one possible interpretation for partial derivatives. Indeed, we saw in theprevious section that f

′x represents the rate of change of the function as we change x, holding y fixed,

while f′y represents the rate of change of f as we change y, holding x fixed. Still, sometimes, we are

interested in finding the rate of change of a given function f in a particular direction, given by avector u. This rate will be expressed by the so-called directional derivative of f .

Definition 4.1 Let E and F be Banach spaces, ∅ 6= U ⊆ E, U open, a ∈ U . Also, let f : U → Fand u ∈ E, u 6= 0. If there exists

δf(a, u) = limt→0

f(a + tu)− f(a)t

∈ F,

then δf(a, u) is called the derivative of the function f , at the point a, in the direction u.

Sometimes, we shall use similar notation, i.e.

δf(a, u) =∂f

∂u(a)

orδf(a, u) = f

′u(a).

Usually, a direction in Rn is specified by a unit vector, i.e. a vector u ∈ Rn with ‖u‖ = 1. So,often, we shall consider that the vectors u are taken to be normalized, although the definition aboveworks for arbitrary (even zero) vectors.

Example. If f : R2 → R is defined by

f(x, y) = 4− 2x2 − y2

17

andu = − 1√

2(1, 1),

then∂f

∂u(1, 1) = 3

√2.

Properties.

1) If f is derivable at the point a in the direction u, then f is derivable at a in the direction −uand

∂f

∂(−u)(a) = −∂f

∂u(a).

More generally, if f is derivable at the point a in the direction u, then f is derivable at a in thedirection αu, α ∈ K and

∂f

∂(αu)(a) = α

∂f

∂u(a).

2) If f is derivable at the point a in the direction u, then f is continuous at a in the direction u.

Theorem 4.2 Let E and F be Banach spaces, ∅ 6= U ⊆ E, U open, a ∈ U . Also, let f : U → Fand u ∈ E, u 6= 0. If f is derivable at the point a, then f is derivable at the point a in any directionu 6= 0 and

δf(a, u) = f′(a)(u), ∀u 6= 0.

Proof. Since f is derivable at the point a, it follows that there exists f′(a) ∈ L(E, F ) such that

limx → ax 6= a

f(x)− f(a)− f′(a) ◦ (x− a)

‖x− a‖ = 0.(4.1)

If we take x = a + tu, with t → 0, we obtain

∃ limt→0

f(a + tu)− f(a)− f′(a) ◦ (tu)

‖tu‖ = 0.

Thus, it follows that

∃ limh→0

f(a + hu)− f(a)h

= f′(a)(u),

for any u 6= 0, which ends the proof of the theorem.The converse implication in Theorem 4.2 is not true. As a counterexample, let us consider thefunction f : R2 → R, defined by:

f(x, y) =

x2y

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0).

It is not difficult to see that this function is not differentiable at the point (0, 0), despite the factthat it possesses directional derivatives at this point with respect to any direction u 6= 0.

18

We have previously used in some contexts the notation −→u for denoting a vector from Rn, insteadof the usual notation u. This was done only to explicitly mark the fact that we are dealing with adirectional limit, along the direction of −→u .

Let us restrict ourselves now to the particular case in which U is an open nonempty set in Rn andf : U ⊆ Rn → R. If a ∈ U , the function f is called Gateaux differentiable (R. Gateaux, 1889-1914) atthe point a if it is derivable at a in any direction u ∈ Rn. The number δf(a, u) is called the Gateauxdifferential of f at the point a in the direction u and the map df(a) : Rn → R, defined by

df(a)(u) = δf(a, u)

is the Gateaux derivative of the function f at the point a.

Let us notice that we cannot establish a connection between the continuity of a given functionand its directional derivability. For instance, on one hand, the function f : R2 → R, defined by

f(x, y) =√

x2 + y2

is continuous at the point (0, 0), but it is not Gateaux differentiable at this point. On the otherhand, if we consider the function f : R2 → R, defined by

f(x, y) =

x2

y, y 6= 0,

0, y = 0,

it is not difficult to see that f is Gateaux differentiable at the point (0, 0), but it is not continuousat this point.

Properties.

1) If U 6= ∅, U open and fi : U ⊆ Rn → R, i = 1, n, are derivable at the point a ∈ U in the

direction u, thenn∑

i=1

cifi, where ci ∈ R, is derivable at the point a in the direction u and

∂

∂u(

n∑

i=1

cifi)(a) =n∑

i=1

ci∂fi

∂u(a).

2) If U 6= ∅, U open and f, g : U ⊆ Rn → R are derivable at the point a ∈ U in the direction u,then fg is derivable at the point a in the direction u and

∂

∂u(fg)(a) =

∂f

∂u(a)g(a) + f(a)

∂g

∂u(a).

3) If U 6= ∅, U open and f : U ⊆ Rn → Rm, f = (f1, . . . , fm), then f is derivable at the pointa ∈ U in the direction u if and only if f1, . . . , fm are derivable at the point a in the direction u.

Let us take now f : U ⊆ Rn → R, U 6= ∅, U open, a = (a1, . . . , an) ∈ U . We shall see inwhat follows another possible way for defining the partial derivatives of f , in terms of directionalderivatives. We know that

∂f

∂xj(a) = lim

xj→aj

f(a1, . . . , xj , ..., an)− f(a1, ..., aj , . . . , an)xj − aj

.

19

If we taket = xj − aj ,

then∂f

∂ej(a) = lim

t→0

f(a + tej)− f(a)t

=∂f

∂xj(a),

whereej = (0, . . . , 1, . . . , 0).

Hence, partial derivatives can be seen as particular directional derivatives along the standard basisvector directions.

Let us suppose that f : Rn → R is derivable at a given point a. Then,

δf(a, u) = f′(a)(u) = ∇f(a) · u, ∀u 6= 0.

So, the slopes in all directions are known from slopes in n directions (given by the partial derivatives).

Exercises.

1) Compute the directional derivative of the function f : R2 → R2, defined by

f(x, y) = (x2 + y, x + y2)

at the point a = (2, 3), in the direction of the vector u = (1, 1).

3) Let f : R2 → R, defined by:

f(x, y) =

xy2

x2 + y2, (x, y) 6= (0, 0),

0, (x, y) = (0, 0),

a) Prove that f is continuous at the point (0, 0).b) Compute the partial derivatives of f at (0, 0).c) Compute the directional derivatives of the function f at the point (0, 0) along any direction

u with ‖u‖ = 1.d) Prove that f is not differentiable at the point (0, 0).

5 Composite Functions

Let us remember now the following results:

Proposition 5.1 Let ∅ 6= U ⊆ Rn, U open, f : U → Rm derivable at the point a ∈ U . The matrixassociated to the linear map f

′(a) : Rn → Rm in the canonical bases of Rn and Rm is

Mf (a) =

∂f1

∂x1(a) · · · ∂f1

∂xn(a)

................................

∂fm

∂x1(a) · · · ∂fm

∂xn(a)

∈M(m,n;R).

20

If m = n, the determinantJf (a) = det Mf (a)

was called the Jacobian of the function f at the point a and it was also denoted by

Jf (a) =∂(f1, . . . , fn)∂(x1, . . . , xn)

(a).

Theorem 5.2 (Chain Rule) Let E,F, G be Banach spaces, U ⊆ E, V ⊆ F , U, V nonempty opensets. If f : U → V is derivable at the point a ∈ U and g : V → G is derivable at the point f(a) ,then g ◦ f is derivable at the point a and

(g ◦ f)′(a) = g

′(f(a)) ◦ f

′(a).

For E = Rn, F = Rm and G = Rk, we obtain the following results:

Proposition 5.3 Let f : U ⊆ Rn → Rm, g : V ⊆ Rm → Rk, U, V nonempty open sets. If f isderivable at a ∈ U and g is derivable at b = f(a) ∈ V , then

Mg◦f (a) = Mg(b) ·Mf (a).(5.1)

Corollary 5.4 Let U, V be nonempty open subsets of Rn, f : U → Rn, f = (f1, . . . , fn) derivable atthe point a ∈ U , g : V → Rn, g = (g1, . . . , gn) derivable at the point b = f(a) ∈ V . Then, the maph = g ◦ f : U → Rn, h = (h1, . . . , hn), is derivable at the point a and

∂(h1, . . . , hn)∂(x1, . . . , xn)

(a) =∂(g1, . . . , gn)∂(f1, . . . , fn)

(b)∂(f1, . . . , fn)∂(x1, . . . , xn)

(a).

Particular Cases.

1) Let U ⊆ R2 and V ⊆ R be nonempty open sets and let u : U → V and g : V → R be C1

functions. IfF (x, y) = g(u(x, y)),

then F is of class C1 and

∂F

∂x=

dg

du

∂u

∂x= g

′ ∂u

∂x,

∂F

∂y=

dg

du

∂u

∂y= g

′ ∂u

∂y.

Example. Let g : R→ R be an arbitrary C1 function and let

F (x, y) = g(x2 + y2).

If we denote by u(x, y) = x2 + y2, then

∂F

∂x=

dg

du

∂u

∂x=

dg

du2x = g

′2x,

∂F

∂y=

dg

du

∂u

∂y=

dg

du2y = g

′2y.

2) Let U and V be nonempty open sets in R2 and let f = (u, v), f : U → V and g : V → R beC1 functions. If

F (x, y) = g(u(x, y), v(x, y)),

21

then F is of class C1 on U and

∂F

∂x=

∂g

∂u

∂u

∂x+

∂g

∂v

∂v

∂x,

∂F

∂y=

∂g

∂u

∂u

∂y+

∂g

∂v

∂v

∂y.

Example. LetF (x, y) = xy + g(x + y, x3 + y3),

where g : R2 → R is an arbitrary given function of class C1. If we denote by

u(x, y) = x + y

and byv(x, y) = x3 + y3,

then∂F

∂x= y +

∂g

∂u

∂u

∂x+

∂g

∂v

∂v

∂x,

i.e.∂F

∂x= y +

∂g

∂u+ 3x2 ∂g

∂v.

In a similar manner, we get∂F

∂y= x +

∂g

∂u+ 3y2 ∂g

∂v.

3) Let U and V be nonempty open sets in R3 and let f = (u, v, w), f : U → V and g : V → Rbe C1 functions. If

F (x, y, z) = g(u(x, y, z), v(x, y, z), w(x, y, z)),

then F is of class C1 on U and

∂F

∂x=

∂g

∂u

∂u

∂x+

∂g

∂v

∂v

∂x+

∂g

∂w

∂w

∂x,

∂F

∂y=

∂g

∂u

∂u

∂y+

∂g

∂v

∂v

∂y+

∂g

∂w

∂w

∂y,

∂F

∂z=

∂g

∂u

∂u

∂z+

∂g

∂v

∂v

∂z+

∂g

∂w

∂w

∂z.

Example. LetF (x, y, z) = xyz + g(x + y + z, x2 + y2 + z2, xyz),

where g : R3 → R is an arbitrary given function of class C1. If we denote

u(x, y, z) = x + y + z,

v(x, y, z) = x2 + y2 + z2

andw(x, y, z) = xyz,

then∂F

∂x= yz +

∂g

∂u

∂u

∂x+

∂g

∂v

∂v

∂x+

∂g

∂w

∂w

∂x,

22

i.e.∂F

∂x= yz +

∂g

∂u+ 2x

∂g

∂v+ yz

∂g

∂w.

In a similar manner, we get∂F

∂y= xz +

∂g

∂u+ 2y

∂g

∂v+ xz

∂g

∂w.

and∂F

∂z= xy +

∂g

∂u+ 2z

∂g

∂v+ xy

∂g

∂w.

4) Obviously, if f = (f1, . . . , fm) : Rn → Rm, g = (g1, . . . , gp) : Rm → Rp, h = g ◦ f : Rn → Rp,then

∂hk

∂xj=

m∑

l=1

∂gk

∂fl

∂fl

∂xj.

In fact, we can consider the partial differential operator:

∂

∂xj(·) =

m∑

l=1

∂

∂fl(·) ∂fl

∂xj.

For second order partial derivatives, if assume the needed regularity, we get:

∂

∂xi(∂hk

∂xj) =

m∑

l=1

m∑

r=1

∂2gk

∂fr∂fl

∂fr

∂xi

∂fl

∂xj+

m∑

l=1

∂gk

∂fl

∂2fl

∂xi∂xj.

Example. LetF (x, y) = xy + g(x2 + y2, xy),

where g : R2 → R is an arbitrary given function of class C2. If we denote by

u(x, y) = x2 + y2

and byv(x, y) = xy,

then∂2F

∂x2= 4x2 ∂2g

∂u2+ 4xy

∂2g

∂u∂v+ y2 ∂2g

∂v2+ 2

∂g

∂u,

∂2F

∂x∂y= 1 + 4xy

∂2g

∂u2+ (2x2 + 2y2)

∂2g

∂u∂v+ xy

∂g

∂v2+

∂g

∂u

and∂2F

∂y2= 4y2 ∂2g

∂u2+ 4xy

∂2g

∂u∂v+ x2 ∂2g

∂v2+ 2

∂g

∂u.

Exercises.

1) Compute the first and the second differential for the function

F (x, y) = (x2 + y2)u(x + y),

where u : R→ R is an arbitrary function of class C2.

2) LetF (x, y) = xyu(x + y, xy),

23

where u : R2 → R is a given function of class C2. Compute the first and the second differential of F .

3) Let F (x, y, z) = (x2 + y2 + z2)u(x + y + z, xyz), where u ∈ C2. Compute the first and thesecond differential of F .

4) Let F (x, y) = u(x + y, xy, x2 + y2), where u ∈ C2. Compute

∆ F =∂2F

∂x2+

∂2F

∂y2.

5) Prove that the functionz(x, y) = f(x + ϕ(y)),

where f, ϕ ∈ C2(R), satisfies the equation:

∂z

∂x

∂2z

∂x∂y=

∂z

∂y

∂2z

∂x2.

6) Prove that the function

g(x, t) = ϕ(x− at) + ψ(x + at),

where ϕ,ψ are given functions of class C2 and a > 0, satisfies the equation:

∂2g

∂t2= a2 ∂2g

∂x2,

called the equation of the vibrating string.

6 Differential Operators

Let us recall the following result:

Remark 6.1 Let ∅ 6= U ⊆ Rn, U open, f : U → R derivable at the point a ∈ U . Then, there existsa unique element y ∈ Rn such that

f′(a)(x) = 〈x, y〉.

The element y, denoted by gradaf , is called the gradient of the function f at the point a.

So, the gradient of the scalar field f at the point a is a vector from Rn, defined by

gradaf = (∂f

∂x1(a), . . . ,

∂f

∂xn(a)).

Usually, it is convenient to work with scalar fields of class C1. So, we obtain the following definition:

Definition 6.2 Let ∅ 6= U ⊆ Rn, U open. Also, let f : U ⊆ Rn → R be a scalar field, f ∈ C1(U)and a ∈ U . The gradient of the function f at the point a is defined as being

grad f : U → Rn, (grad f)(a) = gradaf.

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Thus, the gradient of f at the point a can be written as

(grad f)(a) = (∂f

∂x1(a), . . . ,

∂f

∂xn(a)).

If we introduce the nabla symbol ∇, we have

∇f(a) = (∂f

∂x1(a), · · · , ∂f

∂xn(a))

or

∇f(a) =n∑

i=1

∂f

∂xi(a)ei.

The differential operator

∇ = (∂

∂x1, · · · , ∂

∂xn)

is called the Hamiltonian operator. So, in the n-dimensional Euclidian space Rn, this operator canbe written as

∇ =n∑

i=1

∂

∂xiei

or, using the Einstein summation notation, as

∇ =∂

∂xiei

The name of the symbol ∇ comes from the Greek word for a Hebrew harp, which has a similarshape. The symbol was used for the first time in 1837 by William Rowan Hamilton (1805-1865).Another name for this symbol is atled (delta spelled backwards), because this symbol nabla lookslike an inverted delta. Also, the word del is used for the Hamiltonian operator.

Let us see now which is the connection between the gradient and the directional derivative. Letf : U ⊆ Rn → R be differentiable at the point a ∈ U and let −→u ∈ Rn be a unit vector. Then, f isderivable at the point a in the direction −→u and

∂f

∂−→u (a) = 〈∇f(a),−→u 〉 = |∇f(a)| cos θ,

where θ is the angle between the vectors∇f(a) and −→u . Therefore, the rate of change of the function fin the direction −→u depends on −→u and varies from −|∇f(a)| (when θ = π, i.e. −→u points in a directionopposite to ∇f(a)) to |∇f(a)| (when θ = 0, i.e. −→u points in then same direction as ∇f(a)). Thevector ∇f(a) points in the direction of the greatest rate of increase of the function f at the point aand the greatest rate of change is exactly the magnitude of this vector, i.e. |∇f(a)|.

For instance, let us consider a room in which, at each point (x1, x2, x3), the temperature is givenby the scalar field T = T (x1, x2, x3). If we assume that the temperature does not change in time,then, at each point in the room, the gradient at that point will show the direction in which thetemperature rises most quickly. The magnitude of the gradient will tell us how fast the temperaturerises in that direction.

A generalization of the gradient, for functions which have vectorial values, is the Jacobian matrix.The gradient ∇f is orthogonal to the level surface f(x) = C. Indeed, since

∇f · d−→r = df

25

and f(x) = C, it is not difficult to see that

∇f · d−→r = 0.

Examples. 1) The gradient of the scalar field f : R3 → R, defined by

f(x, y, z) = 2x + y2 − cos z,

is the vector:∇f = (2, 2y, sin z).

2) Let r be the length of the position vector −→r 6= −→0 of a point (x, y, z) in R3, i.e.

r = ‖−→r ‖ 6= 0.

Computing grad (1r), we get

grad (1r) = −

−→rr3

.

More general, it is easy to see that

gradf(r) =f ′(r)

r−→r ,

for any f ∈ C1(U), U ⊆ R∗+.

We shall define now the so-called Laplace operator (Pierre-Simon Laplace, 1749-1827) or theLaplacian, denoted by ∆ or by ∇2. This operator is a second order linear differential operator,arising in the modeling of various physical phenomena, such as wave propagation and heat flowin homogeneous bodies. Also, this operator appears in electrostatics, in dispersion theory, in theso-called Laplace’s and Poisson’s equations, in quantum mechanics, etc.

Definition 6.3 Let ∅ 6= U ⊆ Rn, U open and f : U ⊆ Rn → R be a scalar field such that f ∈ C2(U).The Laplacian of the function f is defined as being:

∆f =∂2f

∂x21

+ · · ·+ ∂2f

∂x2n

.

So, the Laplace operator is a second order differential operator defined as the sum of all the unmixedsecond partial derivatives:

∆ =n∑

i=1

∂2

∂x2i

.

Remark 6.4 A function f is said to be harmonic if ∆f = 0.

Example. Let f : R2 \ {(0, 0)} → R be defined by

f(x, y) = ln(x2 + y2).

Then, f satisfies Laplace’s equation, i.e.∆f = 0.

Indeed, it is not difficult to see that∂2f

∂x2=

2(y2 − x2)(x2 + y2)2

.

26

In a similar manner,∂2f

∂y2=

2(x2 − y2)(x2 + y2)2

.

Hence,

∆f =∂2f

∂x2+

∂2f

∂y2= 0.

This proves that our function f is harmonic.

Let us define now the divergence operator, acting on vector fields, and measuring the magnitudeof a given vector field’s source or sink at a given point. In other words, the divergence of a vectorat a point a will be a signed scalar which measures the ”spreading” of the given vector field at thatpoint, i.e. the extent to which the vector field flow behaves like a source or a sink at the point a.

Definition 6.5 Let ∅ 6= U ⊆ Rn, U open and −→v : U ⊆ Rn → Rn be a vector field such that−→v ∈ C1(U). The divergence of the vector −→v is defined as being

div−→v =∂v1

∂x1+ · · ·+ ∂vn

∂xn.

The divergence of −→v can be interpreted as the symbolic scalar product of the Hamiltonianoperator ∇ and the vector field −→v :

div −→v = ∇ · −→v .

Notice that the divergence of a vector field is a scalar field.

Remark 6.6 A vector field −→v is called solenoidal (or, sometimes, incompressible) if div −→v = 0 (nosources or sinks).

The velocity field of an incompressible fluid is an example of such a solenoidal field, assuming thatthere are no sources or sinks, i.e. there are no points at which the fluid is introduced or removed fromthe system. As a matter of fact, the term solenoidal comes from a Greek word meaning pipe-shaped.So, in this context, solenoidal means constrained in a pipe (tube), so with a fixed volume.

Examples.

1) Let −→r = (x, y, z) ∈ R3, −→r 6= −→0 and

r = ‖−→r ‖,i.e.

r =√

x2 + y2 + z2.

Prove that the divergence of the electrostatic field

−→E =

q−→rr3

is zero. Hence, −→E is a solenoidal field.

Let −→E = (Ex, Ey, Ez), whereEx =

qx

r3, Ey =

qy

r3, Ez =

qz

r3.

Since, by the chain rule,∂r3

∂x= 3xr,

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it is easy to see that∂Ex

∂x= q

r3 − 3x2r

r6.

In a similar manner,∂Ey

∂y= q

r3 − 3y2r

r6

and∂Ez

∂z= q

r3 − 3z2r

r6.

Therefore,div −→E = 0.

2) It is very easy to see thatdiv−→r = 3.

Let us define now the so-called curl operator, acting on vector fields in the Euclidian space R3.The curl of a vector field at a point will measure the amount of swirling or spinning. So, the curlwill be the measure of the rotational strength at a point.

Definition 6.7 Let ∅ 6= U ⊆ R3, U open and −→v : U → R3 be a vector field, −→v ∈ C1(U). Thedifferential operator

rot −→v = (∂v3

∂x2− ∂v2

∂x3,∂v1

∂x3− ∂v3

∂x1,∂v2

∂x1− ∂v1

∂x2)

is called the curl or the rotation of the vector −→v .

We shall also denote by curl −→v the rotation of the vector field −→v .The curl of −→v can be interpreted as the symbolic vector product of the Hamiltonian operator ∇

by the vector field −→v :curl −→v = ∇×−→v ,

i.e.

curl −→v =

∣∣∣∣∣∣∣∣∣∣∣∣∣

−→e1−→e2

−→e3

∂

∂x1

∂

∂x2

∂

∂x3

v1 v2 v3

∣∣∣∣∣∣∣∣∣∣∣∣∣

Notice that the curl of a vector field is also a vector field.

Remark 6.8 A vector field −→v is called irrotational if curl −→v = 0 (irrotational, nonturbulent flow).

The most important physical examples of irrotational vectors are the gravitational and, respec-tively, the electrostatic forces.

Exercises.

1 Prove thatcurl−→r = 0

28

andcurl (f(r)−→r ) = 0,

for any f ∈ C1(U), U ⊆ R∗+.

2) Let −→V = −→V0 +−→ω ×−→r be the velocity field in a rigid solid. It is not difficult to prove that

curl−→V = 2−→ω .

anddiv−→V = 0.

Remark 6.9 If −→v is an irrotational field in a simply connected neighborhood U of a point x, thenin this neighborhood, −→v is a potential field, i.e. −→v is given in terms of the gradient of a scalar fieldΦ:

−→v = −∇Φ.

We shall list now some properties of the above introduced differential operators and some rela-tionships between them. The proof of these properties is very simple and it is left as an exercise tothe reader.

Properties.

Let U ⊆ R3 be a nonempty open set. Prove that:

1) grad (cf) = c grad f, ∀c ∈ R, ∀f : U → R, f ∈ C1(U).

2) grad (f + g) = grad f + grad g, ∀f, g : U → R, f, g ∈ C1(U).

3) grad (fg) = ggrad f + fgrad g, ∀f, g : U → R, f, g ∈ C1(U).

4) grad (f/g) =ggrad f − fgrad g

g2, ∀f, g : U → R, f, g ∈ C1(U),

g(x) 6= 0, ∀x ∈ U .

5) grad F (ϕ) = F′(ϕ) grad ϕ, ∀F, ϕ ∈ C1.

6) div (α−→u + β−→v ) = α div −→u + β div −→v , ∀α, β ∈ R, ∀−→u ,−→v ∈ C1.

7) div (ϕ−→v ) = ϕ div −→v + (grad ϕ) · −→v , ∀ϕ, −→v ∈ C1.

8) div (−→u ×−→v ) = −→v · rot −→u −−→u · rot −→v , ∀−→u ,−→v ∈ C1. As a consequence, if−→u and −→v are irrotatational vector fields, then −→u ×−→v is solenoidal.

9) div (grad ϕ) = 4ϕ, ∀ϕ ∈ C2.

10) rot (grad ϕ) = 0, ∀ϕ ∈ C2.

11) rot (α−→u + β−→v ) = α rot −→u + β rot −→v , ∀α, β ∈ R, ∀−→u ,−→v ∈ C1.

12) rot (ϕ−→v ) = ϕ rot −→v −−→v × grad ϕ, ∀ϕ, −→v ∈ C1.

13) rot (rot −→u ) = grad (div −→u )−4−→u , ∀−→u ∈ C2.

14) div (rot −→v ) = 0, ∀−→v ∈ C2.

15) grad1r

= −−→rr3

, grad ln r =−→rr2

, ∀−→r = (x, y, z) ∈ R3, r =| −→r |6= 0.

29