Partial Derivative

October 1, 2017 | Author: Vishal Vishwakarma | Category: Gradient, Divergence, Derivative, Waves, Euclidean Vector
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The derivative is a way to find out how a function of a single variable will change when the variable changes. We often want to extend this idea to functions of several variables. This requires the introduction of methods for calculating the effect of changing one variable while the other variables are fixed. These ideas are illustrated below. Consider the area enclosed in Figure 7.1. The solid box encloses the original area A0 . The length of the sides of the solid box are denoted by x, y. The dotted lines in the figure represent an expansion of the original area. We wish to find the change in the total area. Let the sides of the original box with area A0 increase by increments Dx and Dy: First note that the original area is A0 ¼ x y and the total area is AT ¼ (x þ Dx ) (y þ Dy) or AT ¼ x y þ xDy þ y Dx þ Dx Dy: The change in area is D A ¼ AT  A0 ¼ y D x þ x D y þ D x Dy ¼ D Ajy þ D Ajx þ D x Dy

Figure 7.1 Expanding area. Math Refresher for Scientists and Engineers, Third Edition By John R. Fanchi Copyright # 2006 John Wiley & Sons, Inc.




where jy and jx signify that the sides of the solid box with lengths y and x are not changed. If the length y is fixed, then DAjy ¼ y Dx or y ¼ DA=Dxjy . If the length x is fixed, DAjx ¼ x Dy or x ¼ DA=Dyjx . Substituting these expressions into DA gives   DA DA Dx þ Dy þ Dx Dy DA ¼ D xy Dy x If we let Dx ! 0 and Dy ! 0, we obtain the differential of A(x, y) expanded in terms of partial derivatives (@[email protected], @[email protected]) and differentials (dx, dy) such that   @A @A dA ¼  dx þ  dy @x y @y x where the second-order term is negligible. A more formal treatment of these ideas follows.



A function y is a function of several variables if it is a function of two or more independent variables. A function of n variables has the form y ¼ f (x1 , x2 , . . . , xn ) where x1 , x2 , . . . , xn are n independent variables and y is a dependent variable. The function f (  ) is a mapping from Rn ! R, where R is the set of real numbers and Rn is an n-dimensional set of real numbers. The partial derivative of y with respect to xi is defined by @y f (x1 , . . . , xi þ Dxi , . . . , xn )  f (x1 , . . . , xi , . . . , xn ) ¼ lim @xi Dxi !0 Dxi and all other {xj } are held constant. Higher-order partial derivatives are found by successive applications of this definition; thus     @ @f @ 2f @ @f @ 2f ¼ fxy ¼ 2 ¼ fxx , ¼ @x @x @x @y @x @[email protected] The symbol fxy denotes the partial differentiation of f with respect to x and then y. The order of differentiation is commutative. If f xy and f yx are continuous functions of x and y, then f xy ¼ f yx .

Example: Let z ¼ ax2 þ bxy þ cy2 þ dx þ ey þ f :



First-Order Partial Derivatives: @z @z ¼ 2ax þ by þ d, ¼ bx þ 2cy þ e @x @y Second-Order Partial Derivatives:   @ 2z @ @z ¼ ¼ 2a @x2 @x @x   @ 2z @ @z ¼ ¼ 2c @y2 @y @y     @ 2z @ @z @ @z ¼ ¼ ¼b @[email protected] @x @y @y @x EXERCISE 7.1: Let f (x, y) ¼ e xy : Find f xy , f yx . The total differential of a function y of variables {xi : i ¼ 1, n} is dy ¼

n X @y i¼1



Example: Let y ¼ x21 þ x32 . In this case n ¼ 2 because there are two independent variables and dy ¼

@y @y dx1 þ dx2 ¼ 2x1 dx1 þ 3x22 dx2 @x1 @x2

EXERCISE 7.2: Let y ¼ x21 þ x21 x2 þ x32 . Calculate @[email protected] , @[email protected] , and dy. Suppose we have a function z of two parameterized variables such that z ¼ f (x(t), y(t)) The total derivative of the parameterized function z of two variables is dz @f dx @f dy ¼ þ dt @x dt @y dt This expression can readily be generalized for a parameterized function of several variables.

EXERCISE 7.3: Let z ¼ x2 þ y2 , x ¼ at, y ¼ bet . Find dz=dt.



Application: Jacobian Transformation. The equation of a coordinate transformation may be written as y ¼ Ax If we consider a two-dimensional system, the total derivative of the transformation equation is dy1 ¼

@y1 @y1 dx1 þ dx2 @x1 @x2

dy2 ¼

@y2 @y2 dx1 þ dx2 @x1 @x2

or dy1 ¼ J11 dx1 þ J12 dx2 dy2 ¼ J21 dx1 þ J22 dx2 where Jij ¼

@yi @x j

Collecting the elements {Jij } in the 2  2 square matrix J gives ½dyi ¼

2 X

½Jij ½dxj


or the matrix equation dy ¼ Jdx The matrix J is called the Jacobian and is given by ½Jif ¼

@yi @xj

EXERCISE 7.4: Find the Jacobian of the coordinate rotation y ¼ ax, where a is the matrix of coordinate rotations introduced in Chapter 4, Section 4.1.



Vector analysis is the study of vectors and their transformation properties. As we show in the following, it is a discipline in which partial differentiation plays a major role.



Scalar and Vector Fields Let x1 , x2 , x3 denote the Cartesian coordinates of a point X in a region of space R. The position vector x of X is x ¼ x1 i 1 þ x2 i 2 þ x3 i 3 where i1 , i2 , i3 are unit vectors defined along the orthogonal axes of the coordinate system. If we can associate a scalar function f with every point in R, then f is the scalar field in R and may be written f (x1 , x2 , x3 ) ¼ f (x) An example of a scalar field is the temperature at each point in a region of space. Instead of a scalar function, suppose we associate a vector v with every point in R. The resulting vector field has the form v(x1 , x2 , x3 ) ¼ v(x) A vector field is exemplified by a velocity field or a magnetic field. The vector field is a function that assigns a vector to every point in a region. Scalar and vector fields commute, that is, f u ¼ uf Vector fields u and v may be multiplied in the usual way as the dot product u  v ¼ (u1 i1 þ u2 i2 þ u3 i3 )  (v1 i1 þ v2 i2 þ v3 i3 ) ¼

3 X m¼1

and the cross product   i1  u  v ¼  u1  v1

i2 u2 v2

 i3  u3  ¼ v  u v3 

Vector fields u, v, w satisfy the triple scalar product u  (v  w) ¼ v  (w  u) ¼ w  (u  v) and the triple vector product u  (v  w) ¼ v(u  w)  (u  v)

um v m



Example: The expansion of the triple scalar product of the vector fields A, B, C gives   ^i  ^   bx A  (B  C) ¼ (ax^i þ ay ^j þ az k)  c x

^j by cy

 k^   bz   c  z

^  ½(by cz  bz cy )^i  (bx cz  bz cx ) ^j ¼ (ax^i þ ay ^j þ az k) ^ þ (bx cy  by cx )k ¼ ax (by cz  bz cy ) þ ay (bz cx  bx cz ) þ az (bx cy  by cx )

Gradient, Divergence, and Curl We can determine the spatial variation of a scalar or vector field by introducing the del operator r defined in Cartesian coordinates as r ; i1

@ @ @ þ i2 þ i3 @x1 @x2 @x3

Applying r to the scalar field f gives a vector field called the gradient of f. It is denoted as grad f ¼ rf ¼ i1

@f @f @f þ i2 þ i3 @x1 @x2 @x3

The gradient of f points in the direction in which the scalar field f changes the most with respect to a change in position. The vector field rf is perpendicular, or normal, to the surfaces corresponding to constant values of f. The arrows in Figure 7.2 illustrate the direction of the gradient at several points around the surface of constant f.

Figure 7.2

Direction of gradient.



Example: Let f (x, y) ¼ x2 y be a scalar field. The gradient of f (x, y) is @f @f rf (x, y) ¼ ^i þ ^j ¼ 2xy^i þ x2 ^j @x @y The del operator can be applied to vector fields in two ways: (1) to create a scalar and (2) to create another vector. Suppose a is a vector field. We create a scalar by defining the divergence of a as the dot product of the operator r and the vector a:   @ @ @ þ i2 þ i3  ða1 i1 þ a2 i2 þ a3 i3 Þ div a ; r  a ¼ i1 @x1 @x2 @x3 ¼

@a1 @a2 @a3 þ þ @x1 @x2 @x3

because im  in ¼ dmn , where dmn is the Kronecker delta. The cross product of the operator r and the vector a is called the rotation of a or the curl of a. It is given by   i1   @ curl a ; r  a ¼   @x1 a 1

i2 @ @x2 a2

 i3   @   @x3  a3

Note that a  r and a  r are not commutative; that is, a  r = r  a and a  r = r  a: The vector products a  r and a  r are given by 

@ @ @ a  r ¼ ða1 i1 þ a2 i2 þ a3 i3 Þ  i1 þ i2 þ i3 @x1 @x2 @x3 ¼

3 X m¼1


@ @xm

and   i1  a a  r ¼  1  @  @x


i2 a2 @ @x2

 i3  a3  @  @x3 

The divergence of the gradient of a scalar field f gives the Laplacian of f ; thus in Cartesian coordinates we have r  (rf ) ¼

3 X @ 2f m¼1


; r2 f



By contrast, the Laplacian of a vector field u is r2 u ¼ r(r  u)  r  (r  u) The curl of the gradient of a scalar field vanishes, r  (rf ) ¼ 0 and the divergence of the curl of a vector field is zero, r  (r  u) ¼ 0 Several other useful relations are summarized as follows: DEL OPERATOR RELATIONS

Let f, g be scalar fields and u, v be vector fields. Sum of fields

r(f þ g) ¼ rf þ rg r  (u þ v) ¼ r  u þ r  v r  (u þ v) ¼ r  u þ r  v r( fg) ¼ f (rg) þ g(rf ) r  ( f u) ¼ f (r  u) þ (rf )  u r  ( f u) ¼ f (r  u) þ (rf )  u

Product of fields

r  (u  v) ¼ v  (r  u)  u  (r  v) r  (u  v) ¼ u(r  v) þ (v  r) u  v(r  u)  (u  r)v r(u  v) ¼ u  (r  v)  v(r  u)þ (v  r) u  (u  r)v r  (rf ) ¼ r2 f r  (r  u) ¼ r(r  u)  r2 u


Example: Suppose A ¼ 2xy^i þ x2^j ¼ Ax^i þ Ay^j: Then the dot product is rA¼

@ @ Ax þ Ay ¼ 2y @x @y

The cross product is   ^i   @ r  A ¼   @x A x

^j @ @y Ay

 k^  @  @z  Az 



Expanding the determinant and simplifying gives       @Az @Ay @Ax @Az @Ay @Ax r  A ¼ ^i    þ ^j þ k^ ¼ 2xk^ @y @z @z @x @x @y The curl of A is a vector transverse to the x –y plane containing the vector A. Also, the vector A is the gradient of f (x, y) ¼ x2 y from a previous example; thus r  A ¼ r  (rf ) ¼ r2 f ¼

@ 2f @ 2f @ 2f þ þ ¼ 2y @x2 @y2 @z2

A field vector v is called irrotational if the curl, or rotation, of v vanishes, that is, rv¼0 The vector field v is said to be solenoidal if rv¼0 A vector field V that is the gradient of a scalar field f is irrotational because the curl of a gradient vanishes; thus r  V ¼ r  (rf) ¼ 0 Similarly, a vector field U that is the curl of a vector field u is solenoidal because the divergence of the curl vanishes; thus r  U ¼ r  (r  u) ¼ 0 ^ Evaluate (a) r ¼ jrj; (b) r  r; (c) r  r; EXERCISE 7.5: Let r ¼ x^i þ y^j þ zk: (d) r(r); and (e) r(1=r): Application: Propagation of Seismic Waves. Seismic waves are vibrations, or displacements from an undisturbed position, that propagate from a source, such as an explosion, through the earth. Seismic wave propagation is an example of a displacement propagating through an elastic medium. The equation for a wave propagating through an elastic, homogeneous, isotropic medium is r

@ 2u ¼ (l þ 2m) r (r  u)  mr  (r  u) @t2


where r is the mass density of the medium, l and m are properties of the elastic medium called Lame´’s constants, and u measures the displacement of the medium from its undisturbed state [Tatham and McCormack, 1991]. If the displacement uI is irrotational, then uI satisfies the constraint r  uI ¼ 0



and Eq. (7.2.1) becomes @ 2 uI (l þ 2m) r(r  uI ) ¼ r @t2


r(r  u) ¼ r  (r  u) þ r2 u


The vector identity

for an irrotational vector is r(r  uI ) ¼ r2 uI so that Eq. (7.2.1) becomes the wave equation @ 2 uI ¼ v2I r2 uI @t2


The speed of wave propagation vI for an irrotational displacement uI is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (l þ 2m) vI ¼ r A solution of Eq. (7.2.4) that is irrotational is the solution for a longitudinal wave propagating in the z direction with amplitude u0 , frequency v, and wavenumber k: uI ¼ u0 ei(kzvt) k^ If the displacement us is solenoidal, then us satisfies the constraint r  us ¼ 0 and Eq. (7.2.1) becomes @ 2 us m ¼  r  (r  us ) r @ t2


The vector identity in Eq. (7.2.3) for a solenoidal vector reduces to r  (r  us ) ¼ r2 us so that Eq. (7.2.5) becomes the wave equation @ 2 us ¼ v2s r2 us @ t2 The speed of wave propagation vs for a solenoidal displacement us is rffiffiffiffi m vs ¼ , vI r




A solution of Eq. (7.2.6) that is solenoidal is the solution for a transverse wave propagating in the z direction: us ¼ u0 ei(kxvt) k^ The irrotational displacement uI represents a longitudinal P (primary) wave, whereas the solenoidal displacement us represents a slower transverse S (secondary) wave. Both types of waves are associated with earthquakes and explosions on or below the earth’s surface. The waves are useful for geophysical studies of the earth’s interior.



A fundamental concept in complex analysis is the concept of analyticity. A function f (z) of a complex variable z is analytic in a domain D if f (z) exists and is differentiable at all points in D. In principle, the complex function f (z) of the complex variable z ¼ x þ iy can be written as the sum of a real function u(x, y) and an imaginary function iv(x, y); thus f (z) ¼ u(x, y) þ iv(x, y)


where u and v are real functions of the real variables x, y: The criteria for establishing the analyticity of f (z) are obtained by extending the definition of partial derivative presented in Section 7.1 from real to complex functions. The derivative of f (z) with respect to z is

f 0 (z) ¼ lim


f (z þ D z)  f (z) Dz


The complex increment Dz ¼ Dx þ iDy may approach 0 along any path in a neighborhood of z. A neighborhood of z is an open set in the complex plane that encloses the point z. One example path for Dz ! 0 is to approach the point z by letting Dx ! 0 and then Dy ! 0. In this case, Dz ! iDy and Eq. (7.3.2) becomes f 0 (z) ¼ lim


f (x, y þ Dy)  f (x, y) iDy


Expanding Eq. (7.3.3) in terms of u, v gives

f 0 (z) ¼ lim


u(x, y þ Dy) þ iv(x, y þ Dy)  u(x, y)  iv(x, y) i Dy




or v(x, y þ Dy)  v(x, y) Dy!0 Dy

f 0 (z) ¼ lim

u(x, y þ Dy)  u(x, y) Dy!0 Dy


 i lim

Applying the definition of partial derivative in Section 7.1 to Eq. (7.3.5) yields f 0 (z) ¼

@v @u i @y @y


An alternative route to an expression for f 0 (z) is provided by letting Dy ! 0 before letting D x ! 0: In this case, D z ! D x and Eq. (7.3.2) becomes f 0 (z) ¼ lim

D x!0

f (x þ D x, y)  f (x, y) Dx


By analogy with the procedure leading from Eq. (7.3.3) to (7.3.6), we obtain f 0 (z) ¼

@u @v þi @x @x


Equations (7.3.6) and (7.3.8) are equivalent expressions for f 0 (z). Equating the real and imaginary parts of Eqs. (7.3.6) and (7.3.8) gives the Cauchy –Riemann equations @u @v ¼ @x @y



@u @v ¼ @y @x


The validity of the Cauchy–Riemann equations depends on the existence of f 0 (z), which depends on the analyticity of f (z). If the function f (z) is analytic, then f 0 (z) exists and the Cauchy–Riemann equations are valid. Conversely, if the Cauchy–Riemann equations apply, then the function f (z) is analytic and f 0 (z) exists [Kreyszig, 1999]. If we differentiate Eq. (7.3.9) with respect to x and Eq. (7.3.10) with respect to y, we obtain @ 2u @ 2v ¼ 2 @x @[email protected]





@ 2u @ 2v ¼ 2 @y @[email protected]


Subtracting Eq. (7.3.12) from (7.3.11) gives Laplace’s equation in two Cartesian dimensions: @ 2u @ 2u þ ¼0 @x2 @y2


Laplace’s equation is an example of a partial differential equation. Partial differential equations are discussed in more detail in Chapter 12.

EXERCISE 7.6: Given f (z) ¼ z2 , show that the Cauchy –Riemann equations are satisfied and evaluate f 0 (z).

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