LECTURE NOTES ON MATHEMATICAL METHODS, Exercises of Mathematical Methods

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LECTURE NOTES ON

MATHEMATICAL METHODS

Mihir Sen

Joseph M. Powers

Department of Aerospace and Mechanical Engineering

University of Notre Dame

Notre Dame, Indiana 46556-

USA

updated

29 July 2012, 2:31pm

10 CONTENTS

Preface

These are lecture notes for AME 60611 Mathematical Methods I, the first of a pair of courses on applied mathematics taught in the Department of Aerospace and Mechanical Engineering of the University of Notre Dame. Most of the students in this course are beginning graduate students in engineering coming from a variety of backgrounds. The course objective is to survey topics in applied mathematics, including multidimensional calculus, ordinary differ- ential equations, perturbation methods, vectors and tensors, linear analysis, linear algebra, and non-linear dynamic systems. In short, the course fully explores linear systems and con- siders effects of non-linearity, especially those types that can be treated analytically. The companion course, AME 60612, covers complex variables, integral transforms, and partial differential equations. These notes emphasize method and technique over rigor and completeness; the student should call on textbooks and other reference materials. It should also be remembered that practice is essential to learning; the student would do well to apply the techniques presented by working as many problems as possible. The notes, along with much information on the course, can be found at http://www.nd.edu/∼powers/ame.60611. At this stage, anyone is free to use the notes under the auspices of the Creative Commons license below. These notes have appeared in various forms over the past years. An especially general tightening of notation and language, improvement of figures, and addition of numerous small topics was implemented in 2011. Fall 2011 students were also especially diligent in identifying additional areas for improvement. We would be happy to hear further suggestions from you.

Mihir Sen [email protected] http://www.nd.edu/∼msen

Joseph M. Powers [email protected] http://www.nd.edu/∼powers

Notre Dame, Indiana; USA ©^ CC © BY: © $ \ © = 29 July 2012 The content of this book is licensed under Creative Commons Attribution-Noncommercial-No Derivative Works 3.0.

Chapter 1

Multi-variable calculus

see Kaplan, Chapter 2: 2.1-2.22, Chapter 3: 3.9,

Here we consider many fundamental notions from the calculus of many variables.

1.1 Implicit functions

The implicit function theorem is as follows:

Theorem For a given f (x, y) with f = 0 and ∂f /∂y 6 = 0 at the point (xo, yo), there corresponds a unique function y(x) in the neighborhood of (xo, yo).

More generally, we can think of a relation such as

f (x 1 , x 2 ,... , xN , y) = 0, (1.1)

also written as f (xn, y) = 0, n = 1, 2 ,... , N, (1.2)

in some region as an implicit function of y with respect to the other variables. We cannot have ∂f /∂y = 0, because then f would not depend on y in this region. In principle, we can write y = y(x 1 , x 2 ,... , xN ), or y = y(xn), n = 1,... , N, (1.3)

if ∂f /∂y 6 = 0. The derivative ∂y/∂xn can be determined from f = 0 without explicitly solving for y. First, from the definition of the total derivative, we have

df =

∂f ∂x 1

dx 1 +

∂f ∂x 2

dx 2 +... +

∂f ∂xn

dxn +... +

∂f ∂xN

dxN +

∂f ∂y

dy = 0. (1.4)

Differentiating with respect to xn while holding all the other xm, m 6 = n, constant, we get

∂f ∂xn

∂f ∂y

∂y ∂xn

14 CHAPTER 1. MULTI-VARIABLE CALCULUS

so that ∂y ∂xn

∂f ∂xn ∂f ∂y

which can be found if ∂f /∂y 6 = 0. That is to say, y can be considered a function of xn if ∂f /∂y 6 = 0. Let us now consider the equations

f (x, y, u, v) = 0 , (1.7) g(x, y, u, v) = 0. (1.8)

Under certain circumstances, we can unravel Eqs. (1.7-1.8), either algebraically or numeri- cally, to form u = u(x, y), v = v(x, y). The conditions for the existence of such a functional dependency can be found by differentiation of the original equations; for example, differen- tiating Eq. (1.7) gives

df =

∂f ∂x

dx +

∂f ∂y

dy +

∂f ∂u

du +

∂f ∂v

dv = 0. (1.9)

Holding y constant and dividing by dx, we get

∂f ∂x

∂f ∂u

∂u ∂x

∂f ∂v

∂v ∂x

Operating on Eq. (1.8) in the same manner, we get

∂g ∂x

∂g ∂u

∂u ∂x

∂g ∂v

∂v ∂x

Similarly, holding x constant and dividing by dy, we get

∂f ∂y

∂f ∂u

∂u ∂y

∂f ∂v

∂v ∂y

∂g ∂y

∂g ∂u

∂u ∂y

∂g ∂v

∂v ∂y

Equations (1.10,1.11) can be solved for ∂u/∂x and ∂v/∂x, and Eqs. (1.12,1.13) can be solved for ∂u/∂y and ∂v/∂y by using the well known Cramer’s^1 rule; see Eq. (8.93). To solve for ∂u/∂x and ∂v/∂x, we first write Eqs. (1.10,1.11) in matrix form:

( (^) ∂f ∂u

∂f ∂g^ ∂v ∂u

∂g ∂v

) ( (^) ∂u ∂x∂v ∂x

−∂f ∂x − ∂g ∂x

(^1) Gabriel Cramer, 1704-1752, well-traveled Swiss-born mathematician who did enunciate his well known rule, but was not the first to do so.

16 CHAPTER 1. MULTI-VARIABLE CALCULUS

Using the formula from Eq. (1.15) to solve for the desired derivative, we get

∂u ∂x =

∣∣ ∣∣^ −^

∂f ∂x

∂f ∂v − ∂g ∂x ∂v∂g

∣∣ ∣∣ ∣∣ ∣∣

∂f ∂u

∂f ∂g^ ∂v ∂u

∂g ∂v

∣∣ ∣∣

. (1.23)

Substituting, we get

∂u ∂x =

∣∣ ∣∣ −^1 −y u

∣∣ ∣∣ ∣∣ ∣∣ 6 u^5 + 1^1 v u

∣∣ ∣∣

= (^) u(6uy (^5) + 1)−^ u − v. (1.24)

Note when v = 6u^6 + u, (1.25) that the relevant Jacobian determinant is zero; at such points we can determine neither ∂u/∂x nor ∂u/∂y; thus, for such points we cannot form u(x, y). At points where the relevant Jacobian determinant ∂(f, g)/∂(u, v) 6 = 0 (which includes nearly all of the (x, y) plane), given a local value of (x, y), we can use algebra to find a corresponding u and v, which may be multivalued, and use the formula developed to find the local value of the partial derivative.

1.2 Functional dependence

Let u = u(x, y) and v = v(x, y). If we can write u = g(v) or v = h(u), then u and v are said to be functionally dependent. If functional dependence between u and v exists, then we can consider f (u, v) = 0. So,

∂f ∂u

∂u ∂x

∂f ∂v

∂v ∂x

∂f ∂u

∂u ∂y

∂f ∂v

∂v ∂y

In matrix form, this is ( (^) ∂u ∂x

∂v ∂u^ ∂x ∂y

∂v ∂y

) ( (^) ∂f ∂u∂f ∂v

Since the right hand side is zero, and we desire a non-trivial solution, the determinant of the coefficient matrix must be zero for functional dependency, i.e. ∣∣ ∣ ∣

∂u ∂x

∂v ∂u^ ∂x ∂y

∂v ∂y

∣ = 0.^ (1.29)

1.2. FUNCTIONAL DEPENDENCE 17

Note, since det J = det JT^ , that this is equivalent to

J =

∂u ∂x

∂u ∂v^ ∂y ∂x

∂v ∂y

∣ =^

∂(u, v) ∂(x, y)

That is, the Jacobian determinant J must be zero for functional dependence.

Example 1. Determine if u = y + z, (1.31) v = x + 2z^2 , (1.32) w = x − 4 yz − 2 y^2 , (1.33) are functionally dependent. The determinant of the resulting coefficient matrix, by extension to three functions of three vari- ables, is

∂(u, v, w) ∂(x, y, z) =

∣∣ ∣∣ ∣∣ ∣

∂u ∂x ∂u ∂y ∂u ∂z ∂v ∂x ∂v ∂y ∂v ∂z ∂w ∂x

∂w ∂y

∂w ∂z

∣∣ ∣∣ ∣∣ ∣

=

∣∣ ∣∣ ∣∣

∂u ∂x

∂v ∂x

∂w ∂u^ ∂x ∂u^ ∂y ∂v^ ∂y ∂w^ ∂y ∂z

∂v ∂z

∂w ∂z

∣∣ ∣∣ ∣∣ ,^ (1.34)

=

∣∣ ∣∣ ∣∣

0 1 1 1 0 −4(y + z) 1 4 z − 4 y

∣∣ ∣∣ ∣∣ ,^ (1.35)

= (−1)(− 4 y − (−4)(y + z)) + (1)(4z), (1.36) = 4 y − 4 y − 4 z + 4z, (1.37) = 0. (1.38) So, u, v, w are functionally dependent. In fact w = v − 2 u^2.

Example 1. Let x + y + z = 0 , (1.39) x^2 + y^2 + z^2 + 2xz = 1. (1.40) Can x and y be considered as functions of z?

If x = x(z) and y = y(z), then dx/dz and dy/dz must exist. If we take f (x, y, z) = x + y + z = 0, (1.41) g(x, y, z) = x^2 + y^2 + z^2 + 2xz − 1 = 0, (1.42) df = ∂f ∂z dz + ∂f ∂x dx + ∂f ∂y dy = 0 , (1.43)

1.3. COORDINATE TRANSFORMATIONS 19

• 0 1

x

-0.

0

y

-0.

0

1

z

• 0 1

x

-0.

0

y

-0.

0

1

-1 -0. (^0) 0. x 1

0

1

2

y

-0.

0

1

z

-1 -0. (^0) 0. x 1

0

1

2

y

-0.

0

1

Figure 1.1: Surfaces of x + y + z = 0 and x^2 + y^2 + z^2 + 2xz = 1, and their loci of intersection.

then the solution matrix (dx/dz, dy/dz)T^ is found as before:

dx dz =

∣∣ ∣∣^ −^

∂f ∂z

∂f ∂y − ∂g ∂z ∂g ∂y

∣∣ ∣∣ ∣∣ ∣∣

∂f ∂x

∂f ∂g^ ∂y ∂x

∂g ∂y

∣∣ ∣∣

=

∣∣ ∣∣ −^1 −(2z + 2x) 2 y

∣∣ ∣∣ ∣∣ ∣∣ 5 1 2 x + 2z 2 y

∣∣ ∣∣

= − 102 yy −+ 2 2 xz + 2− 2 xz , (1.56)

dy dz =

∣∣ ∣∣

∂f ∂x −^

∂f ∂g^ ∂z ∂x −^ ∂g ∂z

∣∣ ∣∣ ∣∣ ∣∣

∂f ∂x

∂f ∂g^ ∂y ∂x

∂g ∂y

∣∣ ∣∣

=

∣∣ ∣∣ 5 −^1 2 x + 2z −(2z + 2x)

∣∣ ∣∣ ∣∣ ∣∣ 5 1 2 x + 2z 2 y

∣∣ ∣∣

= 10 −y 8 −x 2 −x^8 −z 2 z. (1.57)

The two original functions and their loci of intersection are plotted in Fig. 1.2. Straightforward algebra in this case shows that an explicit dependency exists:

x(z) = −^6 z^ ±

√ 2

√ 13 − 8 z^2 26 ,^ (1.58) y(z) = −^4 z^ ∓^5

√ 2

√ 13 − 8 z^2 26.^ (1.59) These curves represent the projection of the curve of intersection on the x, z and y, z planes, respectively. In both cases, the projections are ellipses.

1.3 Coordinate transformations

Many problems are formulated in three-dimensional Cartesian^3 space. However, many of these problems, especially those involving curved geometrical bodies, are more efficiently

(^3) Ren´e Descartes, 1596-1650, French mathematician and philosopher.

20 CHAPTER 1. MULTI-VARIABLE CALCULUS

-0.2 (^0) 0.

-0.

0

1

0

1

z

0

x

-0.

0

1 y

0

1

• -0. 0 0. 1

x

0

1

2

y

-0.

0

1

z

• -0. 0 0. 1

x

0

1

2

y

-0.

0

1

z

Figure 1.2: Surfaces of 5x+y +z = 0 and x^2 +y^2 +z^2 +2xz = 1, and their loci of intersection.

posed in a non-Cartesian, curvilinear coordinate system. To facilitate analysis involving such geometries, one needs techniques to transform from one coordinate system to another. For this section, we will utilize an index notation, introduced by Einstein.^4 We will take untransformed Cartesian coordinates to be represented by (ξ^1 , ξ^2 , ξ^3 ). Here the superscript is an index and does not represent a power of ξ. We will denote this point by ξi, where i = 1, 2 , 3. Because the space is Cartesian, we have the usual Euclidean^5 distance from Pythagoras’^6 theorem for a differential arc length ds:

(ds)^2 =

dξ^1

dξ^2

dξ^3

(ds)^2 =

∑^3

i=

dξidξi^ ≡ dξidξi. (1.61)

Here we have adopted Einstein’s summation convention that when an index appears twice, a summation from 1 to 3 is understood. Though it makes little difference here, to strictly adhere to the conventions of the Einstein notation, which require a balance of sub- and superscripts, we should more formally take

(ds)^2 = dξj^ δjidξi^ = dξidξi, (1.62) (^4) Albert Einstein, 1879-1955, German/American physicist and mathematician. (^5) Euclid of Alexandria, ∼ 325 B.C.-∼ 265 B.C., Greek geometer. (^6) Pythagoras of Samos, c. 570-c. 490 BC, Ionian Greek mathematician, philosopher, and mystic to whom this theorem is traditionally credited.