Lecture I

Physics 501 

Math Models of Physics Problems 

Wednesday, September 4, 13 

Luis Anchordoqui 

1

Bibliography 

L. A. Anchordoqui and T. C. Paul, 

``Mathematical Models of Physics Problems’‘ 

(Nova Publishers, 2013) 

G. F. D. Duff and D. Naylor, 

``Differential Equations of Applied Mathematics’‘ 

(John Wiley & Sons, 1966) 

G. B. Arfken, H. J. Weber, and F. E. Harris 

``Mathematical Methods for Physicists’’ (7th Edition) 

(Academic Press, 2012) 


2

Provisional Course Outline 

(Please note this may be revised during the course 

to match coverage of material during lectures, etc.) 

1st week - Elements of Linear Algebra 

2nd week - Analytic Functions 

3rd week - Integration in the Complex Plane 

4th week - Isolated Singularities and Residues 

5th week - Initial Value Problem (Picard’s Theorem) 

6th week - 

Initial Value Problem (Green Matrix) 

7th week - Boundary Value Problem (Sturm-Liouville Operator) 

8th week - Midterm-exam (October 23) 

9th week - 

Boundary Value Problem (Special Functions) 

10th week - Fourier Series and Fourier Transform 

11th week - 

Hyperbolic Partial Differential Equation (Wave equation) 

12th week - Parabolic Partial Differential Equation (Diffusion equation) 

13th week - Elliptic Partial Differential Equation (Laplace equation) 

14th week - Midterm-exam (December 11) 


3

Elements of Linear Algebra 

1.1 Linear Spaces 

1.2 Matrices and Linear Transformations 


4

Definition 1.1.1. 

field F 

+ 

for which all axioms below hold 8 , µ, ⌫ 2 F : 

(i) closure + µ · µ 

A is a set together with two operations and 

sum and product again belong to F 

(ii) associative law +(µ + ⌫) =( + µ)+⌫ & · (µ · ⌫) =( · µ) · ⌫ 

(iii) commutative law + ⌫ = ⌫ + & · µ = µ · 

(iv) distributive laws · (µ + ⌫) = · µ + · ⌫ 

and ( + µ) · ⌫ = · ⌫ + µ · ⌫ 

(v) existence of an additive identity there exists an element 

0 2 F for which +0= 

(vi) 

Linear Spaces 

existence of a multiplicative identity 

1 2 F with 1 6= 0 for which 1 · = 

(vii) existence of additive inverse to every 2 F 

an additive inverse such that + =0 

there exists an element 

· 

there corresponds 

(viii) existence of multiplicative inverse to every 2 F 

1 

there corresponds a multiplicative inverse such that 1 · =1 


5

Example 1.1.1. 

Underlying every linear space 

is a field 

F 

examples are R 

and 

C 


A linear space 

V 

is a collection of objects 

with a (vector) addition and scalar multiplication defined 

which is closed under both operations 

Such a vector space satisfies following axioms: 

➢ commutative law of vector addition 

➢ associative law of vector addition 

➢ There exists a zero vector 

x + y = y + x, 8 x, y 2 V 

x +(y + w) =(x + y)+w, 8 x, y, w 2 V 

0 

such that 

x + 0 = x, 8 x 2 V 


6

➢ To every element 

x 2 V 

there corresponds an inverse element 

x 

such that 

x +( x) =0 

➢ associative law of scalar multiplication 

( µ) x = (µ x), 8 x 2 V and , µ 2 F 

➢ distributive laws of scalar multiplication 

( + µ) x = x + µ x, 8 x 2 V and , µ 2 F 

(x + y) = x + y, 8 x, y 2 V and 2 F 

➢ 

1 · x = x, 8 x 2 V 


7


Cartesian space 

R n 

is prototypical example 

of real 

n 

-dimensional vector space 

Let x =(x 1 ,...,x n ) be an ordered n tuple of real numbers x i 

to which there corresponds a point 

x 

with these Cartesian 

coordinates and a vector 

x 

with these components 

We define addition of vectors by component addition 

x + y =(x 1 + y 1 ,...,x n + y n ) 

(1.1.1.) 

and scalar multiplication by component multiplication 

x =( x 1 ,..., x n ) 

(1.1.2.) 


8


Given a vector space V over a field F 

W 

if 

W 

is vector space over F 

Corollary 1.1.1 

V 

a subset of is called subspace 

under operations already 

A subset W of a vector space V is a subspace of V , 

defined on 

(i)W is nonempty (ii) if x, y 2 W ☛ then x + y 2 W 

(iii) x 2 W and 2 F ☛ then · x 2 W 

V 

After defining notions of vector spaces and subspaces 

next step is to identify functions that can be used 

to relate one 

vector space to another 

Functions should respect algebraic structure of vector spaces 

so we require they preserve addition 

and scalar multiplication 


9


Let V and W be vector spaces over field F 

A linear transformation from V 

to 

W 

is a function T : V ! W 

such that T ( x + µy) = T (x)+µT (y) (1.1.3.) 

for all vectors x, y 2 V and all scalars , µ 2 F 

If a linear transformation is one-to-one and onto 

it is called vector space isomorphism ☛ or simply isomorphism 


S = x 1 , ··· , x n 

Let be a set of vectors in vector space over field 

Any vector of form y = 

nX 

i=1 

is called linear 

ix i for i 2 F 

combination of vectors in 

V 

S 

F 

S 

Set is said to span if each element of 


V 

can be expressed as linear combination of vectors in 

V 

S 

10


Let be given vectors 

and 

m 

1,... m an equal 

number of scalars 

x 1 ,...,x m 

1x 1 + ···+ k x k + ···+ m x m 

We can form a linear combination or sum 

(1.1.4.) 

which is also an element of the vector space 

Suppose there exist values 1 ... n which are not all zero 

such that above vector sum is the zero vector 

Then the vectors x 1 ,...,x m 

are said to be linearly dependent 

Contrarily vectors x 1 ,...,x m are called linearly independent 

if 

1x 1 + ···+ k x k + ···+ m x m = 0 

(1.1.5.) 

demands scalars 

k must all be zero 


11

Definition 1.1.7 

Dimension of 

V 


V 

☛ maximal number of linearly independent vectors of 

n 

Let be an dimensional vector space 

and 

S = x 1 ,...,x n ⇢ V 

V 

(1.1.6.) 

a linearly independent spanning set for V 

➪ 

S 

is called a basis of 

V 


Let 

S 

be a nonempty subset of vector space 

V 

➪ S is a basis for V if and only if each vector in V 

can be written uniquely as a linear combination of vectors in 


S 

12


An inner product 

(x, y) 

h , i : V ⇥ V ! F 

ordered pair of elements of 

and has following properties: 

➢ conjugate symmetry or Hermiticity 

➢ linearity in second argument 

➢ definiteness 

Corollary 1.1.2. 

is a function that takes each 

V 

to a number 

hx, yi 2F 

hx, yi =(hy, xi) ⇤ 

hx, y + wi = hx, yi + hx, wi and hx, yi = hx, yi 

hx, xi =0, x = 0 

Conjugate symmetry and linearity in second variable gives 

h x, yi =(hy, xi) ⇤ = ⇤ (hy, xi) ⇤ = ⇤ (hx, yi) 

hy + w, xi =(hx, y + wi) ⇤ =(hx, yi) ⇤ +(hx, wi) ⇤ = hy, xi + hw, xi 

Remark 1.1.1. 

In R inner product is symmetric 

whereas in 

C 

is a sesquilinear form 

(i.e. is linear in one argument and conjugate-linear in other) 


13


An inner product h , i is said to be positive definite , 

for all non-zero x in V,hx, xi 0 

A positive definite inner product is usually referred to as 


An inner product space is a vector space V 

equipped with an inner product h , i : V ⇥ V ! F 


V 

F 

over field 

Vector space on endowed with a positive definite inner 

product (a.k.a. scalar product) defines Euclidean space E 


For 

x, y 2 R n 


nX 

hx, yi = x · y = x k y k 

For x, y 2 C n hx, yi = x · y = 

k=1 

(1.1.7.) 

nX 

x ⇤ ky k (1.1.8.) 

k=1 

genuine inner product 


14 

F


C([a, b] ) 

x(t) 

Let 

denote set of continuous functions 

defined on closed interval 1


Axiom of positivity allows one to define a norm or length 

For each vector of an euclidean space 

kxk =+ p hx, xi 

In particular kxk =0, x = 0 

Further ☛ if 

2 C 

k xk = p | | 2 hx, xi = | |kxk 

This allows a normalization for any non-zero length vector 

Indeed ☛ if x 6= 0 then kxk > 0 

Thus ☛ we can take 2 Csuch that | | = kxk 1 y = x 

and 

It follows that kyk = | |kxk =1 


Length of a vector 


x 2 R n is 

kxk = 

Length of a vector x 2C 2 ([a, b]) is 

| 

|xk = 

(1.1.12.) 

(1.1.13.) 

1/2 nX 

xk! 2 (1.1.14.) 

k=1 

( Z b 

|x(t)| 2 dt 

a 

) 1/2 

(1.1.15.) 


16


In a real Euclidean space angle between vectors 

x 

and 

y 

cos cxy = 

|hx, yi| 

kxkkyk 

(1.1.16.) 


Two vectors are orthogonal x y if hx, yi =0 

Zero vector is orthogonal to every vector in E 


In a real Euclidean space 

angle between two orthogonal non-zero vectors is 

⇡/2 

⤶ 

i.e. cos cxy =0 


17

Lemma 1.1.1. 

If 

{x 1 , x 2 , ··· , x k } is a set of mutually orthogonal non-zero vectors 

Proof. 

Assume that vectors are linearly dependent 


k 

then its elements are linearly independent 

Then ☛ there exists numbers i (not all zero) such that 

1x 1 + 2 x 2 + ···+ k x k = 0 

1 6= 0 

(1.1.17.) 

Further ☛ assume that and consider scalar product 

of the linear combination (1.1.17) with vector x 1 

Since 

x i x j 

or equivalently 

for i 6= j ☛ we have 

which contradicts hypothesis 

1hx 1 , x 1 i = hx 1 , 0i 

1kx 1 k 2 =0) x 1 = 0 

If a sum of mutually orthogonal vectors is 


0 

(1.1.18.) 

(1.1.19.) 

then each vector must be 

0 

18


A basis 

x 1 ,...,x n of 

V 

is called orthogonal 

if 

hx i , x j i =0 

for all 

i 6= j 

☛ basis is called orthonormal 


kx i k =1, 8i =1,...,n 

unit length 

Simplest example of an orthonormal basis is standard basis 

e 1 = 

0 

B 

@ 

1 

0 

0 

. 

0 

0 

1 

, e 2 = 

C 

A 

if in addition each vector has 

0 

B 

@ 

0 

1 

0 

. 

0 

0 

1 

, ... e n = 

C 

A 

0 

B 

@ 

0 

0 

0 

. 

0 

1 

1 

C 

A 

(1.1.20.) 


19

Lemma 1.1.2. 

If set of vectors {x 1 , x 2 , ··· , x k } 

* + 

kX 

kX 

y, ix i = 

i=1 

i=1 

Theorem 1.1.1. (Pythagorean theorem) 

If 

then every 

y 2E 

is orthogonal to 

linear combination of this set of vectors 

ihy, x i i (1.1.22.) 

x y 2E then 

kx + yk 2 = hx + y, x + yi = kxk 2 + kyk 2 (1.1.23.) 

In any right triangle ☛ area of square whose side is hypotenuse 

(side opposite right angle) is equal to sum of areas of squares 

whose sides are two legs (two sides that meet at a right angle) 


is also orthogonal to 

If set of vectors {x 1 , x 2 , ··· , x k } are mutually orthogonal 

y 

x i x j with i 6= j then 

(1.1.24.) 


||x 1 + ···+ x k || 2 = ||x 1 || 2 + ···+ ||x k || 2 20

Corollary 1.1.5. (Triangle inequality) 

For 

x, y 2E we have 

kxk 

kyk applekx + yk applekxk + kyk 

(1.1.25.) 

Length of a side of a triangle 

does not exceed sum of lengths of other two sides 

nor is it less than absolute value of difference of other two sides 

Proof. 

Consider scalar product 

kx + yk 2 = hx + y, x + yi = kxk 2 +2


Let 

x =(x 1 ,...,x k ,...) 

be an infinite sequence of real numbers 

such that 

1X 

k=1 

x 2 k converges 

Sequence x defines a point of Hilbert coordinate space 

It also defines a vector with 

which as in 

Addition and scalar multiplication 

x 

k 

with 

k 

-th component x k 

-th coordinate 

Norm of Hilbert vector is Pythagorean expression 

! 1/2 

1X 

kxk = 

By hypothesis 

this series converges if 

k=1 

x 2 k 

R n 

R 1 

x k 

we identify with point 

are defined analogously to (1.1.1) and (1.1.2) 

x is an element of Hilbert space H = E 1 


22

Definition 1.2.0 


Linear Operators on Euclidean Spaces 

An operator A on E is a vector function A : E!E 

Operator is called linear if 

A(↵x + y) =↵Ax + Ay, 8x, y 2E and 8↵, 2 C (or R) 

A n ⇥ n x 

T (x) =Ax 

Let be an 

matrix and a vector 

➪ 

A vector 

the function 


is a linear operator 

x 6= 0 is eigenvector of A if 9 satisfying A x = x 

in such a case (A I) x = 0 ☛ I is identity matrix 

Eigenvalues are given by relation det (A I) =0 

which has 

➪ 

m 

different roots with 

1 apple m apple n 

det(A I) is a polynomial of degree n 

Eigenvectors associated with eigenvalue 

are obtained by 

solving (singular) linear system (A I) x = 0 


23

Remark 1.2.1. 

x 1 x 2 a, b constants 

ax 1 + bx 2 is an eigenvector with eigenvalue because 

If and are eigenvectors with eigenvalue and 

➪ 

A(ax 1 + bx 2 )=aAx 1 + bAx 2 = a x 1 + b x 2 = (ax 1 + bx 2 ) 

It is straightforward to show that: 

(i) eigenvectors associated to a given eigenvalue 

form a vector space 

(ii) two eigenvectors corresponding to different eigenvalues 

are lineraly independent 


A matrix is said to be diagonable (or diagonalizable) 

if the eigenvectors form a base 

i.e. if any vector v can be written as a linear combination 

of eigenvectors 

A matrix 

A 

A 

is said to be diagonable 

if there exists n eigenvectors 

that are linearly independent 

x 1 ,...,x n 


24

In such a case 

☛ 

we can form with 

k 

n eigenvectors an n ⇥ n matrix U 

U 

such that -th column of is -th eigenvector 

In this way 

☛ relations 

can be written in a matrix form 

U 

n Ax k = x k 

AU = UA 0 ☛ A 0 is a n ⇥ n 

The latter can also be written as 

U 1 AU= A 0 , or equivalently A = UA 0 U 1 , 

k 

diagonal matrix such that 

A 0 ij = 

i ij 

which bind diagonal matrix with original matrix 

( is invertible because eigenvectors are linearly independent) 


Transformation (1.2.11.) represents a change of base 

Note that eigenvalues (and therefore matrix A 0 ) 

are independent of change of base 

n ⇥ n 

if B = W 1 AW with W an arbitrary (invertible) 

➪ det (B I) =det(W 1 AW W 1 W)=det(A I) 


such that it has same eigenvalues 

(1.2.11.) 

matrix 

(1.2.12.) 

25


If real function 

f(x) 

f(x) = 

has a Taylor expansion 

1X 

f (n) (0) 

n! 

x n (1.2.13.) 

n=0 

matrix function is defined by substituting argument 

powers become matrix powers, additions become matrix sums 

and multiplications become scaling operations 

If real series converges for |x|


A complex square matrix A is Hermitian if A = A † 

where A † =(A ⇤ ) T 

Remark 1.2.2. 

It is easily seen that if 

is conjugate transpose of a complex matrix 

A 

(i) its eigenvalues are real 

A partially defined linear operator 

is Hermitian then: 

(ii) eigenvectors associated to different eingenvalues 

are orthogonal 

(iii) it has a complete set of orthogonal eigenvectors 

which makes it diagonalizable 


A 

on a Hilbert space 

is called symmetric if hAx, yi = hx,Ayi, 8 x and y in domain of A 

A symmetric everywhere defined operator is called 

self-adjoint or Hermitian 

H 

on this Hilbert space we have ☛ 

H 

C n A 

hx, Ayi = hAx, yi, 8 x, y 2 C n 

Note that if we take as Hilbert space 

with standard dot product 

and interpret a Hermitian square matrix as a linear operator 


27

Example 1.2.1 

A convenient basis for traceless Hermitian 

are Pauli matrices: 

1 = 

They obey following relations: 

(i) 

✓ 0 1 

1 0 

2 

i = I 

(ii) i j = j i 

(iii) i j = i k 

(i, j, k) 

◆ ✓ 0 i 

, 2 = 

i 0 

◆ 

, 3 = 

☛ a cyclic permutation of (1,2,3) 

These three relations can be summarized as 

2 ⇥ 2 matrices 

✓ 1 0 

0 1 

◆ 

(1.1.36.) 

☛ 

✏ ijk 

i j = I ij + i✏ ijk k (1.1.37.) 

is Levi-Civita symbol 


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29

Lecture I

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