Applied Mathematics and Numerical Methodswitek/msmn/handouts.pdfOne identifies tensor σ with...

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Applied Mathematics and Numerical Methods lecture notes by Witold Cecot Contents 1 Motivation 2 2 Modeling of physical quantities 2 2.1 Scalars ............................................ 2 2.2 Vectors ............................................ 2 2.3 Tensors ........................................... 3 2.4 Examples .......................................... 3 3 Review of matrix algebra 4 4 Algebraic eigenproblems 6 5 Solution of systems of linear algebraic equations 7 6 Nonlinear equations 8 7 Approximation 10 8 Errors in numerical computing 12 9 Numerical integration 12 10 Numerical differentiation 15 11 Some examples of differential equations 17 12 Finite difference method (FDM) 18 13 Initial value problems 19 14 Initial boundary value problems 20 15 Galerkin’s and Ritz methods 22 16 Introduction to Optimization 24 17 Elementary concepts of statistics 27 1

Transcript of Applied Mathematics and Numerical Methodswitek/msmn/handouts.pdfOne identifies tensor σ with...

Page 1: Applied Mathematics and Numerical Methodswitek/msmn/handouts.pdfOne identifies tensor σ with matrix σ, that in turn obeys tensor transformation rule σ′ = QσQT Recommended web

Applied Mathematics and Numerical Methods

lecture notes by Witold Cecot

Contents

1 Motivation 2

2 Modeling of physical quantities 22.1 Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Review of matrix algebra 4

4 Algebraic eigenproblems 6

5 Solution of systems of linear algebraic equations 7

6 Nonlinear equations 8

7 Approximation 10

8 Errors in numerical computing 12

9 Numerical integration 12

10 Numerical differentiation 15

11 Some examples of differential equations 17

12 Finite difference method (FDM) 18

13 Initial value problems 19

14 Initial boundary value problems 20

15 Galerkin’s and Ritz methods 22

16 Introduction to Optimization 24

17 Elementary concepts of statistics 27

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1 Motivation

Design of new products demands modeling of materials, processes and phenomena by formulationand solution of appropriate equations. Mathematics answers the question whether a solution tothe problem formulated exists while numerical methods are tools for efficient approximation of thatsolution by transforming a calculus problem to an algebraic one (discretization). At the end of suchan analysis accuracy of numerical results should be assessed with the help of mathematical theorems.

2 Modeling of physical quantities

2.1 Scalars

• Only value (mass, time, density, energy, ...)

2.2 Vectors

• value, direction and point of application (displacement, velocity, force, ...)

• operations: summation (subtraction), dot product (a · b = ||a|| ||b|| cos(ϕ)), vector product(||a× b|| = ||a|| ||b|| sin(ϕ) & the right handed screw rule)

• expansion of a vector in a basis→ coordinates→ identification of a matrix of coordinates withthe vector (not unique)

• change of 2D Cartesian basis by rotating the axes counter-clockwise by an angle θ

[

x′

1

x′

2

]

=

[

q11 q12

q21 q22

]

[

x1

x2

]

or briefly x′

= Qx, Q =

[

cos(θ) sin(θ)

− sin(θ) cos(θ)

]

where x′

,x denote new and old coordinates, Q is a transformation matrix

Q−1 = QT (orthonormal matrix)

• a · b = a1b1 + a2b2 + a3b3

• a× b =

i j k

a1 a2 a3b1 b2 b3

(expansion w.r.t. the first row)

• vector spaces - (V,R,+, ·)

1. (i) u+ v = v + u

2. (ii)(u+ v) +w = v + (u+w)

3. (iii) ∃0 : u+ 0¯= v

4. (iv) ∀v∃ − v : v + (−v) = 0

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5. (v) α(u+ v) = αu+ αv

6. (vi) 1v = v

7. (vii) (α+ β)v = αv + βv

8. (viii)(αβ)v = α(βv)

2.3 Tensors

• Definition of a second order tensor: a linear mapping of vectors to vectors (Rn → Rn)

• A mapping τ is linear if τ(αv1 + βv2) = ατ(v1) + βτ(v2) ∀α, β ∈ IR , v1, v2 ∈ V

• Examples: multiplication by a number, orthogonal projection, stress tensor

• Non-example: addition of a specified non-zero vector

• Every tensor has a corresponding, representing it matrix T . Once a coordinate system is in-troduced we may identify a tensor with its matrix.

• Transformation under change of basis: T′

= QTQT

2.4 Examples

• Example 1. Orthogonal projection onto a horizontal line: IR 2 ∋ v → p = Π(v) ∈ IR 2

x

y

v

p

[

px

py

]

=

[

1 0

0 0

][

vx

vy

]

Figure 1: graphical representation matrix representation

• Example 2. Stress tensor: a mapping σ that relates density of intermolecular forces [Pa], i.e.the stress vector1 t acting on an internal plane, to the unit normal vector n that defines theplane, thus t = σ(n)

stress vector t - internal, shape restoring, distributed (per unit area) force (depends onposition and cross section direction)

unit normal vector n - unit norm, orthogonal, outward pointingstress tensor - determines stress vectors for all cross sections

1Whenever loading is applied to a body the configuration of the body changes and the intermolecular distances vary.As a result extra internal intermolecular forces appear and try to restore the original configuration. These additionalintermolecular forces may be averaged (smeared, for modeling of bodies as continuum, i.e. assuming continuumdistribution of matter) and stress vector is defined as the limit of fractions t = restoring force

area[Pa] as area→ 0.

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x

y

n

t [

tx

ty

]

=

[

σxx σxy

σyx σyy

][

nx

ny

]

Figure 2: graphical representation matrix representation

Typically the matrix that represents the stress tensor σ is symmetric and denoted in 3D by

σ =

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

One identifies tensor σ with matrix σ, that in turn obeys tensor transformation ruleσ

= QσQT

Recommended web page:http://freevideolectures.com/Course/2381/Symmetry-Structure-and-Tensor-Properties-of-Material/28

3 Review of matrix algebra

• Two dimensional matrix - a table of elements of a certain set or more precisely a mapping ofa Cartesian product 1, 2, ..., m × 1, 2, ..., n, m, n ∈ IN to a certain set.Examples:

[

i j k

1 2 3

]

∂∂x

0

0 ∂∂y

∂∂y

∂∂x

[

A B

B C

]

where i, j,k − vectors, A,B,C −matrices

One column and one row matrices are often called vectors

Typically we deal with matrices populated with real numbers

• Special types of matrices: square (An×n), diagonal (D), identity (I), upper(lower)-triangular(U ,L), symmetric, skew-symmetric, sparse, band, orthogonal (Q)

• Operations on number valued matrices: addition (subtraction), multiplication, exponentiation,transposition, inversion; associative, commutative and noncommutative operations, distribu-tivity

• Solution of a system of linear algebraic equations (Ax = b) is one of the most frequently utilizedtasks of algebra. Since computational complexity of the well known Crammer’s formulas is of

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order Cn! (200 years of computation for 20 unknowns with the speed of 1Gflops) we need othermethods.

• Determinant is such a number, associated with every square matrix, that the following condi-tions hold

(i) det(I)=1

(ii) exchange of rows reverses sign of the determinant

(iii) it is linear with respect to each row

• The definition mentioned above implies the following three properties that in turn are the basisof the Gauss elimination - a method which enables solution of a system of e.g. 20 equations inmilliseconds.

– if two rows of A are proportional, then det(A) = 0

– if ri denotes i− th row of a matrix and α ∈ IR , then the following operation ri ← ri+αrj

does not change the determinant.

– for an upper-triangular matrix det(U ) = u11 · u22 · . . . · unn

• Certain useful properties (specify matrices for which these statements hold)

(A+B)T = AT +BT (AB)T = BTAT det(AT ) = det(A)

(A+B)−1 6= A−1 +B−1 (AB)−1 = B−1A−1 det(A−1) = 1

det(A)

det(A+B) 6= det(A) + det(B) det(AB) = det(B)det(A) (A−1)T = (AT )−1 = A−T

AT +A - is symmetric ATA - is symmetric

• Quadratic form (positive definiteness)

F (x) = xTAx

∂F

∂x= (AT +A)x

• Norms of matrices

– vectors (column or row matrices):||x||1 = |x1|+ |x2|+ . . .+ |xn|,

||x||2 =√

x21 + x2

2 + . . .+ x2n,

||x||∞ = max|x1|, |x2|, . . . , |xn|– matrices (in general): ||Apq|| = supx 6=0

||Ax||p||x||q

– matrices (in particular): ||A||2 =√

ρ(ATA)

Recommended web page:http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

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4 Algebraic eigenproblems

• Definition

A nonzero vector x such thatAx = λx (1)

is called an eigenvector of matrix A and the multiplier λ (complex in general) is called theeigenvalue. In other words, multiplication of a matrix by its eigenvector produces a vector inthe same direction whenever the corresponding λ is real.

• Applications in mechanics:

- tensor principal directions and valuescharacteristic equation in 2D: λ2 − tr(A)λ+ det(A) = 0characteristic equation in 3D: λ3 − IAλ

2 + IIAλ− IIIA = 0

- resonant (eigen) frequencies and modesgeneralized eigen problem: Kx = λMx

- critical buckling loading

• Lemmas:

- sp(AT ) = sp(A), sp(P−1AP ) = sp(A)

- A = SΛS−1, where columns of S are eigenvectors of A and diagonal matrix Λ consistsof its eigenvalues (remark: S must be invertible)

- A real, symmetric matrix (e.g. representing a STRESS tensor) has real eigenvalues andorthogonal eigenvectors for distinct eigenvalues

- A real matrix is positive definite iff its symmetric part has all positive eigenvalues

- if sp(A) = λ1, λ2, . . . , λn then sp(A−1) = 1/λ1, 1/λ2, . . . 1/λn- if sp(A) = λ1, λ2, . . . λn then sp(A− dI) = λ1 − d, λ2 − d, . . . , λn − d

- λ1 + λ2 + . . .+ λn = tr(A)

- λ1λ2 . . . λn = det(A)

• Numerical analysis:

- Gershgorin’s theorem: sp(A) ⊂ Uni=1λ ∈ C : |λ− aii| 6 Σj 6=i|aij|

- Power method: iterative procedure for estimation of the dominant eigenvalue and corre-sponding eigenvector. One may use lemmas for inverse and shifted matrices to computean eigenvalue that is closest to a given number.

- Remark: the power method is convergent for matrices with real, distinct eigenvalues

- Other methods: Lanczos (improved PM), Jacobi (all eigen modes for real symmetric ma-trices) and QR algorithm (symmetry not required) that are based on the Givens rotations,...

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• A flow chart of the power method

- Set initial eigenvector x, maximum acceptable: error ε and number of iterations N

- FOR counter k = 1 to N repeat

= normalize vector v = x||x||

= update vector x = Av

= update Rayleigh quotient L = vTx

= compute error estimate e = ||x− Lv||/||x||= IF (e < ε) THEN break loop

CONTINUE LOOP

- Print k, e, L, v = x/||x||

• There is a MATLAB demo (eigshow), ilustrating the eigenvalue problem for a 2 by 2 matrix.

5 Solution of systems of linear algebraic equations

• Example

x − 2y = 0x + y = 1.5

or

[

1 −21 1

] [

xy

]

= x

[

11

]

+ y

[

−21

]

=

[

01.5

]

(2)

b

c2 c1

Figure 3: Row and column pictures of the SLAE

• Numerical methods:

1. direct (Gauss elimination, LU or Cholesky decomposition)

2. iterative (Jacobi, Gauss-Seidel, multigrid)

3. minimization (conjugate gradient, GMRES)

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• Gaussian elimination may be used to compute:

- solution of SLAE

- determinant of a matrix

- rank of a matrix

- inverse matrix

- U factor of the LU decomposition

• Computational complexity of LU decomposition is the same as of Gauss elimination for oner.h.s but may be smaller for multiple r.h.s.

• iterative methods

Ax = b ⇔ Px = Px+ b−Ax P - preconditioner

x(k+1) = x(k) + P−1(b−Ax(k)) - general iteration scheme

A = L +D +U

P = D - Jacobi method

P = L+D - Gauss-Seidel method

error estimation:||b−Ax(k)||||b||

||x(k+1) − x(k)||||x(k+1)||

• Overdetermined systems: x∗ = arg minx||Ax− b||2 → ATAx∗ = ATb

Recommended web page:http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/

6 Nonlinear equations

• Given a continuous function f(x), we seek value x∗ such that

f(x∗) = 0 (3)

The problem is known as equation root finding or function zeros finding.

• Any number of solutions is possible (0,1,2,...,∞).

• Fixed point method (simple iteration) is used for the formulation called fixed point problem:find x∗ such that

x∗ = g(x∗) (4)

Such a form suggests the following iteration scheme:

Assume x0 and iteratively compute xi = g(xi−1) for i = 1, ..., N (5)

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• Bisection method is used for the equation in form (3). Begin with initial bracket, i.e. interval[a, b] for which sign of f differs at endpoints and repeatedly halve its length until solution hasbeen isolated as accurately as desired.

• Newton’s method is also used for the equation in form (3). Differentiability of f is required.Assume x0 and compute

xi+1 = xi + h, h = −fif

i

, for i = 1, ..., N (6)

• Newton’s method for a system of n nonlinear equations

F =

f1(x1, x2, ..., xn)f2(x1, x2, ..., xn)

...fn(x1, x2, ..., xn)

= 0 (7)

Assume x0 and compute for i = 1, ..., N

xi+1 = xi + h, J ih = −F i, J =

∂f1∂x1

∂f1∂x2

. . . ∂f1∂xn

∂f2∂x1

∂f2∂x2

. . . ∂f2∂xn

...∂fn∂x1

∂fn∂x2

. . . ∂fn∂xn

(8)

• Convergence rateDefinition If a sequence x1, x2, ... converges to a value r and if there exist real numbers λ > 0and α > 1 such that

limn→λ

||(xn+1 − r)||||(xn − r)||α = λ (9)

then we say that α is the rate of convergence of the sequence.α = 1 - linear convergence1 < α < 2 - superlinear convergenceα = 2 - quadratic convergence.

Estimation of α for n large enough and the relative error estimate en = ||xn−xn−1||||xn||

en(en−1)α

≈ en+1

(en)α⇒ α ≈ log en+1 − log en

log en − log en−1(10)

• Only the bisection method is always convergent. The asymptotic rates of convergence are 1for the bisection and fixed point methods and 2 for the Newton-Rapson method (the fastestconvergence).

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1 2 3 4 5 6 7 8 9 1010

−20

10−15

10−10

10−5

100

105

Figure 4: Convergence for the fixed point and Newton methods. Error in logarithmic scale versusiteration number.

7 Approximation

1. Polynomial interpolation

• FormulationFor a given data set of n coplanar points

(x1, y1), (x2, y2), ..., (xn, yn), x1 < x2 < ...xn (11)

determine a function f : R→ R such that

f(xk) = yk, ∀k = 1, 2, ...n (12)

• Lagrange formulaDefine the Lagrange basis functions being the following polynomials of order n− 1

l(n−1)1 (x) = (x−x2)(x−x3)·...·(x−xn)

(x1−x2)(x1−x3)·...·(x1−xn)

l(n−1)2 (x) = (x−x1)(x−x3)·...·(x−xn)

(x2−x1)(x2−x3)·...·(x2−xn)

...

l(n−1)n (x) = (x−x1)(x−x2)·...·(x−xn−1)

(xn−x1)(xn−xn)·...·(xn−x(n−1)

(13)

and use the above polynomials to construct the polynomial interpolant of order n − 1 inthe following form

L(n−1)(x) = y1l(n−1)1 (x) + y2l

(n−1)2 (x) + ... + ynl

(n−1)n (x) (14)

• Hermite polynomials are used when not only values but also derivatives are specified atthe nodes

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2. Least squares approximation (best approximation, curve fitting, linear regression)

• FormulationFor a given data set of m points

(x1, y1), (x2, y2), ..., (xm, ym), x1 < x2 < ...xm (15)

and n (n < m) basis functions ϕ1(x), ϕ2(x), ..., ϕn(x) determine coefficients of the followinglinear combination

f(x) =n

k

αkϕk(x) (16)

to minimize the expressionm∑

k

[f(xk)− yk]2 (17)

• SolutionFormulate an overdetermined system of linear equations

Aa = b (18)

where

A =

ϕ1(x1) ϕ2(x1) ... ϕn(x1)ϕ1(x2) ϕ2(x2) ... ϕn(x2)...ϕ1(xm) ϕ2(xm) ... ϕn(xm)

, b =

y1y2...ym

(19)

and find a∗ such that Aa∗ is a projection of b onto the column space of A.

y →b

Aa∗ → f

r

column space of A

Figure 5: Graphical interpretation of the least squares method

In other words one minimizes ||r||2 = ||b−Aa∗||2 by solution of the following (”square”)system of linear equations

ATAa∗ = ATb (20)

Remarks:

- Eq. (20) is a necessary condition for minimum at a∗ in general, but it is also asufficient condition in this particular case.

- If columns of matrix A are linearly independent then ATA is nonsingular.

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8 Errors in numerical computing

1. Steps of modeling

• Real object, phenomenon or process

• Mathematical model

• Numerical model

• Results of computing

2. Error types

• In mathematical modeling

- disregard of certain phenomena

- simplification of data (material properties, geometry,...)

• In numerical modeling

- approximation

- truncation

• In computing

- round-off

- algorithmic

3. Related topics

• Error vs. residuum

• Well and ill posed problems

• Well and ill conditioned problems

• Stable and unstable algorithms

• Stability + approximability → convergence

9 Numerical integration

To derive formulas for numerical (approximate) integration one can use polynomial interpolationof the integrated function. Thus, given a continuous function f(x) on an interval [a, b] such that∫ b

af(x)dx is a proper integral (has neither infinite limit nor the integrand approaches infinity at any

point in the range of integration), we can choose a set of nodes x0, x1, ..., xn ∈ [a, b] to evaluate apolynomial interpolant p. Integration of this polynomial results in the following quadrature

∫ b

a

f(x)dx ≈∫ b

a

p(x)dx =n

i=1

wif(xi) = I (21)

where wi are known coefficients called weights.

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1. Newton Cotes quadratures in the composite versions

For n equally distributed nodes, such that

a = x0 < x1 < ... < xn = b, xi − xi−1 = h ∀i = 2, 3, ..., n (22)

• left, right and midpoint rectangle rules (Riemann sums)

I1 = h(f0 + f1 + ... + fn−1)

I2 = h(f1 + f2 + ... + fn)

I3 = h[f(x0+x1

2) + f(x1+x2

2) + ... + f(xn−1+xn

2)]

(23)

• trapezoid rule

I4 =h2(f0 + 2f1 + 2f2 + ... + 2fn−1 + fn) (24)

• Simpson’s rule - for ODD number of nodes (even n)

I5 =h3(f0 + 4f1 + 2f2 + 4f3 + ... + 4fn−2 + 2fn−1 + fn) (25)

Accuracy

quadrature number of points highest order of polynomialin one subinterval integrated exactly

I1, I2 1 0

I3 1 1

I4 2 1

I5 3 3

2. Gauss-Legendre quadratures

• If we consider any positions of the nodes we can find quadratures that for a given numberof nodes are exact for as high order of polynomial as possible. Eq. assuming 2 nodeswe have 4 parameters to be determined, thus the exact result for 3rd order polynomialis possible. Such rules are called Gaussian quadratures and they are tabulated on thecanonical interval [−1, 1]. The nodes are zeros of the Legendre polynomials of appropriateorders.

• Certain Gaussian quadratures on interval [−1, 1]

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number of integration coordinates weights highest order of exactlypoints integrated polynomial

1 0 2 1

2 − 1√3, 1√

31, 1 3

3 −√

35, 0,

35

59, 8

9, 59

5

• For arbitrary interval [a, b] use change of variable [−1, 1] ∋ t→ x ∈ [a, b] in the form

x =b− a

2t+

b+ a

2(26)

for whichdx

dt=

b− a

2(27)

• Example: using two-point Gauss quadrature one can compute

I =

∫ 1

−1

(x3 +1

4x2 − 3

8x)dx = f1 + f2 =

1

6(28)

where f1, f2 are the two values of the integrand shown in Fig. 6 and 16is exactly equal to

the difference of the red and blue areas, i.e. the value of the integral 28.

Figure 6: Illustration of the Gauss quadrature. Sum of two function values is equal to the differenceof the red and blue areas.

3. Adaptive integration

• Choose n subintervals and integrate

• Divide each subinterval into 2 smaller ones and repeat integration

• Compare results

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• Wherever the results differ by more than assumed tolerance continue the subdivisionprocess

Instead of breaking subintervals one may increase order of quadrature.

10 Numerical differentiation

Given a differentiable function f(x) on an interval [a, b], we can choose a set of nodes x0, x1, ..., xn ∈[a, b] (called a finite difference stencil) to evaluate a polynomial interpolant p of f . Differentiation ofpolynomial p results in the following finite difference formula

dkf

dxk(xk) ≈

dkp

dxk(xi) =

n∑

i=1

wif(xi) (29)

where wi are known coefficients called weights.

1. Derivation of finite difference formulas using the Lagrange interpolationLet’s have 3 nodes

x−1 < x0 < x1, x−1, x0, x1 ∈ [a, b] (30)

Then, the Lagrange interpolant is the following 2nd order polynomial

p(x) = L2(x) = f−1(x− x0)(x− x1)

(x−1 − x0)(x−1 − x1)+ f0

(x− x−1)(x− x1)

(x0 − x−1)(x0 − x1)+ f1

(x− x−1)(x− x0)

(x1 − x−1)(x1 − x0)(31)

therefore,

dp

dx= f−1

x− x0 + x− x1

(x−1 − x0)(x−1 − x1)+ f0

x− x−1 + x− x1

(x0 − x−1)(x0 − x1)+ f1

x− x−1 + x− x0

(x1 − x−1)(x1 − x0)(32)

d2p

dx2= f−1

2

(x−1 − x0)(x−1 − x1)+ f0

2

(x0 − x−1)(x0 − x1)+ f1

2

(x−1 − x−1)(x−1 − x0)(33)

and for a difference stencil consisting of 3 equally spaced nodes (x0 − x−1 = x1 − x0 = h) wehave the following central difference formulas

df

dx(x0) ≈ −f−1

2h+

f12h

=1

h

[

−12

01

2

]

F 0 (34)

d2f

dx2(x0) ≈

f−1

h2− 2f0

h2+

f1h2

=1

h2[1 − 2 1]F 0 (35)

where

F 0 =

f−1

f0f1

(36)

The above derived finite difference formulas are exact for polynomials of orders up to 2.

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x−1 x0 x1

f−1

f0

f1

Figure 7: slope of red line - f ′(x0), slope of green line - central (second order) difference formula,slopes of blue lines - backward and forward (first order) difference formulas

2. First order derivative - graphical interpretation of exact and approximate values

3. Derivation based on the Taylor’s formula (for sufficiently regular functions)

∃ξ ∈ [x0, x0 + h] : f(x0 + h) = f(x0) + hf ′(x0) +1

2h2f ′′

0 +1

6h3f ′′′(ξ) (37)

Neglecting the last term (with unknown ξ)

f−1 ≈ f0 − hf ′0 + 1

2h2f ′′

0

f1 ≈ f0 + hf ′0 + 1

2h2f ′′

0

(38)

Solution of system (38) with respect to f ′0, f

′′0 gives central, finite difference formulas (34), (35).

4. Differentiation of noisy data

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.5 1 1.5 2 2.5 3 3.5−3

−2

−1

0

1

2

3

Figure 8: Direct differentiation of noisy data results in a ”disaster” (blue line in the right hand sidefigure). Differentiation of a polynomial least squares fit gives reasonable result (cyan line).

16

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11 Some examples of differential equations

1. 2nd order ODE, linear IVP

motion of a point-like mass

x = x(t) ∈ C2([0, T ])mx+ cx+ kx = F (t) ∀t ∈ (0, T )x(0) = x0

x(0) = v0

2. 2nd order ODE, linear BVP

uniaxial stress state in a bar

u = u(x) ∈ C2([0, l])−AEu′′ = q(x) ∀x ∈ (0, l)u(0) = 0AEu′(l) = P

3. 2nd order ODE, nonlinear BVP

a suspended inextensible cable

y = y(x) ∈ C2([0, l])

−y′′ +√

1 + (y′)2 = 0 ∀x ∈ (0, l)y(0) = 0y(l) = 0

4. 4th order ODE, linear BVP

deflection of Euler-Bernoulli beam

u = u(x) ∈ C4([0, l])EIwIV = q(x) ∀x ∈ (0, l)w(0) = 0, w(l) = 0w′(0) = 0, EIw′′(l) = −M

5. 2nd order parabolic PDE, linear IBVP

heat transfer

u = u(x, t)

−α∂2u∂x2 +

∂u∂t

= Q(x, t) ∀x ∈ (0, l), ∀tu(x, 0) = u0(x)u(0, t) = g1(t), u(l, t) = g2(t)

6. 2nd order hyperbolic PDE, linear IBVP

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wave equation

u = u(x, t)

−β ∂2u∂x2 +

∂2u∂t2

= f(x, t) ∀x ∈ (0, l), ∀tu(x, 0) = u0(x)u(x, 0) = v0(x)u(0, t) = g1(t), u(l, t) = g2(t)

7. 2nd order elliptic PDE, linear BVP

Poisson (Laplace) equation

u = u(x, y)

−k(

∂2u∂x2 +

∂2u∂y2

)

= q(x, t) ∀x ∈ Ω

u(x, y) = g on ∂Ωu (kinematic,Dirichlet)k ∂u∂n

= t on ∂Ωt (static, Neumann)

8. System of 2nd order elliptic PDE, linear BVP

Linear elasticity (statics)

u = u(x, y, z)divσ = 0σ = Cε

ε = 12(∇u+∇Tu)

u = g on ∂Ωu (kinematic,Dirichlet)σn = t on ∂Ωt (static, Neumann)

12 Finite difference method (FDM)

1. FDM - application to 1D boundary value problemThe method converts BVP (infinite dimensional problem) into a system of algebraic equa-tions (finite dimensional problem) by satisfying approximately the differential equation only atselected points (nodes).

(a) problem formulationGiven q(x) ∈ C0([0, l]), find u = u(x) ∈ C2([0, l]), such that

−u′′ = q(x) ∀x ∈ (0, l)u(0) = 0 Dirichlet b.c.u′(l) = P Neumann b.c.

(39)

(b) discretization

• FDM meshLet’s assume l = 3/4, q = x, P = 0.1 and introduce 4 equally spaced nodes:x1 = 0, x2 = 1/4, x3 = 2/4, x4 = 3/4 (h=1/4).

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• finite difference formulas

u′′i ≈

1

h2[1, −2, 1]

ui−1

ui

ui+1

u′4 ≈

1

h[−1, 1]

[

u3

u4

]

(40)

(c) generation of algebraic equations by collocation methodx = x1 (Dirichlet b.c.) → u1 = 0x = x2 (differential equation) → −(u1 − 2u2 + u3)/h

2 = 0.25x = x3 (differential equation) → −(u2 − 2u3 + u4)/h

2 = 0.5x = x4 (Neumann b.c.) → (−u3 + u4)/h = 0.1

matrix form:

1 0 0 01 −2 1 00 1 −2 10 0 −1 1

u1

u2

u3

u4

=

0−0.015625−0.0312500.025

00.07190.12810.1531

(d.o.f)

(d) solution of sparse (tridiagonal for this particular example), nonsymmetric SLE

(e) post-processing - presentation of results, e.g of approximate solution and its derivative

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.05

0.1

0.15

0.2

0.25

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

(f) assessment of results accuracy and reliabilityMore nodes would be required to achieve acceptable accuracy in this example.

13 Initial value problems

1. Formulation

Find function y(t) : R ⊃ [t0, tF ] ∋ t→ y ∈ R

dy

dt= f(t, y) ∀t ∈ (t0, tF )

y(t0) = y0

(41)

2. Picard’s theorem enables to conclude whether a unique solution exists

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3. The graphical view (for dy

dt= t− y2 )

0 0.5 1 1.5 2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

direction field and 3 integral curves direction field and isoclines (dydt

= const)

4. Euler’s method (1st order, one-step scheme)explicit implicit

ti+1 = ti + h

yi+1 = yi + hfi fi = f(ti, yi)

ti+1 = ti + h

yi+1 = yi + hfi+1 fi+1 = f(ti+1, yi+1)

5. Runge-Kutta 2 method (modified Euler, 2nd order, one-step scheme)explicit

ti+1 = ti + hK1 = fiy = yi + hK1 predictionK2 = f(ti+1, y)

yi+1 = yi + hK1 +K2

2correction

6. Adams methods (two-step schemes are shown below)explicit Adams-Bashforth implicit Adams-Moulton

ti+1 = ti + hyi+1 = yi +

12h(3fi − fi−1)

ti+1 = ti + hyi+1 = yi +

12h(fi + fi+1)

More step schemes (multi-step) are also used.

14 Initial boundary value problems

1. Second order partial differential equations (PDE’s)

auxx + buxy + cuyy + dux + euy + fu+ g = 0 (42)

are classified, by value of discriminant ∆ = b2 − ac, as

• parabolic (∆ = 0) - time dependent, dissipative, evolving towards a steady state (e.g.heat equation)

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• hyperbolic (∆ > 0) - time dep., if conservative then do not evolve towards a steady state(e.q. wave equation)

• elliptic (∆ < 0) - time independent, steady state (e.q. Laplace equation)

For u = u(x1, x2, . . . , xn) the first part of the PDE may be written in the following form

[

∂x1

,∂

∂x2

, . . . ,∂

∂xn

]

a11 a12 . . . a1na21 a22 . . . a2n

...an1 an2 . . . ann

∂u∂x1∂u∂x2

. . .∂u∂xn

(43)

and if A, a matrix of coefficients for the highest order derivatives, is symmetric then thecorresponding PDE is

• parabolic (A singular, zero eigenvalue)

• hyperbolic (A indefinite and nonsingular, nonzero eigenvalues of both signs)

• elliptic (A either positive or negative definite, nonzero eigenvalues of the same sign)

2. 2nd order parabolic PDE, linear IBVP

heat transfer

u = u(x, t), α ∈ R+

−α∂2u∂x2 +

∂u∂t

= Q(x, t) ∀x ∈ (0, l), ∀tu(x, 0) = u0(x)u(0, t) = g1(t), u(l, t) = g2(t)

α∆t

(∆x)26

1

2

t

x

1 initial condition (at every point)

1 boundary condition (at every point)

differential equation

initial and boundary conditions

3. 2nd order hyperbolic PDE, linear IBVP

wave equation

u = u(x, t), β ∈ R+

−β ∂2u∂x2 +

∂2u∂t2

= f(x, t) ∀x ∈ (0, l), ∀tu(x, 0) = u0(x)u(x, 0) = v0(x)u(0, t) = g1(t), u(l, t) = g2(t)

21

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β(∆t)2

(∆x)26 1

t

x

2 initial conditions (at every point)

1 boundary condition (at every point)

differential equation

initial and boundary conditions

4. 2nd order elliptic PDE, linear BVP

Poisson (Laplace) equation

u = u(x, y), k ∈ R+

−k(

∂2u∂x2 +

∂2u∂y2

)

= q(x, t) ∀x ∈ Ω

u(x, y) = g on ∂Ωu (kinematic,Dirichlet)k ∂u∂n

= t on ∂Ωt (static, Neumann)

x

y

1 boundary condition (at every point)

differential equation

15 Galerkin’s and Ritz methods

1. FormulationLet’s consider 1D elliptic problem

−AEu′′ = q ∀x ∈ (0, l)u(0) = 0AEu′(l) = P

(44)

Let’s employ an arbitrary test function v(x), multiply DE by v and integrate over the interval(0, l) to obtain

−∫ l

0

vAEu′′ dx =

∫ l

0

vq dx ∀v ∈ V (45)

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Integrating the first term by parts gives

∫ l

0

v′AEu′ dx− [v(l)AEu′(l)− v(0)AEu′(0)] =

∫ l

0

vq dx ∀v ∈ V (46)

Making use of the static b.c and assuming v(0) = 0 yields variational formulationFind u(x)∈ V0 /u(0) = 0/ such that

∫ l

0

v′AEu′ dx =

∫ l

0

vq dx+ v(l)P ∀v ∈ V0 (47)

2. Approximation (discretization)

u(x) ≈ uh(x) = u0(x) + α1φ1(x) + α2φ2(x) + . . .+ αnφn(x) (48)

where:u0(x) is an arbitrary extension of the Dirichlet b.c. to the whole domain (e.g. u0 = 0 forproblem 44α1, α2, . . . , αn ∈ R are unknown dofφ1(x), φ2(x), . . . , φn(x) are assumed basis (linearly independent) functions

satisfying homogeneous (zero) kinematic b.c

3. SLE that determines dof (entries of column vector a) is obtained by assuming basis functionsto be also test functions and substituting approximation to the variational formulation

Ga = f (49)

where

Gij =

∫ l

0

φ′iAEφ′

j dx fi =

∫ l

0

φiq dx+ φi(l)P i, j = 1, 2, ..., n (50)

4. Two examples of basis functionsx, x2, x3, . . . (this sequence should be complete, i.e containing all powers up to the highest one)sin(x), sin(2x), sin(3x), . . .

5. Numerical exampleLet’s consder BVP (44) with L=0.75, AE=1, q=x, P=-0.1 and 2 basis functions (x, x2). Oneobtains the following results

G =

34

916

916

916

f =

0.0656

0.0229

⇒ a =

73320

− 316

(51)

Thus aproximation of the solution to the considered BVP reads

uh =73

320x− 3

16x2 (52)

This approximation is compared with the exact solution in Fig. 9.

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

exactGalerkin approximationdata

Figure 9: Galerkin method result

6. Minimization of a functionalProblem (47) may be formulated as minimization of a functional, i.e.Find u(x)∈ V0 /u(0) = 0/ such that

J(u) = minV0J(v) (53)

where

J(v) =1

2

∫ l

0

v′AEv′ dx−∫ l

0

vq dx− v(l)P (54)

The same approximation as in the Galerkin method leads to the Ritz approach which in factresults in the same system of linear algebraic equations (49).

7. It is worth mentioning that the weak formulation (47) is known in mechanics as the principleof virtual works and the minimized functional (54) represents the total potential energy of theconsidered mechanical system.

16 Introduction to Optimization

1. Optimization is a large subarea of applied mathematics and numerical methods. Examples ofits applications: design of structures (bridges, airplanes, cars) of minimum weight subject toconstraint on maximum stress and displacement, optimal telecommunication routing, salesmanproblem, encryption and code breaking, optimization aided financing, optimal advertising, ...Typically we can only make one thing best at a time. Eg. is there one airplane which is: thefastest, most efficient, quietest, least expensive,...?)

2. Formulation of the problem

(i) Numbers are decisive variables (discrete)

Find a vector x∗ ∈ D ⊂ Rn such that

f(x∗) 6 f(x) ∀x ∈ D, where f : Rn ⊃ D → R and D is defined by constraints

24

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gi(x∗) 6 0, i = 1, 2, ..., p, gi : R

n → R

hj(x∗) = 0, j = 1, 2, ..., q, hj : R

n → R

equivalently: f(x∗) = minx∈D f(x) orx∗ = argminx∈D f(x)

(ii) Functions are decisive variables (continuous optimization) - optimal control or trajectoryoptimization

Find a function y∗(x) ∈ V = y : Rn → Rn such that x = x(t),y = y(t) and

J(y∗) ≤ J(y) ∀y ∈ V

subject to constraints

(iii) Structural optimization (PDE are constraints): sizing, shape (topology optimization)

3. Basic notions

(a) Objective functions

- linear, nonlinear

- convex (concave), neither convex nor concaveDEF: Function f : Rn ⊃ D → R is convex iff(αx1 + βx2) ≤ αf(x1) + βf(x2) ∀α, β ∈ R,x1,x2 ∈ D

- types of minima: a - strong local, b - weak local, c - strong GLOBAL

- For convex objective function ANY local minimum is a global one

- If function f is continuous on compact (closed and bounded) set D, then f has globalminimum on D

(b) Constraints

- linear, nonlinear

- equality, inequality

4. Optimization without constraints (straight minimization)Example: Least squares

- Primal formulationGiven: Am×n, bm×1, m > n, rank(A) = n, Find: min||Au− b|| → u

- Dual formulation (constrained optimization)Find: min||b− r|| subject to ATr = 0 → rwhere r = b−Au

5. Methods for taking into account equality constraintsExample: minF (x, y); F (x, y) = (x− 3)2+ 1

2(y− 2)2 subject to eq. constraint x+ y+1 = 0

(a) elimination of variables (minF1(x); F1(x) = (x− 3)2 + 12(x+ 3)2 )

(b) Lagrange multipliers (minF2(x, y, λ); F2(x, y, λ) = (x− 3)2 + 12(y − 2)2+λ(x+ y + 1) )

25

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(c) penalty parameter (minF3(x, y); F3(x, y) = (x− 3)2 + 12(y − 2)2+1

ε(x+ y + 1)2 )

6. Kuhn-Tucker necessary (sufficient for convex problems) conditions for a solution to be optimal.

If f, gi, hj are continuously differentiable functions and x∗ is a local minimum then

• without constraints∇f(x∗) = 0

• with inequality constraintsgi(x

∗) 6 0, i = 1, 2, ..., p

∃λ1, λ2, ..., λp > 0, λi gi(x∗) = 0 ∀i = 1, 2, ..., p and

∇f(x∗) +∑p

1 λi∇gi(x∗) = 0

• with both inequality and equality constraintsgi(x

∗) 6 0, i = 1, 2, ..., p

∃λ1, λ2, ..., λp > 0, λi gi(x∗) = 0 ∀i = 1, 2, ..., p

hi(x∗) = 0, j = 1, 2, ..., q

∃µ1, µ2, ..., µq ∈ R and

∇f(x∗) +∑p

1 λi∇gi(x∗) +∑q

1 µj∇hj(x∗) = 0

7. Examples

• Convex f(x, y) = x2 + y2

1− x 6 0

The KT conditions are satisfied at x∗ = (1, 0) with λ = 2. Point (1, 0) is the minimizer.

• Concave f(x, y) = −(x2 + y2)

x− 1 6 0

The KT conditions are satisfied at x∗1 = (1, 0) with λ = 2 and at x∗

2 = (0, 0) with λ = 0.The first point is neither a local minimizer nor a local maximizer. At the second functionf attains a global maximum (not minimum). In fact the problem has no solution.

• A pointwise mass suspended on two springsk

kF

1

2

F1 = k1x, F2 = k2x constitutive relations

F1 + F2 = F equilibrium equation

x∗ = Fk1+k2

solution

– Minimize the total potential energy: J(x∗) = minx J(x)J(x) = 1

2(k1x

2 + k2x2)− Fx

– Minimize the total complementary energy: Jc(F∗1 , F

∗2 ) = minF1,F2 Jc(F1, F2)

Jc(F1, F2) =1

2

(

F 21

k1+

F 22

k2

)

subject to constraint F1 + F2 = F

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−0.2 0 0.2 0.4 0.6 0.8 1 1.2−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25J(x) Jc(F1,F2)

−1 −0.5 0 0.5 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Jc(F1)

17 Elementary concepts of statistics

Statistics deals with analysis of real experimental data which typically are random (depend onchance). Statistics is related to probability theory, i.e. a mathematical theory based on axiomsand used to model randomness of real world.

1. Probability

• basic notions

– a trial - s single performance of an experiment (e.g. a toss of two coins)

– an outcome (elementary event) - each possible result of an experiment (e.g. the pairconsisting of two heads)

– sample space S- the set of all possible outcomes (may be finite or infinite)

– event A - a subset of S

– complement of event - A = S\A• definition of probability

P (A) =k

n(55)

where: k - number of elementary events (cases) favorable to the considered event An - number of all elementary events

• experimental estimation (relative frequency) of probability

Pf(A) =kfnf

(56)

where: kf - number of observed favorable outcomes of an experimentnf - number of all experiment outcomes

2. Random variables - quantities whose values (real numbers) describe events (outcomes of ex-periments), e.g.

• if X is the random variable which represents the roll of a die then it belongs to a finite setX ∈ 1, 2, 3, 4, 5, 6 and X is called a DICRETE random variable, P (X = 1) = 1

6

• if x represents the amount of rain tomorrow then it belongs to an infinite setx[cm] ∈ [0,∞) and x is called a CONTINUOUS random variable, P (x = 1) = 0

27

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3. Statistical parameters describing discrete data

• mean (arithmetic - expected value, geometric, harmonic)

µA = E(X) = 1n

i Xi µG = n√ΠiXi µH = n∑

i1Xi

• median (”middle value” of a sorted set), mode (”most frequent” value in a set)

• standard deviation - average distance from the mean, variance

σ =√

1n

i(xi − µA)2 σ2

• skewness - meassures assymetry about the mean

s =1n

i(xi − µA)3

σ3

4. DISCRETE random variable example: X ∈ 7, 3, 11, 5, 5, 7, 9, 7, 7, 9

1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12data, mean, std

3 5 7 9 110

1

2

3

4

3 5 7 9 110

0.2

0.4

0.6

0.8

1

3 5 7 9 110

0.2

0.4

0.6

0.8

1

Sequence Frequency Probability Cumulativeof data of values (nf ) distribution probability

Pf(X = Xi) C(Xi) = P (X 6 Xi)

5. CONTINUOUS random variable - Gaussian density function

−5 0 5 10 150

0.05

0.1

0.15

0.2

x

Den

sity

−5 0 5 10 150

0.2

0.4

0.6

0.8

1

x

Cum

ulat

ive

dens

ity

ϕ(x) =1

σ√

(2π)e−

12(x−µ

σ)2

∫ b

a

ϕ dx = P (a 6 x 6 b) Φ(x) =∫ x

−∞ ϕ(t) dt

µ =

∫ ∞

−∞ϕ(x)x dx σ =

∫ ∞

−∞ϕ(x)(x− µ)2 dx

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6. Fitting a probability density function to ”experimental” data - exampleLet x ∈ 18.1, 17.5, 20.8, 24.5, 21.5, 22.5, 23.5, 21.2, 18.9 - actually, more results are needed

1 2 3 4 5 6 7 8 917

18

19

20

21

22

23

24

25

16 18 20 22 24 260

0.5

1

1.5

2

2.5

3

16 18 20 22 24 260

0.2

0.4

0.6

0.8

1

16 18 20 22 24 260

0.2

0.4

0.6

0.8

1

Sequence Frequency of Probability Cumulativeof data observations in bins density probability

29