Second Order ODE's (2P) Young Won Lim 7/1/14

Like dokumenter
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Universitetet i Bergen Det matematisk-naturvitenskapelige fakultet Eksamen i emnet Mat131 - Differensiallikningar I Onsdag 25. mai 2016, kl.

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Gradient. Masahiro Yamamoto. last update on February 29, 2012 (1) (2) (3) (4) (5)

Slope-Intercept Formula

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Mathematics 114Q Integration Practice Problems SOLUTIONS. = 1 8 (x2 +5x) 8 + C. [u = x 2 +5x] = 1 11 (3 x)11 + C. [u =3 x] = 2 (7x + 9)3/2

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

UNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Solutions #12 ( M. y 3 + cos(x) ) dx + ( sin(y) + z 2) dy + xdz = 3π 4. The surface M is parametrized by σ : [0, 1] [0, 2π] R 3 with.

Moving Objects. We need to move our objects in 3D space.

Graphs similar to strongly regular graphs

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Ma Flerdimensjonal Analyse Øving 1

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Speed Racer Theme. Theme Music: Cartoon: Charles Schultz / Jef Mallett Peanuts / Frazz. September 9, 2011 Physics 131 Prof. E. F.

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Trigonometric Substitution

Ma Linær Algebra og Geometri Øving 5

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Microsoft Dynamics C5 Version 2008 Oversigt over Microsoft Reporting Services rapporter

Løsningsforslag til utvalgte oppgaver i kapittel 3

Physical origin of the Gouy phase shift by Simin Feng, Herbert G. Winful Opt. Lett. 26, (2001)

Neural Network. Sensors Sorter

TFY4170 Fysikk 2 Justin Wells

UNIVERSITETET I OSLO

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Solution for INF3480 exam spring 2012

Unit Relational Algebra 1 1. Relational Algebra 1. Unit 3.3

SCE1106 Control Theory

TDT4117 Information Retrieval - Autumn 2014

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Eksamen i TMA4190 Mangfoldigheter Onsdag 4 juni, Tid :

Smart High-Side Power Switch BTS730

UNIVERSITETET I OSLO

Splitting the differential Riccati equation

FYSMEK1110 Eksamensverksted 23. Mai :15-18:00 Oppgave 1 (maks. 45 minutt)

0:7 0:2 0:1 0:3 0:5 0:2 0:1 0:4 0:5 P = 0:56 0:28 0:16 0:38 0:39 0:23

2018 ANNUAL SPONSORSHIP OPPORTUNITIES

AC Network Analysis. Learning objectives. Ideell Kondensator (Capasitor) Structure of parallel-plate capacitor

Eksamensoppgave i TMA4135 Matematikk 4D

(ii) x = 0or 4. (ix) x = or 3. (xii) x = or. (vi) x = 4. (xiv) x = Exercise (i) ( x 3) 8. (iii) ( 3q. (iii) 3( x + 2) 16.

Qi-Wu-Zhang model. 2D Chern insulator. León Martin. 19. November 2015

Maple Basics. K. Cooper

TMA4329 Intro til vitensk. beregn. V2017

Information search for the research protocol in IIC/IID

Finite Elements Methods. Formulary for Prof. Estor's exam

Eksamen i FY3466 KVANTEFELTTEORI II Tirsdag 20. mai :00 13:00

TILLEGGSSPØRSMÅL BILLETT- OG ADMINISTRASJONSSYSTEM KINONOR AS COMPLEMENTARY QUESTIONS POINT OF SALE SOFTWARE PACKAGE KINONOR AS

Den europeiske byggenæringen blir digital. hva skjer i Europa? Steen Sunesen Oslo,

ECON3120/4120 Mathematics 2, spring 2004 Problem solutions for the seminar on 5 May Old exam problems

MeijerG1. Notations. Primary definition. Traditional name. Traditional notation. Mathematica StandardForm notation. Generalized Meijer G-function

5 E Lesson: Solving Monohybrid Punnett Squares with Coding

Bærekraftig FM til tiden/ Bærekraftig FM på tid

Oppgavehefte om komplekse tall

Bokmål / Nynorsk / English NORGES TEKNISK- NATURVITENSKAPELIGE UNIVERSITET INSTITUTT FOR FYSIKK. Eksamen TFY4185 Måleteknikk

Ressurser. OpenCV documentation: Eigen documentation : C++: Image Watch: An image debugger plug-in for Visual Studio

Numerical Simulation of Shock Waves and Nonlinear PDE

SVM and Complementary Slackness

Generalization of age-structured models in theory and practice

Løsningsforslag for eksamen i Matematikk 3 - TMA4115

INF Logikk og analysemetoder Forslag til løsning på oppgave fra læreboken

Level Set methods. Sandra Allaart-Bruin. Level Set methods p.1/24

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

1 User guide for the uioletter package

3. Grænseovergange og grænseværdier

KOMPLEKSE TALL KARL K. BRUSTAD

Du må håndtere disse hendelsene ved å implementere funksjonene init(), changeh(), changev() og escape(), som beskrevet nedenfor.

Ma Flerdimensjonal Analyse Øving 2

LØSNINGSFORSLAG EKSAMEN, MAT 1001, HØSTEN (x + 1) 2 dx = u 2 du = u 1 = (x + 1) 1 = 1 x + 1. ln x

Exercise 1: Phase Splitter DC Operation

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Ma Linær Algebra og Geometri Øving 1

Løsningsforslag til prøveeksamen i MAT1050, vår 2019

NKKN typeforslag versjon Definisjon av grunntypene

Innholdsfortegnelse... 1 Endringslogg UD BETALINGSTERMINAL NETS NEW DRIVERS FULL SUPPORT WINDOWS

* très facile ** facile *** difficulté moyenne **** difficile ***** très difficile I : Incontournable

Merak Un-glazed Porcelain Wall and Floor Tiles

Energy Calibration for the Forward Detector at WASA-at-COSY

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

Matematikk Øvingsoppgaver i numerikk leksjon 2 Løsningsforslag

Eksamensoppgave i PSY3100 Forskningsmetode - Kvantitativ

Eksamensoppgave i TMA4123/TMA4125 Matematikk 4M/4N

melting ECMI Modelling week 2008 Modelling and simulation of ice/snow melting Sabrina Wandl - University of Linz Tuomo Mäki-Marttunen - Tampere UT

Kompleks eksponentialform. Eulers inverse formler. Eulers formel. Polar til kartesisk. Kartesisk til polar. Det komplekse signalet

a) The loop transfer function with the process model with a P controller is given by h 0 (s) = h c (s)h p (s) = K p (1 + s)(2 + s) K p

Oversigt [S] 9.6, 11.1, 11.2, App. H.1

Eksamensoppgave i PSY3100 Forskningsmetode - Kvantitativ

Kneser hypergraphs. May 21th, CERMICS, Optimisation et Systèmes

Right Triangle Trigonometry

Eksamen i TMA4123/TMA4125 Matematikk 4M/N

TMA4135 Matematikk 4D Kompendium i numerikk. Eirik Refsdal

Matematikk Øvingsoppgaver i numerikk leksjon 2 Løsningsforslag

Ringvorlesung Biophysik 2016

UNIVERSITETET I OSLO ØKONOMISK INSTITUTT

FYS2140 Kvantefysikk. Løsningsforslag for Oblig 7

Matematikk Øvingsoppgaver i numerikk leksjon 1. Løsningsforslag

Transkript:

Second Order ODE's (2P)

Copyright (c) 2011-2014 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License". Please send corrections (or suggestions) to youngwlim@hotmail.com. This document was produced by using OpenOffice and Octave.

Complex Numbers i = 1 (3, 2) two real numbers 2-d coordinate i 2 i 3 = 1 = i i 2 i = i 3 + i2 one complex number with real part of 3 and imaginary part of 2 i 4 = 1 i 2 i 2 = +1 imaginary (3,2) i i i 3 + i2 1 1 real i i i Second Order ODEs (2P) 3

Coordinate Systems and Complex Numbers (x, y) (a) (b) (x, y) (r, θ) (r cosθ,r sin θ) (r, θ) (c) x + i y r cosθ + i r sin θ (a) Cartesian Coordinate System x + i y = r cos θ + i r sin θ = r(cos θ + i sin θ) imaginary θ r (r,θ) x = r cos y = r sin θ x + i y real (b) Polar Coordinate System (c) Complex Number Second Order ODEs (2P) 4

Euler's Formula (1) e i π/ 2 e iθ = cosθ + isin θ i i i e i π 1 1 e 0 x y R{e iθ } = cos θ I{e i θ } = sin θ i i e i 3π/2 i e 0 = +1 + 0i = +1 = e i 2 π e i π/ 2 = 0 + 1i = +i = e = 1 + 0 i e i π = 0 1i e i 3 π/2 i 3π/ 2 = 1 = e i π i π /2 = i = e Second Order ODEs (2P) 5

Euler's Formula (2) e +iθ = cos θ + isin θ e iθ = cosθ isin θ e i π/2 i3 π/2 e i3 π/ 4 e e i π/ 4 i5 π/ 4 e i7 π/ 4 e e i π e 0 e i π e 0 e i5 π/ 4 i7 π/ 4 e e i3 π/2 e i3 π/ 4 π /4 e e π /2 Second Order ODEs (2P) 6

Absolute Values and Arguments e i π/ 2 r e i θ = r(cos θ + i sin θ) e i π i 1 1 i i i i i e 0 r sin θ r r cos θ e i 3π/2 absolute value argument, phase r θ r e iθ = r cos 2 θ + sin 2 θ arg(r e iθ ) Second Order ODEs (2P) 7

Computing Complex Number Argument y x < 0 y x > 0 atan ( y x ) + π atan ( y x ) [0, +π] y x > 0 y x < 0 atan ( y x ) π atan ( y x ) [0, π] atan ( y atan(y/x) ) x atan2(y, x) π 2 +π π 2 π π atan ( y x ) + π π 2 π π 2 atan ( y x ) π Second Order ODEs (2P) 8

sin and cos e i = cos i sin e +i θ = cosθ + i sinθ e i = cos i sin R {e i } = cos = ei e i 2 I{e i } = sin = ei e i 2i e i θ = cosθ + i sinθ e i θ = cos θ i sinθ e +i θ = cosθ + i sinθ Second Order ODEs (2P) 9

Complex Exponential e j t = cos t j sin t x real Top View z phase θ = ωt time 2π time y imag Front View z y imag x real time z time y imag x Side View real Second Order ODEs (2P) 10

Conjugate Complex Exponential I I R R t t e + j ω 0t = cos (ω 0 t) + j sin (ω 0 t) e j ω t = cos (ω 0 t) j sin (ω 0 t) e + j t = cos (t) + j sin (t) (ω 0 = 1) e jt = cos (t) j sin (t) (ω 0 = 1) Second Order ODEs (2P) 11

Cos(ω 0 t) e + jt = cos (t) + j sin (t) (e + jt + e j t ) = 2cos (t) e jt = cos (t) j sin (t) I R t x(t) = A cos (ω 0 t) = A 2 e+ j ω 0t + A 2 e j ω 0t Second Order ODEs (2P) 12

Sin(ω 0 t) e + jt = cos (t) + j sin (t) (e + jt e j t ) = 2 j sin (t) I e j t = cos (t) + j sin (t ) R t x(t) = A sin (ω 0 t) = A 2 j e+ j ω 0 t A 2 j e j ω 0 t Second Order ODEs (2P) 13

Complex Arithmetic z = a + i b w = c + i d z+w = (a+c) + i(b+d) z = a + i b w = c + i d z w = (a c) + i(b d) z = a + i b w = c + i d z w = (ac bd ) + i(ad+bc) z = a + i b w = c + i d z w = ( ac+bd ) c 2 +d + i ( ad+bc ) 2 c 2 +d 2 ( a + i b c + i d )( c i d c i d ) Second Order ODEs (2P) 14

Complex Conjugate z = x + i y = R{z} + i I{z} z = x i y = R{z} i I{z} z + w = z + w z w = z w R{z} = 1 (z + z) 2 z w = z w I{z} = 1 (z z) 2i ( z w ) = z w z z = (x + i y)(x i y) = x 2 + y 2 1 z = 1 z z z = z z z = z = z x 2 + y 2 z 2 Second Order ODEs (2P) 15

Determinant (1) Determinant of order 2 [ a 1 a 2 b 1 b 2] a 1 a 2 b 1 b 2 = a 1 b 2 a 2 b 1 Determinant of order 3 [a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3] [ + + ] + + + [a1 b 2 b 3] [ 3] [ 3 c 2 c c 1 c a 2 b 1 b 3 a 3 ] b 1 b 2 c 1 c 2 Second Order ODEs (2P) 16

Determinant (2) Determinant of order 3 [a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3] [a1 [ + + ] + + + + + b 2 b 3] [ 3] [ 3 c 2 c c 1 c a 2 b 1 b 3 a 3 ] b 1 b 2 c 1 c 2 a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3 = + a 1 b 2 b 3 3 a 2 b 1 b 3 3 + a 3 b 1 b 2 2 c 2 c c 1 c c c 1 Second Order ODEs (2P) 17

[a1 [a1 Determinant (3) Determinant of order 3 [a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3] [ + + ] + + + [ b 1 b 3] 3 c 1 c a 2 a3 + b 2 c 1 c 3] a3 b 1 b 3 ] c 2 a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3 = a 2 b 1 b 3 3 2 c 1 c + b a 1 a 3 3 c 2 a 1 a 3 3 c 1 c b b 1 Second Order ODEs (2P) 18

a11 Determinant Rule of Sarrus Determinant of order 3 Recursive Method [a1 1 a12 a1 3 a 2 1 a 22 a 2 3 a 3 1 a 32 a 3 3] a1 2 a13 a 2 1 a 22 a 23 a 3 1 a 3 2 a 33 = a 11 a 22 a 23 a 32 a 33 12 a a 21 a 23 a 31 a 33 13 + a a 21 a 22 = a 11 (a 22 a 33 a 23 a 32 ) a 12 (a 21 a 33 a 23 a 31 ) + a 13 (a 21 a 32 a 22 a 31 ) a 31 a 32 Determinant of order 3 only Rule of Sarrus [a11 a1 2 a13 a 21 a 2 2 a 23 a11 a12 a13 a a 31 a 3 2 a 33] 21 a 2 2 a 23 a 31 a 3 2 a 33 [a1 1 a1 2 a13 a 2 1 a 2 2 a 23 a11 a12 a13 a a 3 1 a 3 2 a 33] 21 a 2 2 a 23 a 31 a 3 2 a 33 [a1 1 a12 a13 a 2 1 a 22 a 23 a11 a12 a13 a 3 1 a 32 a 33] a 21 a 2 2 a 2 3 a 31 a 3 2 a 3 3 Second Order ODEs (2P) 19

Solving Linear Equations A set of linear equations If the inverse matrix exists a x + b y = e [ a b d][ x ] c y = [ e ] [ x ] c x + d y = f f y = [ a b c d] 1[ e f ] a d b c = ad bc 0 [ x y ] = 1 [ d b ][ e ] ad bc c a f e f d b = de bf [ x y ] = 1 ad bc [ de bf ce+af ] a c e f = af ce Cramer's Rule x = e f a c d b b d y = a c a c e f b d Second Order ODEs (2P) 20

Cramer's Rule Determinant of order 3 a 1 x + a 2 y + a 3 z = d b 1 x + b 2 y + b 3 z = e c 1 x + c 2 y + c 3 z = f [a1 a2 a3 b 1 b 2 b 3 3][ x ] [ d y = e c 1 c 2 c z f ] x = d a2 a3 e b 2 b 3 f c 2 3 c c 1 c 2 3 c a1 a2 a3 b 1 b 2 b 3 y = d a1 a3 b 1 e b 3 c 1 f 3 c c 1 c 2 3 c a1 a2 a3 b 1 b 2 b 3 z = a1 a2 d b 1 b 2 e c 1 c 2 f a1 a2 a3 b 1 b 2 b 3 c 1 c 2 c 3 Second Order ODEs (2P) 21

Wronskian W (f 1,f 2,, f n ) = f 1 f 2 f n f ' 1 f ' 2 f ' n (n 1) (n 1) f 2 f n f 1 (n 1) f 1 (x) f 2 (x) f n (x) W (f 1 (x), f 2 (x),, f n (x)) = df 1 dx df 2 dx df n dx d (n 1) f 1 dx (n 1) d (n 1) f 2 d dx (n 1) (n 1) f n dx (n 1) Second Order ODEs (2P) 22

Linear Independent Functions and Wronskian C 1 y 1 + C 2 y 2 = 0 C 1 = C 2 = 0 always zero means all coefficients must be zero y 1 and y 2.are linearly independent functions C 1 y 1 + C 2 y 2 = 0 C 1 y 1 ' + C 2 y 2 ' = 0 [ y 1 y 2 y 1 ' y 2 '][ C 1 2] C = [ 0 0] If the inverse matrix exists y 1 y 2 y 1 ' y 2 ' 0 [ C 1 C 2] = [ y 1 y 2 y 1 ' y 2 '] 1[ 0 0] = [ 0 0] the only solution: trivial W ( y 1, y 2 ) 0 Second Order ODEs (2P) 23

Linear Dependent (1) {u, v, w} linearly dependent w = u + v u = w v v = w u w in terms of u & v u in terms of w & v v in terms of w & u v w v w v w u u u u + v w = 0 u + v w = 0 u + v w = 0 Second Order ODEs (2P) 24

Linear Dependent (2) {u, v, w} linearly dependent w1 in terms of u & v w2 in terms of u & v w3 in terms of u & v v w w v v w u u u k 1 u + k 2 v + k 3 w = 0 m 1 u + m 2 v + m 3 w = 0 n 1 u + n 2 v + n 3 w = 0 (k 1 = 0) (k 2 = 0) (k 3 = 0) (k 1 0) (k 2 0) (k 3 0) (m 1 = 0) (m 2 = 0) (m 3 = 0) (m 1 0) (m 2 0) (m 3 0) (n 1 = 0) (n 2 = 0) (n 3 = 0) (n 1 0) (n 2 0) (n 3 0) Second Order ODEs (2P) 25

Linear Dependent (3) {v 1,, v 3, v 4 } linearly dependent v 1 k 1 v 1 + k 2 + k 3 v 3 + k 4 v 4 = 0 (k 1 0) (k 2 0) (k 3 0) (k 4 0) v 3 v 4 0 v 1 + m 2 + m 3 v 3 + m 4 v 4 = 0 (m 1 = 0) (m 2 0) (m 3 0) (m 4 0) m 1 v 1 + m 2 + m 3 v 3 + m 4 v 4 = 0 (m 1 0) (m 2 0) (m 3 0) (m 4 0) Second Order ODEs (2P) 26

Linear Dependent (4) {v 1,, v 3, v 4 } linearly dependent = v 1 v 3 v 4 = = Second Order ODEs (2P) 27

Linear Independent (1) S = {v 1,,, v n } k 1 v 1 + k 2 + + k n v n = 0 non-empty set of vectors in V the solution of the above equation trivial solution: k 1 = k 2 = = k n = 0 if other solution exists if no other solution exists S linearly S dependent linearly independent v 1 v 3 k 2 k 3 v 3 v 1 v 3 v 3 v 1 k 1 v 1 v 1 Second Order ODEs (2P) 28

Linear Independent (2) every point in R 2 can be represented by v 1 k 1 v 1 + k 2 linear combination of v1 and v2 which are one set of linear independent two vectors only points on a line in R 2 v 1 k 1 v 1 + k 2 linear combination of v1 and v2 which are one set of linear dependent two vectors Second Order ODEs (2P) 29

References [1] http://en.wikipedia.org/ [2] M.L. Boas, Mathematical Methods in the Physical Sciences [3] E. Kreyszig, Advanced Engineering Mathematics [4] D. G. Zill, W. S. Wright, Advanced Engineering Mathematics [5] www.chem.arizona.edu/~salzmanr/480a