Quant Found for Sys Engr
Quant Found for Sys Engr SYST 500
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SYST 500CSI 600 Lecture 6 Partial Notes Ariela Sofer Differential Equations 1 FirstOrder Separable Differential Equations A rst order differential equation is said to be separable if it has the form dy 7 h dz gltzgt a If My 31 0 and py lMy7 the equation can be written in the form dy mg 7 9W Solution method employs the equation pltygtdy amen ln the above we assumed that My 31 0 But it is possible that constant function y de ned by My 0 is also a solution7 that was lost77 in the process above This needs to be checked separately Example Consider the equation y 02xy If y 31 0 the equation can be written in the form y y 02x Thus Cl 2 0 112 0 1m2 702d loglyl01x c lyle39 gty0539 y The constant function y 0 for every y is also a solution since it satis ed the differential equation It is not a special case of the general solution for any value of c it was lost by the general solution process Example Consider the equation y yZx If y 31 0 the equation can be written in the form y y2 Thus dy7dx7717z gto71lt D7 7lt1 l W 2727 679x igcx yi ogcy The constant function y 0 for every y is also a solution since it satis ed the differential equation It is not a special case of the general solution for any value of c it was lost by the general solution process Example The equation y 62 can be written in the form 6 4yy 62m e 4ydy 62 dz 76 43 ezm C i 74y 7log 7 262m 5 y iUOgO 252m CD Given the initial value y0 07 we get 0 7 logl 7 2 cl which implies that logl 7 20 07 hence 72c1orc3 2 FirstOrder Linear Equations 2 H 22 A rst order differential equation of the form CL dyc aomy 9W a195 lf g 0 the equation is homogeneous7 otherwise it is nonhomogeneous Dividing this equation by a0z gives the standard form of the equation dy Pltzgty we The Homogeneous Equation The homogeneous equation can be written as Integrating gives log 7Pdz C so that e f pmdwc Cle f pwww39 Thus the general solution to the homogeneous equation is ye ceifPmdm This is a family of solutions7 with different curves corresponding to different values of 0 Example Consider the equation y sinxy 0 Here Pz sinx and fPdz f sinzdx 7 cosx the constant has already been gured into the solution expression7 hence can be ignoredThen ye ceifPmdm Cgcosm Example Consider the equation 2sy y 0 Dividing by 2x gives y y2s 0 Therefore PW 12x and Pdz fdm2z log log lzllZ Then ye ceifPmdm ceiltjglmll2 celoglml l2 C ilZ note that this means that for z gt 07 y cxE7 while for z lt 07 y c 7z check the solutionl At z 0 y is not de ned and the differential equation is not de ned The Nonhomogeneous Equation In analyzing the equation it is convenient to de ne the operater Ly as follows Ly y The homogeneous equation becomes Ly 07 and the nonhomogeneous becomes Ly Suppose that yp is some solution to the nonhomogeneous solution Lyc i on an interval I7 and let yo be the general solution to the homogeneous equation Lyp 0 on I Then Lyc yp Ly0 Lyp 0 fx therefore ye yp is also a solution to the nonhomogeneous equation on I Conversely if yp and 1717 are two solutions to the nonhomogeneous equation then Lyc 7 zip Lyc 7 L7jp f 7 fx 07 so that yp 7 1717 is a solution to the homogeneous equation Thus the general solution to the nonhomogeneous equation on the interval is y yo yp To solve the nonhomogeneous equation we multiply the equation by the integrating factor ef pmdm Multiplying an equation by an appropriate integrating factor simply makes everything easier to integrate Here we get 6fPmdzy P6fPzdmy efpwmfa But the left hand side is the derivative of ef pmdmy with respect to x Thus the particular solution satis es d 6fPmdzy 6f Integrating the derivatives on both sides gives yp 67fPmdm6fPmdmfxdx39 Finally since y yo yp we get y fawn 6 6fPzdzfdgt39 A simpler presentation of the above is as follows Let q 67fPmdm39 y gm 0 dxgt Example Consider the equation y zy cosx Then since Pz x7we get Then yo cqx7 and qltgt IPWW e mz27 6fPmdz 6122 and y e m22 lt6 6m22 cosdzgt The nal integral does not have an explicit form and is generally calculated numerically Equot SYST 500CSI 600 Lecture 1Partial Notes Ariela Sofer Vectors Introduction 0 Scalars are numbers that may be integer a 3 b 75 Rational a 57 b 71007919 irrational a 7139 1 b e o Vectors are a collection of scalars that can convey more information in a compact representation 0 Vectors are presented either geometrically suitable for 2 dimensions R2 or 3 dimensions 133 or algebraically suitable for any dimension Geo metric Representation 0 Vector viewed as a quantity that has both a magnitude and a direction 0 Examples are Force operating on body velocity of a body acceleration etc Vectors are represented graphically by arrows For example the velocity vector of a particle moving northeast at 10 mph is represented by an arrow of length 10 in the appropriate scale in 450 to the horizontal axis 0 We denote the length of a vector a by a b ltgt Hall HbH and directionadirectionb Example The vector starting from the point 00 and terminating at the point 11 is equal to the vector starting from the point 10 and terminating at the point 0 1 since both have the same direction and the same magnitude 0 In the two dimensional space the vector emanating from the origin and ending at the point P with coordinates x y is called the position vector of P and is denoted by z 0 Various operations can be de ned on a vector a such as 7a ka and on two vectors ab such as 11 ail 2a3b etc 0 Angle between two vectors refers to the angle between them when they are arranged tail to tail 0 Angles may be represented in degrees or in radians Note that the default in MATLAB is radians In general 27139 radians 3600 so that 1800 1 radian 7 7139 o Dot Product of two vectors a I cos0 where 9 is angle between a and b o It is possible to show from geometric considerations that say in 3 dimensions a I albl agbg agbg Thus ifa 123 and b 71174 then a I 711 9 0 When 9 7r2 9007 or 9 37r2 2700 a I 0 Two vectors a and b are said to be orthogonal if and only a I O o abgt0 ltgt 0is acute7 andablt0 ltgt 0isobtuse o a a HallHaH cos0 Hallz so that Hall xa a ab 0 7 MlMW Example ifa 712719 27 71 0050 708 Therefore 9 arccos708 2498 radians 2498 gtk 1807139 1430 Algebraic Representation o Advantage we can extend the concept of a vector beyond two or three dimensions Vector notation allows us to represent large amount data sets concisely o A vector in n space Rn is an n tuple of numbers a 011012 quot701 As long as we don t work with matrices7 it doesn t matter very much whether we work with row vectors or column vectors Later when we work with matrices it will be convenient to work with column vectors We will then denote 7 T x 7 m m17m27 ln Lecture 1 of the notes however we will stick to row vectors for notational ease o If a and b are two n vectors then ka ka17ka27kan 10 ltgt MO7 i1n ab a1b1a2b2anbn a ltgt 041 7 i1n o Dot product of two n vectors 71 a I Zajbj j1 o Norm 2 norm in this course of a vector Hwwmmz j1 M30 10 ltgt HaH0 o For any scalar k HkaH t o A vector a is said to be a unit vector if Hall 1 A normalized vector is a vector scaled to be a unit vector Normalizing a gives the vector i aHaH For example normalizing a 3 4 gives the unit vector i 35 45 0 Two vectors a and b are orthogonal if and only if a I 0 o Triangular inequality Ha bH Hall HbH with equality only if a kb with k gt 0 Vector Spaces 0 A vector space V is a set of elements with two operations addition and ii multiplication by a scalar that satis es the following Axioms for vector addition Closure under addition Commutative law Associative law Zero vector Negative of a vector zerzy V myymey V myzxyzVxyz V Hauniquevector 0 such that Ozm0sz V Vz E V3 aunique vector ix such that z 7m 0 Axioms for scalar multiplication vi z E V km E V V scalar k Closure under scalar multiplication 1m kz y km kg Vm y E V and scalar k Distributive law um k1 k2m klz kgm Vz E V and scalars k1 k2 Distributive law k1k2z k1k2z Vz E V and scalars k1 k2 Distributive law m m m Vm E 0 Examples V R2 V R3 V R V1 04012 a1 0 But V2 011012 a1 2 0 is not a vector space since it violates the negative vector rule77 and V3 a1a2 a1 a2 1 is not because it is not closed under addition The set Pnall polynomials of degree n is a vector space o If a subset W of a vector space V is also a vector space then W is a subspace subspace of V o A subset W of a vector space V is a subspace of V if and only if it satis es the closure properties 0 Note that any set that does not contain a zero element is not a vector space 0 Linear independence A set of n vectors m1 mn is said to be linearly independent Ll if the only constants 041 an satisfying alxl 04ng l anzn are 041 042 an O A set that is not linearly independent is said to be linearly dependent LD 0 Examples 1 1 2 2 1 then 0411 0422 0 implies 041 0420 2a14a20 ia1az07 so the vectors are Ll SYST 500CSI 600 Lecture 4Partial Notes Ariela Sofer Eigenvectors and Eigenvalues 1 A Digression Complex numbers 0 A complex number is a number of the form 2 x iy where x and y are real numbers and 1392 l The real part of z is Rezx and the imaginary part is lmzy o The complex conjugate of z x iy is 2 x iy 0 Some operations on complex numbers If 21 x1 z39yl and 22 x2 iyg then 0 21i22 1i2iy1iy2 39 3132 951Z3931X952 1112 951952 31132 Z95132 95231 0 The norm or magnitude of z is xZ V052 32 0 Division of two complex numbers is performed using 2122 2122 22 22 thus giving a real denominator 2 Another Digression Roots of Polynomials o If a0 a1 an are n 1 numbers with an y 0 then the function 1395 anx an1x 1 alx a0 is a polynomial of degree n The following discusses polynomials with real coef cients 0 The numbers that solve 1305 0 are its roots 0 A polynomial of degree n has n roots some of which may be complex o If x is a complex root of a polynomial then so is a complex roots come in pairs 0 The polynomial 1395 can be be written as 1305 anx x1x x2 05 xn where 051 7 are its roots Some roots of a polynomial may have multiplicity greater than 1 For example 05 0 is a root of multiplicity 3 of 1305 x4 x3 05305 1 The roots of the quadratic polynomial P ax bx c are bl Vb2 4ac 9512 2a 3 Eigenvalues and Eigenvectors Eigenvalues and eigenvectors are useful in many problems including dif ferential equations and difference equations They may determine the physical properties of a system Eigenvalues for example may represent the fundamental frequencies of oscillations in an electrical system system They may also affect how accurately we can solve the system Ax I Let A be a square matrix The scalar A is said to be an eigenvalue of A if there exists some nonzero vector x such that Ax A05 The vector x is said to be an eigenvector of A corresponding to the eigen value A Alt Then 05 1 1T is an eigenvector of A and A 5 is the corresponding eigenvalue since 1 5 M M 5596 Notice that v 2x 2 2Tis also an eigenvector for A 5 since Av 51 Example Let 4 1 1 4 41 14 If x is an eigenvector of A corresponding to A then so is v am for all nonzero scalar oz This follows since Av Aozx ozAx ozAx Aozx Av Finding the eigenvalues AxAxgtAx Ax0gtA A1x0 gdetA AI0 2 The equation CA detA AI 0 is called the characteristic equa tion of A To nd the eigenvalues we nd the n roots of the polynomial C To nd the corresponding eigenvectors we solve A AIM 0 by applying Gaussian elimination to the augmented matrix A A 0 for each eigenvalue Example Let Then 4 l comA Mj 1 4A AI M24 A2 815 0 A1 mg 3 The eigenvectors for A1 5 satisfy law Choosing x1 1 gives the eigenvector mm The eigenvectors for A2 3 satisfy 110 gt 110 Choosing x1 1 gives the eigenvector we Both v1 and 112 were only determined upto a scalar multiple It is sometimes common to normalize the eigenvectors this is What MATLAB produces Normalization of a vector u yields the vector aa which has norm 1 The normalization of U1 and 112 gives me w 1 10 0 0 0 1952 110 m Mmlt 000 gt gt1 2 52 452 l 0 Example Let 2 A i 1 1 Then 2 A detA AI 1 A 2 A1 A2 0 A1 2 A23 1 1 A The eigenveotors for A 2 satisfy 0 0 A 2I0 1 0 gtx1 is free x2x30 1 0 Choosing say 951 1 gives the eigenveotor 1 U1 0 0 The eigenveotors for the double root A 1 satisfy 1 0 A 10 0 0 gt 051 0952 and 953 free 0 0 Choosing x2 1 x3 0 on one hand and x2 0 x3 1 on the other gives two L1 vectors 0 0 U2 1 7 U3 0 0 1 0 Example Let 1 0 1 A i 2 1 1 Then 1 A 0 1 detA AI 2 A 1 2 A1 A2 0 gt A1 2 A23 1 1 A The eigenvectors for A 2 satisfy 1 1 0 A 20 0 1 0 1 0 gt 952 is free 951 x3 0 Choosing say 952 1 gives the eigenvector 1 U1 0 0 The eigenvectors for A 1 satisfy 0 1 0 A I0 1 1 0 gtx30x20andx1free 0 0 We have therefore only one L1 vector for this double root Choosing x1 1 we get the eigenvector 0 U2 1 0 Example Let 2 2 A lt 1 0 Then 2 A 2 2 detA I 2 A 2 A 22 0 gt A1 12 A2 1 1 Note that A2 X1 The eigenvectors for A1 satisfy A 1 z I0 111 12Z 0 0 Notice that the rst equation is 1 times the second since 1 1 z39 1 z392 1 2 Thus as expected the equations are LD We can use the second equation to set 951 1 239 x2 1 giVing 1z39 U1 1 5 The eigenvectors for A2 satisfy l 139 2 0 A l z0ilt 1 1Z 0 The rst equation is z39 1 times the second Therefore we can use the second equation to set 951 l 7052 l giving Note that 112 171Both v1 and 112 were only determined upto a scalar multi ple This scalar might be a complex number MATLAB gives eigenvectors equal to 13 31 1 i 31z39 1 3i 7 2 3239 Check to verify that these are eigenvectors of norm 1 that are scalar mul tiples of U1 and Ugl 0 Some properties of eigenvalues o Eigenvalues corresponding to distinct eigenvalues are Ll 0 An eigenvalue of multiplicity k gt 2 may have less than k Ll eigenvectors o If x1 and x2 are eigenvectors corresponding to the same eigenvalue A then so is any combination 01051 02x2 0 detA A1A2 An This follows from setting Setting A 0 into the equation detA AI co mo A1 A A o A is singular ltgt some AZ 0 o traceA A1A2An o The eigenvalues of a triangular matrix are its diagonal terms In partic ular this holds for a diagonal matrix 0 eigAeigATgt o If A is a complex eigenvalue of A with corresponding eigenvector x then its conjugate A is also an eigenvalue of A with corresponding eigenvector 0 That is complex eigenvalues come in conjugate pairs SYST 500CSI 600 Lecture 7 Partial Notes Ariela Sofer HighOrder Linear Differential Equations 1 Preliminary Theory An nthorder linear differential equation has the form d n d 1 d My am mg W gm lfthe problem has conditions on the value ofthe function and its rst 7171 derivatives at a given point7 ie7 9amp0 107 y o 917 quot397 yn710yn717 the problem is called an initialvalue problem if it has conditions on the function value at n distinct points7 ie7 y1y17 We y2 Wen ym it is called a boundaryvalued problem If g 0 the equation is said to be ho mogeneous7 otherwise it is nonhomogeneous lf ll al for all 239 and x then the equation is said to have constant coe icients A set of functions f17 f27 is linearly dependent LD on an interval I if there exist constants 0102 0 not all zeros such that 01f1 02f2 cnfnx 0 for all z in I If it is not LD it is linearly independent Ll Example Consider the functions f1 17 f2z 7 and f3z 2 on the real line 7007 00 Suppose that there exist constants 010203 so that 01 ng 03x2 0 for all x Plugging in 0 yields 01 07 and plugging in z 1 and z 71 yields 02 03 0 and 702 0307 respectively7 which implies that 02 03 0 Hence cl 0 and the functions are Ll ln constrast7 the functions f1 7 f2 2 and f3 2x2 7 z are LD on the real line since f1 2f2x 7 f3 0 for all x Finally7 the functions f1 em f2z 62m and f3z 64 are Ll on the real line7 since the only constants 010203 for which 016m 026 036 m 0 for all x are 01 02 03 0 Consider now the n th order homogeneous equation Ly 0 It can be shown that there exist 71 Ll functions y17y27 yn called a fundamental set of solutions satisfying 0 Furthermore7 if yhyg yn is a fundamental set of solutions to Ly 0 then the general solution to the equation is y Gil1W UziXi quot39 Cnyn7 where 01 are arbitrary constants o The question whether the functions y17y27 yn are Ll can be resolved by analyzing the determinant in 12 yn 13 9 y Wy13y23m3yn 3 3 3 71 71 yin gt ygn gt y n 1 lfy17y27 y are 71 solutions to Ly 0 on an interval I7 then they are Ll on I if and only if Wy1y2 7y 31 0 for all z in I 0 Example The functions yl 62m and y2 e zw are solutions to the differential equation y 7 4y 0 on 007 00 This can be veri ed by plugging into the equation Now 21 72m 5 e Kl17 262m 726721 4 7 0 for every x on the real line Therefore yl and y2 are Ll and they constitute a fundamental set of solutions to the equation The general solution the equation is y 016 02672 2 Homogeneous Linear Equations with Constant Coe icients 0 Consider the linear differential equation with constant coef cients My any an71y 1 a1y aoy 0 where an 31 0 If n 1 the equation is my aoy 07 and we can show from last class that the general solution is cle WO m Following this pattern we will try to guess a solution of the form y 6 to the equation Ly 0 Using the fact that the k th order derivative of our guessed solution satis es Mk Aka we obtain7 when substituting into the equation Lfy 6 lankn anilAnil a1 a0 0 3 so that anAn l anilAnil l 11 10 0 This is called the characteristic equation for the ODE Suppose that the characteristic equation CA 0 has n distinct roots A17A27 A Then the functions 6quot 6quot 7 6quot are a fundamental set of solutions to the homoge neous equation7 and the general solution of the equation is 016A 026A 0736A Example Consider the equation y 7 2y 7 3y 0 The characteristic equation gives A272A730 A137 A2771 m are a fundamental set of solutions7 and the Then the functions yl 63m and y2 6 general solution to the equation is y 016 026 If we wish to solve the equation with initial value conditions y0 1y 0 2 we have 90 01 62 1 y 0 301 7 02 2 7 cl 34 02 14 7 y 34e3m 1205 A special case is when some of the roots are complex Suppose that A1 a 2 is a complex root of the characteristic equation Then its conjugate A2 a 7 2b is also a root of the equation Together the two roots contribute the solutions emiim email e cosb i z sinb Now7 any real combination of these two solutions can also be written as a combination of y1 e cosb and y2 e sinbx7 and furthermore7 yl and y2 are Ll solutions to the differential equation It is more conve nient to use yl and y2 as part of the fundamental set7 rather than 601i The complex roots A1 and A2 contribute therefore the term y 016 cosbz 026 sinb to the general solution to the equation Example Consider the equation y 2y By 0 The characteristic equation gives A22A50 A71i221 Then the functions yl e w cos2x and y2 e w sin2 are a fundamental set of solutions7 and the general solution to the equation is y 016 cos2m 026 w sin2 Finally suppose now that A is a repeated root of order k Then the functions yl e yg x6 7 yk zk le are k Ll solutions to Ly 0 These are the k Ll solutions contributed to the fundamental set by the root A Example Consider the equation M4 7 2y y 0 The characteristic equation gives A472A3A07A2A7120 A1A20 A3A41 Then the functions yl 60m 1 and y2 meow z are Ll solutions contributed by the double root A 07 and the functions yg em and y4 mm are Ll solutions contributed by the double root A 1 The general solution to the equation is c Ccewcmm 1 2 3 4 3 The Nonhomogeneous Linear Equation 0 De ne Ly is de ned as in the previous section The homogeneous equation becomes Ly 07 and the nonhomogeneous becomes Ly 0 Suppose that yp is some solution to the nonhomogeneous solution Lyp on an interval I7 and let yo be the general solution to the homogeneous equation Ly0 0 on I Then Lyc yp Ly0 Lyp 0 fx therefore ye yp is also a solution to the nonhomogeneous equation on I Conversely if yp and 1717 are two solutions to the nonhomogeneous equation then Lyc 7 zip Lyc 7 L7jp f 7 fx 07 so that yp 7 1717 is a solution to the homogeneous equation Thus the general solution to the nonhomogeneous equation on the interval is yycyp 0 Our challenge therefore is to nd a particular solution yp to the nonhomogeneous equation We will discuss only one approach called the the method of undetermined coef cients The method is appropriate for functions f so for which the set x f 7 f77 s7 consists of a nite number of Ll functions 0 Example The function f cosx has f 7sinz7 f x 7cosx etc7 so f and its derivtives cconsist of only two Ll functions The same is true for f sinx The function f 6 satis es f 16 so f and its derivatives are made out of one Ll function For all of these functions the method of undetermined coef cients may be used On the other hand for fx 4 f 7x and f x 71 Lz quotl 7 so that f and its derivatives are not made out of a nite number of Ll functions7 and the method does not apply 0 The method uses the structure of f to guess the general form for yp Once the correct form is guessed77 the exact values of the undetermined coef cients denoted below by A1 A2 and so forth is obtained by plugging into the differential equation The exaxt form of yp depends also on the roots of the characteristic equation CA 0 o The guessed solutions have the following forms y k 7 am y Aleam y Alx 60m fa 7 k6 if A a is not a root of CA if A a is a root of order k fzkn ypA0A1zAn ypxkA0A1zAn if A 0 is not a root of CA if A 0 is a root of order k f k sinbz or yp A1 cosb A2 sinbz yp kA1 cosb A2 sinbx f k cosbz if A 2b is not a root of CA if A 2b is a root of order k f k6 sinbx7 yp A16 cosbz A26 sinb yp zke A1 cosbz A2 sinbz f k6 cosb if A 1 21 is not a root of CA if A 2b is a root of order k SYST 500CSI 600 Lecture 3Partial Notes Ariela Sofer Determinants and Inverses 1 Determinants 0 Every square matrix has a quantity associated with it called its determinant denoted detA The determinant is de ned recursively on the size of the matrix lfAis1gtlt1 detA deta a lfAis2gtlt2 thidt a b e a b e deb e i e C d i C d 7a C For general 71 we will rst need two other de nitions First the minor determinant of 17 is the determinant M of the submatrix obtained by deleting rowz and column j The cofactor of 17 is its minor determinant multiplied by its sign factor that is CH 71 7 MH The sign factors correspond to the following checkerboard pattern 7 signz j 71 7 a f 1 Given these detA 2791 11701 i It can be shown that detA can be expanded by any of As rows or columns ZaijClj 1 71 j1 ZaijClj 1 n i1 0 Example 1 3 2 A 75 2 1 2 0 3 2 1 75 1 75 2 011 0 3 67 0127 2 3 7177 013 2 0 7 47 so that detA 6 3 17 274 4 Alternatively expanding by the second column 75 1 1 2 1 2 012 l 2 3quot 7 022 l 2 3l 71 032 l 51l 7117 so that detA 3 17 271 4939 V L A is lower triangular upper triangular or diagonal detA H ail i1 In general if B is obtained from A via elementary row operations then R H 013 c 31 0 i detB cdetA R e R CR detB detA R lt gt Rj detB idetA The same properties apply if the operations are performed on columns rather than rows Computing the determinant of a matrix via its de nition is computationally prohibitive The work is of order 711 A more e ective approach is to triangularize the matrix via row operations 0 1 m 1 o detA 0 ltgt rows and also columns of A are D 0 Thus detA 0 ltgt rankA lt n o detAT detA o detAB detA detB 0 In general detA B 31 detA detB Example 13 0 13 A 72 31 m 0 9 011 01 Some other properties OOH O W 0 1 gt i detA 8 89 2 Inverse of 1 Matrix 0 Suppose A is a square matrix If there exists another matrix B such that AB BA I where I is the identity matrix then B is said to the inverse of A denoted by A l and A is said to be invertible 0 Example The matrix A lt 2 0 gt is invertible since B 0395 0 satis es 1 5 701 02 AB BA I Checkl The 1 gtlt 1 matrix A 0 is not invertible because no other 1 gtlt 1 matrix B satis es AB BA o If Aquot1 exists then since detA detA 1 detAA 1 detI 1 it follows that detA 7E 0 A is said to be nonsingular Conversely if detA 0 A is said to be singular A singular matrix has no inverse A invertible ltgt rows of A are Ll ltgt columns of A are Ll ltgt detA 31 0 0 Some properties of the inverse A7571 A7 AB71 B71A717 A571 A71T In general there is no closed formula for A B 1 directly in terms of A 1 and B l 0 Consider the linear system Axb7 where A is n gtlt n If A is nonsingular then Ax b i A lAm A lb i z A lb 0 Conclusion The n gtlt n homogeneous system Ax 0 has a nontrivial solution ltgt detA 31 0 0 Computing the inverse A useful formula for n 2 If ad 7 b0 31 07 then 7 a b 1 i 1 d 7b Ailtcdgt A fadibcltic a i For general 71 Let C Clj be the matrix of cofactors of A De ne the adjoint matrix adjA CT Then 211 adjA detA Example 1 1 0 A 1 2 1 1 0 1 we have 2 0 72 T 1712 12 adjAOT i1 1 1 detA2 A 1 0 12 712 1 71 1 71 12 12 7 In general the formula is computationally inef cient7 and is not practical7 even for small 71 i A more ef cient alternative is via Gauss Jordan Elimination If A is nonsingular we can transform All egg IlA l Example 1 SYST 500CSI 600 Lecture 8 Partial Notes Ariela Sofer Systems of Linear Differential Equations Introduction We are concerned with the system of differential equations 3 a11 a12M quot a1nn f1t W2 a21 122 quot3912nn f2t 96 an i an z quot annn fnt where 1 2 xn are functions of t This is a system of linear differential equations Setting f1t f t 7 2 Mt 95100 Mt s 7 9M0 we can write the system as an aln A s s anl 1 ann Mt AM f or x Ax f in short If f 0 the system is called homogeneous7 otherwise it is nonhomogeneous 2 Solution by Elimination Consider the homogeneous system We can eliminate one of the variable from the equation For example we can use the rst equation to set 2 4x1 7 3 Plugging this into the second equation yields lzf1L 7 x 7x1 16L 7 4 V we 412 Collecting terms we get 78x315x1 07 CA A278A150 A53 z1 0163t0265t Therefore x2 4x1 7 3 40165 40265 7 30165 7 30265 0165 7 0265 7 1 7 31 5t 1 z7ltmgt7wlt1gt626 4 Thus 0 Consider now a nonhomogeneous version of the system above 3 7 4x1 7 2 t z z 7z14x27t For example we can use the rst equation to eliminate 2 2 4x1 7 3 t Plugging this into the second equation yields lzf1L 7 x 1 7z1 16L 7 lzf1L 4t 7t v v mg 4122 Collecting terms we get x 7 8xf1L 15L 73t 1 Now the solution to the homogeneous equation for this is as above Q10 6163t6265t To get a solution to the nonhomogeneous equation we guess zlp A Bt we have 1PABt zipB z p 17 15A78B1 A7125B715 73 1537 and 1 01 0262 7 t5 7125 Moving to 2 952 495179591 40153 4cze5 74157425730153 730255 715H 0163t70265tt57925 Thus 1 961 1 3t 1 5t 1 715 7125 ltz2gt 015lt1gt1025lt71gttlt 15 1 7925 A more systematic representation of the substitution method is obtained by using the notation Dz dmdt7 collecting the resulting coef cients of each of the variables in each equation and then solving As an example if we consider our homogeneous problem it can be written us 374z120 D74x1x20 gt x1z274z20 x1D74x20 We can then use Gaussian elimination To shorten the presentation here7 we note that we can eliminate x2 by multiplying the rst row by D 4 and then subtracting the second row This gives D74 217x1 0 D278D15 1 0 x 78x 15L 0 1 1 This yields the same solution as we found earlier 3 A Direct Approach Consider the homogeneous system t Axt A solution vector on an interval I is a vector 1 2 Mt X163tlt1gtandX265tlt71gt are solution vectors for our eargier homogeneous example satisfying the system Example The set of solution vectors XLXZ7 Xn be a is linearly dependent LD on an interval I if there exist constants 0102 6 not all zeros such that 01X102X2Can0 for all t in I If it is not LD it is linearly independent Ll Any set of n Ll solution vectors of the homogeneous system on an interval I is said to be a fundamental set of solutions on the interval lf X17X27Xn are a fundamental set of solutions Then the general solution of the system x Ax on the interval is Cle Cng Can Our previous example suggest that a fundamental set of solutions may exist of solution vectors of the form t e Vq where q is some vector To verify this we plug our suggested solution into the equation We obtain Ae vq Ae vq i q Aq Aq Aq It follows that A must be an eigenvalue of A with corresponding eigenvalue q We therefore have the following Let A1 A2 7 An be the n eigenvalues of A not necessarily distinct Then if q1 q2 qn are n Ll eigenvectors corresponding to M then 6 qu 92912 m 6mg H SYST 500CSI 600 Lecture 9 Partial Notes Ariela Sofer Laplace Transforms and Fourier Series The Laplace Transform The idea of transforms in mathematics is to assign to a function or7 in some cases7 to an in nite sequence of numbers some unique companion function in a different domain This for example the Laplace transform of the function ft will be a function The beauty of transforms is that in many cases dif cult problems involving ft may be easy to solve with respect to The most famous transforms are the Laplace transform7 the Fourier transform7 and the Z transform Given a function ft7 t 2 07 its Laplace transform is de ned as M amt LO e tmdt provided that the integral converges Example For ft 1 F5 ft A gistdt 7679 0 7 1 1 for s 2 0 The integral diverges blows up for s lt 07 and so the transform is not de ned Example For ft 6 co co 1 gisteatdt 67siatdt 7 t0 t0 s 7 for s 2 a The integral diverges blows up for s lt 17 and so the transform is not de ned 1 1 07 57a Fs 7siat a 6 7 7 0 8 1 Example For ft 263 Fs 15 7 3 Example For the piecewise function 07 0 S t S 1 1 co 00 00 115 e ftdt e Stftdt 0 e stdt 77679 7 0 t1 t1 1 t 0 Some transforms ft1 Fs 520 farce Fs8ia52a ftt n12 F 8771 gt0 m coskt 115 s gt 0 m sinkt 115 s gt 0 o if Fs is the transform of ft then we say that ft is the inverse transform of Fs7 and write ft 1Fs 0 Examples 0 Properties of transforms Linearity of transform aft bgt aFs bGs Linearity of inverse transform 1aFs bGs aft bgt Transform of the derivative e f m From this we get recursively 6 5 ftl s 6 ftdt 40 5115 Law smai m awe W f wwSWVWwWV 7 7mi WWW Example 3 6 15 24 i 3 i i 3 7 5s1n3t 4t 75 s1n3t4 t 75829 4 84 7 829 84 7 Example 1 2 71 674 2673 s 2 Using Transforms to Solve Differential Equations Suppose we have a differential equation in We can take the Laplace transform of the equation and get from that an algebraic equation for Ys We can solve for Ys and nally apply the inverse transform to get the solution The key idea here is that instead of working in original space t we work in transformed space 5 and in certain cases this is easier Example Consider the equation y 2y 4t with y0 0 Then taking the Laplace transform of this equation we get y 2y we 7 mg 7 110 215 7 Si 7 55 52s 2 To nd the inverse transform of Ys we will need to break it into partial fractions 4 i 0 7A552B52052 5252Ts 52 52 7 52s2 52A O 52A B 2B 7 52s2 Comparing coef cients we get A C 0 2A B 0 and 2B 4 This yields B 2 A 71 C 1 Therefore 1 2 l YS m 8 872 7 yt712te 2 Example Consider the equation y 7 5y 6y 6 with y0 0y 0 1 Taking the Laplace transform of this equation we get 6 My 7 52 6y M6 8255 8110 40 5l8Y8 90 6Y8 We now solve for Ys 6 5 55 7 2s 7 3 To nd the inverse transform of Ys we will need to break it into partial fractions Ys52 7 55 6 6 glgtY8 65 A B L A527556B5273s052725 5572s73 5 572 573 5572573 i52ABOs75A73B72O6A T 5572573 Comparing coef cients we get A B C 0 75A 7 3B 7 2C 1 and 6A 6 This yields A 1 B 74 C 3 Therefore i yt 717 462 365 03 Fourier Series Preliminary Many physical phenomena have a cyclical nature Annual weather patters electrical current Vibration of a string optical phenomena etc Fourier analysis represents com plicated cyclical structures by combinations of simpler cyclical fucntions such as sines cosines From this we can get new insights and the representation is easier to handle A function f is said to be periodic with period T if fT f for all x For example 27139 is a period of cos2x since cos2 2 0052r The smallest possible period of a periodic function is said to be the fundamental period of f The fundamental period of cos2 is 7139 A function is de ned to be even if f f7x It is de ned to be odd if f 7f7x Examples of even functions include 2 cosz coshz eme w 2 Examples of odd functions include x x3sinz sinhz em 7 e m2 Some functions are neither odd nor even for example f em and some can be both eg fx 0 Every function can be written as a sum of an even and odd function f96 f95 2 7 W rm 7 ch where f0z 2 fex f07 SYST 500CSI 600 Lecture 2Partial Notes Ariela Sofer Matrices and Linear Systems 1 Introduction 0 A matrix is a rectangular array of quantities an 012 am 021 022 12n A aml am2 amn We say A dzj Here the dimension A is of size m x n We say A aijmgtltn 0 Example The matrices 5 7 3 3 7T2 Alt2 44gt Blt65gt C 1040 have size 2 x 3 2 x 2 and 1 x 4 respectively Here B is a square matrix m If m 1 as in C the matrix is a row vector and if n 1 the matrix is a column vector 0 In general we can de ne several operations on matrices If A aijmgtltn7 B 927an then 7614 aij 527an A B lt gt aij bij A B aij bijmgtltn If A and B do not have the same size they cannot be equal and their sum is not de ned 0 Examples 41 1 2 3 3 6 9 3lt0 4 7 0 12 ml 33 l HlJkOT H010 H D H o Matrixvector product For an m x 71 matrix A and an n Vector 05 TL 2 01m j1 Ax 39 TL 2 am j j1 0 Examples 1 2 3 1 0 5 0 1 5 54 726 45 11 1 1 1 0 o If we denote the j th column of A by AJ then A A1 A2 An 0 Product of matrices If A is m x n and B is n x k then k AB AB17AB27 7ABk7 SO Zailblj 11 Note the rule concerning the sizes involved in the product A B vv mxnngtltk ka oExamples 1 0 1 2 1 2 3 0 14 lt5 44 2 1 lt632gt 321 0 1 5 5 1 4 2 1 0 In general AB 74 BA in fact it may not be de ned When the product is de ned ABC ABC o If A is square n x n then A2 A x A A3 A2 x A etc Examples for some special square matrices here with 713 5 0 0 1 0 0 5 3 7 5 0 0 D0 10I010U091L1 10 0 0 4 0 0 1 0 0 4 2 2 4 v diagonal identity lower upper matrix matrix triangular trlangular The identity matrix I has the property that A I A I 2 l0 0 The transpose of the m x 71 matrix A is the n x m matrix AT obtained by interchanging the rows of A with its columns Thus A1 ajZ7Vz39 j 0 Example 1 5 HA i j thenAT 2 4 3 7 0 Some properties of the transpose ATV A A BT AT BT ABT BTAT MDT kAT Note that in general ABT7 ATBT and in fact the product on the right may not be de ned even if the produce AB is de ned Why o If x and y are two column vectors of equal size then 05 3 xTy o A square matrix A is symmetric if AT A Thus For example a diagonal matrix is always symmetric Systems of Linear Equations 0 A system of m linear equations in n unknowns has the general form 011951 012952 39 39 39 ain in b1 0211 0221 agnfn b2 amlxl amgxg amnxn bm Denoting A dzj b 117 written in matrix form bmT and x x1 xnT the system is Axb If I 0 the system is called homogeneous otherwise it is nonhomoge neous Solution of linear systems of equations is based on transforming them to an equivalent systems with a structure that is easy 7 to solve The idea uses the fact that the following elementary operations on a linear system transform it into an equivalent system that has the same solution Solving Linear Equations Solution is often done using elimination using elementary row operations ERO s The following row operations on a system of equations give an equivalent system Multiply an equation by a nonzero constant Add a multiple of one equation to another Switch two equations To avoid lots of writing or for more ef cient coding we work on the arrays of numbers involved in the system speci cally on the augmented matrix A l b Performing elementary operations on the equations is equivalent to per forming the following elementary row operations on the augmented matrix 1312 lt cRi c 74 0 Multiply row 1 by a nonzero constant 1312 lt RZ ch7 Add a multiple of row j to row 1 1312 lt gt Rj Switch rows 1 and j Gaussian elimination Ab ell 08 UE Here the matrix U has a row echelon form in which the rst nonzero element in each row is to the right by at least one of the rst nonzero element in any row above it Note that in contrast to the book we will not require the rst nonzero to be 1 Gaussian elimination example 1 what is the original system of equations 1 3 0 4 l 3 0 4 2 3 l 2 gt 0 9 l 10 0 l l 2 0 l l 2 l 3 0 4 89x389 gt 9131 l gt 0 9 l 10 9x2110 gtx21gtx l 0 089 89 x134 gtx1l 1 Check the solutionl Gaussian elimination example 2 2 2 4 2 2 4 gt 1 l 3 0 0 l 4 The second equation gives 0951 0952 1 which cannot be satis ed by any x1 952 so the system has no solution The system is said to be inconsistent Gaussian elimination example 3 2 2 4 2 2 4 gt 1 1 2 0 0 0 The second equation is redundant The rst equation states 2951 2952 4 which has lots of solutions they are all on the line 951 x2 2 If we let x2 t then the set of all solutions can be written as the parametric family 952 t 951 2 t for all possible values oft The system is said to have Gaussian elimination example 4 2 2 1 multiple solutions 2 2 1 4 4 1 1 1 3 0 0 12 1 Here 953 2 If we let x2 t then the set of all solutions can be written as the parametric family 953 2 x2 t 951 1 t for all possible values of t Gaussian elimination example 5 Here we will have to interchange rows 1 and 2 0 1 2 1 lt1 13 cl 2 At each step of Gaussian elimination the element that will eventually be the rst nonzero of its row is called the pivot Performing the elementary row operations is called pivoting Note that when a row operation of the form RZ lt RZ ch is performed to zeroize 7 the element under the pivot in the i th row j should generally be the pivot row otherwise the row echelon structure may be lost 113 012 gt gtxg2x11 gtxlt GaussJordan elimination Ab er Ols Here T has a reduced row echelon form in which the rst nonzero element in each row is equal to 1 it is to the right by at least one of the rst nonzero element in any row above it and all other elements in its column are zero Example 130 4 A 231b2 011 2 5 00 2 3 1 2 gt 0 9 1 10 0 1 1 2 0 1 1 2 1 0 13 23 1 0 0 1 gt 0 1 19 109 gt 0 1 0 1 gtx 1 0 0 89 89 0 0 1 1 0 In general if A is m x n then the system Ax b is referred to as an m x 71 system If m lt n the system is called underdetermined because there are usually but not always too few equations to determine a unique solution If m gt n the system is called overdetermined because there are usually but not always too many equations to afford a feasible solution The homogeneous system A95 0 always has a trivial solution x 0 If there are any other solutions they are called nontrivial If x is a nontriVial solution then so is km for any scalar k so that there are in nitely many solutions An m x n homogeneous system will always have nontriVial solutions of m lt n Rank of a Matrix 0 The row rank of a matrix A is the maximum number of LI rows in A o The column rank of a matrix A is the maximum number of LI columns inA oExample l l 0 0 A 0 0 l 2 l l l 2 has row rank 2 The rst two rows are LI and last row sum of rst two rows The column rank is also 2 since cols l and 3 say are Ll and cols 2 and 4 are LC s of these columns 0 In general row rankAcolumn rank This common number is call the rank of A
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