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Systems of First Order Linear Differential Equations X1

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Systems of First Order Linear Differential Equations X1 Systems of First Order Linear Differential Equations We will now turn our attention to solving systems of simultaneous homogeneous first order linear differential equations. The solutions of such systems require much linear algebra (Math 220). But since it is not a prerequisite for this course, we have to limit ourselves to the simplest instances: those systems of two equations and two unknowns only. But first, we shall have a brief overview and learn some notations and terminology. A system of n linear first order differential equations in n unknowns (an n × n system of linear equations) has the general form: x1′ = a11 x1 + a12 x2 + … + a1n xn + g1 x2′ = a21 x1 + a22 x2 + … + a2n xn + g2 x3′ = a31 x1 + a32 x2 + … + a3n xn + g3 (*) : : : : : : xn′ = an1 x1 + an2 x2 + … + ann xn + gn Where the coefficients aij ’s, and gi’s are arbitrary functions of t. If every term gi is constant zero, then the system is said to be homogeneous. Otherwise, it is a nonhomogeneous system if even one of the g’s is nonzero. © 2008, 2012 Zachary S Tseng D-1 - 1 The system (*) is most often given in a shorthand format as a matrix-vector equation, in the form: x′ = Ax + g ′ x a a ... a x g 1 11 12 1n 1 1 ′ a a ... a x g x2 21 22 2n 2 2 ′ x a31 a32 ... a3n x3 g3 3 = + : : : : : : : : : : : : : : ′ an1 an2 ... ann xn g n xn x′ A x g Where the matrix of coefficients, A, is called the coefficient matrix of the system. The vectors x′, x, and g are ′ x1 x1 g1 ′ x g x2 2 2 ′ x g x3 3 3 x′ = , x = , g = . : : : ′ x g xn n n For a homogeneous system, g is the zero vector. Hence it has the form x′ = Ax . © 2008, 2012 Zachary S Tseng D-1 - 2 Fact : Every n-th order linear equation is equivalent to a system of n first order linear equations. (This relation is not one-to-one. There are multiple systems thus associated with each linear equation, for n > 1.) Examples: (i) The mechanical vibration equation m u″ + γ u′ + k u = F(t) is equivalent to x1′ = x2 − k γ F(t) x′ = x − x + 2 m 1 m 2 m Note that the system would be homogeneous (respectively, nonhomogeneous) if the original equation is homogeneous (respectively, nonhomogeneous). (ii) y″′ − 2 y″ + 3 y′ − 4 y = 0 is equivalent to x1′ = x2 x2′ = x3 x3′ = 4 x1 − 3 x2 + 2 x3 © 2008, 2012 Zachary S Tseng D-1 - 3 This process can be easily generalized. Given an n-th order linear equation (n) (n−1) (n−2) any + an−1 y + an−2 y + … + a2 y″ + a1 y′ + a0 y = g(t). (n−1) (n) Make the substitutions: x1 = y, x2 = y′, x3 = y″, … , xn = y , and xn′ = y . The first n − 1 equations follow thusly. Lastly, substitute the x’s into the original equation to rewrite it into the n-th equation and obtain the system of the form: x1′ = x2 x2′ = x3 x3′ = x4 : : : : : : xn−1′ = xn ′ − a a a a g(t) x = 0 x − 1 x − 2 x − ... − n−1 x + n 1 2 3 n an an an an an Note : The reverse is also true (mostly) *. Given an n × n system of linear equations, it can be rewritten into a single n-th order linear equation. * The exceptions being the systems whose coefficient matrices are diagonal matrices. However, our Eigenvector method will nevertheless be able to solve them without any modification. © 2008, 2012 Zachary S Tseng D-1 - 4 Exercises D-1.1 : 1 – 3 Convert each linear equation into a system of first order equations. 1. y″ − 4y′ + 5y = 0 2. y″′ − 5y″ + 9y = t cos 2 t 3. y(4) + 3y″′ − πy″ + 2πy′ − 6 y = 11 4. Rewrite the system you found in (a) Exercise 1, and (b) Exercise 2, into a matrix-vector equation. 5. Convert the third order linear equation below into a system of 3 first order equation using (a) the usual substitutions, and (b) substitutions in the reverse order: x1 = y″, x2 = y′, x3 = y. Deduce the fact that there are multiple ways to rewrite each n-th order linear equation into a linear system of n equations. y″′ + 6y″ + y′ − 2y = 0 Answers D-1.1 : 1. x1′ = x2 2. x1′ = x2 x2′ = −5x1 + 4 x2 x2′ = x3 x3′ = −9x1 + 5 x3 + t cos 2 t 3. x1′ = x2 x2′ = x3 x3′ = x4 x4′ = 6x1 − 2πx 2 + πx 3 − 3 x4 + 11 0 1 0 0 0 1 0 0 1 0 4. (a) x′ = x (b) x′ = x + −5 4 −9 0 5 tcos2t 5. (a) x1′ = x2 (b) x1′ = −6x1 − x2 + 2 x3 x2′ = x3 x2′ = x1 x3′ = 2x1 − x2 − 6x3 x3′ = x2 © 2008, 2012 Zachary S Tseng D-1 - 5 A Crash Course in (2 ××× 2) Matrices Several weeks worth of matrix algebra in an hour… (Relax, we will only study the simplest case, that of 2 × 2 matrices.) Review topics: 1. What is a matrix ( pl . matrices)? A matrix is a rectangular array of objects (called entries ). Those entries are usually numbers, but they can also include functions, vectors, or even other matrices. Each entry’s position is addressed by the row and column (in that order) where it is located. For example, a52 represents the entry positioned at the 5th row and the 2nd column of the matrix A. 2. The size of a matrix The size of a matrix is specified by 2 numbers [number of rows ] × [number of columns ]. Therefore, an m × n matrix is a matrix that contains m rows and n columns. A matrix that has equal number of rows and columns is called a square matrix . A square matrix of size n × n is usually referred to simply as a square matrix of size (or order) n. Notice that if the number of rows or columns is 1, the result (respectively, a 1 × n, or an m × 1 matrix) is just a vector. A 1 × n matrix is called a row vector , and an m × 1 matrix is called a column vector . Therefore, vectors are really just special types of matrices. Hence, you will probably notice the similarities between many of the matrix operations defined below and vector operations that you might be familiar with. © 2008, 2012 Zachary S Tseng D-1 - 6 3. Two special types of matrices Identity matrices (square matrices only) The n × n identity matrix is often denoted by In. 1 0 0 1 0 0 1 0 I = , I = , etc. 2 0 1 3 0 0 1 Properties (assume A and I are of the same size): AI = IA = A In x = x, x = any n × 1 vector Zero matrices – matrices that contain all-zero entries. Properties: A + 0 = 0 + A = A A 0 = 0 = 0 A 4. Arithmetic operations of matrices (i) Addition / subtraction a b e f a ± e b ± f ± = c d g h c ± g d ± h © 2008, 2012 Zachary S Tseng D-1 - 7 (ii) Scalar Multiplication a b ka kb k = , for any scalar k. c d kc kd (iii) Matrix multiplication a b e f ae + bg af + bh = c d g h ce + dg cf + dh The matrix multiplication AB = C is defined only if there are as many rows in B as there are columns in A. For example, when A is m × k and B is k × n. The product matrix C is going to be of size m × n, and whose ij -th entry, cij , is equal to the vector dot product between the i- th row of A and the j-th column of B. Since vectors are matrices, we can also multiply together a matrix and a vector, assuming the above restriction on their sizes is met. The product of a 2 × 2 matrix and a 2- entry column vector is a b x ax + by = . c d y cx + dy Note 1: Two square matrices of the same size can always be multiplied together. Because, obviously, having the same number of rows and columns, they satisfy the size requirement outlined above. Note 2: In general, AB ≠ BA . Indeed, depending on the sizes of A and B, one product might not even be defined while the other product is. © 2008, 2012 Zachary S Tseng D-1 - 8 5. Determinant (square matrices only) For a 2 × 2 matrix, its determinant is given by the formula a b det = ad − bc c d Note : The determinant is a function whose domain is the set of all square matrices of a certain size, and whose range is the set of all real (or complex) numbers. 6. Inverse matrix (of a square matrix) Given an n × n square matrix A, if there exists a matrix B (necessarily of the same size) such that AB = BA = In, then the matrix B is called the inverse matrix of A, denoted A−1. The inverse matrix, if it exists, is unique for each A. A matrix is called invertible if it has an inverse matrix. a b Theorem : For any 2 × 2 matrix A = , c d its inverse, if exists, is given by 1 d − b A−1 = . ad − bc − c a Theorem : A square matrix is invertible if and only if its determinant is nonzero. © 2008, 2012 Zachary S Tseng D-1 - 9 1 − 2 2 − 3 Examples : Let A = and B = . 5 2 −1 4 1 − 2 2 − 3 2 − 2 − 4 − (− )3 (i) 2 A − B = 2 − = 5 2 −1 4 10 − (− )1 4 − 4 0 −1 = 11 0 1 − 2 2 − 3 2 + 2 − 3 − 8 4 −11 (ii) AB = = = 5 2 −1 4 10 − 2 −15 + 8 8 − 7 On the other hand: 2 − 3 1 − 2 2 −15 − 4 − 6 −13 −10 BA = = = −1 4 5 2 −1+ 20 2 + 8 19 10 (iii) det( A) = 2 − (−10) = 12, det( B) = 8 − 3 = 5.
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