MEC702_Wk_8_Lect_1

# MEC702_Wk_8_Lect_1 - MEC702 Wk 8 Lect 1 Background on...

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Unformatted text preview: MEC702 Wk 8 Lect 1 Background on Matrices Given are the constant c and the matrices of size 2 × 2  a A = 11 a21 Matrix Addition  A+B = a11 a21 a12 a22   b B = 11 b21   a12 b + 11 a22 b21 b12 b22    b12 a + b11 = 11 b22 a21 + b21 ,  a12 + b12 . a22 + b22 • Note that for matrix addition to be possible, both matrices should be of equal size m rows × n columns. Scalar Multiplication cA = c  a11 a21   a12 ca11 = a22 ca21 ca12 ca22  • Note that for scalar multiplication, the constant c goes inside the matrix to all entries. 1 Matrix Multiplication Given two matrices A of size m × p and B of size p × n, we use the so-called method of Run and Dive (run across the row of rst matrix and dive along the column of second matrix). Examples given below: AB =  a11 a21 a12 a22  b11 b21   b12 a b + a12 b21 = 11 11 b22 a21 b11 + a22 b21 a11 b12 + a12 b22 a21 b12 + a22 b22  Note that the matrix multiplication is also about the compatibility of the matrix sizes: AB = (m × p)(q × n) • If p = q , the matrix multiplication is possible and the result matrix is of size m × n. Examples given below:  1 3  2 9 4 7   8 4 3 2 6 3 4 5 9  8 7 6 5 4 3 3 4 9 5  8 6 7 5 5  2 5  (1)(9) + (2)(7) (3)(9) + (4)(7)   68 53 = 93 66 =   (1)(8) + (2)(6) 23 = (3)(8) + (4)(6) 55 20 48  = not possible. Linear System Given a system of linear equations, we obtain a linear system A~x = ~b: a1 x + b1 y = c1 a2 x + b2 y = c2 =⇒  a1 a2 b1 b2     x c = 1 y c2 =⇒ A~x = ~b where A is the coecient matrix, ~x is the variable matrix and ~b is the constant matrix. Similarly, for a system of 3 linear equations in 3 variables a1 x + b1 y + d1 z = c1 a2 x + b2 y + d2 z = c2 a3 x + b3 y + d3 z = c3 =⇒ a1 a2 a3 b1 b2 b3 2 d1 x c1 d2 y = c2 d3 z c3 =⇒ A~x = ~b. Eigenvalues and Eigenvectors Eigenvalues is one of the most important topics in Linear Algebra. Some applications of eigenvalues are studying populations growth, solving systems of dierential equations, in quadratic forms, and in engineering and science. The concept of eigenvalues comes from the eigenvalue problem. Eigenvalue Problem If A is a n × n matrix, can we nd a scalar λ such that it can replace A in the linear system? For what vectors x is the λ satised? The fundamental equation to the eigenvalue problem is Ax = λx. where λ is the called the eigenvalue of A and x is the eigenvector of A.       1 4 1 2 Example. Let A = , and x1 = and x2 = . 2 3 1 −1 Verify that λ1 = 5 is an eigenvalue of A corresponding to x1 and λ2 = −1 is an eigenvalue of A corresponding to x2 . Solution.      1 4 1 5 Ax1 = = 2 3 1 5      1 4 2 −2 Ax2 = = 2 3 −1 1   1 = λ 1 x1 1   2 = (−1) = λ2 x2 . −1 = (5) Finding Eigenvalues and Eigenvectors The procedure to nd the eigenvalues and its corresponding eigenvectors is given below.We have to solve for the equation |λI − A| = 0. Example. Find the eigenvalues and eigenvectors of the matrix A = Solution. We have     1 0 1 4 0 = |λI − A| = λ − 0 1 2 3     λ 0 1 4 0 = − 0 λ 2 3 λ − 1 −4 0 = −2 λ − 3 (λ − 1)(λ − 3) − (−2)(−4) 0 = 0 = λ2 − 4λ − 5 0 = (λ − 5)(λ + 1). 3  1 2  4 . 3 This yields two eigenvalues, λ1 = 5 and λ2 = −1. We now go on to nd the eigenvectors of each of the eigenvectors. For λ1 = 5, substitute into the homogenous system, we have  λ−1 −2  0 0 −4 λ−3   (5) − 1 −4 0 −2 (5) − 3 0   4 −4 0 . −2 2 0 =⇒ = The rst equation is 4x1 − 4x2 = 0 =⇒ x1 = x2 . Let x2 = t, t ∈ R. Then x1 = x2 = t as well. Then the eigenvectors ~x is given by ~x =   x1 x2 =    t 1 =t , t ∈ R. t 1 Thus, all eigenvectors corresponding to λ1 = 5 are non-zero scalar multiples of   1 . 1 For λ1 = −1, substitute into the homogenous system, we have  λ−1 −2 −4 λ−3  0 0  (−1) − 1 −4 −2 (−1) 3   −2 −4 0 . −2 −4 0 =⇒ =  0 0 The rst equation is −2x1 − 4x2 = 0 =⇒ x1 = −2x2 . Let x2 = t, t ∈ R. Then x1 = −2x2 = −2t as well. Then the eigenvectors x is given by x=   x1 x2 =     −2t −2 =t , t ∈ R. t 1 Thus, all eigenvectors corresponding to λ2 = −1 are non-zero scalar multiples of   −2 . 1 4 Systems of ODEs as Models in Engineering Applications Mixing Problem involving Two Tanks Tank T1 and T2 contain initially 100 gallons of water each. In T1 , the water is pure, whereas 150 pounds of fertilizer are dissolved in T2 . By circulating liquid at a rate of 2 gallons/min and stirring (to keep the mixture uniform), the amounts of fertilizer y1 (t) in T1 and y2 (t) in T2 changes with time t. How long should we let the liquid circulate so that T1 will contain at least half as much fertilizer as there will be left in T2 ? Step 1: Setting up the Model The variables y1 (t) = the amount of fertilizer in T1 in time t y2 (t) = the amount of fertilizer in T2 in time t y10 (t) = the rate of change of amount of fertilizer in T1 y20 (t) = the rate of change of amount of fertilizer in T2 y10 (t) = Inow rate − Outow rate = y20 (t) = 2 y2 − 100 2 Inow rate − Outow rate = y1 − 100 2 y1 100 2 y2 . 100 Hence the mathematical model of our mixture problem is the system of rst order ODEs y10 = −0.02y1 + 0.02y2 y20 = 0.02y1 − 0.02y2 Step 2: General Solution The system of ODEs can be written as a linear system (A~x = ~b) as → −0 y = A~y 5 where  0 → −0 y y = 10 , y2 A=  −0.02 0.02 0.02 −0.02  ~y =   y1 y2 Looking at the system as a single equation, we try an exponential function of t (same as λ-method done in Chapter 2). So suppose ~y is one of the solutions of the system = ~xeλt ~y → −0 y λ~xeλt = and equating to the linear system → −0 y λ~xe = A~y λt = A~xeλt λ~x = A~x. Hence we see that we have to nd the eigenvalues and eigenvectors of A to nd the solutions. −0.02 − λ 0.02 |A − λI| = 0 0.02 = 0 −0.02 − λ (−0.02 − λ)(−0.02 − λ) − 0.0004 = 0 λ + 0.04λ = 0 λ(λ + 0.04) = 0 2 λ1 = 0, For λ1 = 0:  −0.02 0.02 0.02 −0.02 λ2 = −0.04.  The rst equation is −0.02x + 0.02y = 0 implies x = y , hence parameterizing, ~x1 = For λ2 = −0.04:      x t 1 = =t . y t 1   0.02 0.02 0.02 0.02 The rst equation is 0.02x + 0.02y = 0 implies y = −x, hence parameterizing,       x t 1 ~x2 = = =t . y −t −1 Since there are two eigenvalues, we have the general solution as ~y = ~y = c ~x eλ1 t + c2 ~x2 eλ2 t 1 1     y1 1 0 1 = c1 e + c2 e−0.04t . y2 1 −1 6 Step 3: Applying initial conditions       y1 = c1 + c2 e−0.04t y1 1 1 = c1 + c2 e−0.04t =⇒ y2 1 −1 y2 = c1 − c2 e−0.04t . Applying initial conditions, y1 (0) = 0 lbs of fertilizer in Tank 1 and y2 (0) = 150 lbs of fertilizer in Tank 2. y1 = 0 = c1 + c2 y2 = 150 = c1 − c2 =⇒ c1 = 75 c2 = −75. Therefore the particular solution to the IVP is     1 1 ~y = 75 − 75 e−0.04t 1 −1 or component-wise, y1 (t) = 75 − 75e−0.04t y2 (t) = 75 + 75e−0.04t . The above shows the plots of y1 and y2 . Notice that as time goes on, the two functions approach a common limit of 75 lbs of fertilizer in each tank. Step 4: Answer to the question As for the question asked, T1 should contain half of the fertilizer thats left in T2 . Hence y1 = 50 and y2 = 100. So y1 = t = 50 = 75 − 75e−0.04t 27.5. Hence the uid should circulate for at least half an hour. 7 ...
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