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2 Characteristic-Value Problems and Quadratic Forms SEC. 2-1. INTRODUCTION 47 Assuming a solution of the form °t y' = Ale wt' Y2 = A2eic (b) and substituting in (a) lead to the following set of algebraic equations relating the frequency, co, and the amplitudes, AI, A2: 2 2 (k1 + k2)A - k2A2 = n1co A 1 2 c -k 2 Al + k 2A2 = 2co A 2 We can transform (c) to a form similar to that of(2-1) by defining new amplitude Characteristic-Value measures,* = )2 Problems and A = A 12, (d) A.2 = A 2 I and the final equations are Forms kL + k2- k2 - 1 Quadratic 712 = 4A1 --- - A1 inYZ1n nrz (e) A, + A--2 =A2 1/1-1I1n /'2 2-1. INTRODUCTION The characteristic values and corresponding nontrivial solutions of (e) are Consider the second-order homogeneous system, related to the natural frequencies and normal mode amplitudes by (d). Note in (e) is symmetrical. This fact is quite significant, (all - )x + a,2x-2 0 (2-1) that the coefficient matrix as we shall see in the following sections. a21x1 + (a2 2 - )X2 0 where A is a scalar. Using matrix notation, we can write (2-1) as ax = Ax (2-2) Y21 M2 or (2-3) (a - 21 2)X 0 k2 The values of 2 for which nontrivial solutions of (2-1) exist are called the characteristicvalues of a. Also, the problem of finding the characteristic values and corresponding nontrivial solutions of(2-) is referred to as a second-order characteristic-value problem.* The characteristic-value problem occurs naturally in the free-vibration analysis of a linear system. We illustrate for the system shown in Fig. 2-1. The equations of motion for the case of no applied forces (the free-vibration case) are Fig. 2-1. A system with two degrees of freedom. m2y 2 + k2(y2 - Y) = 0 Although the application to dynamics is quite important, our primary reason - k (y - Y) = 0 m2 + kjy2 2 2 for considering the characteristic-value problem is that results obtained for the characteristic value problem provide the basis for the treatment of quadratic The term "eigenvalue" is a hybrid of the * Also called "eigenvalue" problem in some texts. * See Prob. 2-1. German term Eigenwerte and English "value." 46 SEC. 2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM 49 CHAP. 2 CHARACTERISTIC-VALUE PROBLEMS 48 Solving (a), Al = +6 = +1 forms which are encountered in the determination of the relative extrema of a 2 function (Chapter 3), the construction of variational principles (Chapter 7), and (2) stability criteria (Chapters 7, 18). This discussion is restricted to the case where al -2 a is real. Reference 9 contains a definitive treatment of the underlying theory all -1] ± x2 = 1 and computational procedures. A1 ° ;11, = +i where i = -I- 2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM By definition, nontrivial solutions of (2-4) exist only when = 21 or 1. know from Cramer's rule that nontrivial solutions of 2 We In what follows, we suppose the characteristic values are real. We consider (all - ))Xi + a1z2 2 = 0 (2-4) first the case where a = . Equation (2-4) becomes a2 1 xl + (a2 2 -- )X 2 = 0 + al x - 0 (all - l)xl 2 2 (a) are possible only if the determinant of the coefficient matrix vanishes, that is, a ,xl + (a 2 - ; 1l)x 2 = 0 when I _ I I 2 2 a l l -- U12 (2->) The second equation is related to the first by Ia21 a22 -_ a 21 (2-5) results in the following equation (usually called the charac- second eq. = ( ) times the first eq. (b) Expanding - Al teristic equation) for A: \'-O 2 0 I-- o} This follows from the fact that the coefficient matrix is singular. ' - (all + a2 2)A + (alla2 2 - a2 1a 12 ) We let (all - l)(a2 2 - 1) - a1 2 a21 = 0 (c) f11 = all + a2 2 (2-7) Since only one equation is independent and there are two unknowns, the solu­ 1f2 al a - a 2a2 1 = al 1 2 2 tion is not unique. We define x(?, x21) as the solution for A = Al. Assuming* reduces to and the characteristic equation that a 2 # 0, the solution of the first equation is 2 - 1A + /32= (2-8) X C 1 (d) by ~~1)011 -~~~~ The roots of (2-8) are the characteristic values of a. Denoting the roots X(t) _ all - A l il, A2,the solution is a1 2 A , = (WIt + I- 42)2 (2-9) 1 2 where cl is an arbitrary constant. Continuing, we let symmetrical, al = a2 1, and 1 When a is 2 X(1 = {x('), x( } (e) 2 2 pi - 4 = (all - a ) + 4(a 12) 2 2 and take cl such that (X(1))Tx(l) = 1. This operation is called normalization, the charac­ Since this quantity is never negative, it follows that the characteristic values for a and the resulting column matrix, denoted by Q 1, is referred to as symmetrical second-order matrix are always real. teristic vector for A1. Example 2-1 Q, = c,{+1 -la - (2-10) a12 (I) 2 2511 + 1 1 fI = 2 - (2)(2)= 67 By definition, QTQ = 1 fl (2)(5) - (2)(2) 6 (2-11) The characteristic equation for this matrix is *If a2 =O, we work with the second equation. (a) A2 - 7 + 6 = 0 50 CHARACTERISTIC-VALUE PROBLEMS CHAP. 2 SEC. 2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM 51 Since Q1 is a solution of (2-4) for = Al, we see that Example 2-2 aQ = Q (2-12) 1 (1) Following the same procedure for =A2, we obtain [2 2] Q2 = c {1 a - 2b2} (2-13) 2 1 A1 = +6 2 = + where The equations for = 1 = + 6 are (i\2 =1 + all -+- 22 -4x, + 2x 2 = 0 at2 Also, 2x - x = 0 Q2Q = 1 We see that the second equation is - times the first equation. Solving the first equation, 2 (2-14) we obtain aQ 2 = i;2Q2 1 X(?) = C1 X1) = 2X( ) = 2c1 It remains to discuss the case where A1 = iA2. If a is symmetrical, the char­ Then, ( acteristic values will be equal only when al 1 = a22 and a = a = 0. Equa­ x )= cl{1, 2} 1 2 21 and the normalized solution is tion (2-4) takes the form (al - )xl + ()x 2 = 0 (a) Q, = {1,2} (0)x1 + (all - )x 2 = 0 Repeating for = 2 = + 1, we find These equations are linearly independent, and the two independent solutions (2)= c2{1, -} are and x(l) = {c , 0} 1 (b) 2 1 Q X(2) = {0, C2} 2 = / {1, -} = -/5 {2, - 1} One can easily verify that The corresponding characteristic vectors are aQj =2 and j Qj i = 1,2 Q = {+1,0} (2-15) QIQ = Q2Q = 0 Q2 {0, + 1} (2) 2 1 If a is not symmetrical, there is only one independent nontrivial solution when the characteristic values are equal. a:[1 3] It is of interest to examine the product, QrQ2. From (2-10) and (2-13), The characteristic values and corresponding normalized solutions for this matrix are we have 21 = +5 2 = -1 QTQ2 -c 2 (all -- A,)(a -A2) a (a) 12 1 5 Now, when a is symmetrical, the-right-hand term vanishes since Q = {4 , - 1} al - 22 = -(a - 2 2 ) = -a 1 (b) We see that QTQ 2 # 0. Actually, all - 1 and we see that Q Q2 = 0. This result is also valid when the roots are equal. Q 7 In general, QQ2 #¢ 0 when a is unsymmetrical. Two nth order column vectors QTQ 2 = -5 U, V having the property that (3) UT V=VTU = 0 (2-16) are said to be orthogonal. Using this terminology, Q1 and Q2 are orthogonal = +i )2 = -i for the symmetrical case. We have included this example to illustrate the case where the characteristic values are 2 A nrTRI .lTIC.A_\ A I IF PROBLEMS CHAP. SEC. 2-3. SIMILARITY AND ORTHOGONAL TRANSFORMATIONS 53 52 rV, cI I1v-., ,.. ­ We have shown that the characteristic vectors are always linearly indepen­ The equations corresponding to ). = ). are complex. dent when a is symmetrical. They are also independent when a is unsym- - i)xt - 2x = 0 (1 2 metrical, provided that Al A2. Then, ql # 0 except for the case where a is xl- (1 + i)x 2 = 0 unsymmetrical and the characteristic values are equal. If ql # 0, q-t exists and we can express (2-18) as Note that the second equation is (1 - i) times the first equation. The general solution is I- i q-laq = x(l) {1, 2j} (2-19) The matrix operation, p- ( )p where p is arbitrary, is called a similaritytrans- Repeating for = .2, we find formation. Equation (2-19) states that the similarity transformation, q-l'( )q, = C2 {1, W X(2) reduces a to a diagonal matrix whose elements are the characteristic values · ~~-· ·~·L'2 of a. c. When the roots are complex, 22 is the complex conjugate of At,.Now, we take c2 = If a is symmetrical, the normalized characteristic vectors are orthogonal, ( Then, x(2) is the complex conjugate of x 1) We determine c, such that that is, 2 (X( t))Tx( )= 1 QfQ 2 = Q2Q1 = 0 Also, by definition, Finally, the characteristic values and characteristic vectors are QtQ = QQ2 = 1 Al,2-- +i 1 Using these properties, we see that Q1 2- q = LQtJfI [Qt Q2] = In general, the characteristic values are complex conjugate quantities when the elements vectors are complex conjugates. of a are real.
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