2 Characteristic-Value Problems and Quadratic Forms 2-1. INTRODUCTION Consider the second-order homogeneous system, (all -)x + a,2x-2 0 (2-1) -)X2 0a21x1 + (a2 2 where A is a scalar. Using matrix notation, we can write (2-1) as ax = Ax (2-2) or (a -212)X 0 (2-3) 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 is referred to as a second-orderand corresponding nontrivial solutions of(2-) 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 m2y 2 + k2(y2 -Y) = 0 m2 + kjy2 -k2(y2 -Y) = 0 * Also called "eigenvalue" problem in some texts. The term "eigenvalue" is a hybrid of the German term Eigenwerte and English "value." SEC. 2-1. INTRODUCTION 47 Assuming a solution of the form ' y' = Alewt Y2 = A2eic °t (b) and substituting in (a) lead to the following set of algebraic equations relating the frequency, co, and the amplitudes, AI, A2: (k1 + k2)A -k2A2 = n1co2A 1 c-k 2Al + k 2A2 = 2co2A 2 We can transform (c) to a form similar to that of(2-1) by defining new amplitude measures,* = )2 A = A 12, (d) A.2 = A 2 I and the final equations are kL + k2-k2 -1 ----A1 712 = 4A1 inYZ1n nrz (e) 2 =A2A, + A--1/1-1I1n /'2 The characteristic values and corresponding nontrivial solutions of (e) are related to the natural frequencies and normal mode amplitudes by (d). Note that the coefficient matrix in (e) is symmetrical. This fact is quite significant, as we shall see in the following sections. Y21 M2 k2 Fig. 2-1. A system with two degrees of freedom. Although the application to dynamics is quite important, our primary reason for considering the characteristic-value problem is that results obtained for the characteristic value problem provide the basis for the treatment of quadratic * See Prob. 2-1. 46SEC. 2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM 49 CHAP. 2 48 CHARACTERISTIC-VALUE PROBLEMS Solving (a), forms which are encountered in the determination of the relative extrema of a Al = +6 2= +1 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] A1 ±° x2 = 1and computational procedures. ;11, = +i where i = -I-2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM We know from Cramer's rule that nontrivial solutions of By definition, nontrivial solutions of (2-4) exist only when = 21 or 12. In what follows, we suppose the characteristic values are real. We consider (all -))Xi + a1z22 = 0 (2-4) first the case where a = . Equation (2-4) becomes a2 1xl + (a22 --)X2 = 0 (all -l)xl + al 2x2 -0 (a)are possible only if the determinant of the coefficient matrix vanishes, that is, a2,xl + (a22 -;1l)x2 = 0 when I _ I Ia l l Ia21 --U12 (2->) The second equation is related to the first by a22 -_ a 21 ) times the first eq.Expanding (2-5) results in the following equation (usually called the charac-second eq. = ( (b) teristic equation) for A: \'-O -Al ' 2 -(all + a22)A + (alla2 2 -a21a12) 0 I--o} This follows from the fact that the coefficient matrix is singular. We let f11 = all + a2 2 (2-7) (all -l)(a22 -1) -a1 2 a21 = 0 (c) 1f2 al1a2 2 -a 2a2 1 = al Since only one equation is independent and there are two unknowns, the solu-tion is not unique. We define x(?, x21) as the solution for A = Al. Assuming* and the characteristic equation reduces to that a 2 # 0, the solution of the first equation is 2 -1A + /32= (2-8) 1 -~~~~ X C~~1)011 (d) The roots of (2-8) are the characteristic values of a. Denoting the roots by X(t) _ all -A l il, A2,the solution is a1 2 A1,2 = (WIt + I-42)2 (2-9) where cl is an arbitrary constant. Continuing, we let When a is symmetrical, al2 = a2 1, and X(1 = {x('), x(1 } (e) pi -4 = (all -a2 2)2 + 4(a12)2 and take cl such that (X(1))Tx(l) = 1. This operation is called normalization, 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 the charac-symmetrical second-order matrix are always real. teristic vector for A1. Example 2-1 Q, = c,{+1 -la -(2-10) a12 (I) 2 2511 + 11 fI = 2 -(2)(2)= 67 By definition, fl (2)(5) -(2)(2) 6 QTQ = 1 (2-11) The characteristic equation for this matrix is *If a2 =O, we work with the second equation. A2 -7 + 6 = 0 (a)50 51 CHARACTERISTIC-VALUE PROBLEMS CHAP. 2 Since Q1 is a solution of (2-4) for = Al, we see that aQ = Q1 (2-12) Following the same procedure for =A2, we obtain Q2 = c2 {1 1 a -2b2} (2-13) where (i\2 =1 + all -+-22 at2 Also, Q2Q 2 = 1 (2-14) aQ 2 = i;2Q2 It remains to discuss the case where A1 = iA2. If a is symmetrical, the char-acteristic values will be equal only when al 1 = a22 and a1 2 = a21 = 0. Equa-tion (2-4) takes the form (al -)xl + ()x2 = 0 (a)(0)x1 + (all -)x2= 0 These equations are linearly independent, and the two independent solutions are x(l) = {c1, 0} (b)X(2) = {0, C2} The corresponding characteristic vectors are Q = {+1,0} (2-15) Q2 {0, + 1} If a is not symmetrical, there is only one independent nontrivial solution when the characteristic values are equal. It is of interest to examine the product, QrQ2. From (2-10) and (2-13), we have QTQ2 -c 2 (all --A,)(a -A2) (a)a 12 Now, when a is symmetrical, the-right-hand term vanishes since al -22 = -(a 2 2 -) = -a 1 (b)all -1 and we see that Q Q2 = 0. This result is also valid when the roots are equal. In general, QQ2 #¢ 0 when a is unsymmetrical. Two nth order column vectors U, V having the property that UT V=VTU = 0 (2-16) are said to be orthogonal. Using this terminology, Q1 and Q2 are orthogonal for the symmetrical case. SEC. 2-2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM Example 2-2 (1) [2 2] A1 = +6 2 = + The equations for = 1 = + 6 are -4x, + 2x2= 0 2x -x = 0 We see that the second equation is -times the first equation. Solving the first equation, we obtain X(?) = C1 X1) = 2X(1) = 2c1Then, x( )= cl{1, 2}and the normalized solution is Q, = {1,2} Repeating for = 2 = + 1, we find (2)= c2{1, -}and 2 1 Q2 = / {1, -} = -/5 {2, -1} One can easily verify that and aQj =2 …