SEC 2 1 47 INTRODUCTION Assuming a solution of the form Y2 A2eic t y Ale wt 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 k 2 A k 2A2 n1co A 1 2co k 2 Al k 2A2 Characteristic Value Problems and Quadratic Forms x a 2x 2 0 X2 0 kL k2 2 1 where A is a scalar Using matrix notation we can write 2 1 as ax Ax 2 2 or 2 3 a 21 2 X 0 A1 inYZ1n kjy 2 k 2 y2 Y 0 is a hybrid of the Also called eigenvalue problem in some texts The term eigenvalue German term Eigenwerte and English value 46 712 nrz A 1 k2 2 4A1 e A 2 A2 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 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 m2 y 2 k2 y 2 Y 0 m2 d I and the final equations are Consider the second order homogeneous system a 2 1x1 a2 2 c 2 A A 12 A 2 A 2 INTRODUCTION all A2 We can transform c to a form similar to that of 2 1 by defining new amplitude measures 1 1 1I1n 2 1 2 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 CHARACTERISTIC VALUE PROBLEMS 48 SEC 2 2 CHAP 2 SECOND ORDER CHARACTERISTIC VALUE PROBLEM Solving a Al 6 forms which are encountered in the determination of the relative extrema of a function Chapter 3 the construction of variational principles Chapter 7 and stability criteria Chapters 7 18 This discussion is restricted to the case where a is real Reference 9 contains a definitive treatment of the underlying theory and computational procedures 2 1 2 2 al all 1 x2 A1 11 2 2 49 i 1 I where i SECOND ORDER CHARACTERISTIC VALUE PROBLEM 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 first the case where a Equation 2 4 becomes We know from Cramer s rule that nontrivial solutions of 2 4 all Xi a1z2 2 0 a2 1 xl a2 2 X 2 0 are possible only if the determinant of the coefficient matrix vanishes that is I I Ia l l when U12 2 Ia21 a22 all 0 22 2 7 f11 all a 2 2 al1 a2 2 a 2 a2 1 al and the characteristic equation reduces to 2 1A 32 The second equation is related to the first by all all a 2 2 4 a 12 l a2 2 1 a1 2 a21 X C 1 1 011 X t d l all A a1 2 X 1 x x 1 Q c 1 la a12 I 2511 The characteristic equation for this matrix is 6 0 By definition a e and take cl such that X 1 Tx l 1 This operation is called normalization and the resulting column matrix denoted by Q 1 is referred to as the charac teristic vector for A1 Example 2 1 2 2 67 fI 2 6 2 5 2 2 fl c where cl is an arbitrary constant Continuing we let 2 Since this quantity is never negative it follows that the characteristic values for a symmetrical second order matrix are always real 0 b 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 that a 2 0 the solution of the first equation is When a is symmetrical al 2 a2 1 and 2 a 21 times the first eq Al This follows from the fact that the coefficient matrix is singular second eq 2 8 The roots of 2 8 are the characteristic values of a Denoting the roots by il A2 the solution is 2 9 42 2 A1 2 WIt I 2 A 7 a a 2 xl a2 2 1l x 2 0 We let pi 4 2 Expanding 2 5 results in the following equation usually called the characteristic equation for A O 2 I o 0 all a A alla2 2 a 2 1a 12 1f2 l xl al 2x QTQ 1 If a2 O we work with the second equation 2 10 2 11 2 11 50 CHARACTERISTIC VALUE PROBLEMS Since Q1 is a solution of 2 4 for SEC 2 2 Al we see that aQ Following the same procedure for A2 SECOND ORDER CHARACTERISTIC VALUE PROBLEM 1 we obtain 1 a 2 2 13 2b2 The equations for 1 all 4x 2x 2 0 22 2x Also Q2Q 2 1 2 2 14 These equations are linearly independent and the two independent solutions are x l c 1 0 b X 2 0 C2 The corresponding characteristic vectors are a 12 A2 1 2 2 1 we find and 2 c2 1 2 1 Q2 1 5 2 1 One can easily verify that aQj 2 i 1 2 jQj QIQ 2 Q2Q 1 0 2 all A a 2 X1 2X 1 2c 1 Q Repeating for 0 x cl 1 2 and the normalized solution is 2 15 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 x times the first equation Solving the first equation X C1 Then and Q 1 0 Q2 0 1 c We see that the second equation is we obtain 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 x 2 0 a 0 x1 all x 2 0 QTQ2 2 1 6 are at2 aQ 2 A1 6 where i 2 a 1 3 The characteristic values and corresponding normalized solutions for this matrix are 21 5 a 2 1 15 Now when a is symmetrical the right hand term vanishes …