Algebraic Properties of Solutions of Linear Systems

Transcription

1 Algebraic Properties of Solutions of Linear Systems In this chapter we will consider simultaneous first-order differential equations in several variables, that is, equations of the form f 1t,,,x n d f 2t,,,x n 1 dx n f n t,,,x n In addition to equation 1, we will often impose initial conditions on the functions t,,x n t These will be of the form t 0 x 0 1, t 0 x 0 2,, x n t 0 x 0 n 1 Equation 1, together with the initial conditions 1, is referred to as an initial-value problem First-order systems of differential equations also arise from higher-order equations for a single variable yt Every nth-order differential equation for the single variable y can be converted into a system of n first-order equations for the variables EXAMPLE: Convert the differential equation t y, t dy,, x nt dn 1 y n 1 into a system of 2 first-order equations 4 d2 y 2 + dy +3y 0 Solution: Let From this it immediately follows that But the original equation implies therefore So t y and t dy dy t and d d 4 d2 y 2 dy 3y dy d 2 y +3y 2 4 d +3 4 d dy d2 y 2

2 EXAMPLE: Convert the differential equation into a system of n first-order equations Solution: Let From this and 2 it follows that a n t dn y n +a n 1t dn 1 y n 1 ++a 0y 0 2 t y t dy x 3 t d2 y 2 d x 4 t d3 y 3 d dy d d 2 y dx 3 2 x n t dn 1 y d d n 2 y n 1 n 2 d n y d d n 1 y n n 1 a n t dn y n +a n 1t dn 1 y n 1 ++a 0y 0 dx n 1 dx n a n t dn y n a n 1t dn 1 y n 1 a n 2t dn 2 y n 2 a 0y d n 1 y d n y a n 1t +a n 2t dn 2 y n 1 ++a 0y n 2 n a n t dx n a n 1tx n +a n 2 tx n 1 ++a 0 a n t So d x 3 dx n 1 dx n x n a n 1tx n +a n 2 tx n 1 ++a 0 a n t 2

3 EXAMPLE: Convert the initial-value problem d 3 2 y dy + +3y e t ; 3 y0 1, y 0 0, y into an initial-value problem for the variables y, dy/, and d 2 y/ 2 Solution: Let t y From this and 3 it follows that t dy x 3 t d2 y 2 d d 3 y 3 d dy d d 2 y dx 3 2 d 3 y dy +3y e t d 3 y 3 et 2 dy 3y So dx 3 et 2 3 d x 3 dx 3 et 2 3 Moreover, the functions,, and x 3 satisfy the initial conditions 0 y0 1 0 y 0 0 x 3 0 y 0 0 The most general system of n first-order linear equations has the form a 11t ++a 1n tx n +g 1 t dx n a n1 t ++a nn tx n +g n t Ifeachofthefunctionsg 1,g n isidenticallyzero, thenthesystem4issaiobehomogeneous; otherwise it is nonhomogeneous 3 4

4 Now, even the homogeneous linear system with constant coefficients a 11 ++a 1n x n dx n a n1 ++a nn x n is quite cumbersome to handle This is especially true if n is large Therefore, we seek to write these equations in as concise a manner as possible To this end we introduce the concepts of vectors and matrices 5 DEFINITION: A vector x is a shorthand notation for the sequence of numbers,,x n The numbers,,x n, are called the components of x If t,,x n x n t, then t t xt x n t is called a vector-valued function Its derivative dxt/ is the vector-valued function t d t dx n t x n DEFINITION: A matrix A a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn is a shorthand notation for the array of numbers a ij arranged in m rows and n columns 4

5 DEFINITION:LetAbeann nmatrixwithelementsa ij andletxbeavectorwithcomponents,,x n We define the product of A with x, denoted by Ax, as the vector whose ith component is a i1 +a i2 ++a in x n, i 1,2,,n In other words, the ith component of Ax is the sum of the product of corresponding terms of the ith row of A with the vector x Thus, a 11 a 12 a 1n a 11 +a 12 ++a 1n x n a 21 a 22 a 2n a 21 +a 22 ++a 2n x n Ax a n1 a n2 a nn x n a n1 +a n2 ++a nn x n For example, Finally, we observe that the left-hand sides of 5 x 3 x x 3 +17x x 3 +46x x 3 +31x 4 a 11 ++a 1n x n dx n a n1 ++a nn x n are the components of the vector dx/, while the right-hand sides of 5 are the components of the vector Ax Hence, we can write 5 in the concise form where x x n ẋ dx Moreover, if t,,x n t satisfy the initial conditions Ax 6 a 11 a 12 a 1n a 21 a 22 a 2n and A a n1 a n2 a nn t 0 x 0 1, t 0 x 0 2,, x n t 0 x 0 n then xt satisfies the initial-value problem ẋ Ax, xt 0 x 0, where x 0 x 0 1 x x 0 n 5

6 For example, the system of equations can be written in the concise form where x x x 3 d 15 + x 3 dx x 3 ẋ Ax and A DEFINITION: Let c be a number and x a vector with n components,,x n We define cx to be the vector whose components are c,,cx n, that is c c cx c x n cx n DEFINITION: Let x and y be vectors with components,,x n and y 1,,y n respectively We define x+y to be the vector whose components are +y 1,,x n +y n, that is y 1 +y 1 y 2 +y 2 x+y + x n y n x n +y n THEOREM 1: Let xt and yt be two solutions of 6 Then a cxt is a solution for any constant c b xt+yt is again a solution An immediate corollary of Theorem 1 is that any linear combination of solutions of 6 is again a solution of 6 6

7 EXAMPLE: Solve the system d 4 Solution: We first note that 8 can be rewritten in the following matrix form or 8 where x x1 ẋ Ax and A This system of equations can be derived from the second-order differential equation by setting d 2 y +4y y and dy To find two linearly independent solutions of 9 we note that the characteristic equation is with the roots r 1,2 ±2i Consequently, r y 1 t e 0t cos2t cos2t and y 2 t e 0t sin2t sin2t are two solutions of 9 It follows that x1 t y1 t y2 t t y 1t or y 2t is a solution of 8 Hence by Theorem 1, y1 t y2 t xt c 1 +c y 1t 2 y 2t cos2t sin2t c 1 +c 2 2sin2t 2cos2t c1 cos2t+c 2 sin2t 2c 1 sin2t+2c 2 cos2t is a solution of 8 again REMARK: Another way to solve 8 will be discussed in Section 39 7

8 EXAMPLE: Solve the system 10 d 2 + Solution 1: The system can be derived from the second-order differential equation by setting d 2 y 2dy +y y and 1 2 Indeed, if we rewrite the second equation of system 10 as and plug it into the first equation, we get 1 2 x 2 dy y 1 2 x x 2 x 2 x 2 x 2 x 2 2x To find two linearly independent solutions of 11 we note that the characteristic equation is Thus, r 1 is a repeated root Consequently, r 2 2r +1 0 y 1 t e t and y 2 t te t are two solutions of 11 It follows that x1 t y 1 t y 1 t/2 y 2 t y 2 t/2 t y 1 t or y 2 t is a solution of 10 Hence by Theorem 1, y 1 t y 1 t/2 y 2 t y 2 t/2 xt c 1 +c 2 y 1 t y 2 t e t e t /2 e t +te t te t /2 c 1 +c e t 2 te t 0 e t /2 c 1 +c e t 2 te t c2 e t /2 c 1 e t +c 2 te t is a solution of 10 again 8

9 Solution 2: We solve the first equation of system 10 We have dx1 ln t+c e ln e t+c e t+c e c e t so ±e c e t We can easily verify that the function 0 is also a solution of So we can write the general solution in the form c 1 e t where c 1 is an arbitrary constant c 1 e c, or c 1 e c, or c 1 0 We now substitute c 1 e t into the second equation of system 10: x x 2 2c 1 e t + x 2 2c 1 e t This is a first-order linear differential equation see Section 12 Here at 1 so that µt exp at exp e t Multiplying both sides of the equation x 2 2c 1 e t by µt we obtain the equivalent equation Hence e t x 2 2c 1 e t or 2c 1 2c 1 t+c 2 d e t 2c1 so Therefore 2c 1 t+c 2 e t x1 t xt t c 1 e t 2c 1 t+c 2 e t REMARK: Another way to solve 10 will be discussed in Section 310 9

10 EXAMPLE: Solve the system { x x Solution: The system can be derived from the second-order differential equation by setting d 2 y 6dy +5y y and 1 4 Indeed, if we rewrite the first equation of system 12 as and plug it into the second equation, we get dy 7y x x 1 +7 x 1 7x 1 12 x 1 +7 x 1 6x To find two linearly independent solutions of 13 we note that the characteristic equation is r 2 6r+5 0 with the roots r 1 1 and r 2 5 Consequently, y 1 t e t and y 2 t e 5t are two solutions of 13 It follows that x1 t y 1 t t y 1t 7y 1 t/4 or y 2 t y 2t 7y 2 t/4 is a solution of 12 Hence by Theorem 1, y 1 t y 2 t xt c 1 +c y 1t 7y 2 1 t/4 y 2t 7y 2 t/4 e t e c 1 +c 5t e t 7e t 2 /4 5e 5t 7e 5t /4 e t e 5t c 1 +c 3e t 2 /2 e 5t /2 c 1 e t +c 2 e 5t 3c 1 e t /2 c 2 e 5t /2 is a solution of 12 again REMARK: Another way to solve 12 will be discussed in Section 38 Also, in Section 34 we will show that the solutions that we found in the last three examples are the general solutions of the systems 10