System of Linear Differential Equations

Transcription

1 Sysem of Linear Differenial Equaions In "Ordinary Differenial Equaions" we've learned how o solve a differenial equaion for a variable, such as: y'k5$e K2$x =0 solve DE yx = K 5 2 ek2 x C_C1 2$y''C7$y =0 solve DE yx = _C1 sin x C_C2 cos x In many engineering problems, we ofen have a siuaion where we need o solve more han one differenial equaion a he same ime for wo or more unknown variables. For example, a 10-sory building can be modeled as a sysem of 10 equaions, each describing a moion of a sory. We can have 2 or more reserviors in which here are consan in-flow and ou-flow beween hese reserviors and we wan o know he amoun of waer (or a concenraion of some subsances in he waer) a any ime. Le's firs define he erm "Linear sysem". A Linear Sysem is he sysem in he form of: y'= Ay C g y' 1 a 11 a 12 a 13 g 1 where y'= y' 2 y' 3, A = a 21 a 22 a 23, y = a 31 a 32 a 33 y 3, and g = g 2 g 3 if we wrie each equaion separaely (insead of he marix form above), we would ge: y' 1 x = a 11 Ca 12 Ca 13 y 3 Cg 1 y' 2 x = a 21 Ca 22 Ca 23 y 3 Cg 2 y' 3 x = a 31 Ca 32 Ca 33 y 3 Cg 3 I is a "Sysem" because we need o solve 3 unknowns, namely x, x, y 3 x. I is "Linear" because each variables are separae (i.e. we do no have erms like x $ x. If g s 0 (noe ha i is a vecor, no scalar), he sysem is called nonhomogenous. If g =0, he sysem

2 is called homogenous and is of he form y'= Ay. Soluion for Homogeneous and Consan Coefficien Linear Sysem This is he soluion for y'= Ay where a ij of A are jus numbers, no funcions of. From he previous course, we recall ha he ODE y'= ky has he soluion y = C$e k$ where k=consan. Similar o ha, he soluion of y'= Ay has he form of y = xe λ$ where x is a vecor of consans (no a funcion of or y) Subsiue he soluion ino he differenial equaion, we ge y' = λxe λ$ = Ay = Axe λ$ λx = Ax We can see ha his is an eigenvalue problem. λ is he eigenvalue of A wih associaing eigenvecor x. In many cases (no all!!), A n # n will have n linearly independen se of eigenvecors. Therefore, we have n soluions saisfying y'= Ay. λ λ λ = x 1 1 e, = x 2 2 e,, = x n n e A general soluion is a linear combinaion of hese soluions. λ λ λ y = C 1 x 1 1 e +C 2 x 2 2 e C CC n x n n e Remember ha x i are eigenvecors and C i are consans. We can also wrie a marix Y ha conains all he y i as column vecors. Y= If he soluions,,, are all linearly independen, hey form a basis. Then Y is called a Fundamenal Marix and we can wrie: y = Yc

3 c 1 = c 2 c n The quesion is how do we know if hese soluions are linearly independen? To answer his, we define Wronskian as Deerminan of Y. Tha is: 1 n W = De 1 n Noe ha y i i are no jus numbers! They are funcions of The soluion is linearly independen if and only if W is no zero a any one value of (need jus one value) in he inerval considered.

4 Soluion for Nonhomogeneous Linear Sysem A nonhomogeneous linear sysem is he sysem in he form of: y'= Ay C g where g s 0 (zero vecor) The soluion of nonhomogenous sysem is he combinaion of he soluion of homogenous sysem (assuming g = 0, callled homogeneous soluion, and a paricular soluion which occurs due o he effec of g. y = y h C y p where y h is he homogeneous soluion and y p is he paricular soluion. To find he soluion, we use he mehod of Variaion of Parameers. We assume he soluion in he form of: y p = Y u where Y is he fundamenal marix (known from he soluion of homogeneous sysem as disscussed in he previous par), u is a vecor o be deermined. Noe ha u is no a consan. Consequenly y = y h C y p = Y c C Y u Differeniaing o ge y'= Y' c C Y' u CY u' Subsiue his ino y'= Ay C g o ge Y' c C Y' u CY u' = A Y c C A Y u +g We should recognize ha Y is a fundamenal marix conaining all he soluions y i of he homogeneous sysem as column vecors, i.e. Y=. Then

5 Y'= y' 1 y' 2 y' n bu y' 1 = A, y' 2 = A, Y'= y' 1 y' 2 y' n = A A A = A = AY Consequenly, Y' c = A Y c and Y' u = A Y u. And we now have: Y u' = g And since all columns of Y are linearly independen vecors (because we form Y from linearly independen soluions!), we can always find is inverse, Y K1. Muliply his inverse on boh side of he equaion above: Y K1 Y u' = Y K1 g u' = Y K1 g To find u we inegrae he above equaion elemen-by-elemen. u = 0 Y K1 g d To find he paricular soluion, we recall ha y p = Y u = Y 0 Y K1 g d

6 Summary of Seps for a Soluion of y'= Ay C g 1. Find he soluion of homogeneous equaion y'= Ay 1.1 Find he eigenvalues and eigenvecors of A 1.2 The homogenous soluion is of he form y h = c 1 x 1 e λ 1 Cc 2 x 2 e λ 2 CCc n x n e λ n = Y c, where Y is a fundamenal marix 2. Find Y K1 3. Find vecor u = 0 Y K1 g d 4. The paricular soluion is y p = Yu 5. The general soluion is y = y h C y p = YcCYu 6. The only unknown now is c which can be solved from he iniial condiions Noes: Please observe he following noaions: scalars are in normal leer, UPPERCASE or lowercase vecors are in bold lowercase MATRICES ARE IN BOLD UPPERCASE

7 Conversion of Higher Order Differenial Equaions o a Sysem of Firs Order Differenial Equaions We can conver a higher order ODE ino a sysem of firs order ODEs. In some cases (no always!), i is easier o obain a soluion. Suppose we have an n-h order ODE = F, y, y', y'', K1 we can se = y = y' y 3 = y'' = K1 Warning: Please be careful wih he subscrips (equaion number) and he superscrips (order of he differeniaion)! From his, we can se he equaions in he following form: y' 1 = y' 2 = y 3 y' n K1 = y' n = F, y, y', y'', K1 Noe ha if we have an n-h order ODE, we will ge an n n marix. From his, we can solve he sysem of equaions using he mehod oulined above and we will have he soluions in his form: = y = y' y 3 = y'' = K1 The firs soluion is he soluion y ha we wan!!!