hapter 6 Higher Dimensional Linear Systems Linear systems in R n : Observations: X x. x n X X a a n..... a n a nn ; Let T the the coordinate change matrix such that T T is in canonical form ;then Y T X solves Y T X T X T T T X Y For block-diagonal matrix... the system Y Y is reduced to total of k smaller linear systems k Yj j Y j Y it su ces to solve Y Y for in the following two forms 2 I 2 2 I 2 (i). ; (ii)... I.. 2 pp Y. Y k 2 2q2q where 2 ; I 2
ase (i): Y Y is each is a linear rst-order DE y y + y 2 y p y p + y p y p y p we start with solving the last equation y p c p e t and substitute it into the one above it: and solve this linear DE: y p y p + y p y p + c p e t e t y p e t y p e t y p e t y p y p cp e t y p c p t + c p y p (c p t + c p ) e t In the same manner, we can "move upward" to solve y p 2 ;then y p 3 ; :::;till nally solve y ase (ii): q ; 2 : We know from planar system (chapter 3), for + i a cos t + b sin t Y e t a sin t + b cos t for q > ;we write Y Y. Y q ; Y j yj y j2 2
We again start with the last DE and move backwards. lving the last equation: Yq 2 Y q ) Y q e t aq cos t + b q sin t a q sin t + b q cos t Substitute into the next one above Y q 2 Y q + Y q This is a planar system of linear nonhomogeneous DEs. it may be solved using the method of "variation of parameters" by looking for solution in the form (optional homework) Y e t aq (t) cos t + b q (t) sin t a q (t) sin t + b q (t) cos t once we solve this system, we can then move upward to solve for Y q 2 ; Y q 3 ; :::; Y successively. We shall introduce another approach to solve this nonhomogeneous system In summary, to solve X X;. we rst nd its canonical form T T : 2. next, we solve Y Y by solving several subproblems in case (i) and/or case (ii) 3. Finally, X T Y is the desired solution. Example lve X X Example 2 lve X X 3
l: T ; The Exponential of Matrix Recall that in solving case (ii), we need to solve nonhomogeneous system Y 2 Y + Y q (t) ; Y q (t) is a given vector function The method of variation of parameter is used then but better methods are need Recall the Taylor series expansion e x X k x k k! it is convergent for all x: for diagonal matrix diag ( ; :::; n ) k diag k ; :::; k n as N! NX k k k! diag NX k N! k! ; :::; X k n! diag e ; :::; e n e k! k k De nition of e exp () e X k k k! Theorem: The above series convergent if we de ne metric in L (R n ) as, for (a ij ) kk max (ja ij j) 4
Example 3. Find e if Example 4. Find e if l: Note that E 2 I; E 3 E; E 4 I: cos sin e sin cos Example 5. Find e if l: e e t t te t e t Example 6. Guess what is e if Properties of exponential of matrices:. If T T; then e T e T? 2. If ; then exp ( + ) e e 3. exp ( ) (exp ()) E 4. If is an eigenvalue of and V is an associated eigenvector, then e is an eigenvalue of e and V is an eigenvector of e associated with e 5. e t e t e t Theorem: e t X is the only solution of X X; X () X 5
Example7: Find general solutions for X X with in Example 3-5:. In Example 3. e ; e t t e t e X e t X t x e t y 2. In Example 4. cos t sin t ; e t sin t cos t X e t X x cos t + y sin t x sin t + y cos t 3. In Example 5. e ; e t t te t e t Nonhomogeneous X e t x e X t + y te t y e t X X + G (t) Variation of Parameters: onsider solution in the form X e t Y Note that when G ; Y (t) constant. function. Substituting it into the equation Now consider Y (t) is a LHS X e t Y e t Y + e t Y e t Y + e t Y RHS X + G (t) e t Y + G (t) 6
and e t Y G (t) ; Y X + lution of nonhomogeneous IVP is X e t X + or Y e t G (t) e s G (s) ds e s G (s) ds Example 8: Find solution for systems in case (ii) of canonical forms X 2 X + G (t) ; G (t) (g (t) :g 2 (t)) is a given vector function l: To nd e t 2 ; we rst try: 2 ; 2 2 2 2 2 2 2 2 3 2 3 3 2 3 + 3 2 3 3 2 3 3 2 It seems not so easy! We try something else. Notice that X e t 2 X is the solution of a X 2 X; X () X : b On the other hand, we know the solution of this planar system is a cos t + b sin t X e t a sin t + b cos t e t 2 a b e t a cos t + b sin t a sin t + b cos t In particular, if we choose a ; b ; and a ; b ; respectively, then we shall see e t 2 cos t sin t e t : sin t cos t ::: 7
Recall a b c d e t 2 cos t sin t e t sin t cos t cos t sin t e t sin t cos t d b ad bc c a : cos t e t sin t sin t cos t and thus X e t 2 X + cos t sin t e t sin t cos t e 2s G (s) ds Example 9: Harmonic oscillators x + R t (g (s) cos s g 2 (s) sin s) e s ds x + R t (g (s) sin s + g 2 (s) cos s) e s ds x + x cos!t : or X X + Recall that cos t sin t e t sin t cos t e t e ( t) cos!t ; cos ( t) sin ( t) sin ( t) cos ( t) ; cos t sin t sin t cos t e s G (s) ds cos s sin s sin s cos s sin s cos!s cos s cos!s cos!s ds 2 ds cos (! + ) t (! + ) sin (! + ) t (! + ) cos (! ) t + 2 (! )! 2 sin (! ) t (! ) 8
X e t X + e t e s G (s) ds Homework: (h), 4, 6, 7, 2(c)(g)(j) 9