Ma2aPracLectures 16-18

# Ma2aPracLectures 16-18 - Lecture 16 For any m × m-matrix M...

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Unformatted text preview: Lecture 16 For any m × m-matrix M = ( m ij ) , 1 ≤ i, j ≤ m, the exponential of M is defined by exp( M ) = e M = ∞ ∑ n =0 M n n ! , when the right hand side converges, with M = I . Notation: ∞ ∑ n =0 ( a n b n c n d n ) = ∑ ∞ n =0 a n ∑ ∞ n =0 b n ∑ ∞ n =0 c n ∑ ∞ n =0 d n Why is this relevant for us? Reason : If we have x ′ = A x , where x = x 1 . . . x m , x ′ = d x dt = dx 1 dt . . . dx m dt , and A an m × m-matrix with constant coeﬃcients, a basic set of solutions is given by the columns of the matrix Φ( t ) = e At Note : Φ ′ ( t ) = Ae At = A Φ( t ). A set of basic solutions are given by the columns of Φ( t ) = ( x (1) x (2) . . . x ( m ) ) , with x ( i ) ( t ) = x ( i ) 1 ( t ) x ( i ) 2 ( t ) . . . x ( i ) m ( t ) , ∀ i ≤ m. We have mostly looked so far at the case where A has distinct eigenvalues. A modification is necessary for repeated eigenvalues ! Now we will look at the exponential of some simple 2 × 2-cases: 1 (i) A is diagonal : A = ( λ 1 λ 2 ) A 2 = ( λ 2 1 λ 2 1 ) , . . . , A n = ( λ n 1 λ n 2 ) So e A = A + A 1 1! + A 2 2! + ··· + A n n ! + . . . = ( 1 0 0 1 ) + ( λ 1 λ 2 ) + ( λ 2 2! λ 2 2 2! ) + ··· + ( λ n 1 n ! λ n 2 n ! ) + . . . = ( 1 + λ 1 + λ 2 1 2! + ··· + λ n n n ! 1 + λ 2 + λ 2 2 n ! + ··· + λ n 2 n ! + . . . ) = ( e λ 1 e λ 2 ) Conclusion : exp ( λ 1 λ 2 ) = ( e λ 1 e λ 2 ) Note : exp ( a b c d ) ̸ = ( e a e b e c e d ) ! (ii) A : upper triangular A = ( 1 α 0 1 ) A 2 = ( 1 2 α 1 ) A 3 = ( 1 3 α 1 ) , . . . , A n = ( 1 nα 1 ) (by induction) Thus e A = ∞ ∑ n =0 A n n ! = ∞ ∑ n =0 ( 1 n ! nα n ! 1 n ! ) = ∑ ∞ n =0 1 n ! ∑ ∞ n =0 nα n ! ∑ ∞ n =0 1 n ! 2 ∞ ∑ n =0 1 n ! = e α ∞ ∑ n =1 n n ! = α ∞ ∑ n =1 1 ( n − 1)! = α ∞ ∑ k =0 1 k ! = αe = ⇒ e A = ( e αe e ) Check : For A = ( 1 α 0 1 ) , e At = ( e t αe t e t ) . (iii) A = ( λ 1 λ ) A 2 = − ( λ 2 2 λ λ 2 ) A 3 = ( λ 3 3 λ 2 λ 3 ) . . . A n = ( λ n nλ n − 1 λ n ) Remembering that A = I , e A = ∑ ∞ n =0 λ n n ! ∑ ∞ n =0 nλ n − 1 n ! ∑ ∞ n =0 λ n n ! Note that for any t , ∞ ∑ n =0 nλ n − 1 t n n ! = ∞ ∑ n =1 λ n − 1 t n ( n − 1)! = t ∞ ∑ k =0 λ k t k k ! = te λt Putting t = 1, e A = ( e λ e λ e λ ) Check : For A = ( λ 1 λ ) , e At = ( e λt te λt e λt ) . 3 It appears hopeless to get a nice expression for e M for an arbitrary m × m matrix M = ( m ij ), even for m = 2. We tackle this problem by appealing to similarity of matrices (also called conjugacy )....
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Ma2aPracLectures 16-18 - Lecture 16 For any m × m-matrix M...

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