lecture_32 (dragged) 2

lecture_32 (dragged) 2 - C such that W ³ x(1 x n ´ t = C...

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MA 36600 LECTURE NOTES: MONDAY, APRIL 13 3 It suffices then to show that x ( c ) = Ψc for some constant c . Consider an initial value problem in the form d dt x ( c ) = P ( t ) x ( c ) , x ( c ) ( t 0 )= x 0 . According to the Existence and Uniqueness Theorem, we know that a unique solution exists. Hence if we Fnd a solution, it must be the solution. We have seen that x ( c ) ( t )= Ψ ( t ) c is a solution to the homogeneous equation for any constant c . We choose c subject to the initial conditions. When t = t 0 ,wehavethesystem of linear algebraic equations Ψ ( t 0 ) c = x 0 . As Ψ ( t ) is a nonsingular matrix for all t , we see that its inverse exists when t = t 0 : c = Ψ ( t 0 ) 1 x 0 . We conclude that x ( t ) x ( p ) ( t )= x ( c ) ( t )= Ψ ( t ) c = n ° k =1 c k x ( k ) = x ( t )= n ° k =1 c k x ( k ) + x ( p ) ( t ) . Wronskian. Consider the homogeneous system d dt x = P ( t ) x . Say ± x (1) ( t ) ,..., x ( n ) ( t ) ² is a set of solutions – not necessarily a fundamental set of solution. DeFne the Wronskian of this set as the determinant W ³ x (1) ,..., x ( n ) ´ ( t )=det x 11 ( t ) x 12 ( t ) ··· x 1 n ( t ) x 21 ( t ) x 22 ( t ) ··· x 2 n ( t ) . . . . . . . . . . . . x n 1 ( t ) x n 2 ( t ) ··· x nn ( t ) where x ( k ) ( t )= x 1 k ( t ) x 2 k ( t ) . . . x nk ( t ) . Then there exists a constant
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Unformatted text preview: C such that W ³ x (1) , . . . , x ( n ) ´ ( t ) = C exp µ tr ¶ t P ( τ ) dτ · in terms of the trace of an n × n matrix tr µ p ij ( t ) · = p 11 ( t ) + p 22 ( t ) + · · · + p nn ( t ) . This is known as Abel’s Formula . We sketch the proof. It suffices to show that dW dt = ( p 11 + p 22 + · · · + p nn ) W. ±irst, recall that the determinant of a matrix is deFned as det ³ x ij ´ = ° σ ° ( σ ) x 1 σ (1) x 2 σ (2) · · · x n σ ( n ) where ° ( σ ) = ± 1 is a sign that depends on the permutation σ . We wish to di²erentiate this expression. Recall the Product Rule: d dt ³ y 1 y 2 ´ = dy 1 dt y 2 + y 1 dy 2 dt d dt ³ y 1 y 2 y 3 ´ = dy 1 dt y 2 y 3 + y 1 dy 2 dt y 3 + y 1 y 2 dy 3 dt . . . d dt ¸ n ¹ k =1 y k º = n ° i =1 dy i dt · ¹ k ° = i y k...
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