Matrix Solutions to Linear Systems of ODEs
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1 Matrix Solutions to Linear Systems of ODEs James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University November 3, 216
2 Outline 1 Symmetric Systems of ODEs 2
3 Symmetric Systems of ODEs Example For the symmetric system below [ [ x (t) 2 12 y = (t) 12 5 [ [ x() 1 = y() 2 [ x(t) y(t) Find the characteristic equation Find the general solution Solve the IVP
4 Symmetric Systems of ODEs Solution The characteristic equation is ( [ 1 det r 1 [ ) = Thus ([ r = det 12 r 5 = (r + 2)(r 5) 144 = r r 244 ) Hence, eigenvalues or roots of the characteristic equation are r 1 = 9.83 and r 2 =
5 Symmetric Systems of ODEs Solution For eigenvalue r 1 = 9.83, substitute the value into [ [ [ [ r V V1 12 r V 2 V 2 Picking the top row, we get 29.83V 1 12V 2 = implying V 2 = 6.18V 1. Letting V 1 = a, we find V 1 = a and V 2 = 6.18a: so [ [ V1 1 V = = a 6.18 V 2 Let a = 1/ [ 1 eigenvector 6.18 T = 1/6.26 =.16 and choose the unit [ E 1 = = [.16.99
6 Symmetric Systems of ODEs Solution So one of the solutions is [ x1 (t) = E y 1 (t) 1 e 9.83t = [ e 9.83t. For eigenvalue r 2 = 24.83, substitute the value into [ [ [ [ r V V1 12 r 5 V V 2 Picking the top row, we get 4.83V V 2 = implying V 2 =.16V 1. Letting V 1 = b, we find V 1 = b and V 2 =.16b: so [ [ V1 1 V = = b V 2.16
7 Symmetric Systems of ODEs Solution Let b = 1/ [ 1 eigenvector.16 T = 1/1.1 =.99 and choose the unit [ E 2 = [ = Note < E 1, E 2 >= (.16)(.99) + (.99)(.16) =. So the other solution is [ [ x2 (t) = E y 2 (t) 2 e 29.83t = The general solution [ x(t) = A E y(t) 1 e 9.83t + B E 2 e 29.83t [ [.16 = A e 9.83t.99 + B e 29.83t. e 29.83t
8 Symmetric Systems of ODEs Solution Find A and B: use the ICs. [ [ [ x() 1.16 = = A e + B y() 2.99 [ [ = A + B [ e Thus, [ [ [ A 1 E1 E 2 = [ E B 2 1 T E2 T [ A = B [ [ E1 T E2 T 1 2 [ [ A E1 E 2 = [ E B 1 T E2 T [ 1 2 Letting D = [ 1 2 T, A =< E1, D > and B =< E 2, D >.
9 The coefficient matrix is A = [ and the eigenvalues of A are λ 1 = 9.83 and λ 2 = Then we know if P = [ E 1 E 2 then P T A P = [ λ1 λ 2 [ λ1 = A = P P T λ 2 Let Λ = [ λ1 λ 2 = A = PΛP T
10 A 2 = (PΛP T ) (PΛP T ) ( ) = PΛ P T P ΛP T = PΛ 2 P T ) ) ) ) A 3 = (PΛP (A T 2 = (PΛP (PΛ T 2 P T = PΛ 3 P T It is easy to see by POMI that A n = PΛ n P T. ( ) = PΛ P T P Λ 2 P T Recall for the scalar t, At = ta simply multiplies each entry of A by the number t. Hence, from the above we have A n t = PΛ n P T t. Consider [ [ [ λ1 At = P P T λ1 E11 t E t = P 12 t λ 2 λ 2 E 21 t E 22 t [ λ1 t = P P T λ 2 t
11 Thus, [ [ λ1 t E11 E At = P 12 λ 2 t E 21 E 22 A similar calculation shows [ [ A 2 t 2 λ 2 = P 1 t 2 E11 E 12 λ 2 2 t2 E 21 E 22 And the function [ λ 2 = P 1 t 2 λ 2 P T 2 t2 W 3 (t) = I + At + A 2 t 2 /2! + A 3 t 3 /3! [ 1 + λ1 t + λ = P 2 1 t2 /2 + λ 3 1 t3 /6 1 + λ 2 t + λ 2 2 t2 /2 + λ 3 2 t3 /6 [ 3 = P k= (λk 1 tk )/k! 3 P T k= (λk 2 tk )/k! P T
12 In general W n (t) = [ n P k= (λk 1 tk )/k! n k= (λk 2 tk )/k! P T
13 In general W n (t) = [ n P k= (λk 1 tk )/k! n k= (λk 2 tk )/k! We know e λ1t = lim n n k= (λk 1 tk )/k! and e λ2t = lim n n k= (λk 2 tk )/k! P T
14 In general W n (t) = [ n P k= (λk 1 tk )/k! n k= (λk 2 tk )/k! We know e λ1t = lim n n k= (λk 1 tk )/k! and e λ2t = lim n n k= (λk 2 tk )/k! This suggests [ n lim W limn n(t) = P k= (λk 1 tk )/k! n n lim n k= (λk 2 tk )/k! [ e λ 1t = P e λ2t P T P T P T
15 In general W n (t) = [ n P k= (λk 1 tk )/k! n k= (λk 2 tk )/k! We know e λ1t = lim n n k= (λk 1 tk )/k! and e λ2t = lim n n k= (λk 2 tk )/k! This suggests [ n lim W limn n(t) = P k= (λk 1 tk )/k! n n lim n k= (λk 2 tk )/k! [ e λ 1t = P e λ2t P T P T P T Define the matrix e Λt = [ e λ 1t e λ2t
16 The system of ODEs X = AX can be written as X = PΛP T X. Let Y = P t X.
17 The system of ODEs X = AX can be written as X = PΛP T X. Let Y = P t X. Then the system becomes Y = ΛY. with general solution [ [ [ αe λ 1t e λ 1t α Y (t) = βe λ2t = e λ2t β for arbitrary α and β. Define Θ = [ α β
18 The system of ODEs X = AX can be written as X = PΛP T X. Let Y = P t X. Then the system becomes Y = ΛY. with general solution [ [ [ αe λ 1t e λ 1t α Y (t) = βe λ2t = e λ2t β for arbitrary α and β. Define Θ = [ α β Then the general solution is Y (t) = P T X = e Λt Θ
19 The system of ODEs X = AX can be written as X = PΛP T X. Let Y = P t X. Then the system becomes Y = ΛY. with general solution [ [ [ αe λ 1t e λ 1t α Y (t) = βe λ2t = e λ2t β for arbitrary α and β. Define Θ = [ α β Then the general solution is Y (t) = P T X = e Λt Θ This gives X (t) = Pe Λt Θ
20 The system of ODEs X = AX can be written as X = PΛP T X. Let Y = P t X. Then the system becomes Y = ΛY. with general solution [ [ [ αe λ 1t e λ 1t α Y (t) = βe λ2t = e λ2t β for arbitrary α and β. Define Θ = [ α β Then the general solution is Y (t) = P T X = e Λt Θ This gives X (t) = Pe Λt Θ Now define the matrix e At = Pe Λt P T. Then e At P = Pe Λt and so X (t) = e At PΘ.
21 Note PΘ = [ [ α E 1 E 2 = [ αe β 1 βe 2
22 Note PΘ = [ [ α E 1 E 2 = [ αe β 1 βe 2 So for initial data, X = [ x 1 x 2 T, we have [ x 1 x 2 = [ αe 1 βe 2 implying α =< X, E 1 >= x 1 and β =< X, E 1 >= x 2
23 Note PΘ = [ [ α E 1 E 2 = [ αe β 1 βe 2 So for initial data, X = [ x 1 x 2 T, we have [ x 1 x 2 = [ αe 1 βe 2 implying α =< X, E 1 >= x 1 and β =< X, E 1 >= x 2 Thus, the solution to the initial value problem is X (t) = e At X.
24 Note PΘ = [ [ α E 1 E 2 = [ αe β 1 βe 2 So for initial data, X = [ x 1 x 2 T, we have [ x 1 x 2 = [ αe 1 βe 2 implying α =< X, E 1 >= x 1 and β =< X, E 1 >= x 2 Thus, the solution to the initial value problem is X (t) = e At X. And the general solution to the dynamics has the form X (t) = e At C where C is an arbitrary vector.
25 The general solution to the scalar ODE x = λx is x(t) = e λt c and we now know we can write the general solution to the vector system X = AX as X (t) = e At C also as long as we interpret the exponential matrix e At right.
26 The general solution to the scalar ODE x = λx is x(t) = e λt c and we now know we can write the general solution to the vector system X = AX as X (t) = e At C also as long as we interpret the exponential matrix e At right. Our argument was for 2 2 symmetric matrices, but essentially the same argument is used in the general n n case but the canonical form of A we need is called the Jordan Canonical Form which is discussed in more advanced classes.
27 Homework Prove that A n = PΛ n P T by POMI. 4.2 Show via POMI 4.3 Calculate e At for A = [ A n t n λ n = P 1 t n λ n P T 2 tn [
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