Solving Systems of Linear Differential Equations with Real Eigenvalues

Transcription

1 Solving Systems of Linear Differential Equations with Real Eigenvalues David Allen University of Kentucky February 18, 2013

2 1 Systems with Real Eigenvalues This section shows how to find solutions to linear systems of differential equations when the eigenvalues of the system matrix are all real. Finding solutions when there are complex eigenvalues is considerably more difficult. But in an instructional setting, most of the concepts can be conveyed within the real eigenvalue context. Back 2

3 A Quasi-triangular Matrix A real matrix that is lower block-triangular with each diagonal block being either a 1 1 matrix or 2 2 matrix with complex eigenvalues is said to be a lower quasi-triangular matrix. A schematic representation is A A 21 A A 31 A 32 A A n1 A n2 A n3 A nn Back 3

4 An Existence Theorem The following is Theorem 3.9 in Stewart [1, page 285]. Fact 1.1. Let A R n n. Then there exists an orthogonal matrix R such that R t AR is quasi-triangular. Moreover, R may be chosen so that any 2 2 diagonal block of R t AR has only complex eigenvalues (which must therefore be conjugates). This decomposition is called the Real Schur form. When all eigenvalues of A are real (as assumed in this presentation), Fact 1.1 implies R t AR is triangular. Back 4

5 A Transformation to Lower Triangular Form The system to be solved is (t) = A (t) (1) with (0) given. If A in (1) is not lower triangular, by Fact 1.1 one may compute an orthogonal matrix R such that R t AR is lower triangular. The algorithm for finding R is far too complicated to present in this class. The algorithm is an intermediate step in solving non-symmetric eigen systems. Refer to the collection of papers in Wilkinson and Reinsch [2]. Back 5

6 Recasting to a Triangular System The system (1) is recast (t) = A (t) (R t (t)) = (R t AR)(R t (t)) (t) = A (t) The algorithm presented in proof of Fact 1.2 that follows, shows how to solve the triangular system. The solution is then back transformed (t) = R (t). Back 6

7 Solving the Triangular System Fact 1.2. Assume a lower triangular matrix A has real elements. The eigenvalues of A are its diagonal elements, so let λ =, = 1,, n. Construct a vector e(t) such that e (t) = t k exp(λ t) where k is the number of occurrences of λ j = λ, j = 1,, 1. Consider the equations (t) = A (t) (2) with (0) given. The solution of is of the form where B is lower triangular matrix. (t) = Be(t) (3) Back 7

8 Construction of e(t) Illustrated To illustrate the construction of e(t), suppose the diagonal elements of A, the eigenvalues of A, are 0, -2, -3, -2, -3. Then 1 exp( 2t) e(t) = exp( 3t) t exp( 2t) t exp( 3t) Consider = 4; λ = 2 and -2 has occurred one time previously, hence, k = 1. Back 8

9 The Proof The proof of Fact 1.2 is constructive in that it gives the recipe for computing B. The proof is by induction. Begin by initializing all elements of B to zero. For = 1 let b 1,1 = 1 (0). Now, with > 1 assume (t) = j=1 b,j e j (t) for = 1,, 1. The th equation is where c j = 1 =1, b,j. 1 (t) λ (t) =,j j (t) j=1 1 = c j e j (t) (4) j=1 Back 9

10 Multiply Equation (4) by exp( λ t): 1 exp( λ t)( (t) λ (t)) = c j exp( λ t)e j (t) (5) Substitute the definition of e j (t) into Equation (5) and integrate both sides: 1 exp( λ t) (t) = c j exp((λ j λ ) t) t k j d t + d (6) j=1 j=1 where d is a constant of integration. Back 10

11 Multiply Equation (6) by exp(λ t): 1 (t) = c j exp(λ t) exp((λ j λ ) t) t k j d t + d exp(λ t). j=1 (7) The integral in each term in equation (7) is to be evaluated. They take a different form depending on the equality or not of λ j and λ. Back 11

12 Case where λ j = λ When λ j = λ, the jth term in Equation (7) evaluates to c j k j + 1 tk j+1 exp(λ t). Let be the index of the next term for which > j and λ = λ j. Add c j k to b,. Back 12

13 Case where λ j = λ When λ j = λ, the jth term in Equation (7) evaluates to k j c j h=0 With h = 0 add k j! (k j h)!(λ λ j ) h+1 tk j h exp(λ j t). (8) c j λ λ j to b,j. If k j > 0 set = j. For each h > 0, find the next smaller value of such that λ = λ j, c and add j k j! to b (k j h)!(λ λ j ) h+1,. Back 13

14 Finding the value of d Let represent the index of the first term where λ = λ. Let c = (k j = 0) b,j j=1 and add (0) c to b,. Proof of Fact 1.2 is complete. Back 14

15 A Numerical Illustration The transformation to triangular form is illustrated numerically here. Consider the kinetic diagram θ 2 θ 1 GI tract Plasma Other θ 4 θ 3 Back 15

16 The System Matrix With θ 1 = 4, θ 2 = 2, θ 3 = 3, and θ 4 = 1, the system matrix is A = The initial conditions are (0) = (100, 0, 0) t. Back 16

17 The Orthogonal Matrix The orthogonal matrix that will transform A to Schur form is R = The Schur matrix is and (0) = R t (0). A = R t AR = Back 17

18 Solution of the Transformed System Apply the algorithm presented above. The solution of the transformed system is (t) = B e(t) exp( 4 t) = exp( t) exp( t) Back 18

19 Solution to the Original System The solution of the original system is (t) = R (t) exp( 4 t) = exp( t) exp( t) Back 19

20 An Exercise Exercise 1.1. Assume the same model and initial conditions as given in the numerical example above, but with θ 1 = 2, θ 2 = 1, θ 3 = 3, and θ 4 = 4. Find the solution for (t) expressed as Be(t). Back 20

21 References [1] G. W. Stewart. Introduction to Matrix Computations. Academic Press, New York, [2] J. H. Wilkinson and C. Reinsch, editors. Handbook for Automatic Computation, volume 2, Linear Algebra. Springer-Verlag, New York, NY, Back 21