Homogeneous Linear Systems of Differential Equations with Constant Coefficients


 Mildred Cooper
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1 Objective: Solve Homogeneous Linear Systems of Differential Equations with Constant Coefficients dx a x + a 2 x a n x n, dx 2 a 2x + a 22 x a 2n x n,. dx n = a n x + a n2 x a nn x n. This is a system of n differential equations for n unknowns x (t),, x n (t). We can put the system in an equivalent matrix form: d x A x, x (t) with unknown being the vector function x(t) =. x n (t) Solution Method: Supposethat A is diagonalizable; that is, there are an invertible matrix P and a diagonal matrix D = λ... The system for u(t) becomes d u P AP u = D u, λ n such that A = PDP. In this case, set x(t) = P u(t). or, equivalently, du λ u, du n. = λ n u n. The last system is a completely decoupled system, which we can easily solve to get solutions u (t) = C e λ t,, u n (t) = C n e λnt. Going back to x we get the solutions to the given system d x/ A x: C e λ t x(t) = P. C n e λnt An equivalent but even simpler formulation of the solution method: First construct a basis of R n consisting of eigenvectors of A: v, v 2,, v n,
2 corresponding to the eigenvalues λ, λ 2,, λ n. Then the solutions of the system of differenttial equations are: x(t) = C e λ t v + C 2 e λ 2t v C n e λnt v n, where C, C 2,, C n are free parameters. What if not diagonalizable? In the case A is not diagonalizable (i.e., you cannot find n linearly independent eigenvectors of A), the system cannot be completely decoupled as above and the problem is a little harder. But even in such cases, the eigenvectors provide a partial but crucial help. In general, if λ is an eigenvalue of A with multiplicity k, we can obtain k linearly independent solutions by considering solutions of the following form: x(t) = e λt v + te λt v t k e λt v k. Plugging this into the system of differential equations and comparing the coefficients, we will get a system of linear equations for vectors v,, v k. The solution space of this linear system turns out to be a k dimensional subspace and thus yields a kparameter family of solutions of the differential equations. See Examples 2 and. EXAMPLE (Simple Eigenvalues). Solve d x = A x for A = The eigenvalues of A are λ =, λ 2 = 2, λ =. For each eigenvalue we can find a corresponding eigenvector: v =, v 2 = 2, v = /2 / Here, v j is an eigenvector associated to eigenvalue λ j (j =, 2, ). This shows matrix A is diagonalizable: A = PDP with /2 λ P = [ v v 2 v ] = 2 /2, D = 0 λ 2 0 = λ 0 0 Now set x(t) = P u(t). The system for u(t) becomes du / u, d u D u, or, equivalently, du 2 / 2u 2, du / u. Solving this decoupled system, we obtain u (t) = C e t, u 2 (t) = C 2 e 2t, u (t) = C e t. Finally, the solutions to the given system are x(t) = P C e t C 2 e 2t = C e t + C 2 e 2t 2 + C e t /2 /2, C e t 2
3 where C, C 2, C are free parameters. EXAMPLE 2 (Repeated Eigenvalues, Diagonalizable). Solve the initial value probelm d x = A x, x(0) = for A = 5 5 The eigenvalues of A are λ =, λ 2 = λ = 2. Set solutions in the following form: a b c x(t) = e t a + e 2t b + te 2t c = e t a 2 + e 2t + te 2t c 2 a b c Plug this into d x/ A x: e t a + 2e 2t b + e 2t c + 2te 2t c = e t A a + e 2t A b + te 2t A c. A comparison of coefficients gives: A a = a A b = 2 b + c A c = 2 c (A + I) a [ ][ ] A 2I I b 0 A 2I c a 6 a 2 6 a b b c c 2 c Solve this system of linear equations for vectors a, b, c: a = a, b = b2 + b 0, c 0 where a,, b are arbitrary constants. The general solutions to the given system are x(t) = a e t + e 2t + b e 2t 0, ( ) 0 where a,, b are free parameters. Finally, let s get the solution to the initial value problem. Let t in equation ( ): a + + b 0 = 2 0
4 This gives a = 0 b 0 2 = 2 2 Plugging this back in ( ), we obtain the answer 9e t + 2e 2t x(t) = e t 5e 2t e t + 7e 2t 2 EXAMPLE (Repeated Eigenvalues, Not Diagonalizable). 5 8 Solve d x/ A x for A = The eigenvalues of A are λ =, λ 2 = λ =. Set solutions in the following form: a b x(t) = e t a + e t b + te t c = e t a 2 + e t + te t a Plug this into d x/ A x: b = 5 7 e t a e t b + e t c te t c = e t A a + e t A b + te t A c. c c 2 c A comparison of coefficients gives: A a = a A b = b + c A c = c (A + I) a [ ][ ] A + I I b 0 A + I c 8 a 2 2 a 2 6 a b b c c 2 c Solve this system of linear equations for vectors a, b, c: 2 2 a = a, b = b + c /2, c = c, 0 where a, b, c are arbitrary constants.
5 The general solutions of the system of differential equations are 2 2 x(t) = e t a + e t b + c /2 + te t c 0 2 2t = a e t + b e t + c e t /2 + t, t where a, b, c are arbitrary constants. EXAMPLE (Complex Eigenvalues). Solve d x = A x for A = The eigenvalues of A are λ = 2 + i, λ 2 = 2 i, λ = 5. For each eigenvalue we can find a corresponding eigenvector: 2i + 2i v = + i, v 2 = i, v = 2 Here, v j is an eigenvector for eigenvalue λ j (j =, 2, ), obtained by solving (A λ j I) v. The solutions to the given system are 2i + 2i x(t) = C e λt v +C 2 e λ2t v 2 +C e λt v = C e ( 2+i)t + i +C 2 e ( 2 i)t i +C e 5t 2, where C, C 2, C are free parameters. Via Euler s formula, we can write down an alternative expression of general solutions: } } x(t) = C Re {e λt v + C 2 Im {e λt v + C e λt v = C e 2t cos(t) sin(t) 2 + C 2 e 2t sin(t) + cos(t) 2 + C e 5t 2, 0 0 where C, C 2, C are free parameters. 5
6 [] Solve d x [2] Solve d x [] Solve d x [] Solve d x [5] Solve d x [6] Solve d x EXERCISES [ ] [ ] 5 5 x, x(0) =. [ ] 2 x x, x(0) = x x x. 0 7 Answers: [ 9e [] x(t) = t e 2t ] e t e 2t [ ] [ ] 2 / + 2t [2] x(t) = C e 2t + C 2 e 2t where C t,c 2 are arbitrary constants [] x(t) = e t + e 2t 6 + e t 6 /2 2/ / [] x(t) = C e t +C 2 e t +C e t 0 where C,C 2,C are arbitrary constants t [5] x(t) = C e t 2 + C 2 e t + C e t t where C,C 2,C are arbitrary constants 2 t 2 + i 2 i [6] x(t) = C e t + C 2 e (+i)t 2i + C e ( i)t + 2i where C,C 2,C are arbitrary constants, or equivalently, 2 2 x(t) = C e t +C 2 e t cos(t) sin(t) 2 +C e t sin(t) + cos(t)
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