Systems of Linear Differential Equations CHAPTER 10 Systems of Linear Differential Equations

Contents 10.1 Preliminary Theory 10.2 Homogeneous Linear Systems 10.3 Solution by Diagonalization 10.4 Nonhomogeneous Linear Systems 10.5 Matrix Exponential

10.1 Preliminary Theory Introduction Recall that in Sec 3.11, we have the following system of linear DEs: (1)

We have also the normal form (2)

Linear Systems When (2) is linear, we have the normal form as (3)

Matrix Form of a Linear Systems If we let then (3) becomes (4) If it is homogeneous, (5)

Example 1 (a) If , the matrix form of is

(b) If , the matrix form of is

A solution vector on an interval I is any column DEFINITION 10.1 A solution vector on an interval I is any column matrix whose entries are differentiable functions satisfying (4) on the interval. Solution Vector

Example 2 Verify that on (−, ) are solutions of (6)

Example 2 (2) Solution From we have

Initial Value Problem (IVP) Let Then the problem Solve: Subject to: X(t0) = X0 (7) is an IVP.

Let the entries of A(t) and F(t) be functions continuous on a common interval I that contains t0, Then there exists a unique solution of (7) on I. THEOREM 10.1 Existence of a Unique Solution Let X1, X2,…, Xk be a set of solution of the homogeneous system (5) on I, then X = c1X1 + c2X2 + … + ckXk is also a solution on I. THEOREM 10.2 Superposition Principles

Example 3 Please verify that are solutions of (8)

Example 3 (2) then is also a solution.

Let X1, X2, …, Xk be a set lf solution vectors of the DEFINITION 10.2 Let X1, X2, …, Xk be a set lf solution vectors of the homogeneous system (5) in an interval I. We say the set is linear dependence in the interval of there exist constants c1, c2, …, ck, not all zero, so that c1X1 + c2X2 + … + ckXk = 0 for every t in the interval. If the set of vectors is not linearly dependence on the interval, it is said to be linearly independence. Linear Dependence/Independence

Let be n solution vectors of the homogeneous system (5) on THEOREM 10.3 Let be n solution vectors of the homogeneous system (5) on an interval I. Then the set of solution vectors is linearly Independent on I is and only if the Wronskian Criterion for Linearly Independence Solution

(continued) (9) for every t in the interval. THEOREM 10.3 (9) for every t in the interval. Criterion for Linearly Independence Solution

Example 4 We saw that are solutions of (6). Since they are linearly independent for all real t.

Any set X1, X2, …, Xn of n linearly independent DEFINITION 10.3 Any set X1, X2, …, Xn of n linearly independent solution vectors of the homogeneous system (5) on an Interval I is said to be a fundamental set of solutions on the interval. Fundamental Set of Solution

There exists a fundamental set of solutions for the homogeneous system (5) on an interval I. THEOREM 10.4 Existence of a Fundamental Set Let X1, X2, …, Xn be a fundamental set of solutions of the homogeneous system (5) on an interval I. Then the general solution of the system in the interval is X = c1X1 + c2X2 + … + cnXn where the ci, i = 1, 2,…, n are arbitrary constants. THEOREM 10.5 General Solution—Homogeneous Systems

Example 5 We saw that are linearly independent solutions of (6) on (−, ). Hence they form a fundamental set of solutions. Then general solutions is then (10)

Example 6 Consider the vectors

Example 6 (2) Their Wronskian is

Example 6 (2) Then the general solution is

homogeneous system (5). Then the general solution of the THEOREM 10.6 Let Xp be a given solution of the nonhomogeneous system (4) on the interval I, and let Xc = c1X1 + c2X2 + … + cnXn denote the general solution on the same interval of the associated homogeneous system (5). Then the general solution of the nonhomogeneous system on the interval is X = Xc + Xp. The general solution Xc of the homogeneous system (5) is called the complementary function of the nonhomogeneous system (4). General Solution—Nonhomogeneous Systems

Example 7 The vector is a particular solution of (11) on (−, ). In Example 5, we saw the solution of is Thus the general solution of (11) on (−, ) is

10.2 Homogeneous Linear Systems A Question We are asked whether we can always find a solution of the form (1) for the homogeneous linear first-order system (2)

Eigenvalues and Eigenvectors If (1) is a solution of (2), then X = Ket then (2) becomes Ket = AKet . Thus we have AK = K or AK – K = 0. Since K = IK, we have (A – I)K = 0 (3) The equation (3) is equivalent to

If we want to find a nontrivial solution X, we must have If we want to find a nontrivial solution X, we must have det(A – I) = 0 The above discussions are similar to eigenvalues and eigenvectors of matrices.

Let 1, 2,…, n be n distinct eigenvalues of the matrix THEOREM 10.7 Let 1, 2,…, n be n distinct eigenvalues of the matrix A of (2), and let K1, K2,…, Kn be the corresponding eigenvectors. Then the general solution of (2) is General Solution—Homogeneous Systems

Example 1 Solve (4) Solution we have 1 = −1, 2 = 4.

Example 1 (2) For 1 = −1, we have 3k1 + 3k2 = 0 2k1 + 2k2 = 0 Thus k1 = – k2. When k2 = –1, then For 1 = 4, we have −2k1 + 3k2 = 0 2k1 − 2k2 = 0 Thus k1 = 3k2/2. When k2 = 2, then

Example 1 (3) We have and the solution is (5)

Example 2 Solve (6) Solution

Example 2 (2) For 1 = −3, we have Thus k1 = k3, k2 = 0. When k3 = 1, then (7)

Example 2 (3) For 2 = −4, we have Thus k1 = 10k3, k2 = − k3. When k3 = 1, then (8)

Example 2 (4) For 3 = 5, we have Then (9)

Example 2 (5) Thus

Example 3: Repeated Eigenvalues Solve Solution

Example 3 (2) We have – ( + 1)2(– 5) = 0, then 1 = 2 = – 1, 3 = 5. For 1 = – 1, k1 – k2 + k3 = 0 or k1 = k2 – k3. Choosing k2 = 1, k3 = 0 and k2 = 1, k3 = 1, in turn we have k1 = 1 and k1 = 0.

Example 3 (3) Thus the two eigenvectors are For 3 = 5,

Example 3 (4) Implies k1 = k3 and k2 = – k3. Choosing k3 = 1, then k1 = 1, k2 = –1, thus Then the general solution is

Second Solution Suppose 1 is of multiplicity 2 and there is only one eigenvector. A second solution will be of the form (12) Thus X = AX becomes we have (13) (14)

Example 4 Solve Solution First solve det (A – I) = 0 = ( + 3)2,  = －3, －3, and then we get the first eigenvector Let

Example 4 (2) From (14), we have (A + 3 I) P = K. Then Choose p1 = 1, then p2 = 1/6. However, we choose p1 = ½, then p2 = 0. Thus

Example 4 (3) From (12), we have The general solution is

Eigenvalue of Multiplicity 3 When there is only one eigenvector associated with an eigenvalue of multiplicity 3, we can find a third solution as and

Example 5 Solve Solution (1 – 2)3 = 0, 1 = 2 is of multiplicity 3. By solving (A – 2I)K = 0, we have a single eigenvector

Example 5 (2) Next, solving (A – 2I) P = K and (A – 2I) Q = P, then Thus

Let A be the coefficient matrix having real entires of THEOREM 10.8 Let A be the coefficient matrix having real entires of the homogeneous system (2), and let K1 be an eigenvector corresponding to the complex eigenvalue 1 =  + i ,  and  are real. Then and are solutions of (2). Solution Corresponding to a Complex Eigenvalue

Let 1 =  + i be a complex eigenvalue of the THEOREM 10.9 Let 1 =  + i be a complex eigenvalue of the coefficient matrix A in the homogeneous system (2), and let B1 and B2 denote the column vectors defined in (22). Then (23) are linearly independent solutions of (2) on (－, ). Real Solutions Corresponding to a Complex Eigenvalue

Example 6 Solve Solution First, For 1 = 2i, (2 – 2i)k1 + 8k2 = 0, – k1 + (–2 – 2i)k2 = 0, we get k1 = –(2 + 2i)k2.

Example 6 (2) Choosing k2 = –1, then Since  = 0, then

Fig 10.4

10.3 Solution by Diagonalization Formula If A is diagonalizable, then there exists P, such that D = P-1AP is diagonal. Letting X = PY then X = AX becomes PY = APY, Y = P-1APY, that is, Y = DY, the solution will be

Example 1 Solve Solution From det (A – I) = – ( + 2)(– 1)(– 5), we get 1 = – 2, 2 = 1 and 3 = 5. Since they are distinct, the eigenvectors are linearly independent. For i = 1, 2, 3, solve (A –iI)K = 0, we have

Example 1 (2) then and Since Y = DY, then Thus

10.4 Nonhomogeneous Linear Systems Example 1 Solve Solution First solve X = AX,  = i, −i,

Example 1 (2) Now let we have Thus Finally, X = Xc + Xp

Example 2 Solve Solution First we solve X = AX. By similar procedures, we have 1 = 2, 2 = 7, and Then

Example 2 (2) Now let After substitution and simplification, or

Example 2 (3) Solving the first equations , and substitute these values into the least two equations, we obtain The general solution of the system on (－, ) is X = Xc + Xp or

Example 3 Determine the form pf Xp for dx/dt =5x + 3y – 2e-t + 1 dy/dt =−x + y + e-t – 5t + 7 Solution Since then

Fundamental Matrix If X1, X2,…, Xn is a fundamental set of solutions of X = AX on I, its general solution is the linear combination X = c1X1 + c2X2 +…+ cnXn, or (1)

(1) can be also written as. X = Φ(t)C (1) can be also written as X = Φ(t)C (2) where C is the n  1 vector of arbitrary constants c1, c2,…, cn, and is called a fundamental matrix. Two Properties of (t): (i) nonsingular; (ii) (t) = A(t) (3)

Variation of Parameters In addition, we to find a function such that Xp = Φ(t)U(t) (4) is a particular solution of (5) Since (6) then (7)

Using (t) = A(t), then or (8) Thus and so Since Xp = Φ(t)U(t), then (9) Finally, X = Xc + Xp (10)

Example 4 Find the general solution of on (−, ). Solution First we solve the homogeneous system The characteristic equation of the coefficient matrix is

Example 4 (2) We can get  = −2, −5, and the eigenvectors are Thus the solutions are Thus

Example 4 (3) Now

Example 4 (4) Hence

Example 5 Solve Solution From the same method, we have Then

Example 5 (2) Let X = PY, We have two DEs:

Example 5 (3) The solutions are y1 = 1/5 et + c1 and y2 = –7/20 et + c2 e5t. Thus

10.5 Matrix Exponential Matrix Exponential For any n  n matrix A, (3) DEFINITION10.4 For any n  n matrix A, (3) Matrix Exponential Also A0 = I, A2 = AA, ….

Derivative of eAt (4) Since

Besides, (5) then eAt is also the solution of X = AX.

Computation of eAt

Using the Laplace Transform X = AX, X(0) = I (7) If x(s) = L{X(t)} = L{eAt}, then sx(s) – X(0) = Ax(s) or (sI – A)x(s) = I Now x(s) = (sI – A)–1I = (sI – A)–1. Thus L{eAt} = (sI – A)–1 (8)

Example 1 Use the Laplace Transform to compute eAt, where Solution

Example 1 (2)

Using Powers Am (10) where the coefficient cj are the same in each and the last expression is valid for the eigenvalues 1, 2, …, n of A. Assume that the eigenvalues of A are distinct. By setting  = 1, 2, …n in second expression in (10), we found the cj in first expression by solving n equations in n unknown. From (3) and (2), we have

Replacing Ak and k as finite sums followed by an interchange of the order of summations

Example 2: Using Powers Am Compute eAt, where Solution We already know 1= −1 and 2 = 2, then eAt = b0I + b1A and (14) Using the values of , then we have b0 = (1/3)[e2t + 2e– t], b1 = (1/3)[e2t – e–t].

Example 2 (2) Thus

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