LECTURE 3 Nonhomogeneous Linear Systems We now turn our attention to nonhomogeneous linear systems of the form dx (1) = A (t) x (t) + g (t) dt where A (t) is a (potentially t-dependent) matrix and g (t) is some prescribed vector function of t. As in the last lecture, we shall concentrate on 2 × 2 linear systems; as they are simple to compute and yet they still retain the essential features of the general case. Just as in the case a single nonhomogenous linear ODE, the general solution of (1) will be of the form x (t) = xp (t) + xo (t) where xp (t) is a particular solution of (1) and xo (t) is the general solution of the corresponding homogeneous equation dx = A (t) x (t) . (2) dt (This afterall is a consequence of the linearity of the system, not the number of equations.) And so, just as in the case of a single ODE, we will need to know the general solution of homogeneous system (2) in order to solve the nonhomogeneous system (1). 1. Diagonalizable Systems with Constant Coefficients Let’s begin with the simple 2 × 2 system of the form dx = Ax (t) + g (t) (3) dt where A is a constant (i.e. t-independent) diagonalizable matrix. Let (r1 , ξ), and (r2 , η) be the eigenvalue/eigenvector pairs for A. We then have x(1) (t) = er1 t ξ , x(2) (t) = er2 t η as two fundamental solutions of (2). Recall that if we form a matrix C by using the eigenvectors ξ and η as, respectively, the first and second columns then r1 0 C−1 AC = ≡D 0 r2 And so if we define (4) y = C−1 x h (t) = C−t g (t) , and multiply both sides of (3) from the left by C−1 we get dy d dx = C−1 x = C−1 dt dt dt C−1 (Ax (t) + g (t)) = C−1 Ax (t) + C−1 g (t) = C−1 ACC−1 x (t) + C−1 g (t) = Dy (t) + h (t) 1 2. GENERAL DIAGONALIZABLE SYSTEMS 2 Thus, we get (5) or or dy1 dt dy2 dt dy = Dy (t) + h (t) dt r1 0 y1 (t) h1 (t) = + 0 0 y1 (t) h2 (t) dy1 − r1 y1 = h1 (t) dt dy2 (6b) − r2 y2 = h2 (t) dt Thus, the change of variable (4) allows us to convert the original system into an uncoupled pair of inhomogeneous ODEs. (6a) Recall that the general solution of (7) 0 y + p (t) y = g (t) =⇒ 1 y= µ (t) Z Z C µ (t) g (t) + µ (t) where µ (t) = exp p (t) dt In the cases at hand p (r) −→ g (t) −→ −r1 −r2 , , a constant a constant , a constant h1 (t) h2 (t) and accordingly the solutions of (6a) and (6b) are Z r1 t y1 (t) = e e−r1 t h1 (t) dt + c1 er1 t Z y2 (t) = er2 t e−r2 t h2 (t) dt + c2 er2 t Hence, we can write R er1 t R e−r1 t h1 (t) dt + c1 er1 t y (t) = er2 t e−r2 t h2 (t) dt + c2 er2 t as the general solution of the auxiliary, decoupled system (5). To recover the general solution of the original system, all we have to do is multiply the solution y (t) from the left by C, as (8) x (t) = CC−1 x (t) = C C−1 x (t) = Cy (t) 2. General Diagonalizable Systems Recall that the general solution of a single linear ODE y 0 + p (x) y = g (x) (9) is given by (10) y= 1 µ (x) Z µ (x) g (x) dx + C µ (x) Z , µ (x) = exp Let me state this result a little differently. First, note that if g (x) = 0, then C y= µ (x) and so −1 µ (x) p (x) dx 2. GENERAL DIAGONALIZABLE SYSTEMS 3 is interpretable as a fundamental solution for the corresponding homogeneous problem y 0 + p (x) y = 0. Let me denote by ψ (x) this fundamental solution: ψ (x) ≡ µ (x) −1 Z = exp − p (x) dx Then in terms of the fundamental solution ψ (x) of the corresponding homogeneous problem, the general solution of nonhomogeous equation (9) is Z −1 (100 ) y = ψ (x) ψ (x) g (x) dx + Cψ (x) In fact, the general solution of a nonhomgeneous system dx = A (t) x (t) + g (t) dt (11) can also be expressed in terms of the fundamental solutions of the corresponding homogeneous problem dx = A (t) x (t) dt (12) Suppose that the homogeneous system (12) has been solved in such a way that we can express its general solution in terms of a fundamental matrix Ψ (t)1 x0 (t) = Ψ (t) c =⇒ dx0 = A (t) x0 (t) dt I claim that Z x (t) ≡ Ψ (t) (13) Ψ−1 (t) g (t) dt + Ψ (t) c is then the general solution of dx = A (t) x (t) + g (t) dt Before demonstrating this claim, however, note that (13) is a very straightforward generalization of the solution(100 ) of a single nonhomogeneous linear ODE to a system of nonhomogeneous linear ODEs. 1In other words, suppose we have found n independent solutions ψ (1) (t) , . . . , ψ (n) (t) of an n × n homogeous linear system dxo dt = A (t) xo and have rewritten the right hand side of expression of the general solution as a linear combination of the fundamental solutions x0 (t) = c1 ψ (1) (t) + · · · + cn ψ (n) (t) as a matrix product | x0 (t) = ψ (1) (t) | c1 . ψ (n) (t) .. | cn | ··· ≡ Ψ (t) c 2. GENERAL DIAGONALIZABLE SYSTEMS 4 As for the demonstration that (13) is a solution of 11, that’s easy Z d d −1 x (t) = Ψ (t) Ψ (t) g (t) dt + Ψ (t) c dt dt Z Z d dΨ d −1 −1 Ψ (t) g (t) dt + = Ψ (t) g (t) dt + Ψ (t) Ψ (t) c dt dt dt Z = (A (t) Ψ (t)) Ψ−1 (t) g (t) dt + Ψ (t) Ψ−1 (t) g (t) + A (t) Ψ (t) c Z = A (t) Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c + g (t) = A (t) x (t) + g (t) The step Z d Ψ−1 (t) g (t) dt → Ψ (t) Ψ−1 (t) g (t) dt is just the application of the fundamental theorem of calculus. Ψ (t) Example 3.1. dx1 = 2x1 − x2 + et dt dx2 = 3x1 − 2x2 + e−t dt This set of differential equations corresponds to an inhomogeneous system x˙ = Ax + g (t) with t 2 −1 e A= , g (t) = 3 −2 e−t We shall first find the fundamental matrix for the corresponding homogeneous system x˙ = Ax. This means finding the eigenvalues r1 , r2 of A, the correponding eigenvectors ξ1 , ξ2 and then forming the matrix | | Ψ (t) = er1 t ξ1 er2 ξ2 | | each column of which being a fundamental solution. Now 0 = det (A − λI) = det 2−λ 3 −1 −2 − λ = λ2 − 4 + 3 = (λ − 1) (λ + 1) λ=1 =⇒ =⇒ vλ=1 N ullSp (A − λI) = N ullSp 1 = 1 λ = −1 =⇒ =⇒ vλ=−1 N ullSp (A − λI) = N ullSp 1 = 3 1 3 −1 −3 −1 −1 3 3 and so when we can take r1 = 1 , r2 = −1 ξ1 = , ξ2 = 1 1 1 3 =⇒ = N ullSp = N ullSp 1 0 3 0 λ = 1, −1 −1 0 −1 0 2. GENERAL DIAGONALIZABLE SYSTEMS 5 the fundamental matrix for the homogenous x˙ = Ax is t e e−t Ψ (t) = et 3e−t Using the formula −1 1 d −b = −c a ad − bc we find −t 1 3e−t 1 3e −e−t −1 = Ψ (t) = t −et et e (3e−t ) − et e−t 2 −et We can now plug into the formula Z x (t) = Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c a c b d −e−t et for the general solution of x˙ = Ax + g (t). We have t t e e−t c1 e c1 + e−t c2 Ψ (t) c = = et 3e−t c2 et c1 + 3e−t c2 and t −e−t e dt et e−t 1 −2t 3 2 − 2e = dt − 12 e2t + 21 R 3 1 −2t − 2 e dt ≡ R 2 1 2t − 2 e + 12 dt 3 1 −2t 2t + 4e = − 14 e2t + 12 t Z Ψ−1 (t) g (t) = Z 1 2 Z 3e−t −et And so Z Ψ (t) Ψ−1 (t) g (t) dt = = et et 1 t 4e 1 t 4e e−t 3e−t 1 −2t 3 2t + 4e 1 2t − 4 e + 12 t −2t −2t 6t + e − 1 + 2te 6t + e−2t − 3 + 6te−2t Thus, finally we have Z x (t) = Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c t 1 t e c1 + e−t c2 e 6t + e−2t − 1 + 2te−2t 4 + = 1 t −2t et c1 + 3e−t c2 − 3 + 6te−2t 4 e 6t + e : :

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