Section 9.8 Higher Order Linear Equations

Transcription

1 Section 9.8 Higher Order Linear Equations Key Terms: Higher order linear equations Equivalent linear systems for higher order equations Companion matrix Characteristic polynomial and equation

2 A linear equation of order n is of the form y (n) + a 1 (t)y (n 1) + +a n 1 (t)y + a n (t)y = F(t). An example is the equation of the vibrating spring that we developed earlier. my'' +μy' + ky = F(t) We have seen that we can replace the single higher-order DE with an equivalent first-order system of dimension n. To do so, we introduce new variables x 1 = y, x 2 = yꞌ,..., x n = y (n 1). We differentiate each of the new variables with respect to obtain the system of differential equations

3 Converting to matrix form we let f 0 = 0 Ft () and we can express the system of DEs as x' = Ax + f. Thus, the equation is homogeneous if and only if F(t) = 0. x 1(t 0) x 2(t 0) = x n(t 0) If there is an initial condition we express it in vector form as xt ( 0 ). Notice, in particular, that if x(t) is a solution to x' = Ax + f then it is the first component of vector x, y(t) = x 1 (t), which is the solution to the higher-order differential equation.

4 Next, we can use the results about systems to find an appropriate existence and uniqueness theorem for a higher-order linear equation. All the ideas about linear independence, fundamental sets, Wronskian, etc apply since we can characterize the higher order DE as a linear system of differential equations. For example we have the following. Suppose that y 1 (t), y 2 (t),..., and y n (t) are linearly independent solutions to equation n th order homogeneous DE y (n) + a 1 (t)y (n 1) + +a n 1 (t)y + a n (t)y = 0. Then every solution to this DE is a linear combination of y 1 (t), y 2 (t),..., and y n (t).

5 The solution strategy for a higher-order equation is almost the same as that for a system. To find the general solution to the nth-order linear, homogeneous equation y (n) + a 1 (t)y (n 1) + +a n 1 (t)y + a n (t)y = 0, we need to find a fundamental set of solutions; that is, a set of n solutions y 1 (t), y 2 (t),..., y n (t) that are linearly independent. Then the general solution is a linear combination y(t) = C 1 y 1 (t) + C 2 y 2 (t)+ +C n y n (t). To find the particular solution to homogeneous DE to y (n) + a 1 (t)y (n 1) + +a n 1 (t)y + a n (t)y = 0, that satisfies the initial conditions y(t 0 ) = y 0, yꞌ(t 0 ) = y 1,..., and y (n 1) (t 0 )= y n 1, we substitute the initial conditions into y(t) = C 1 y 1 (t) + C 2 y 2 (t)+ +C n y n (t). The resulting system of n linear equations to solve for the n unknown constant C 1, C 2,..., C n.

6 Finding a fundamental set of solutions Since we know that y is a solution to Note the constant coefficients. if and only if is a solution to the system xꞌ = Ax, where A is the matrix, we could just find a fundamental set of solutions for the system xꞌ = Ax. Then use that set to find the fundamental set for DE However, there is an easier procedure, that can be derived from information gained from the solutions to the system. We will state the results and give a short explanation.

7 The matrix of the system of DEs is This matrix has a special structure and is called the companion matrix of the polynomial since det(a λi) = (-1) n p(λ). n n n-2 n-1 n p(λ) =λ +a λ + +a λ +a λ+a It follows that the characteristic equation of the matrix A is p(λ) = 0. Hence the eigenvalues of matrix A are the roots of characteristic polynomial p(λ). Note that homogeneous DE is so the coefficients of the DE lead directly to the characteristic polynomial. Once we have the eigenvalues we can proceed in a manner similar to that for systems with a general coefficient matrix. However, we do not need to directly use the matrix exponential. Again it breaks into the three cases: distinct, repeated, and complex roots.

8 Real roots Corresponding to the real root λ of p(λ), we have the solution y(t) = e λt. If the roots are distinct, this is all we need to know. Example: Find a fundamental set of solutions to the equation yꞌꞌꞌ yꞌ = 0. The characteristic equation is p(λ) = λ 3 λ = 0. The roots are 1, 0, and 1. Thus, y 1 (t) = e t, y 2 (t) = e 0t = 1, and y 3 (t) = e t are solutions. Since the roots of the characteristic polynomial are distinct, these solutions are linearly independent. Hence, they form a fundamental set of solutions so the general solution of the 3 rd order DE is y(t) = C e -t t 1 + C 2 + C3e

9 Repeated real root Suppose λ is a root of the characteristic polynomial of algebraic multiplicity q. From previous results we know that we can find q linearly independent solutions of the system xꞌ = Ax of the form Recall that a solution x(t) of xꞌ = Ax is a vector function in the form so the first component is a solution of the higher order DE Now lets look carefully at Each term of this expression is an n-vector so the first component is e λt times a polynomial of degree q 1. So we have y(t) = P(t) e λt where P(t) is some polynomial of degree at most q - 1. Because of algebraic multiplicity q we need q linearly independent expressions of this form. Question: Is there an easy way to get q such functions?

10 Here is where we can use some linear algebra. Recall that the vector space of polynomials of degree q 1 or less has a natural basis {1, t, t 2,..., t q-1 }. So is it possible that the q functions λt λt 2 λt q -1 λt e, te, t e,, t e which are linearly independent are a set of q solutions corresponding to eigenvalue λ? The answer is YES! So we have If λ is a real root to the characteristic polynomial of algebraic multiplicity q, then y 1 (t) = e λt, y 2 (t) = te λt,..., and y q (t) = t q 1 e λt are q linearly independent solutions. Example: Find a fundamental set of solutions to the equation y (4) 2yꞌꞌꞌ + 2yꞌ 1 = 0. The characteristic polynomial is λ 4 2λ 3 + 2λ 1 = (λ 1) 3 (λ + 1). Thus, the roots are ± 1, and 1 has multiplicity 3. From the root λ = 1, we get the solution y 1 (t) = e t. From the root λ = 1, we get three linearly independent solutions y 2 (t) = e t, y 3 (t) = te t, and y 4 (t) = t 2 e t. Thus the general solution is y(t) = C 1 e t + C 2 e t + C 3 te t + C 4 t 2 e t.

11 Complex roots Since the homogeneous DE has real coefficients, the characteristic polynomial does as well. Hence, complex roots come in complex conjugate pairs, λ = α + iβ and λ = α iβ. We have the corresponding solutions Example: Find a fundamental set of solutions to yꞌꞌꞌ + 7yꞌꞌ+ 19yꞌ + 13y = 0. The characteristic polynomial is λ 3 + 7λ λ + 13 = (λ + 1)(λ 2 + 6λ + 13). Thus, the roots are 1, and 3 ± 2i. The real root 1 leads us to the solution y 1 (t) = e t. The complex conjugate pair 3 ± 2i yields the real solutions y 2 (t) = e 3t cos 2t and y 3 (t) = e 3t sin 2t.

12 Repeated complex roots We follow the ideas from repeated real roots. Example: Find a fundamental set of solutions to y (4) + 4yꞌꞌꞌ + 14yꞌꞌ + 20yꞌ + 25y = 0. The characteristic polynomial is λ 4 + 4λ λ λ + 25 = (λ 2 + 2λ + 5) 2. Consequently, we have roots 1 ± 2i, each of multiplicity 2. Thus, we have solutions y 1 (t) = e t cos 2t, y 2 (t) = e t sin 2t, y 3 (t) = te t cos 2t, and y 4 (t) = te t sin 2t.