Vector Differential Equations: Defective Coefficient Matrix and Higher Order Linear Differential Equations

Transcription

1 Vector Differential Equations: Matrix and Higher Order Differential Equations Calculus III Summer 2013, Session II Thursday, August 1, 2013

2 Agenda 1. Vector equations: defective coefficient matrix 2. equations of order n A taste of what s to come

3 Introduction We ve learned how to find a matrix S so that S 1 AS is almost a diagonal matrix. Recall that diagonalization allows us to solve linear systems of diff. eqs. because we can solve the equation y = ay. Jordan form will give us small systems that look like Is there an obvious solution? A nontrivial one? Yes! y 1 = ay 1 + y 2, y 2 = ay 2. y 1 (t) = e at and y 2 (t) = 0. y 1 (t) = te at and y 2 (t) = e at. Write this in the vector form [ ] ( y1 = e at t y 2 [ ] [ ]) 0. 1

4 2 2 defective systems Switching back to the standard basis, these are the solutions x 1 (t) = e at v 1 and x 2 (t) = e at (tv 1 + v 2 ) where v 2, v 1 is a chain of generalized eigenvectors. Example Find the general solution to x = Ax, A = [ ] The single eigenvalue is λ = Chain of generalized e-vectors is v 1 = (1, 3), v 2 = (0, 1). (A 3I)v 1 = 0 and (A 3I)v 2 = v Fundamental set of solutions is therefore x 1 (t) = e 3t v 1 and x 2 (t) = e 3t (tv 1 + v 2 ).

5 Longer chains What about chains of generalized eigenvectors longer than 2? If A is an n n matrix with eigenvalue λ and chain of generalized eigenvectors v 1 = (A λi) p 1 v, v 2 = (A λi) p 2 v,... v p 1 = (A λi)v, v p = v, check that the following are solutions to x = Ax: x 1 (t) = e λt v 1 x 2 (t) = e λt (v 2 + tv 1 ). x p (t) = e λt ( v p + tv p (p 1)! tp 1 v 1 )

6 Longer chains We should also check that {x 1 (t),..., x p (t)} is independent. We know that {v 1,..., v p } is independent, that is, det ([ v 1 v 2 v n ]) 0. Theorem The set {x 1 (t),..., x p (t)} is a linearly independent subset of V n (I). Thus, we can construct a fundamental set of solutions by applying the foregoing construction to each chain of generalized eigenvectors.

7 Example Find the general solution to x = Ax if A = Only eigenvalue is λ = Yesterday we found the chain v 1 = ( 2, 0, 1), v 2 = (0, 1, 0), v 3 = ( 1, 0, 0). 3. Thus, solutions are x 1 (t) = e t v 1, x 2 (t) = e t (v 2 + tv 1 ), x 3 (t) = e t ( v 3 + tv t2 v 3 ).

8 Example Find the general solution to x = Ax if A = Eigenvalues are λ 1 = 2 and λ 2 = Eigenvectors and generalized eigenvectors are Ae 1 = 2e 1, Ae 2 = 2e 2 + e 1, Ae 3 = 5e 3, Ae 4 = 5e 4, Ae 5 = 5e 5 + e 4, Ae 6 = 5e 6 + e Our fundamental set of solutions is x 1 (t) = e 2t e 1, x 2 (t) = e 2t (e 2 + te 1 ), x 3 (t) = e 5t e 3, x 4 (t) = e 5t e 4, x 5 (t) = e 5t (e 5 + te 4 ), x 6 (t) = e 5t ( e 6 + te t2 e 6 ).

9 Introduction We now turn our attention to solving linear equations of order n. The general form of such an equation is a 0 (x)y (n) + a 1 (x)y (n 1) + + a n 1 (x)y + a n (x)y = F (x), where a 0, a 1,..., a n, and F are functions defined on an interval I. The general strategy is to reformulate the above equation as Ly = F, where L is an appropriate linear transformation. In fact, L will be a linear operator.

10 Recall that the mapping D : C 1 (I) C 0 (I) defined by D(f) = f is a linear transformation. This D is called the derivative operator. Higher order derivative D k : C k (I) C 0 (I) are defined by composition: so that D k = D D k 1, D k (f) = dk f dx k. A linear operator of order n is a linear combination of derivative of order up to n, defined by L = D n + a 1 D n a n 1 D + a n, Ly = y (n) + a 1 y (n 1) + + a n 1 y + a n y, where the a i are continous functions of x. L is then a linear transformation L : C n (I) C 0 (I). (Why?)

11 Examples Example If L = D 2 + 4xD 3x, then We have Ly = y + 4xy 3xy. L (sin x) = sin x + 4x cos x 3x sin x, L ( x 2) = 2 + 8x 2 3x 3. Example If L = D 2 e 3x D, determine 1. L ( 2x 3e 2x) = 12e 2x 2e 3x + 6e 5x 2. L ( 3 sin 2 x ) = 3e 3x sin 2x 6 cos 2x

12 Homogeneous and nonhomogeneous equations Consider the general n-th order linear equation a 0 (x)y (n) + a 1 (x)y (n 1) + + a n 1 (x)y + a n (x)y = F (x), where a 0 0 and a 0, a 1,..., a n, and F are functions on an interval I. If a 0 (x) is nonzero on I, then we may divide by it and relabel, obtaining y (n) + a 1 (x)y (n 1) + + a n 1 (x)y + a n (x)y = F (x), which we rewrite as Ly = F (x), where L = D n + a 1 D n a n 1 D + a n. If F (x) is identically zero on I, then the equation is homogeneous, otherwise it is nonhomogeneous.

13 The general solution If we have a homogeneous linear equation Ly = 0, its solution set will coincide with Ker(L). In particular, the kernel of a linear transformation is a subspace of its domain. Theorem The set of solutions to a linear equation of order n is a subspace of C n (I). It is called the solution space. The dimension of the solutions space is n. Being a vector space, the solution space has a basis {y 1 (x), y 2 (x),..., y n (x)} consisting of n solutions. Any element of the vector space can be written as a linear combination of basis vectors y(x) = c 1 y 1 (x) + c 2 y 2 (x) + + c n y n (x). This expression is called the general solution.

14 The Wronskian We can use the Wronskian y 1 (x) y 2 (x) y n (x) y 1 W [y 1, y 2,..., y n ](x) = (x) y 2 (x) y n(x) y (n 1) 1 (x) y (n 1) 2 (x) y n (n 1) (x) to determine whether a set of solutions is linearly independent. Theorem Let y 1, y 2,..., y n be solutions to the n-th order equation Ly = 0 whose coefficients are continuous on I. If W [y 1, y 2,..., y n ](x) = 0 at any single point x I, then {y 1, y 2,..., y n } is linearly dependent. To summarize, the vanishing or nonvanishing of the Wronskian on an interval completely characterizes the linear dependence or independence of a set of solutions to Ly = 0.

15 The Wronskian Example Verify that y 1 (x) = cos 2x and y 2 (x) = 3(1 2 sin 2 x) are solutions to the equation y + 4y = 0 on (, ). Determine whether they are linearly independent on this interval. W [y 1, y 2 ](x) = cos 2x 3(1 2 sin 2 x) 2 sin 2x 12 sin x cos x = 6 sin 2x cos 2x + 6 sin 2x cos 2x = 0 They are linearly dependent. In fact, 3y 1 y 2 = 0.

16 Nonhomogeneous equations Consider the nonhomogeneous linear equation Ly = F. The associated homogeneous equation is Ly = 0. Theorem Suppose {y 1, y 2,..., y n } are n linearly independent solutions to the n-th order equation Ly = 0 on an interval I, and y = y p is any particular solution to Ly = F on I. Then every solution to Ly = F on I is of the form y = c } 1 y 1 + c 2 y c {{ n y n + y } p, = y c + y p for appropriate constants c 1, c 2,..., c n. This expression is the general solution to Ly = F. The components of the general solution are the complementary function, y c, which is the general solution to the associated homogeneous equation, the particular solution, y p.

17 Something slightly new Theorem If y = u p and y = v p are particular solutions to Ly = f(x) and Ly = g(x), respectively, then y = u p + v p is a solution to Ly = f(x) + g(x). Proof. We have L(u p + v p ) = L(u p ) + L(v p ) = f(x) + g(x). Q.E.D.

18 A taste of what s to come Example Determine all solutions to the equation y + y 6y = 0 of the form y(x) = e rx, where r is a constant. Substituting y(x) = e rx into the equation yields e rx (r 2 + r 6) = r 2 e rx + re rx 6e rx = 0. Since e rx 0, we just need (r + 3)(r 2) = 0. Hence, the two solutions of this form are y 1 (x) = e 2x and y 2 (x) = e 3x. Could this be a basis for the solution space? Check linear independence. Yes! The general solution is y(x) = c 1 e 2x + c 2 e 3x.

19 A taste of what s to come Example Determine the general solution to the equation y + y 6y = 8e 5x. We know the complementary function, y c (x) = c 1 e 2x + c 2 e 3x. For the particular solution, we might guess something of the form y p (x) = ce 5x. What should c be? We want 8e 5x = y p + y p 6y p = (25c + 5c 6c)e 5x. Cancel e 5x and then solve 8 = 24c to find c = 1 3. The general solution is y(x) = c 1 e 2x + c 2 e 3x e5x.