Complex Eigenvalues. 1 Complex Eigenvalues

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1 Complex Eigenvalues Today we consider how to deal with complex eigenvalues in a linear homogeneous system of first der equations We will also look back briefly at how what we have done with systems recapitulates what we did with second der equations Complex Eigenvalues Second Order Equations as Systems Complex Eigenvalues We know that to solve a system of n equations written in matrix fm as Ax, we must find n linearly independent solutions x,,x n In the case where A has n real and distinct eigenvalues, we have already solved the system by using the solutions e λ it v i, where λ i and v i are the eigenvalues and eigenvects of A We now consider the case where A has complex eigenvalues We will assume that A has only real entries Then the characteristic polynomial A ri has real coefficients, and therefe any eigenvalues occur in conjugate pairs: r = a + bi and r = a bi Only slightly me surprising is the fact that the eigenvects also occur in conjugate pairs F example, suppose we have eigenvalue r with eigenvect v Then they satisfy the equation A riv = 0 Now if we take the complex conjugate of both sides, and note that both A and I have only real entries, we get A riv = 0 Therefe, an eigenvect associated with r is v! If we have a solution e rt v, we also have its conjugate e rt v, and this means that we also have its real and imaginary parts, since Rex = x + x and Imx = i x x

2 Now let us write the eigenvect split into real and imaginary parts, as v = a + bi Note that a and b are real vects If we also write our eigenvalues with real and imaginary parts as r = λ + µi, then one solution can be rewritten as follows: a + bie λ+µit = a + bie λt cosµt + i sinµt = e λt a cosµt b sinµt + ie λt a sinµt + b cosµt Of course, we also have the complex conjugate of this solution Therefe, we can get both the real and imaginary parts as solutions So we have found two real solutions: ut = e λt a cosµt b sinµt and vt = e λt a sinµt + b cosµt Solve the system 6 0 First we find the eigenvalues of the matrix A in Ax: x 6 λ λ = 6 λ λ + = λ 6λ + = 0 Solving f λ yields λ = 6 ± 6 5 = 6 ± 4i = ± i We only need to find the eigenvect associated with one of these eigenvalues Let s find the eigenvect f λ = + i by solving A λiv = 0 We row-reduce the augmented matrix 6 + i 0 + i 0 i 0 i 0

3 A useful trick to convert a complex value into a real value is to multiply by the complex conjugate, so to get rid of the complex number in the first column of row one, let us multiply by the conjugate + i Then i + i = 9 4i = =, and we get i 0 i 0 9 6i 0 i 0 i 0 i 0 after also dividing through by on row one Then we can subtract row one from row two, and we end the row reduction with: i We note that we now have v + iv = 0 in the first row, and nothing in the second row Note that as expected, we have eliminated at least one row in solving f our eigenvects So we have v = + iv, and v is a free variable Let s assign v =, and then we have the eigenvalue/eigenvect pair λ = + i, and + i So we get a solution of the fm + i + i e +it = e t e it = e t cost + i sint + i Remember: e it = Multiplying through and separating into real and imaginary parts yields e t cost e t sint + i [e t sint + e t cost] e t cost + ie t = sint e t cost e t sint e e t + i t sint + e t cost cost e t sint + i We know that the real and imaginary parts are both solutions, so our general solution is e c t cost e t sint e e t + c t sint + e t cost cost e t sint

4 If we wish to set an initial condition, such as x0 = c : c e 0 cos0 e 0 sin0 e 0 cos0 which gives us the following augmented matrix: 5 0 Row reduction leads to 0 0 so c = and c = are the required constants 5 e + c 0 sin0 + e 0 cos0 e 0 sin0 c, we can solve f c and + c 0 = = 5 Solve the system Eigenvalues: x, x0 = Eigenvects: General Solution: x = Solving the initial condition: x0 = Solution: 4

5 x = x t = x t = Second Order Equations as Systems We know that any der n equation can be converted to a system of n first der equations Let s see what happens when we use this approach to solve a second der equation Solve y + y y = 0 We know the characteristic equation is r + r = 0, which has roots r = and r = Thus we know the general solution is yt = c e t + c e t If we first convert to a system, we set x = y, x = y, and get the following: x t = x x t = x x 0 We find our eigenvalues: λ λ = λ λ + = λ + λ = 0 Thus λ = and λ = are the eigenvalues 5 x

6 We find our eigenvects: x = 0 We have the single relation x = x, so we can use, T F λ =, we solve x = 0 Here we get x = x, so we use, T Thus, our general solution is c e t + c x t = c e t + c e t x t = c e t + c e t Since x = y, we see that we have obtained the same solution as we did befe Solve y + 4y = 0 We know that the characteristic equation is r +4 = 0, so r = ±i Thus our general solution is yt = c cost + c sint If we convert this to a system, we let x = y and x = y to get We get our eigenvalues: λ 4 λ x = x x = 4x x = λ + 4 = 0 So our eigenvalues are λ = ±i We can then find eigenvects: If λ = i, we get i x = 0 4 i e t 6

7 Solving this, we get eigenvect i, T The eigenvect f λ = i is then the conjugate, i, T So expanding the solution cresponding to λ = i and i, T into real and imaginary parts yields So our solution is cost + i sint i c sint cost = sint cost + i + c cost sint cost sint The first row gives c sint c cost Since c could be any value, this is equivalent to the answer we would get from solving the second der equation previously 7