G. NAGY ODE October, 8.. Homogeneous Linear Differential Systems Section Objective(s): Linear Di erential Systems. Diagonalizable Systems. Real Distinct Eigenvalues. Complex Eigenvalues. Repeated Eigenvalues. Non-Diagonalizable Systems. Repeated Eigenvalues. Remarks: We introduce systems of linear di erential equations. We focus on homogeneous systems with constant coe cients. If the homogeneous linear di erential system is diagonalizable, then we have a formula for all the solutions. If the homogeneous linear di erential system is not diagonalizable, then the formula above give only half the solutions. The other half second order scalar of the solutions can be found generalizing ideas from equations with repeated roots of their characteristic polynomial.
G. NAGY ODE october, 8... Linear Di erential Systems. Definition. A first order linear di erential system is the equation x (t) =A(t) x(t)+b(t), where the coe cient matrix A, the source vector b, and the unknown vector x are A(t) = 4 a (t) a (t) 7 5, b(t) = 4 b (t) 7 5, x(t) = 4 x (t) 7 a (t) a (t) b (t) x (t) The system above is called: homogeneous i b =, of constant coe cients i A is constant, diagonalizable i A is diagonalizable. Remarks: In this class we focus on homogeneous systems with constant coe cients. Diagonal systems are very simple to solve. Example : Find functions x, x solutions of the first order,, constant coe homogeneous di erential system x =x, x =x. cients, Solution: In this case, the system is decoupled, so we are just solving (independent) scalar equations. Recall that x (t) =c e t and x (t) =c e t. In vector notation we get x(t) = 4 x (t) 7 5 = 4 c e t 7 5 = c 4 7 5 e t + c 4 7 5 e t x (t) c e t C
G. NAGY ODE October, 8... Diagonalizable Systems: Real Eigenvalues. Example : Now, we consider a system where the equations are coupled. Find functions x, x solutions of the follwoing system of ODEs x = x +x, x =x + x. Solution: We saw that solving a decoupled system (diagonal matrix) is easy. If we have a diagonalizable matrix, A = PDP,i.e.,D = P AP. Multiply the di erential equation x = Ax by P, (P x) =(P AP )(P x), so introduce y = P x, and the equation of y is then y = Dy. To find P and D for the given matrix, find the eigenpairs of A. The solution is =4, v = 4 7 5, and - =, v = 4 7 Therefore, matrix A is diagonalizable with P = 4 7 5, D = 4 4 7 5, P = 4 7 That is, y = 4 y 7 5 = 4 7 5 4 x 7 5 = y x 4 x + x x + x The di erential equation for y is y = D y, hence y 4 y 7 5 = 4 4 7 5 4 y 7 5 ) y 7 5 ) 8 >< y =4y ) >: y = y 8 >< y = (x + x ) >: y =. ( x + x ) 8 >< y = c e 4t. >: y = c e t
4 G. NAGY ODE october, 8 We now transform back to x = P y, x x = 4 x 7 5 = 4 7 5 4 c e 4t 7 5 = 4 c e 4t c e t 7 5 c e t c e 4t + c e t that is,. x(t) = 4 x (t) 7 5 = c 4 7 5 e 4t + c 4 7 5 e t x (t) C
G. NAGY ODE October, 8 5 Theorem. (Homogeneous Diagonalizable Systems) If an n n constant matrix A is diagonalizable, with eigenpairs (, v ),, ( n, v n ), then the general solution of x = A x is x(t) =c e t v + + c n e nt v n. Remark: Each function x k (t) =e kt v k is solution of the system x = A x, because x k = k e kt v k, Ax k = A v k e kt = k v k e kt = k e kt v k. Example : Use the theorem above to find the general solution of the IVP apple apple x = A x, A =, x() =. Solution: Find the eigenpairs of A. The solution is apple apple + =, v + =, and - =, v - = So the general solution is x(t) =c + e t apple + c - e t apple Now we find the coe cients c + and c - that satisfy the initial condition apple apple apple apple apple apple = x() = c + + c c+ - ) = 5 5 c - The inverse of the coe apple 5 We conclude that c + = cient matrix is apple 5 = 5 4 and c - =, hence x(t) = e t apple + e t apple 5 ) apple apple c+ 5 = c - apple, x(t) = apple apple = e t +e t e t +5e t. C
G. NAGY ODE october, 8... Diagonalizable Systems: Complex Eigenvalues. Remarks: A real matrix can have complex eigenvalues. But in this case, the eigenpairs come in conjugate pairs, = +, and v = v +. Theorem. (Complex and Real Solutions) If a matrix A has eigenpairs where, ± = ± i, v ± = a ± ib,, a, and b real, then the equation x = A x has fundamental solutions x + (t) =e +t v +, x - (t) =e -t v -, but it also has real-valued fundamental solutions x (t) = a cos( t) b sin( t) e t, x (t) = a sin( t)+b cos( t) e t. Proof of Theorem : We know that the solutions x ± are linearly independent. Now, ( ±i )t x ± =(a ± ib) e = e t (a ± ib) e ±i t = e t (a ± ib) cos( t) ± i sin( t) = e t a cos( t) b sin( t) ± ie t a sin( t)+b cos( t). Therefore, we get x + = e t a cos( t) b sin( t) + ie t a sin( t)+b cos( t) x = e t a cos( t) b sin( t) ie t a sin( t)+b cos( t). Since the di erential equation x = Ax is linear, the functions below are also solutions, x = x+ + x - = a cos( t) b sin( t) e t, x = i x+ x - = a sin( t)+b cos( t) e t.
G. NAGY ODE October, 8 7 Example 4: Find real-valued fundamental solutions to the di erential equation apple x = Ax, A =. Solution: Fist find the eigenvalues of matrix A above, = ( ) ( ) =( ) +9 ) ± =± i. Then find the respective eigenvectors. The one corresponding to homogeneous linear system with coe cients given by + is the solution of the ( + i) 7 4 5 = 4 i 7 5! 4 i 7 5! 4 i 7 5! 4 i 7 ( + i) i i i Therefore the eigenvector v (+) = 4 v+ v + 7 5 is given by v (+) = iv (+) ) v (+) =, v (+) = i, ) v (+) = 4 i 7 5, + =+i. The second eigenvector is the complex conjugate of the eigenvector found above, that is, v (-) = 4 i 7 5, - = i. Notice that v (±) = 4 7 5 ± 4 7 5 i. Then, the real and imaginary parts of the eigenvalues and of the eigenvectors are given by =, =, a = 4 7 5, b = 4 7
8 G. NAGY ODE october, 8 So a real-valued expression for a fundamental set of solutions is given by x () = x () = 4 7 5 cos(t) 4 7 5 sin(t) e t ) x () = 4 sin(t) 7 5 e t, cos(t) 4 7 5 sin(t)+ 4 7 5 cos(t) e t ) x () = 4 cos(t) 7 5 e t. sin(t) C
G. NAGY ODE October, 8 9..4. Diagonalizable Systems: Repeated Eigenvalues. Remark: linear di erential systems with a diagonalizable coefficient matrix with a repeated eigenvalue are very simple to solve. Theorem. (Diagonalizable with Repeated Eigenvalues) Every diagonalizable matrix with a repeated eigenvalue must have the form A = I. Proof of Theorem : Since matrix A diagonalizable, there exists a matrix P invertible such that A = PDP.SinceAis with a repeated eigenvalue, then Put these two fatcs together, D = 4 7 5 = I. A = P IP = PP = I. Remark: : The di erential equation x = I x is already decoupled. ) x = x x ) too simple. = x
G. NAGY ODE october, 8..5. Non-Diagonalizable Systems: Repeated Eigenvalues. Example 5: Find fundamental solutions to the system apple x 4 = A x, A = Solution: We start computing the eigenvalues of A. p( )= 4 =( + )( + ) + 4 = +8 + = ( + 4). We have a repeated eigenvalue = 4. The eigenvector v is the solution of (A +4I)v =, 4 +4 4 7 5 4 v 7 5 = 4 7 ) +4 So we have only one equation v v =v ) v = 4 v 7 5 = 4 7 5 v v 4 4 7 5 4 v 7 5 = 4 7 and choosing v = we get the eigenpair = 4, v = 4 7 5. So one fundamental solution is v x = 4 7 5 e 4t. However, we do not know what is a second fundamental solution in this case. C
G. NAGY ODE October, 8 Theorem 4. (Non-Diagonalizable with a Repeated Eigenvalue) If a matrix A has a repeated eigenvalue with only one eigen direction determined by v,thenx (t) =A x(t) has the linearly independent solutions x (t) =e t v, x (t) =e t v t + w, where the vector w is one solution of the algebraic linear system (A I)w = v. Example 5-continued: Find the fundamental solutions of the di erential equation apple x 4 = Ax, A =. Solution: We already know that an eigenpair of A is =, v = 4 7 Any other eigenvector associated to = is proportional to the eigenvector above. The matrix A is not diagonalizable, so we solve for a vector w the linear system (A +4I)w = v, 4 4 7 5 4 w 7 5 = 4 7 5 ) w +w = ) w =w. w Therefore, w = 4 w 7 5 = 4 w 7 5 ) w = 4 7 5 w + 4 7 w w So, given any solution w, thecv + w is also a solution for any c R. We choose w =, w = 4 7
G. NAGY ODE october, 8 Therefore, a fundamental set of solutions to the di erential equation above is formed by x (t) =e 4t 4 7 5, x (t) =e 4t t 4 7 5 + 4 7 C