Calculus 2 - Course Notes 4

Calculus 2 - Course Notes 4 - MAT 1332: Frithjof Lutscher...

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MAT 1332: Frithjof Lutscher 77 Systems of diFerential equations - the linear case Recall that a single linear di±erential equation d dt x ( t )= ax ( t ) ,x (0) = x 0 has the solution x ( t e at x 0 . In particular, if a> 0 , the solution grows to in²nity, and if a< 0 then the solution approaches zero. We now want to generalize this result to two coupled equations d dt x 1 ( t ax 1 ( t )+ bx 2 ( t ) 1 (0) = x 10 , d dt x 2 ( t rx 1 ( t sx 2 ( t ) 2 (0) = x 20 . We can also write this in matrix notation (suppressing the argument t for the moment) as d dt ! x 1 x 2 " = ! ab rs "! x 1 x 2 " =: A ! x 1 x 2 " , ! x 1 (0) x 2 (0) " = ! x 10 x 20 " ,A = ! " . Observation I: Eigenvalues and eigenvectors provide solutions Suppose we are looking for solutions of the form ! x 1 ( t ) x 2 ( t ) " = e λt ! v 1 v 2 " . Plugging this expression into the equation above, we get d dt ! x 1 ( t ) x 2 ( t ) " = λe λt ! v 1 v 2 " = e λt A ! v 1 v 2 " . Canceling the exponential term on both sides, we ²nd the condition λ ! v 1 v 2 " = A ! v 1 v 2 " , which means that λ has to be an eigenvalue of A and ! v 1 v 2 " has to be the corresponding eigenvector. Fact: If λ is an eigenvalue of A and v is the corresponding eigenvector, then x ( t e λt v is a solution of the linear system of di±erential equations d dt x ( t Ax ( t ) , where x =( x 1 ,...x n ) T , and A is an n × n -matrix. 77

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MAT 1332: Frithjof Lutscher 78 Example 1 Take the system d dt x 1 ( t )= x 1 ( t )+4 x 2 ( t ) , d dt x 2 ( t )=2 x 1 ( t ) x 2 ( t ) , with matrix A = ! 14 2 1 " . The eigenvalues of A are given by the equation (1 λ )( 1 λ ) 8= λ 2 9=0 . Hence, the eigenvalues are λ = 3 and μ = 3 . The corresponding eigenvectors are v = ! v 1 v 2 " = ! 2 1 " ,w = ! w 1 w 2 " = ! 1 1 " , respectively. Hence, we have the two solutions ! x 1 ( t ) x 2 ( t ) " = e λt v = e 3 t ! 2 1 " , ! x 1 ( t ) x 2 ( t ) " = e μt w = e 3 t ! 1 1 " . We have two solutions, but we don’t have any constants in the equation yet with which we could match the initial condition. This is the next topic. Observation II: Sums and multiples of solutions are solutions Suppose we have two solutions w ( t ) and z ( t ) for the same system, i.e., d dt w ( t Aw ( t ) , d dt z ( t Az ( t ) . Now pick two numbers C 1 ,C 2 and form x ( t C 1 w ( t )+ C 2 z ( t ) . Then x ( t ) is also a solution since d dt x ( t C 1 d dt w ( t C 2 d dt z ( t C 1 Aw ( t C 2 Az ( t A [ C 1 w ( t C 2 z ( t )] = Ax ( t ) . Example 1, continued Find the solution of the system d dt x 1 ( t x 1 ( t x 2 ( t ) , d dt x 2 ( t x 1 ( t ) x 2 ( t ) , 78
MAT 1332: Frithjof Lutscher 79 with initial values x 10 =3 ,x 20 . We already know the eigenvalues and eigenvectors, and hence, we get the general solution as ! x ( t ) y ( t ) " = C 1 e 3 t ! 2 1 " + C 2 e 3 t ! 1 1 " . Substituting the initial conditions, i.e., setting t =0 , we get ! 3 3 " = C 1 ! 2 1 " + C 2 ! 1 1 " . Hence we solve the system ! 21 | 3 1 1 | 3 " −→ ! 1 1 | 3 03 |− 3 " ! 10 | 2 01 1 " . The solution is C 1 =2 ,C 2 = 1 , and therefore the solution to the di±erential equation is ! x 1 ( t ) x 2 ( t ) " e 3 t ! 2 1 " e 3 t ! 1 1 " , or, equivalently x 1 ( t )=4 e 3 t e 3 t , x 2 ( t )=2 e 3 t + e 3 t .

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This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.

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Calculus 2 - Course Notes 4 - MAT 1332: Frithjof Lutscher...

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