Differential Equations and Models

Abstract

At The system of equations below describes how the values of variables u 1 and u 2 affect each other over time: du 1 dt = −u 1 + 2u 2 du 2 dt = u 1 − 2u 2 . Just as we applied linear algebra to solve a difference equation, we can use it to solve this differential equation. For example, the initial condition u 1 = 1, 1 u 2 = 0 can be written u(0) = 0 . Differential equations d dt u = Au By looking at the equations above, we might guess that over time u 1 will de­ crease. We can get the same sort of information more safely by looking at the eigenvalues of the matrix A = − 1 1 −2 2 of our system d dt u = Au. Because A is singular and its trace is −3 we know that its eigenvalues are λ 1 = 0 and λ 2 = −3. The solution will turn out to include e −3t and e 0t . As t increases, e −3t vanishes and e 0t = 1 remains constant. Eigenvalues equal to zero have eigenvectors that are steady state solutions. 2 x 1 = is an eigenvector for which Ax 1 = 0x 1 . To find an eigenvector 1 corresponding to λ 2 = −3 we solve (A − λ 2 I)x 2 = 0: 2 2 1 x 2 = 0 so x 2 = 1 1 −1 and we can check that Ax 2 = −3x 2 . The general solution to this system of differential equations will be: u(t) = c 1 e λ 1 t x 1 + c 2 e λ 2 t x 2 . Is e λ 1 t x 1 really a solution to d dt u = Au? To find out, plug in u = e λ 1 t x 1 : du = λ 1 e λ 1 t x 1 , dt which agrees with: Au = e λ 1 t Ax 1 = λ 1 e λ 1 t x 1 . The two " pure " terms e λ 1 t x 1 and e λ 2 t x 2 are analogous to the terms λ k i x i we saw in the solution c 1 λ 1 k x 1 + c 2 λ 2 k x 2 + + c n λ n k x n to the difference equation · · · u k+1 = Au k .