University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

287

Pages:

2
Author:

ovolodrizzliest

Solve the linear first order system x0 = Ax Given x01 = ax1 + bx2 x02 = cx1 + dx2 ⇔ x01 x2 = a b c d x1 x2 x0 = Ax ⇔ Then we could solve the system in three ways: • Convert the system to a single second order DE and use Chapter 3 methods. Note that we should also be able to go backwards- Given a second order DE, convert it into a system of first order. There are some questions in the review that will extend this to nonlinear DEs and perhaps third order. • Could try to write the system in implicit form: dy dy/dt cx1 + dx2 = = dx dx/dt ax1 + bx2 Then we could try to use a method from Chapter 2. We can also do this with many nonlinear systems of equations (there are a couple of them in the review). • Use the eigenvalues and eigenvectors from the matrix A. This is the topic of Ch 7. Eigenvalues and Eigenvectors • Definition: Given an n × n matrix A, if there is a constant λ and a non-zero vector v so that av1 + bv2 = λv1 or Av = λv cv1 + dv2 = λv2 then λ is an eigenvalue, and v is an associated eigenvector for matrix A. Note that an eigenvector is not uniquely determined; we usually choose the simplest vector (integer valued if possible). • If we try to solve our equation: av1 +bv2 = λv1 cv1 +dv2 = λv2 ⇒ (a − λ)v1 +bv2 = 0 cv1 +(d − λ)v2 = 0 or (A − λI)v = 0 (1) This system has a non-trivial (non-zero) solution for v1 , v2 only if the determinant of coefficients is 0: a−λ b det =0 c d−λ And this is the characteristic equation. We solve this for the eigenvalues: λ2 − (a + d)λ + (ad − bc) = 0 ⇔ λ2 − Tr(A)λ + det(A) = 0 where Tr(A) is the trace of A (which we defined as a + d). For each λ, we must go back and solve Equation (1). 1 • Given a λ and v, the generalized eigenvector w is computed as below, and is used when we are solving the associated system of differential equations. (a − λ)w1 + bw2 = v1 cw1 + (d − λ)w2 = v2 (2) Summary To solve x0 = Ax, find the trace, determinant and discriminant for the matrix A. The eigenvalues are found by solving the characteristic equation: √ ∆ Tr(A) ± λ2 − Tr(A)λ + det(A) = 0 λ= 2 The solution is one of three cases, depending on ∆: • Real λ1 , λ2 give two eigenvectors, v1 , v2 : x(t) = C1 eλ1 t v1 + C2 eλ2 t v2 • Complex λ = a + ib, v (we only need one): x(t) = C1 Real eλt v + C2 Imag eλt v • One eigenvalue, one eigenvector v. Compute the generalized eigenvector w using Equation 2. Then x(t) = eλt (C1 v + C2 (tv + w)) You might find this helpful- Below there is a chart comparing the solutions from Chapter 3 to the solutions in Chapter 7: Chapter 3 Form: ay + by 0 + cy = 0 Ansatz: y = ert Char Eqn: ar2 + br + c = 0 Real Solns y = C1 er1 t + C2 er2 t Complex y = C1 Re(ert ) + C2 Im(ert ) Repeated Root y = ert (C1 + C2 t) 00 Chapter 7 x0 = Ax x = eλt v λ2 − Tr(A)λ + det(A) = 0 x(t) = C1 eλ1 t v1 + C2 eλ2 t v2 x(t) = C1 Re eλt v + C2 Im eλt v x(t) = eλt (C1 v + C2 (tv + w)) Poincaré Diagram Because we have the trace, determinant and discriminant, we can classify the origin as to its stability using the Poincaré diagram (which will be provided). 2

Solve the linear first order system - Study Guide

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