X 1 d x 2 g 2 once we have found the critical point

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x 1 + d x 2 + g 2 Once we have found the critical point, say it is the point ( x 1 , x 2 ) = ( α , β ), it then could be moved to (0, 0) via the translations χ 1 = x 1 α and χ 2 = x 2 β . The result after the translation would be the homogeneous linear system χ ′ = . The two systems (before and after the translations) have the same coefficient matrix.* Their respective critical points will also have identical type and stability classification. Therefore, to determine the type and stability of the critical point of the given nonhomogeneous system, all we need to do is to disregard b , then take its coefficient matrix A and use its eigenvalues for the determination, in exactly the same way as we would do with the corresponding homogeneous system of equations.
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© 2008 Zachary S Tseng D -2 - 17 * (Optional topic) Note : Here is the formal justification of our ability to discard the vector b when determining the type and stability of the critical point of the nonhomogenous system x ′ = Ax + b . Suppose ( x 1 , x 2 ) = ( α , β ) is the critical point of the system. That is 0 = a α + b β + g 1 0 = c α + d β + g 2 Now apply the translations χ 1 = x 1 α and χ 2 = x 2 β . We see that x 1 = χ 1 + α , x 2 = χ 2 + β , χ 1 ′ = x 1 ′, and χ 2 ′ = x 2 ′. Substitute them into the system x ′ = Ax + b : x 1 ′ = χ 1 ′ = a x 1 + b x 2 + g 1 = a ( χ 1 + α ) + b ( χ 2 + β ) + g 1 = a χ 1 + b χ 2 + ( a α + b β + g 1 ) = a χ 1 + b χ 2 x 2 ′ = χ 2 ′ = c x 1 + d x 2 + g 2 = c ( χ 1 + α ) + d ( χ 2 + β ) + g 2 = c χ 1 + d χ 2 + ( c α + d β + g 2 ) = c χ 1 + d χ 2 That is, with the new variables χ 1 and χ 2 the given system has become χ 1 ′ = a χ 1 + b χ 2 χ 2 ′ = c χ 1 + d χ 2 Notice it has the form χ ′ = . That is, it is a homogeneous system (with the critical point at the origin) whose coefficient matrix is exactly A , the same as the original system’s. Hence, b could be disregarded, and a determination of the type and stability of the critical point of the system (with or without the translations) could be based solely on the coefficient matrix A .
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© 2008 Zachary S Tseng D -2 - 18 Example : x 1 ′ = x 1 − 2 x 2 − 1 x 2 ′ = 2 x 1 − 3 x 2 − 3 The critical point is at (3, 1). The matrix A has characteristic equation r 2 + 2 r + 1 = 0. It has a repeated eigenvalue r = − 1 that has only one linearly independent eigenvector. Therefore, the critical point at (3, 1) is an asymptotically stable improper node . The phase portrait is shown below.
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© 2008 Zachary S Tseng D -2 - 19 Example : x 1 ′ = −2 x 1 − 6 x 2 + 8 x 2 ′ = 8 x 1 + 4 x 2 − 12 The critical point is at (1, 1). The matrix A has characteristic equation r 2 − 2 r + 40 = 0. It has complex conjugate eigenvalues with positive real part, λ = 1. Therefore, the critical point at (1, 1) is an unstable spiral point . The phase portrait is shown below.
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© 2008 Zachary S Tseng D -2 - 20 Exercises D-2.1 : 1 – 8 Determine the type and stability or the critical point at (0,0) of each system below.
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