Matrix a 1 1 a for the zero equilibrium to be stable

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Matrix A = 0 1 - 1 a . For the zero equilibrium to be stable spiral, the discriminant and the trace τ must be negative. Tr ( A ) = τ = a < 0 , det ( A ) = δ = 1 , so that γ = τ 2 - 4 δ = a 2 - 4 < 0 . The two inequalities give the conditions for stable spiral: - 2 < a < 0. J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 20 / 41
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Outline 1. Phase Plane Analysis 2. Eigenvalues. Real Eigenvalues Complex Eigenvalues Examples 3. Gershgorin’s Theorem 4. An Example: Pharmacokinetics Model 5. References J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 21 / 41
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4.9 Gershgorin’s Theorem When the zero equilibrium is an isolated equilibrium for the system dX/dt = AX , then all solutions converge to the origin iff the eigenvalues of A are negative or have negative real parts (lie in the left half of the complex plane). The Routh-Hurwitz criteria give necessary and sufficient conditions for the eigenvalues to lie in the left half of the complex plane. Another result which provides sufficient conditions for the eigenvalues to lie in the left half of the complex plane in known as Gershgorin’s Theorem [Noble, B., 1969]. J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 22 / 41
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Gershgorin’s Theorem Theorem (4.5) Let A be an n × n matrix. Let D i be the disk in the complex plane with center at a ii and radius r i = n j =1 ,j 6 = i | a ij | . Then all eigenvalues of the matrix A lie in the union of the disks D i , i = 1 , 2 , . . . , n , S n i =1 D i . In particular, if λ is an eigenvalue of A , then for some i = 1 , 2 , . . . , n , | λ - a ii | ≤ r i . A disk D i is graphed in Figure (4). a ii r i Imaginary axis Real axis Figure 4: Gershgorin’s disk in the complex plane. J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 23 / 41
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Proof. Let λ be any eigenvalue of A and V = ( v 1 , . . . , v n ) T an eigenvector corresponding to this eigenvalue. Then AV = λV , which implies λv i = n j =1 a ij v j or ( λ - a ii ) v i = n X j =1 ,j 6 = i a ij v j , i = 1 , . . . , n. Let v k denote the element of V with greatest magnitude ( i.e. | v k | ≥ | v j | , j 6 = k . Then | v j /v k | ≤ 1 for all j = 1 , . . . , n , and | λ - a kk | ≤ n X j =1 ,j 6 = k | a kj | v j v k n X j =1 ,j 6 = k | a kj | . Hence, λ lies in the disk D k . The conclusion of the theorem follows. J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 24 / 41
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Gershgoring’s Theorem applies to real and complex matrices A . When the entries of A are real, it follows from Gershgorin’s Theorem that the disks lie in the left half of the complex plane if r i + a ii < 0 or a ii < - r i for i = 1 , . . . , n. The strict inequality guarantees the Gershgorin disk lies entirely in the left half of the complex plane. Corollary (4.3) Let A be an n × n with real entries. If the diagonal elements of A satisfy a ii < - r i , where r i = n X j =1 ,j 6 = i | a ij | for i = 1 , . . . , n , then the eigenvalues of A are negative or have negative real parts. J. D. Flores ( USD ) Math-735: Math Modeling Spring 2013 25 / 41
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Example 4.15 Let the matrix A = a - 1 3 1 / 2 b - 2 0 - 5 c .
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