u 1 u 2 14 which has the equilibrium solution u e 16 16 Taking v t u t u e we

U 1 u 2 14 which has the equilibrium solution u e 16

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u 1 u 2 + 14 0 , which has the equilibrium solution u e = 16 16 Taking v ( t ) = u ( t ) - u e , we examine the translated model ˙ v 1 ( t ) ˙ v 2 ( t ) = - 13 8 3 4 1 4 - 1 4 ! v 1 ( t ) v 2 ( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equations: — (17/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 2 Example Model: Try a solution v ( t ) = ξe λt with ξ = [ ξ 1 , ξ 2 ] T , so the DE can be written λ ξ 1 ξ 2 e λt = - 13 8 3 4 1 4 - 1 4 ! ξ 1 ξ 2 e λt Dividing by e λt , we obtain the eigenvalue problem - 13 8 - λ 3 4 1 4 - 1 4 - λ ! ξ 1 ξ 2 = 0 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equ — (18/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 3 Eigenvalue Problem: Eigenvalues for the problem ( A - λ I ) ξ = 0 solve det | A - λ I | = 0, so det - 13 8 - λ 3 4 1 4 - 1 4 - λ = 0 The characteristic equation is λ 2 + 15 8 λ + 7 32 = 0 , which has solutions λ 1 = - 1 8 and λ 2 = - 7 4 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equations: — (19/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 4 Eigenvalue Problem: For λ 1 = - 1 8 , we solve - 3 8 3 4 1 4 - 1 8 ! ξ 1 ξ 2 = 0 0 , which gives a corresponding eigenvector , ξ (1) = 1 2 1 For λ 2 = - 7 4 , we solve 1 8 3 4 1 4 - 3 2 ! ξ 1 ξ 2 = 0 0 , which gives a corresponding eigenvector , ξ (2) = - 6 1 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equ — (20/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 5 Solution v ( t ) : The eigenvalue problem shows that there are two solutions to the Greenhouse example, ˙ v = Kv v 1 ( t ) = 1 2 1 e - t/ 8 and v 2 ( t ) = - 6 1 e - 7 t/ 4 along with any constant multiples of these solutions We combine results above to obtain the general solution u ( t ) = c 1 v 1 ( t )+ c 2 v 2 ( t )+ u e = c 1 1 2 1 e - t/ 8 + c 2 - 6 1 e - 7 t/ 4 + 16 16 The solution exhibits the property of exponentially decaying to the steady-state solution Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equations: — (21/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 6 Unique Solution: Suppose that the rockbed stored heat during the day, so we start with an initial condition of u 20 (0) = 25 C, while the cool night air comes into the greenhouse with u 10 (0) = 5 C.
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