To solve the IVP we solve u 0 c 1 1 2 1 c 2 6 1 16 16 5 25 Equivalently solve 1

# To solve the ivp we solve u 0 c 1 1 2 1 c 2 6 1 16 16

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To solve the IVP, we solve: u (0) = c 1 1 2 1 + c 2 - 6 1 + 16 16 = 5 25 Equivalently, solve 1 2 - 6 1 1 c 1 c 2 = - 11 9 or c 1 = 86 13 , c 2 = 31 13 Thus, the solution to the IVP is u ( t ) = 86 13 1 2 1 e - t/ 8 + 31 13 - 6 1 e - 7 t/ 4 + 16 16 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equ — (22/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 7 Greenhouse/ Rockbed Solution: Graph shows temperature in each compartment u 1 ( t ) ( greenhouse ) and u 2 ( t ) ( rockbed ) 0 1 2 3 4 5 6 7 8 9 10 0 5 10 15 20 25 30 u 1 u 2 t hrs Temperature ( C) Greenhouse/Rockbed Air (numeric) Rockbed (numeric) Air (exact) Rockbed (exact) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equations: — (23/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 8 Greenhouse/ Rockbed Solution Observations Both solutions tend toward the equilibrium solution of 16 C There is more heat capacitance in the rock (high mass), so solution changes more slowly in this compartment The air of the greenhouse responds more quickly (low heat capacitance) The air of the greenhouse heats above steady state before returning toward the equilibrium solution This simplified model assumes a constant external temperature of 16 C rather than the more interesting dynamics of solar power and nocturnal heat loss - significantly more complicated model Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equ — (24/32)

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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Greenhouse Example Revisited MatLab and Maple Summary Direction Fields and Phase Portraits Definition (Autonomous System of Differential Equations) Let x 1 and x 2 be state variables , and assume that the functions, f 1 ( x 1 , x 2 ) and f 2 ( x 1 , x 2 ) are dependent only on the state variables. The two-dimensional autonomous system of differential equations is given by: ˙ x 1 = f 1 ( x 1 , x 2 ) ˙ x 2 = f 2 ( x 1 , x 2 ) Definition (Autonomous Linear System of Differential Equations) Let x 1 and x 2 be state variables with x = [ x 1 , x 2 ] T , and assume that A is a constant matrix. The autonomous linear system of differential equations is given by: ˙ x = Ax . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Equations: — (25/32) Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Greenhouse Example Revisited MatLab and Maple Summary Direction Fields and Phase Portraits The state variables , u 1 = u 1 ( t ) and u 2 = u 2 ( t ), are parametric equations depending on t Define the vector, u ( t ) = u 1 ( t ) i + u 2 ( t ) j The u 1 u 2 -plane is called the state plane or phase plane As t varies, the vector u ( t ) traces a curve in the phase plane called a trajectory or orbit An autonomous system of differential equations describes the dynamics of the orbit The functions, f 1 ( x 1 , x 2
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