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# c35 - 18.03 Class 35 May 8 The companion matrix and its...

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The companion matrix and its phase portrait; The matrix exponential: initial value problems. [1] We spent a lot of time studying the second order equation x" + bx' + kx = 0 and if b and k are nonnegative we interpreted them as the damping constant and spring constant (divided by the mass). The companion system is obtained by setting x' = y y' = - kx - by whose matrix of coefficients is the "companion matrix" A = [ 0 1 ; -k -b ] Note that tr(A) = -b , det(A) = k, and the characteristic polynomial of A is lambda^2 + b lambda + k, i.e., it is the same as the characteristic polynomial of the original second order equation. If lambda_1 is an eigenvalue of a companion matrix, then to find an eigenvector we look for v such that [ -lambda_1 , 1 ; * , * ] v = 0 v = [ 1 ; lambda_1 ] does nicely. This makes sense: x = e^{lambda_1 t} has derivative x' = lambda_1 e^{lambda_1 t} , so [ x ; x' ] = e^{lambda t} [ 1 ; lambda_1 ] is a solution to the companion system. QUESTION 1: What region in the (tr,det) plane corresponds to c > 0, k > 0? Ans: the upper left quadrant. QUESTION 2: What region in the (tr,det) plane corresponds to overdamping? Ans: The part of the upper left quadrant which is below the critical parabola, where there are stable nodes. For example x" + (3/2)x' + (1/2)x = 0 is overdamped. We saw long ago that solutions to overdamped equations decay as time increases, can cross the x = 0 slope). axis at most once, and can have at most one critical point (zero

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c35 - 18.03 Class 35 May 8 The companion matrix and its...

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