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# lecture_8 (dragged) - MA 36600 LECTURE NOTES FRIDAY JANUARY...

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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30 Population Dynamics Logistic Equations. We discuss an application of autonomous equations by considering population growth. Say that we have a population which has a size P = P (t) at time t. Assume that the rate of change of the size is proportional to both the size of the population and the diﬀerence of the size from some maximal sustainable size K . That is ￿ ￿ dP P ∝ P (K − P ) = K · P 1 − . dt K Hence there exists a positive constant r such that ￿ ￿ dP P = rP 1− . dt K This is known as the logistic equation. The equilibrium solutions P = PL can be found by computing the critical points of the function ￿ ￿ P f (P ) = r P 1 − . K Clearly these are PL = 0 and PL = K . Sometimes we refer to K as the saturation level or the environmental carrying capacity. Today we will study the solution to this problem. Solution to Initial Value Problem. We solve this problem explicitly. To this end, it will be useful to consider the “normalized” function and the “normalized” initial condition P (t) P0 y (t) = and y0 = . K K We substitute P = K y into the diﬀerential equation above: ￿ ￿ dP P = rP 1− dt K ￿ ￿ d Ky [K y ] = r K y 1 − dt K dy dy K = r K y (1 − y ) =⇒ = r y (1 − y ). dt dt Hence we ﬁnd the equivalent initial value problem dy = r y (1 − y ) , y (0) = y0 . dt Fortunately this diﬀerential equation is a separable equation: dy = r y (1 − y ) dt =⇒ 1 dy = r. y (1 − y ) dt We wish to compute the antiderivative of the function on the left-hand side. To this end, we assume that it can be expressed in a partial fraction expansion : 1 A B =+ y (1 − y ) y 1−y for some constants A and B yet to be determined. The right-hand side can be simpliﬁed by ﬁnding a common denominator: A B A (1 − y ) By A (1 − y ) + B y A + ( B − A) y + = + = = . y 1−y y (1 − y ) y (1 − y ) y (1 − y ) y (1 − y ) 1 ...
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