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Unformatted text preview: 4 MA 36600 MIDTERM #1 REVIEW is an equilibrium solution. In general, an equilibrium solution is a constant which is a solution, so it
satisﬁes the equation
f (yL ) = 0.
Any constant yL such that f (yL ) = 0 is called a critical point for the function f (y ). A critical
point yL is a stable equilibrium solution if f (yL ) < 0; and yL is an unstable equilibrium solution if
f (yL ) > 0.
• 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
environmental carrying capacity K . That is
∝ P (K − P )
= rP 1−
P (0) = P0 ;
for some positive constant r called the intrinsic growth rate. This is known as the logistic equation.
The solution is
P (t) =
P0 + (K − P0 ) e−rt
• More generally, say that the rate of change of the size of the population is proportional to three
– the size P ,
– the diﬀerence K − P of a carrying capacity K from the size P , and
– the diﬀerence P − T of the size P from a threshold T .
A logistic equation with threshold is an initial value problem in the form
∝ P (K − P ) (P − T )
= −r P 1 −
, P (0) = P0 .
We call K the environmental carrying capacity, T the threshold, and r the intrinsic growth rate.
– When 0 < P0 < T we have f (P ) < 0, so that P = P (t) is a decreasing function. This is because
P < T is below the threshold so we expect the population to become extinct.
– When T < P0 < K we have f (P ) > 0, so that P = P (t) is an increasing function. This is
because P < K so the population is below the carrying capacity of the population.
– When P0 > K we have f (P ) < 0, so that P = P (t) is a decreasing function. This is because P
is above the carrying capacity of the population. §2.6: Exact Equations and Integrating Factors.
• Consider an ordinary diﬀerential equation in the form
= G(x, y )
dx where G(x, y ) = − M (x, y )
N (x, y ) We can write
M (x, y )
N (x, y ) =⇒ M (x, y ) dx + N (x, y ) dy = 0. We call this an exact equation if there exists a function f (x, y ) such that
= M (x, y )
∂x and ∂f
= N (x, y ).
∂y If this is the case, the solution is f (x, y ) = C .
• A diﬀerential equation M (x, y ) dx + N (x, y ) dy = 0 is an exact equation if and only if
∂x (Note how M = M (x, y ) is the function multiplied by the diﬀerential dx, yet we take the partial
with respect to y !) If this is the case, we explain how to construct the function f (x, y ):
#1. Choose a function g (x, y ) according to the partial diﬀerential equation
= M (x, y )
g (x, y ) =
M (σ, y ) dσ.
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.
- Spring '09
- Critical Point