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Notes-1st order ODE pt2.pdf

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Mathematics
Course
MATH 251
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Spring

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Autonomous Equations / Stability of Equilibrium Solutions First order autonomous equations, Equilibrium solutions, Stability, Long▯ term behavior of solutions, direction fields, Population dynamics and logistic equations Autonomous Equation: A differential equation where the independent variable does not explicitly appear in its expression. It has the general form of y′ = f(y). 2y 3 Examples: y′ = e − y y′ = y − 4y 4 y′ = y − 81 + siny Every autonomous ODE is a separable equation. Because, assuming that f(y) ≠ 0, dy dy dy dt = f (y) → f (y)= dt → ∫ f (y)= d∫ . Hence, we already know how to solve them. What we are interested now is to predict the behavior of an autonomous equation’s solutions without solving it, by using its direction field. But what happens if the assumption that f (y) ≠ 0 is false? We shall start by answering this very question. © 2008, 2012 Zachary S Tseng A▯2 ▯ 1 Equilibrium solutions Equilibrium solutions (or critical points) occur whenever y′ = f(y) = 0. That is, they are the roots of f(y). Any root c of f(y) yields a constant solution y = c. (Exercise: Verify that, if c is a root of fhen y = c is a solution of y′ = f(y.) Equilibrium solutions are constant functions that satisfy the equation, i.e., they are the constant solutions of the differential equation. Example: Logistic Equation of Population  y  r 2 y = r 1 − y = ry − y  K  K Both r and K are positive constants. The solution y is the population size of some ecosystem, r is the intrinsic growth rate, and K is the environmental carrying capacity. The intrinsic growth rate is the natural rate of growth of the population provided that the availability of necessary resource (food, water, oxygen, etc) is limitless. The environmental carrying capacity (or simply, carrying capacity) is the maximum sustainable population size given the actual availability of resource. Without solving this equation, we will examine the behavior of its solution. Its direction field is shown in the next figure. © 2008, 2012 Zachary S Tseng A▯2 ▯ 2 Notice that the long▯term behavior of a particular solution is determined solely from the initial condition y(t ) =0y . Th0 behavior can be categorized by the initial value y : 0 If y < 0, then y → −∞ as t→ ∞. 0 If y = 0, then y = 0, a constant/equilibrium solution. 0 If 0 < y < K, then y → K as t→ ∞. 0 If y = K, then y = K, a constant/equilibrium solution. 0 If y > K, then y → K as t→ ∞. 0 © 2008, 2012 Zachary S Tseng A▯2 ▯ 3 Comment: In a previous section (applications: air▯resistance) you learned an easy way to find the limiting velocity without having to solve the differential equation. Now we can see that the limiting velocity is just the equilibrium solution of the motion equation (which is an autonomous equation). Hence it could be found by setting v′ = 0 in the given differential equation and solve for v. Stability of an equilibrium solution The stability of an equilibrium solution is classified according to the behavior of the integral curves near it – they represent the graphs of particular solutions satisfying initial conditions whose initial values, y 0, differ only slightly from the equilibrium value. If the nearby integral curves all converge towards an equilibrium solution as t increases, then the equilibrium solution is said to be stable, or asymptotically stable. Such a solution has long▯term behavior that is insensitive to slight (or sometimes large) variations in its initial condition. If the nearby integral curves all diverge away froman equilibrium solution as t increases, then the equilibrium solution is said to be unstable. Such a solution is extremely sensitive to even the slightest variations in its initial condition − as we can see in the previous example that the smallest deviation in initial value results in totally different behaviors (in both long▯ and short▯terms). Therefore, in the logistic equation example, the solution y = 0 is an unstable equilibrium solution, while y = K is an (asymptotically) stable equilibrium solution. © 2008, 2012 Zachary S Tseng A▯2 ▯ 4 An alternative graphical method: Plotting y′ = f(y) versus y. This is a graph that is easier to draw, but reveals just as much information *s the direction field. It is rather similar to the First Derivative Test for local extrema in calculus. On any interval (they are separated by equilibrium solutions / critical points, which are the horizontal▯intercepts of the graph) where f (y) > 0, y will be increasing and we denote this fact by drawing a rightward arrow. (Because, y in this plot happens to be the horizontal axis; and its coordinates increase from left to right, from −∞ to ∞.) Similarly, on any interval where f(y) < 0, y is decreasing. We shall denote this fact by drawing a leftward arrow. To summarize: f (y) > 0, y goes up, therefore, rightward arrow; f(y) < 0, y goes down, therefore, leftward arrow. The result can then be interpreted in the following way: Suppose y = c is an equilibrium solution (i.e. f (y) = 0), then (i.) If f(y) < 0 on the left of c, and f(y) > 0 on the right of c, then the equilibrium solution y = c is unstable. (Visually, the arrows on the two sides are moving away from c.) (ii.) If f(y) > 0 on the left of c, and f(y) < 0 on the right of c, then the equilibrium solution y = c is asymptotically stable. (Visually, the arrows on the two sides are moving toward c.) Remember, a leftward arrow means y is decreasing as t increases. It corresponds to downward▯sloping arrows on the direction field. While a rightward arrow means y is increasing as t increases. It corresponds to upward▯sloping arrows on the direction field. * All the steps are really the same, only the interpretation of the result differs. A result that would indicate a local minimum now means that the equilibrium solution/critical point is unstable; while that of a local maximum result now means an asymptotically stable equilibrium solution. © 2008, 2012 Zachary S Tseng A▯2 ▯ 5 As an example, let us apply this alternate method on t2e same logistic equation seen previously: y′ = ry − (r/K)y , r = 0.75, K = 10. The y′▯versus▯y plot is shown below. As can be seen, the equilibrium solutions y = 0 and y = K = 10 are the two horizontal▯intercepts (confusingly, they are the y▯intercepts, since the y▯axis is the horizontal axis). The arrows are moving apart from y = 0. It is, therefore, an unstable equilibrium solution. On the other hand, the arrows from both sides converge toward y = K. Therefore, it is an (asymptotically) stable equilibrium solution. © 2008, 2012 Zachary S Tseng A▯2 ▯ 6 Example: Logistic Equation with (Extinction) Threshold y′= − r 1− y1− y y  T  K  Where r, T, and K are positive constants: 0 < T < K. The values r and K still have the same interpretations, T is the extinction threshold level below which the species is endangered and eventually become extinct. As seen above, the equation has (asymptotically) stable equilibrium solutions y = 0 and y = K. There is an unstable equilibrium solution y = T. © 2008, 2012 Zachary S Tseng A▯2 ▯ 7 The same result can, of course, be obtained by looking at the y′▯versus▯y plot (in this example, T = 5 and K = 10): We see that y = 0 and y = K are (asymptotically) stable, and y = T is unstable. Once again, the long▯term behavior can be determined just by the initial value y 0 If y < 0, then y → 0 as t→ ∞. 0 If y 0 0, then y = 0, a constant/equilibrium solution. If 0 < y 0 T, then y → 0 as t→ ∞. If y = T, then y = T, a constant/equilibrium solution. 0 If T < y 0 K, then y → K as t→ ∞. If y 0 K, then y = K, a constant/equilibrium solution. If y 0 K, then y → K as t→ ∞. Semistable equilibrium solution A third type of equilibrium solutions exist. It exhibits a half▯and▯half behavior. It is demonstrated in the next example. © 2008, 2012 Zachary S Tseng A▯2 ▯ 8 3 2 Example: y′ = y − 2y The equilibrium solutions are y = 0 and 2. As can be seen below, y = 2 is an unstable equilibrium solution. The interesting thing here, however, is the equilibrium solution y = 0 (which corresponding a double▯root of f (y). Notice the behavior of the integral curves near the equilibrium solution y = 0. The integral curves just above it are converging to it, like it is a stable equilibrium solution, but all the integral curves below it are moving away and diverging to −∞, a behavior associated with an unstable equilibrium solution. A behavior such like this defines a semistable equilibrium solution. © 2008, 2012 Zachary S Tseng A▯2 ▯ 9 An equilibrium solution is semistable if y′ has the same sign on both adjacent intervals. (In our analogy with the First Derivative Test, if the result would indicate that a critical point is neither a local maximum nor a minimum, then it now means we have a semistable equilibrium solution. (iii.) If f(y) > 0 on both sides of c, or f(y) < 0 on both sides of c, then the equilibrium solution y = c is semistable. (Visually, the arrows on one side are moving toward c, while on the other side they are moving away from c.) Comment: As we can see, it is actually not necessary to graph anything in order to determine stability. The only thing we need to make the determination is the sign of y′ on the interval immediately to either side of an equilibrium solution (a.k.a. critical point), then just apply the above▯ mentioned rules. The steps are otherwise identical to the first derivative test: breaking the number line into intervals using critical points, evaluate y′ at an arbitrary point within each interval, finally make determination based on the signs of y′. This is our version of the first derivative test for classifying stability of equilibrium solutions of an autonomous equation. (The graphing methods require more work but also will provide more information – unnecessary for our purpose here – such as the instantaneous rate of change of a particular solution at any point.) Computationally, stability classification tells us the sensitivity (or lack thereof) to slight variations in initial condition of an equilibrium solution. An unstable equilibrium solution is very sensitive to deviations in the initial condition. Even the slightest change in the initial value will result in a very diff
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