Autonomous Equations / Stability of Equilibrium Solutions. y = f (y).

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1 Autonomous Equations / Stabilit of Equilibrium Solutions First order autonomous equations, Equilibrium solutions, Stabilit, Longterm behavior of solutions, direction fields, Population dnamics and logistic equations Autonomous Equation: A differential equation where the independent variable does not explicitl appear in its expression. It has the general form of = f (. Examples: = e 3 = 3 4 = sin Ever autonomous ODE is a separable equation. Because, assuming that f ( 0, d d = f ( dt = f ( dt d f ( = dt. Hence, we alread know how to solve them. What we are interested now is to predict the behavior of an autonomous equation s solutions without solving it, b using its direction field. But what happens if the assumption that f ( 0 is false? We shall start b answering this ver question. 008, 01 Zachar S Tseng A- - 1

2 Equilibrium solutions Equilibrium solutions (or critical points) occur whenever = f ( = 0. That is, the are the roots of f (. An root c of f ( ields a constant solution = c. (Exercise: Verif that, if c is a root of f (, then = c is a solution of = f (.) Equilibrium solutions are constant functions that satisf the equation, i.e., the are the constant solutions of the differential equation. Example: Logistic Equation of Population r = r 1 = r K K Both r and K are positive constants. The solution is the population size of some ecosstem, r is the intrinsic growth rate, and K is the environmental carring capacit. The intrinsic growth rate is the natural rate of growth of the population provided that the availabilit of necessar resource (food, water, oxgen, etc) is limitless. The environmental carring capacit (or simpl, carring capacit is the maximum sustainable population size given the actual availabilit of resource. Without solving this equation, we will examine the behavior of its solution. Its direction field is shown in the next figure. 008, 01 Zachar S Tseng A- -

3 Notice that the long-term behavior of a particular solution is determined solel from the initial condition (t 0 ) = 0. The behavior can be categorized b the initial value 0 : If 0 < 0, then as t. If 0 = 0, then = 0, a constant/equilibrium solution. If 0 < 0 < K, then K as t. If 0 = K, then = K, a constant/equilibrium solution. If 0 > K, then K as t. 008, 01 Zachar S Tseng A- - 3

4 Comment: In a previous section (applications: air-resistance) ou learned an eas wa to find the limiting velocit without having to solve the differential equation. Now we can see that the limiting velocit is just the equilibrium solution of the motion equation (which is an autonomous equation). Hence it could be found b setting v = 0 in the given differential equation and solve for v. Stabilit of an equilibrium solution The stabilit of an equilibrium solution is classified according to the behavior of the integral curves near it the represent the graphs of particular solutions satisfing initial conditions whose initial values, 0, differ onl slightl from the equilibrium value. If the nearb integral curves all converge towards an equilibrium solution as t increases, then the equilibrium solution is said to be stable, or asmptoticall stable. Such a solution has long-term behavior that is insensitive to slight (or sometimes large) variations in its initial condition. If the nearb integral curves all diverge awa from an equilibrium solution as t increases, then the equilibrium solution is said to be unstable. Such a solution is extremel 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 totall different behaviors (in both long- and short-terms). Therefore, in the logistic equation example, the solution = 0 is an unstable equilibrium solution, while = K is an (asmptoticall stable equilibrium solution. 008, 01 Zachar S Tseng A- - 4

5 An alternative graphical method: Plotting = f ( versus. This is a graph that is easier to draw, but reveals just as much information as the direction field. It is rather similar to the First Derivative Test * for local extrema in calculus. On an interval (the are separated b equilibrium solutions / critical points, which are the horizontal-intercepts of the graph) where f ( > 0, will be increasing and we denote this fact b drawing a rightward arrow. (Because, in this plot happens to be the horizontal axis; and its coordinates increase from left to right, from to.) Similarl, on an interval where f ( < 0, is decreasing. We shall denote this fact b drawing a leftward arrow. To summarize: f ( > 0, goes up, therefore, rightward arrow; f ( < 0, goes down, therefore, leftward arrow. The result can then be interpreted in the following wa: Suppose = c is an equilibrium solution (i.e. f ( = 0), then (i.) If f ( < 0 on the left of c, and f ( > 0 on the right of c, then the equilibrium solution = c is unstable. (Visuall, the arrows on the two sides are moving awa from c.) (ii.) If f ( > 0 on the left of c, and f ( < 0 on the right of c, then the equilibrium solution = c is asmptoticall stable. (Visuall, the arrows on the two sides are moving toward c.) Remember, a leftward arrow means is decreasing as t increases. It corresponds to downward-sloping arrows on the direction field. While a rightward arrow means is increasing as t increases. It corresponds to upward-sloping arrows on the direction field. * All the steps are reall the same, onl 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 asmptoticall stable equilibrium solution. 008, 01 Zachar S Tseng A- - 5

6 As an example, let us appl this alternate method on the same logistic equation seen previousl: = r (r / K), r = 0.75, K = 10. The -versus- plot is shown below. As can be seen, the equilibrium solutions = 0 and = K = 10 are the two horizontal-intercepts (confusingl, the are the -intercepts, since the -axis is the horizontal axis). The arrows are moving apart from = 0. It is, therefore, an unstable equilibrium solution. On the other hand, the arrows from both sides converge toward = K. Therefore, it is an (asmptoticall stable equilibrium solution. 008, 01 Zachar S Tseng A- - 6

7 Example: Logistic Equation with (Extinction) Threshold r = 1 1 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 eventuall become extinct. As seen above, the equation has (asmptoticall stable equilibrium solutions = 0 and = K. There is an unstable equilibrium solution = T. 008, 01 Zachar S Tseng A- - 7

8 The same result can, of course, be obtained b looking at the -versus- plot (in this example, T = 5 and K = 10): We see that = 0 and = K are (asmptoticall stable, and = T is unstable. Once again, the long-term behavior can be determined just b the initial value 0 : If 0 < 0, then 0 as t. If 0 = 0, then = 0, a constant/equilibrium solution. If 0 < 0 < T, then 0 as t. If 0 = T, then = T, a constant/equilibrium solution. If T < 0 < K, then K as t. If 0 = K, then = K, a constant/equilibrium solution. If 0 > K, then K as t. Semistable equilibrium solution A third tpe of equilibrium solutions exist. It exhibits a half-and-half behavior. It is demonstrated in the next example. 008, 01 Zachar S Tseng A- - 8

9 Example: = 3 The equilibrium solutions are = 0 and. As can be seen below, = is an unstable equilibrium solution. The interesting thing here, however, is the equilibrium solution = 0 (which corresponding a double-root of f (. Notice the behavior of the integral curves near the equilibrium solution = 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 awa and diverging to, a behavior associated with an unstable equilibrium solution. A behavior such like this defines a semistable equilibrium solution. 008, 01 Zachar S Tseng A- - 9

10 An equilibrium solution is semistable if has the same sign on both adjacent intervals. (In our analog 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 ( > 0 on both sides of c, or f ( < 0 on both sides of c, then the equilibrium solution = c is semistable. (Visuall, the arrows on one side are moving toward c, while on the other side the are moving awa from c.) Comment: As we can see, it is actuall not necessar to graph anthing in order to determine stabilit. The onl thing we need to make the determination is the sign of on the interval immediatel to either side of an equilibrium solution (a.k.a. critical point), then just appl the abovementioned rules. The steps are otherwise identical to the first derivative test: breaking the number line into intervals using critical points, evaluate at an arbitrar point within each interval, finall make determination based on the signs of. This is our version of the first derivative test for classifing stabilit of equilibrium solutions of an autonomous equation. (The graphing methods require more work but also will provide more information unnecessar for our purpose here such as the instantaneous rate of change of a particular solution at an point.) Computationall, stabilit classification tells us the sensitivit (or lack thereof) to slight variations in initial condition of an equilibrium solution. An unstable equilibrium solution is ver sensitive to deviations in the initial condition. Even the slightest change in the initial value will result in a ver different asmptotical behavior of the particular solution. An asmptoticall stable equilibrium solution, on the other hand, is quite tolerant of small changes in the initial value a slight variation of the initial value will still result in a particular solution with the same kind of long-term behavior. A semistable equilibrium solution is quite insensitive to slight variation in the initial value in one direction (toward the converging, or the stable, side). But it is extremel sensitive to a change of the initial value in the other direction (toward the diverging, or the unstable, side). 008, 01 Zachar S Tseng A- - 10

11 Exercises A-.1: 1 8 Find and classif all equilibrium solutions of each equation below. 1. = = = ( 1)( )( 3) 4. = sin 5. = cos (π / ) 6. = 1 e 7. = (3 1)e 8. = ( 1) (3 ( 5) 9. For each of problems 1 through 8, determine the value to which will approach as t increases if (a) 0 = 1, and (b) 0 = π. 10. Consider the air-resistance equation from an earlier example, 100v = v. (i) Find and classif its equilibrium solutions. (ii) Given (t 0 ) = 0, determine the range of (t). (iii) Given (8) = 60, determine the range of (t). 11. Verif the fact that ever first order linear ODE with constant coefficients onl is also an autonomous equation (and, therefore, is also a separable equation). 1. Give an example of an autonomous equation having no (real-valued) equilibrium solution. 13. Give an example of an autonomous equation having exactl n equilibrium solutions (n 1). 008, 01 Zachar S Tseng A- - 11

12 Answers A-.1: 1. = 0 (unstable), = ±10 (asmptoticall stable). = 0 (asmptoticall stable), = ± (unstable) 3. = 0 and = (asmptoticall stable), = 1 and = 3 (unstable) 4. = 0, ±π, ±4π, (unstable), = ±π, ±3π, ±5π, (asmptoticall stable) 5. = ±1, ±3, ±5, (all are semistable) 6. = 0 (asmptoticall stable) 7. = 1/3 (asmptoticall stable), = 1 (unstable) 8. = 0 (unstable), = 1 and = 5 (semistable), = 3 (asmptoticall stable) 9. (1) 10, 10; () 0, ; (3) 0, ; (4) π, π; (5) 1, 5; (6) 0, 0; (7) 1/3, ; (8), (i) = 50 (unstable) and = 50 (asmptoticall stable); (ii) ( 50, 50); (iii) (, 50) 11. For an constants α and β, + α = β can be rewritten as = β α, which is autonomous (and separable). 1. One example (there are infinitel man is = e. 13. One of man examples is = ( 1)( )( 3) ( n). 008, 01 Zachar S Tseng A- - 1

13 Exact Equations An exact equation is a first order differential equation that can be written in the form M(x, + N(x, = 0, provided that there exists a function ψ(x, such that x = M ( x, and N( x, =. Note 1: Often the equation is written in the alternate form of M(x, dx + N(x, d = 0. Theorem (Verification of exactness): An equation of the form M(x, + N(x, = 0 is an exact equation if and onl if M = N x. Note : If M(x) is a function of x onl, and N( is a function of onl, then M N triviall = 0 =. Therefore, ever separable equation, x M(x) + N( = 0, can alwas be written, in its standard form, as an exact equation. 008, 01 Zachar S Tseng A- - 13

14 The solution of an exact equation Suppose a function ψ(x, exists such that ( x, x = M and N ( x, =. Let be an implicit function of x as defined b the differential equation M(x, + N(x, = 0. (1) Then, b the Chain Rule of partial differentiation, d d ψ ( x, ( x)) = + = M ( x, + N( x,. dx x dx As a result, equation (1) becomes d ψ ( x, ( x)) = 0 dx. Therefore, we could, in theor at least, find the (implicit) general solution b integrating both sides, with respect to x, to obtain ψ(x, = C. Note 3: In practice ψ(x, could onl be found after two partial integration steps: Integrate M (= ψ x ) respect to x, which would recover ever term of ψ that contains at least one x; and also integrate N (= ψ ) with respect to, which would recover ever term of ψ that contains at least one. Together, we can then recover ever non-constant term of ψ. Note 4: In the context of multi-variable calculus, the solution of an exact equation gives a certain level curve of the function z = ψ(x,. 008, 01 Zachar S Tseng A- - 14

15 Example: Solve the equation ( 4 ) + 4x 3 = 0 First identif that M(x, = 4, and N(x, = 4x 3. Then make sure that it is indeed an exact equation: M 3 N = 3 4 and = 4 x Finall find ψ(x, using partial integrations. First, we integrate M with respect to x. Then integrate N with respect to. 4 4 M ( x, dx= ( ) dx= x x+ ψ x, = C ( ( N ( x, d= 4x d= x + ψ x, = C ( x) ( Combining the result, we see that ψ(x, must have non-constant terms: x 4 and x. That is, the (implicit) general solution is: x 4 x = C.., Now suppose there is the initial condition ( 1) =. To find the (implicit) particular solution, all we need to do is to substitute x = 1 and = into the general solution. We then get C = 14. Therefore, the particular solution is x 4 x = , 01 Zachar S Tseng A- - 15

16 Example: Solve the initial value problem ( cos( x + + x) dx+ ( xcos( x + ln x+ e ) d= 0, (1) = 0. x First, we see that M ( x, = cos( x + + x and x N ( x, = xcos( x + ln x+ e. Verifing: M = xsin( x + cos( x + 1 x = N = xsin( x + cos( x + 1 x x Integrate to find the general solution: ψ ( x, = cos( x + + x dx= sin( x + ln x+ x + C1(, x as well, ( x cos( x ) ln x e d sin( x ) ln x e = + + = C ( ) ψ ( x, ) x. Hence, sin x + ln x + e + x = C. Appl the initial condition: x = 1 and = 0: C = sin ln (1) + e = The particular solution is then sin x + ln x + e + x =. 008, 01 Zachar S Tseng A- - 16

17 Example: Write an exact equation that has general solution x 3 e + x = C. We are given that the solution of the exact differential equation is ψ(x, = x 3 e + x = C. The required equation will be, then, simpl M(x, + N(x, = 0, such that ( x, x = M and N( x, =. Since = x 3 = x e 3 4 3x e + 4x + 4x 4 3, and 6. Therefore, the exact equation is: (3 x e + 4 x 3 4 ) + (x 3 e + 4 x ) = , 01 Zachar S Tseng A- - 17

18 Summar: Exact Equations M(x, + N(x, = 0 Where there exists a function ψ(x, such that x = M ( x, and N( x, =. 1. Verification of exactness: it is an exact equation if and onl if M = N x.. The general solution is simpl ψ(x, = C. Where the function ψ(x, can be found b combining the result of the two integrals (write down each distinct term onl once, even if it appears in both integrals): ψ ( x, = M ( x, dx, and ψ ( x, = N( x, d. 008, 01 Zachar S Tseng A- - 18

19 Exercises A-.: 1 Write an exact equation that has the given solution. Then verif that the equation ou have found is exact. 1. It has the general solution x tan + x 3 3x 4 = C.. It has a particular solution x ln x + 5 = For each equation below, verif its exactness then solve the equation. 3. x + x cos(x ) + = x 4 3 x 4. 4x x+ (4x + + 5) = 0 5. (x + ( x) = 0, (10) = 5 6. (3x e x ) + (x 3 + 3x x e x ) = 0, () = 0 7. (5 e x ) + ( 5 e x ) = 0, (0) = 4 sin x x cos x x 8. ( + ) + ( ) = 0 3, (0) = 1 9. ( x 1 x + ) + (arctan( x ) ) = 0 4, (1) = x sin(x)sin( + cos(x) + (cos(x)cos( + sin(x)) = 0, (π/) = π 11. Rewrite the equation into an exact equation, verif its exactness, and then solve the initial value problem. e =, (1) = 0. xe sin( 1 14 Find the value(s) of λ such that the equation below is an exact equation. Then solve the equation (λ x ) + (3x λ) = 0 x 008, 01 Zachar S Tseng A- - 19

20 13. (λ sec (x λ x ) + (x sec (x λ x = (10 4 6x + 6x sin(x 3 )) + (40x 3 3x + λ cos(x 3 )) = 0 Answers A-.: 1. (x tan + 3x 1x 3 ) + (x sec 6x 4 ) = ( ) + (x + 5) = 0 x 3. x + + sin(x ) = C 4 4 x 4. x x + 5= C 5. x x + = 5 6. x 3 + x 3 + 4x e x = x 5 e x = 4 cos x x 8. + = 1 x π 1 9. arctan( x ) + = cos(x)sin( + sin(x) = π 11. The equation is e + (x e sin = 0; x e + cos = 1. λ = 3; x x 1 3 = C 13. λ = ; tan(x x = C 14. λ = 0; 10x 4 3x cos(x 3 ) = C 008, 01 Zachar S Tseng A- - 0

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