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3.2 Sources, Sinks, Saddles, and Spirals

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3.2 Sources, Sinks, Saddles, and Spirals 3.2. Sources, Sinks, Saddles, and Spirals 161 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay00 By0 Cy 0. These are linear equations with constant coefficients A; B; and C .C The graphsC showD solutions y on the horizontal axis and their slopes y0 dy=dt on the vertical axis. These pairs .y.t/;y0.t// depend on time, but time is not inD the pictures. The paths show where the solution goes, but they don’t show when. Each specific solution starts at a particular point .y.0/;y0.0// given by the initial conditions. The point moves along its path as the time t moves forward from t 0. D We know that the solutions to Ay00 By0 Cy 0 depend on the two solutions to 2 C C D As Bs C 0 (an ordinary quadratic equation for s). When we find the roots s1 and C C D s2, we have found all possible solutions : s1t s2t s1t s2t y c1e c2e y c1s1e c2s2e (1) D C 0 D C The numbers s1 and s2 tell us which picture we are in. Then the numbers c1 and c2 tell us which path we are on. Since s1 and s2 determine the picture for each equation, it is essential to see the six possibilities. We write all six here in one place, to compare them. Later they will appear in six different places, one with each figure. The first three have real solutions s1 and s2. The last three have complex pairs s a i!. D ˙ Sources Sinks Saddles Spiral out Spiral in Center s1 > s2 > 0 s1 < s2 < 0 s2 <0<s1 a Re s>0 a Re s<0 a Re s 0 D D D D In addition to those six, there will be limiting cases s 0 and s1 s2 (as in resonance). D D Stability This word is important for differential equations. Do solutions decay to zero ? The solutions are controlled by es1t and es2t (and in Chapter 6 by e1t and e2t ). We can identify the two pictures (out of six) that are displaying full stability : the sinks. A center s i! is at the edge of stability (ei!t is neither decaying or growing). D ˙ 2: Sinks are stable s1 <s2 <0 Then y.t/ 0 ! 5: Spiral sinks are stable Re s1 Re s2 <0 Then y.t/ 0 D ! Special note. May I mention here that the same six pictures also apply to a system of two first order equations. Instead of y and y0, the equations have unknowns y1 and y2. Instead of the constant coefficients A;B;C , the equations will have a 2 by 2 matrix. Instead of the roots s1 and s2, that matrix will have eigenvalues 1 and 2. Those 2 eigenvalues are the roots of an equation A B C 0, just like s1 and s2. We will see the same six possibilities forC the ’s,C andD the same six pictures. The eigenvalues of the 2 by 2 matrix give the growth rates or decay rates, in place of s1 and s2. y a b y y .t/ v 10 1 has solutions 1 1 et : y D c d y2 y2.t/ D v2 20 t The eigenvalue is and the eigenvector is v .v1; v2/. The solution is y.t/ ve . D D 162 Chapter 3. Graphical and Numerical Methods The First Three Pictures We are starting with the case of real roots s1 and s2. In the equation Ay00 By0 Cy 0, this means that B2 4AC . Then B is relatively large. The square rootC in theC quadraticD formula produces a real number pB2 4AC . If A;B;C have the same sign, we have overdamping and negative roots and stability. The solutions decay to .0;0/ : a sink. If A and C have opposite sign to B as in y 3y 2y 0, we have negative damping 00 0 C D and positive roots s1; s2. The solutions grow (this is instability : a source at .0;0/). Suppose A and C have different signs, as in y 3y 2y 0. Then s1 and s2 also 00 0 D have different signs and the picture shows a saddle. The moving point .y.t/;y0.t// can start in toward .0;0/ before it turns out to infinity. The positive s gives est . 2 ! 1 Second example for a saddle : y00 4y 0 leads to s 4 .s 2/.s 2/ 0. D D 2t C 2t D The roots s1 2 and s2 2 have opposite signs. Solutions c1e c2e grow D D C unless c1 0. Only that one line with c1 0 has arrows inward. D D In every case with B2 4AC , the roots are real. The solutions y.t/ have growing exponentials or decaying exponentials. We don’t see sines and cosines and oscillation. s1t s2t The first figure shows growth : 0 < s2 < s1. Since e grows faster than e , the larger number s1 will dominate. The solution path for .y;y0/ will approach the straight s1t s1t line of slope s1. That is because the ratio of y c1s1e to y c1e is exactly s1. 0 D D If the initial condition is on the “s1 line” then the solution (y; y0) stays on that line: c2 0. If the initial condition is exactly on the “s2 line” then the solution stays on that D s1t secondary line : c1 0. You can see that if c1 0, the c1e part takes over as t . D ¤ !1 Reverse all the arrows in the left figure: Paths go in toward .0; 0/ 0<s2 <s1 s1 <s2 <0 s2 <0<s1 Source : Unstable Sink : Stable Saddle : Unstable Figure 3.6: Real roots s1 and s2. The paths of the point .y.t/;y0.t// lead out when roots are positive and lead in when roots are negative. With s2 <0<s1, the s2-line leads in but all other paths eventually go out near the s1-line : The picture shows a saddle point. 3.2. Sources, Sinks, Saddles, and Spirals 163 2 Example for a source : y00 3y0 2y 0 leads to s 3s 2 .s 2/.s 1/ 0. The roots 1 and 2 are positive. TheC solutionsD grow and e2t dominates.C D D 2 Example for a sink : y00 3y0 2y 0 leads to s 3s 2 .s 2/.s 1/ 0. The roots 2 and 1 areC negative.C TheD solutions decayC and eCt dominates.D C C D The Second Three Pictures We move to the case of complex roots s1 and s2. In the equation Ay00 By0 Cy 0, this means that B2 < 4AC . Then A and C have the same signs and BCis relativelyC smallD (underdamping). The square root in the quadratic formula (2) is an imaginary number. The exponents s1 and s2 are now a complex pair a i! : ˙ Complex roots of B pB2 4AC s1; s2 a i!: (2) As2 Bs C 0 D 2A ˙ 2A D ˙ C C D at The path of .y;y0/ spirals around the center. Because of e , the spiral goes out if a>0 : spiral source. Solutions spiral in if a<0 : spiral sink. The frequency ! controls how fast the solutions oscillate and how quickly the spirals go around .0;0/. In case a B=2A is zero (no damping), we have a center at (0, 0). The only terms Di!t i!t left in y are e and e , in other words cos !t and sin !t. Those paths are ellipses in the last part of Figure 3.7. The solutions y.t/ are periodic, because increasing t by 2=! will not change cos !t and sin !t. That circling time 2=! is the period. Reverse all the arrows in the left figure: Paths go in toward .0; 0/: a Re s>0 a Re s<0 a Re s 0 Spiral sourceD : Unstable SpiralD sink : Stable Center :D NeutraDlly stable Figure 3.7: Complex roots s1 and s2. The paths go once around .0;0/ when t increases by 2=!. The paths spiral in when A and B have the same signs and a B=2A is negative. They spiral out when a is positive. If B 0 (no damping) andD4AC > 0, we have a center. The simplest center is y sin t; y D cos t (circle) from y y 0. D 0 D 00 C D 164 Chapter 3. Graphical and Numerical Methods First Order Equations for y1 and y2 On the first page of this section, a “Special Note” mentioned another application of the same pictures. Instead of graphing the path of .y.t/;y0.t// for one second order equation, we could followthe path of .y1.t/;y2.t// for two first order equations. The two equations look like this: dy1=dt ay1 by2 First order system y 0 Ay D C (3) D dy2=dt cy1 dy2 D C The starting values y1.0/ and y2.0/ are given. The point .y1; y2/ will move along a path in one of the six figures, depending on the numbers a;b;c;d. Looking ahead, those four numbers will go into a 2 by 2 matrix A. Equation (3) will become dy=dt Ay. The symbol y in boldface stands for the vector y .y1; y2/. D D And most important for the six figures, the exponents s1 and s2 in the solution y.t/ will be the eigenvalues 1 and 2 of the matrix A.
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