MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.LECTURE 30. PHASE PLANES II Isoclines. We continue study the behavior of the trajectories of � �� � �� � x ab x(30.1) y = cd y. By forming the quotient of the two equations, we obtain dy dy/dt cx + dy c + d(y/x)(30.2) = = = ,dx dx/dt ax + by a + b(y/x)provided that x .=0and ax + by =0It is readily seen from (30.2) that if y/x is constant then dy/dx is constant. That means, at all points on y = mx, where m is a constant, the orbits passing through the points have the same slope. Such a line y = mx is called an isocline of (30.1). The word “iso” means “same”. On the x-axis (y =0), the slope of the orbits of (30.1) is c/a. Therefore, the x-axis is an isocline of (30.1). Similarly, the y-axis is an isocline of (30.1). Indeed, on the y-axis (x =0), the slope of the orbits is d/b. Other useful isoclines are the line ax + by =0, on which the slope is ∞, that is, the orbit is vertical, and the line cx + dy =0, on which the orbits are horizontal. An isocline is not a solution of (30.1). But, if c + dm m = , a + bmthen y = mx is a solution of (30.1). The condition is equivalent to that m is a root of (30.3) bm2 +(a − d)m + c =0. Assume b . (If b =0then x =0is an orbit which corresponds to m = ∞. See the exercise =0below.) The discriminant =(a − d)2 − 4bc of (30.3) is the discriminant of pA(λ)=λ2 − (a + d)λ + (ad − bc). Hence, there are two straight-line orbits y = mx in case of a node or a saddle ( > 0) but no straight-line orbits in case of a focus ( < 0). Exercise. If m solves bm2 +(a − d)m + c =0, then show that a + bm solves the characteristic equation λ2 − (a + d)λ +(ad − bc)=0. If y = mx is a solution of (30.1) then x� = x(a + bm), and hence, x = ce(a+bm)t . Therefore, the sign of a + bm determines whether the orbit y = mx moves toward the origin or away from the origin. If a + bm < 0 then the orbit y = mx is oriented toward the origin and if a + mb > 0 then it is away from the origin. Exercise. For the system x � = ax, y � = cx + dy with a = d, show that x = c1e at , cx +(d − a)y = c2e dt . Hence, u = x, v = cx +(d − a)y satisfy |u|d = k|v|a. Sketch the trajectories. 1Summary. We summarize our discussion so far in the diagram below. p q p2 =4q unstable nodes unstable focus vortex stable focus stable nodes saddles Figure 30.1. Degenerate cases. We now consider the limiting or degenerate cases which fall outside of the six regions in Figure 30.1. Degenerate nodes. Suppose p2 =4q and p =0 , and suppose it does not happen that a = d and b = c =0. Then, p(λ) has only one (double) root, −p/2. This case may be thought of two roots with the same sign, but it happens that the two roots coincide. Hence, we expect that the behavior c + dmof the orbits are similar to the case of a node. The difference, though, is that m = has one a + bm solution and hence the two straight-line orbits coincide. Thus, the orbits are tangent to a line at the origin and almost parallel to the same line at the distant points. In this case, the origin is called a degenerate node. If p> 0, then the origin is a stable (degenerate) node, and if p< 0, it is a unstable (degenerate) node. Figure 30.2. A degenerate node. Lecture 30 2 18.034 Spring 2009� � Star points. Suppose p2 =4q, and suppose a = d and b = c =0. Then, (30.1) reduces to x = ax, y = ay. The solution is given as x(t)=c1e at , y(t)=c2e at , yielding y/x is constant. Hence, the orbits are radial lines throught the origin. In this case the origin is called a star point or a singular node.If p> 0 (a> 0), the origin is a stable star point and if p< 0, then it is unstable. Figure 30.3. A star point. Vanishing determinant. Next, we consider q =0, that is, ad−bc =0. It then follows that cx�−ay� =0, and hence cx − ay is constant. It gives a family of parallel lines, unless a = c =0. Similarly, dx − by is constant, and it gives a family of parallel lines unless b = d =0. Zero coefficient. Finally, if p = q =0, then a = b = c = d =0. In this case, every solution of (30.1) is a critical point. This case is uninteresting. Lecture 30 3 18.034 Spring
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