Click for printer friendely version of this HowTo Graphical Solutions: Phase Plane Analysis Oftentimes we can gather a good deal of qualitative information about a solution to a differential equation without going through the trouble of finding an analytic or numerical solution. Instead, we can simply look for equilibrium points, or points where the derivative is zero, and determine whether the function moves toward or away from these points as time passes giving us the asymptotic behavior of the function without having to solve for it. One Stable Pointno_title Example 2.10.1.1   Example 2.10.1.2 (One Stable Point)   Consider the differential equation: $$\frac{\textrm{d}y}{\textrm{d}t} = -2y - 4.$$ (2.10.1) When $y(t) = -4/2 = -2$, $\textrm{d}y / \textrm{d}t = 0.$ Thus, if $y(0) = -2$, then for any value of $t$, $y(t) =-2$, since the derivative will always be zero. Thus $y(t) =-2$ is called an equilibrium since it will not change. However if $y(0) > -2$, then the derivative will be negative, and thus, as $t$ grows larger and larger, $y(t)$ will converge to $-2$. This is easily seen by simply plugging in different values for $y$ that are greater than $-2$. For example, if $y = 0$, then $\textrm{d}y / \textrm{d}t = -4$. If $y = 100$, then $\textrm{d}y / \textrm{d}t = -196.$ Likewise, if $y(0) < -2$, than the derivative will be positive for all values of $t$ and $y(t)$ will approach $-2$ from below as $t$ goes to infinity. Since the line $y(t) =-2$ is approached from above when $y(t) > -2$ and below when $y(t) < -2$, it is called a node or a stable state. This is because small perturbations to the system at this point will only lead back to it. That is, if the system is at $y(t) =-2$ and some outside force knocks it to $y(t) = -1.98$ or $y(t) = -2.02$, it will asymptotically return to $y(t) =-2$. Figure 2.10.1 shows an actual plot of the phase lines, or slopes for various values of $t$ and $y(t)$. Due to the fact that there are no free instances of $t$ in Equation 2.10.1, the slopes are the same for each value of $t$. In this illustration, it is easy to see the stable point $y(t) =-2$ and how the slope of any point above or below this line points toward it. Figure 2.10.2 demonstrates how that regardless of the initial condition, as $t$ gets larger, the solution will converge on the stable equilibrium. Figure 2.10.3 shows how a nullcline graph represents the same information. Figure: A Phase Plot for Equation 2.10.1. There is a single, stable equilibrium at $y(t) =-2$. Figure: A Phase plane for Equation 2.10.1, with the solutions for the initial conditions, $t=0$, $y(t) = 1$ and $t=0$, $y(t) = -3$. Notice how, regardless of whether or not the initial condition puts $y(t)$ above or below the stable equilibrium, as $t$ grows, they both converge to it. Figure: A nullcline for Equation 2.10.1. When $t < -2$, the derivative is positive. When $t = -2$, the derivative is zero, and thus, this is an equilibrium point. For $t > -2$, the derivative is negative. Since the derivative is positive on the left and negative on the right, $y(t) =-2$ is a stable equilibrium $\vert\boldsymbol{\vert}$ 1 Stable and 2 Unstableno_title Example 2.10.1.3   Example 2.10.1.4 (1 Stable and 2 Unstable)   Consider the cubic differential equation: $$\frac{\textrm{d}y}{\textrm{d}t} = (y - 1)(y - 2)(y - 3),$$ (2.10.2) where $\textrm{d}y/\textrm{d}t = 0$ when $y(t) = 1$, $y(t)=2$ and $y(t)=3$. By plugging in different values for $y$, we can determine the slope at different points. In this case, we end up with one stable point, $y(t)=2$ and two unstable points, $y(t) = 1$ and $y(t)=3$. By unstable, we mean that if the system is at $y(t) = 1$ or $3$, and it is perturbed slightly, it will not return to its original state. Instead, it will either move toward $y(t)=2$ or $\pm \infty$. This is illustrated in Figure 2.10.4. Figure 2.10.5 shows the equivalent phase information contained in a plot of the nullcline. When the initial conditions are known, specific solutions can be plotted and this is shown in Figure 2.10.6. Figure: A Phase plane for Equation 2.10.2, equilibria at $y(t) = 1$, $y(t)=2$ and $y(t)=3$. The equilibrium at $y(t)=2$ is stable since the slopes immediately above and below it converge to it. Figure: A nullcline for Equation 2.10.2. There is one stable equilibrium at $t = 2$, where the derivative to the left is positive and the derivative on the right is negative. The other two equilibria are unstable. Figure: A Phase plane for Equation 2.10.2, with solutions with the initial conditions set to $y(0) = 3.1$, $y(0) = 3$, $y(0) = 2.5$, $y(0) = 1.5$ and $y(0) = 0.9$. Notice how even though $y(t)=3$ is an unstable equilibrium, and thus, solutions will not converge on it, if that is where your solution begins, it will not deviate from it without some other force acting on it. $\vert\boldsymbol{\vert}$ 2 Stable and 1 Unstableno_title Example 2.10.1.5   Example 2.10.1.6 (2 Stable and 1 Unstable)   Consider the cubic differential equation: $$\frac{\textrm{d}y}{\textrm{d}t} = (y - 1)(y - 2)(3 - y),$$ (2.10.3) where $\textrm{d}y/\textrm{d}t = 0$ when $y(t) = 1$, $y(t)=2$ and $y(t) = -3$. Again, by plugging in different values for $y$, we can determine the slope at different points. In this case, we end up with two stable points and one unstable point. This is illustrated in Figure 2.10.7. Figure: A Phase Plane for Equation 2.10.3. There are two stable equilibria at $y=1$ and $y=3$ and an unstable equilibrium at $y=2$. $\vert\boldsymbol{\vert}$ Click for printer friendely version of this HowTo

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