# Phase Planes and Solutions Curves: Qualitative Analysis

Updated: Jan 12, 2020

When I took MAT 22B, one of the topics I heard the most complaints about was Qualitative Analysis. But I thought Qualitative Analysis was one of the easier—or at least more fun—parts of the course. Let’s take a look at one such problem now, and you’ll see what I mean.

Personally, I like these kinds of problems because they don’t require you to perform any complex mathematics! Of course, qualitative analysis is incredibly useful, but it doesn’t require solving advanced differential equations. Let’s jump right in with Part A.

## Part A

What’s an equilibrium solution? What does equilibrium mean? Well you already know the answer to that—equilibrium occurs when physical properties no longer change with time. So it should come as no surprise that equilibrium occurs when (frac{d y}{d t}=0). That means we’re looking at the following equation:

$$y(1-y)(2-y)=0$$

Conveniently, this equation is already factored for us, so the equilibrium solutions are pretty straightforward. The above equation is only satisfied for (y) values of 0, 1, and 2; therefore, **the three equilibrium solutions are (y^* = 0, 1, 2).**

Notice the other question in Part A: “classify their stability”. We’re going to skip this step. Stability is something that becomes much more obvious when looking at a phase plane, which we will draw in Part B.

## Part B

The phase plane is the scariest part of qualitative analysis—I trust that lots of people are able to make it through Part A, but get stumped here. So to clarify things: **a phase plane is a graph of (frac{d y}{d t}) versus (y)***.* The most important part of your phase plane is the fact that your equilibrium solutions are the “roots” of the graph. This means that your phase plane graph only passes through the horizontal axis at your equilibrium points. Take a look:

This is the first thing you should draw on your phase plane. Yes, I know, it’s not done yet, but we have to start somewhere.

Something you probably noticed about the original function, [(y(1-y)(2-y))], by now is that it’s a cubic function. If you’re a graphing expert, you may have realized that a cubic function with three roots can only be drawn in two ways:

Possibility 1

Possibility 2

How do I know that these are the only two possibilities of my graph’s shape? The **multiplicity **of each equilibrium solution is 1. (Recall that the multiplicity of a root is the number of times it is represented as a root in a fully factored polynomial equation.) When the multiplicity of a root is odd, the graph crosses over the horizontal axis at that root. When the multiplicity of a root is even, the graph “bounces” back from the horizontal axis at that root.

Did I confuse you? I hope not. If all that stuff went over your head, don’t worry about it. There are other ways for you to reach all of my same conclusions without knowing anything about multiplicity.

Now let’s move on to the next logical step: which of the two possible graphs are we dealing with here? If we’re dealing with the first graph, then (frac{d y}{d t}) will always be negative for (y leq 0). Alternatively, if we’re dealing with the second graph, (frac{d y}{d t}) will be positive for all values of (y leq 0). All we need to do is plug in some sample value of (y) (let’s pick -1), to find out:

$$frac{d y}{d t}=y(1-y)(2-y)=(-1)(1+1)(2+1)=-6$$

This test confirmed that (frac{d y}{d t}) is negative for all negative values of (y), and therefore that __the solution should look like the first graph above__.

If all the multiplicity stuff from earlier confused you, or if you couldn’t convince yourself that the two graphs above were the only possible solutions, you can continue to plug in numbers until you’re satisfied with the shape of the graph. You will find that plugging in any value for (y) between 0 and 1 will yield a positive (frac{d y}{d t}), while any value for y between 1 and 2 yields a negative (frac{d y}{d t}). Finally, testing a value of greater than 3 always results in a positive (frac{d y}{d t}).

Here is our phase plane:

Now we need to return to Part A, where we will identify the stability of each equilibrium solution. In order to do so, we need to recognize an interesting feature of our phase plane: by definition, (frac{d y}{d t}) will cause changes in (y) over time. Whenever (frac{d y}{d t}) is positive, (y) will increase until it reaches an equilibrium solution. Similarly, when (frac{d y}{d t}) is negative, (y) will decrease until it reaches an equilibrium solution. Let’s represent this property on our phase plane:

Notice that two of our equilibrium solutions (0 and 2) now have arrows pointing away from them, while one solution (1) has arrows pointing toward it. These arrows—which represent the direction that (y) tends to change over time—reveal which solutions are stable, and which are unstable. In this case, we see that the solution **[(y = 1)] is stable, and [(y = 0)] and [(y = 2)] are unstable.**

## Part C

There is nothing left to do in our phase plane, so we can move on to the final part of the problem: the solution curves. This graph is not the same thing is a phase plane—in fact, it’s on an entirely different axis! In contrast to the phase plane, **a solution curve is located on a graph of (y) versus (t)**. We draw this graph with our equilibrium solutions clearly labeled—they will come into play shortly. Typically, equilibrium solutions are labeled as (y^*)

We went through so much trouble to figure out which solutions were stable, and which were unstable, so let’s draw that information onto our graph:

And we’re nearly done! All we have to do is draw some solution curves on our graph. Technically we have already drawn 3 solution “curves”: the ones originating at [(y = 0)], [(y = 1)], and [(y = 2)]; however, we need to draw 6 more curves to complete the picture. Each of these other curves must originate “nearby” one of our 3 equilibrium solutions. Note that each will gravitate away from unstable equilibrium, and move toward stable equilibrium. Below is the graph with all the major curves drawn in:

A general rule of thumb is that these curves will look smooth, much like exponential growth/decay curves. If you’re confused by how slow or fast any given curve moves toward/away from equilibrium, check out the phase plane. The solution curves move faster vertically (through the y-axis) whenever (frac{d y}{d t}) on the phase plane has a local maximum or minimum.

And we’re done! We found the equilibrium solutions, investigated some important properties that set them apart from one another, and then drew two graphs. It’s not nearly as scary as it looked, don’t you think?