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0:00
Hi, in this video I'll explain the concept of equilibrium solutions and their stability. Let's take a differential equation that can be written in this form. We call this an autonomous first order equation.
0:25
It's first order because there's only a first derivative of y in this equation. And it's called autonomous because the t, the independent variable in the equation, it does not appear explicitly. It's implied inside of this derivative, but otherwise it does not appear anywhere else in the equation. It's an equation that only depends on the value of y, the dependent variable, not of t, not of the independent variable. For an autonomous first
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order equation, an equilibrium solution occurs when the derivative dy/dt is equal to zero. It's best to visualize a slope field. Some of you may call it a direction field. Let's take a simple equation. dy/dt = y -1. Right? Our definition - this is a definition - our definition is that the equilibrium solution is when dy/dt equals zero. So let's figure out when is dy/dt equals zero. When dy/dt equals 0, we put zero right here and we solve this equation. This gives us y = 1. And
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therefore the line - this is not just a single value, this is a function, right? y = 1 as a function of t. And therefore we're talking about a whole line y = 1. Let's take a look at this on a graph. Suppose that at some time t1, y happens to be equal to 1. Then we have that point there. What's the slope there? The slope there is zero. dy/dt is equal to zero at that point. And here's where being autonomous comes into play. Because the slope does not depend on t, then doesn't
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matter what the value of t is, the slope stays constant, stays the same. And in this case, it stays zero. And therefore the slope field would have all of these horizontal lines all lining up at y = 1. Right now we have the value of y right here at t = t1 and y = 1. What happens when t increases by a little bit? If t increases by a little bit to, let's say t2 over here, how does the y curve move? Well, it moves along the slope, yeah? And the slope is horizontal.
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The slope is equal to zero. And so that means this moves horizontally to that point. So now we have the point (t2, 1). And what do you think happens after that? As t changes, as t changes, y follows the same slope of zero. And therefore it stays equal to one the whole time. That's why we call it an equilibrium solution. It's an equilibrium solution because it hits the number 1 and it stays number 1. So the horizontal line y = 1 is a equilibrium solution. And you can see that
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that's always true, right? Because an equilibrium solution is defined with dy/dt equals zero. And because dy/dt does not change with t, that means an equilibrium solution is a horizontal line. That means we can use the slope field to just find the equilibrium solutions. We just look for the horizontal lines. Let's take a look at the slope field of this equation. Here is the slope field,
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and I'm using the slope field generator hosted at the University of Arkansas. You can see that there is a horizontal line right here. So if I have - if I put in an initial value t = 0 and y = 1 and I trace that path I get a horizontal line and that is the equilibrium point, or the equilibrium solution. So let's summarize what we got so far. An equilibrium solution is defined as when the derivative dy/dt is equal to zero.
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That's also how we would find an equilibrium solution. To find it, we set dy/dt = 0. Well, the equation is dy/dt = f(y). Therefore, it leads to this equation f(y) =0 which is not a differential equation, which is an algebraic equation which you can solve algebraically. Right? That's how you find the equilibrium solution. Here are some examples. Go ahead and try to find the equilibrium solutions for these differential equations.
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Press pause before I show the answer. All right, welcome back. Here are the answers. The algebra is pretty straightforward, so I'm just giving you the answer. All right. That's the concept of equilibrium. Now, let's talk about stable and unstable equilibrium.
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Let's take a look at this slope field again. The equilibrium point or solution is at y = 1. If we accurately measure our initial value at t = 0, y = 1, we will get the correct answer of y = 1. And if people ask us to predict what will happen in the future, we will correctly follow this line and say that in the future the value of y will be equal to 1. Great. But what if our measuring instruments are not exact? And most measuring instruments are a little bit off.
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So instead of arriving at y = 1 when t equals 0, we may have some measurement error and we measure it out to be 1 point, let's say 1.05. Put it in. So at time t equals z, we think the value of y is 1.05 because that's what our instrument says. Then we draw this solution curve. And check this out. We thought that the starting point is 1.05. It's a very small error right here. But because
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the slope grows this way, by the time we need to predict what happens in the future, we are very far off. We're way off. And vice versa. What if our instrument is a little bit below the reality and instead of getting y = 1, we get y = 0.95. And we draw the solution curve for that on the slope field and we get this other error going the other way. Our prediction will also be very far off. This is a situation where we say that the equilibrium solution is unstable.
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We say it's unstable because we start out with a very tiny error and it causes a huge difference later on. Looking at the slope field, could you have predicted this to happen? Yes, right? I could have seen that because all the slopes are pointing away from the equilibrium solution. So I could have predicted that if you're off from the solution by a little bit, the slope will pull you away from it. Right? So that's actually a rule. Let's write it down.
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An equilibrium is unstable if the slopes point away from it. A small error leads to a large difference. And what happens to all the solution curves here? We take a solution curve that's off by a little bit. It moves away from the equilibrium.
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Another way of saying this is that nearby solutions or specifically nearby solution curves diverge away from the equilibrium. And you can pretty much predict that the opposite of this makes it a stable equilibrium.
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Let's take a different slope field. This is dy/dt = 1 - y. What's the equilibrium here? Again, I put zero for the slope and I get y = 1. It's the same equilibrium as the previous equation. If I start out with the initial value of t = 0 and y = 1, I get a horizontal line. It's an equilibrium. Now, what happens if my measurement is off by a little bit and I get 1.05? I get this curve. So, my initial value is at t equals zero. So, I'm at time t. And then if I need to predict
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what's going on in the future, I will be okay. My measurement is a little bit off at the beginning, but eventually the curve merges into y=1. So if anyone wants a prediction at this point, I'll get a pretty good answer. And likewise, if the measurement is a little bit too low, 0.95 for y, I have the same thing. I have the measurement a little bit off at t equals 0, but if I predict things in the future, I'm okay again. How did this work? This
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works because all the slopes are pointing toward the equilibrium solution. And that makes this a stable solution or a stable equilibrium. Let's write it down.
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An equilibrium is stable if the slopes point toward it and that the nearby solution curves converge into it.
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Then it's considered to be stable because if we're off by a little bit we fall back in. All right, let's take a new example. dy/dt = (2 – y) * (y ^ 2 – 1). Where are the equilibrium solutions? Go ahead and press pause and do this and come back when you're done. All right, welcome back. To find the equilibrium, we just put zero where dy/dt is and therefore it gives us the answers of y = 2, y = 1 and y = -1. We have three equilibrium solutions. Are
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these stable or unstable? Let's take a look at the slope field. Using this slope field, can you tell? Press pause and think about it. Answer before continuing on. Welcome back. Let's take a look at one equilibrium solution at a time. The first one is y = 2. And if you're off of it by a little bit, you fall back in because all the slopes are pointing toward the um the equilibrium. So if you're off by a little bit, if you're here, then you just fall back in.
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If you're off by a little bit, you just fall back in. And the same thing at the bottom. You just merge up into it. Merge in. Merge in. For the solution y = 2, it’s a stable equilibrium. Now let's take a look at y = 1. Also an equilibrium. But if we're off by a little bit then the curve will send it out, send us away from the equilibrium. Above. Yes. below also it moves away and therefore it is an unstable equilibrium.
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That's the line y = 1. Now let's take a look at y = -1. What happens if I'm off of it by a little bit? Let's say that my measurement comes up down here. Well I just follow the curve and the slope point me toward the equilibrium. Same thing on top. So this makes this a stable equilibrium. This equation has three equilibrium solutions. y = 2 is stable, y = 1 is unstable, and y = -1 is stable. All right, are we good? Now I want to show you a way to figure out the stability
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without drawing a whole slope field. I'll introduce the concept of phase diagram. Let's take a look at just the y axis. Actually, since the slope field does not depend on t, because the equation is autonomous, I will take a um I'll take a copy of the y axis but over here instead. Over on the left edge so I can have all the numerical values. I know I have an equilibrium at y= 2, y = 1 and y = -1. I mark those points. Now let's take a look at the slopes. Well,
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at y down here below -1, all the slopes are pointing toward -1. That means I know if I'm below negative one off by a little bit and below -1, the slope will push me up toward -1. What happens in this segment between -1 and 1? Well, all the slopes are pointing down. That means on the one hand if I'm off from -1 by a little bit I will get sent back down to it from both direction. From below it I got pushed up to it. From above it I got pushed down
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to it. That means from both direction I get pushed toward y=-1. That means -1 is stable. But up here at y equals 1, if I'm a little bit below it, I’ll got pulled away from it. Same thing up here, all the slopes are going up. If I'm off of y= 1 by a little bit, I'll get pulled away from it up there as well. And therefore, from this, I know this equilibrium is unstable. What about at 2? Well, from below, if I'm off by a little bit below, I get pushed up back to 2. So, it indicates
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stability. But what about the other side? On the other side, all the slopes are pointing down. And so, if I'm off a little bit above y = 2, I got pushed back down to it. Yes, this is completely stable. It's stable from below. I got p pushed back up. It's stable from above, I got pushed back down. This is what's called a phase diagram. Just by looking at the y-axis and looking which way the
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slope is pointing, I can tell the stability of the equilibrium solutions without having to draw an entire slope field. The moral of the story is we only need to check the direction of the slopes around the equilibrium points or equilibrium solutions. Then we can figure out the stability. To determine stability, we only need to check the direction of the slopes. We don't need to check that is exactly slanted this way or slanted this way. All we need to know is that it going up.
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Great. How do we determine the direction of the slope? What does it mean when a solution curve is increasing? To be more precise, the slope here is actually of the solution curves. And to check whether a slope is going this way or that way, all I really need to do is check the sign of the slope. And the sign of slope means the sign of dy/dt. And what is dy/dt supposed to be equal to? It is supposed to be equal to f(y). And that
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means at the end of the day, all I really need to do is to check the sign of f(y) and I can figure out the stability of the equilibrium solutions. So let's do it as an example. Let's take the uh equation we've been working with. dy/dt = (2 – y) * (y^2 – 1). We already found equilibrium is 0 = (2 – y) * (y^2 – 1). y = 2, y = 1, y = -1. And now I'm going to check the signs of f from negative infinity
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to infinity. That's what I did over here. I check the sign of dy/dt throughout the entire y-axis. Have we had any kind of problems where we need to check the sign of dy/dt or sign of y’ throughout a number line? Have we done that? Yes, we have. We did it in Calc 1. In Calculus 1, when we were doing graphing, analyzing a graph of a function, we check the sign of y’ on a sign table. So that we're going to do the same thing here. Make a sign table. I have my three equilibrium points.
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And I know that that is where my y prime is equal to zero. Well, y prime is also equal to (2 – y) * (y^2 – 1). And it's equal to zero at these three points.
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Let's take a look at this part. On this side of the number line, y is less than -1. And I want to know what the sign of this expression is. Is it positive or is negative? Well, I could do complicated algebra with inequalities and all that, but if you remember what we did in Calc 1, we didn't need to go through all the algebra. All we needed was a test value. So over here all I need is a test value. Um let's say I test it with
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y = -2. I plug y = -2 into here. I have 2 minus -2, times -2^ 2 -1 like that. This is equal to 4. This is equal to 4 minus one which is positive. Positive * positive = positive. The sign here is positive. Now this interval. In this interval I can test y equals 0. Have 2 - 0 that's a 2. 0^ 2 - 1 that's a -1. So it's a negative number. Over here I test with y equals - what? - 1.5 maybe. Then I have (2 - 1.5)* (1.5^ 2 – 1). I don't really need to compute
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this out. I just know that 2 - 1.5 is a positive number. 1.5^ squared is going to be bigger than 1. and you have a number bigger than 1, I subtract 1. I get a positive number. Positive * positive is positive. So, I know it's positive here. And finally, for this interval, I'm going to test something like y = 3. Then what do I have? I have (2 – 3) times (3^ 2 – 1). This is -1. This is some
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positive number. -1 * a positive number makes a negative number. It’s negative here. I have all the signs, now what do I do. For this interval y prime is positive. It means the solution curves over there are increasing and on a horizontal number line increasing means going this way going to the right. So on the number line here, I'm going to draw an arrow that goes to the right. Now the interval between -1 and 1. y prime is negative and therefore the
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solution curve is decreasing and on, again, a horizontal number line, decreasing means you're going that way to the left. So I'm going to draw an arrow going to the left.
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Between 1 and 2, positive increasing I'm drawing my arrow to the right. And after y = 2, for that interval the solution curves are decreasing so I'm drawing my arrow to the left. That horizontal line that you see right here with the equilibrium and all the arrows. That is called a phase diagram. And this phase diagram allows me to see which equilibrium is stable. y = -1 all the arrows are pointing toward it. It's stable. y = 1 the arrows are pointing away from it. It's unstable.
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And y equals 2 the arrows are pointing toward it. It's stable. Now just one more concept before I summarize. It's a concept of a semistable equilibrium. Let me introduce it using an exercise. Find the equilibrium solutions
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and determine their stability for this equation. dy/dt = (y – 2)(y^2 – 4). Go ahead and try this. Press pause and come back when you're done or when you're stuck. I'll show the answer. All right, welcome back. You should be able to arrive at these solutions for equilibrium and this table of signs. Correct? If you didn't get the same answer, uh, press pause and double check your work and try again. Let's now draw the phase diagram.
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In the interval before y = -2, I have y prime is negative and therefore the arrow is pointing to the left. In the interval between y = -2 and y = 2, y prime is positive pointing to the right toward the positive direction, the increasing direction.
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And for the interval of y greater than 2, y prime is positive. Arrow pointing to the right. To the increasing direction. What does this say about stability? At y= -2, all the solution curves are moving away from it. So that's unstable. But what about at y = 2? At y = 2, if I'm off of this measurement by a little bit down here, then the curve merges back to 2. So on this side, it seems to say that the equilibrium point is stable. But on the side greater than 2, if I'm off
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by a little bit greater than 2, the slope field will pull me away from this equilibrium curve. It indicates unstable. So the equilibrium solution y= 2 is stable on one side and unstable on the other side. It’s considered to be semistable. Where it is stable for y less than two and unstable for y greater than two. That's a semistable equilibrium point. Let's take a look at its slope field. Here is the slope field for the equation we've been looking at.
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Let's take a look at the initial value t=0, y = -2. That's the equilibrium down there. And at positive 2, that's the other equilibrium up there. Let's take a look at the issue at y= -2. If I'm off by a little bit above it, I got pulled away. If I'm off by a little bit below it, I got pulled away. So definitely that's an unstable solution. So let's draw some of that in here. If I end up with -2.05 instead of -2, I got dropped away from it pretty quickly.
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Well on the other side if I'm at - 1.95 instead I also got pulled away from it but then I merge into y = 2 so as far as the line y = 2 is concerned if you're off a little bit from y = 2 on the low side let's say 1.95, then I got pulled into 2. It’s a stable kind of error. But if I'm off a little bit on the high side, if I'm at 2.05, then I got pulled away from it. And that makes it an unstable situation. So for y= 2, it's stable on the low side and it's unstable on the high side.
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All right, here's a summary. In this video, we're talking about autonomous first order equations only, but the concept can apply to other equations as well. This is just an intro. Autonomous means the independent variable t does not appear explicitly. The form of the equation is dy/dt = f(y). The equilibrium solution is the value of y such that dy/dt = 0. Meaning f of y equals zero. In the slope field, the equilibrium solution is a horizontal line. Stable equilibrium
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means all of these things. Solutions nearby converge to it. The slope field flows toward that horizontal line. And for that matter, let's go back to a slope field. When you look at the slope field, always go from left to right. Flow with the uh independent variable. Stability also means phase diagram arrows point toward that point. Unstable equilibrium means the exact opposite. So solutions nearby diverge from it. The slope field flows away from that horizontal line and the
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phase diagram arrows point away from that point. Semistable equilibrium means an equilibrium that's stable on one side and unstable on the other. Now the summary of methods. Given the equation dy/dt = f(y). To find equilibrium, set dy/dt equals z and solve the equation f(y) = z algebraically. To determine the stability of the equilibrium solutions, make a number line and a table of sign. A positive d dy/dt means an arrow that points from left to right.
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A negative dy/dt, draw an arrow from right to left and look at where the arrows are pointing. Arrows pointing toward the equilibrium is stable. Arrows pointing away from an equilibrium is unstable. One of each, one pointing toward and one pointing away, it's semistable. It's stable on the side where it's pointing toward and it's unstable on the side that pointing away. All right, hope that helps. Like, share, and subscribe for more contents. Any questions,
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ask in the comments section below. Thanks for watching. Bye.
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