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Question 44 ptsRecall the following bifurcation diagram discussed in the course:This shows the location of the critical points x* versus the bifurcation parameter B...

Question

Question 44 ptsRecall the following bifurcation diagram discussed in the course:This shows the location of the critical points x* versus the bifurcation parameter B The solid blue lines indicate stable critical points, and the dashed red lines indicate unstable critical points.0.5-0.50.5Briefly discuss the phenomenon of hysteresis in relation to this diagramFurther; discuss the physical interpretation of hysteresis in the context of the Spruce Budworm Model"?Enter your discussion in the box

Question 4 4 pts Recall the following bifurcation diagram discussed in the course: This shows the location of the critical points x* versus the bifurcation parameter B The solid blue lines indicate stable critical points, and the dashed red lines indicate unstable critical points. 0.5 -0.5 0.5 Briefly discuss the phenomenon of hysteresis in relation to this diagram Further; discuss the physical interpretation of hysteresis in the context of the Spruce Budworm Model"? Enter your discussion in the box below: HTML Editorna] B I 4 A ` A - I E = 3 3 2 * * = E B - 0 8 8 @ v 6 0 v D M T 12pt Paragraph



Answers

in Exercises 1-8,
a. Identify the equilibrium values. Which are stable and which are unstable?
b. Construct a phase line. Identify the signs of $y^{\prime}$ and $y^{\prime \prime}$ .
c. Sketch several solution curves.
$$\frac{d y}{d x}=y^{2}-4$$

So this question we have do I x or Y prime? It's equal to y squared, biased for cameras to do three different things and part a were asked to identify the equilibrium values and then decide which are stable. Wench Run, Stay well, Part B. We're going to make our phase line and see we're going to sketch several solution curves. So for part A, we're asked to find equilibrium, points equally their values and determine if they're stable or unstable. So to find these values, I'm just going to factor this so I get why, plus two and why minus to you, So that means my equilibrium points are going to be when the city got zero site. Get y equals negative, too. And why's he would've positive to him? And now it was determined whether they're stable or unstable, and you look at the phase line and see whether they're going, the solution's around. They are going towards it or away from it. So, looking at this phase line, I have negative two into right here. And so I see when, Why prime when I plug in anything less than negative, too. So let's say, like native three I get now you three scores nine minus for which is five says may be greater than zero My employees and anything in between these values. So say, like, zero I get why primacy will negative forces We mean less than zero. And then when I plug in, anything greater through is going again be greater than zero. So as we learned in the book, when Why prime screens here on the side? That means this is going to go towards solution. This way. Here it is just you kind of just follow where is pointed. So this means this one's gonna go this way and then again that way. So I see that for the native to it's going towards its that means this one stable for the to both sides of its going away system unstable. So the answer for part A for part, me asked us constructive phase line and and by the science of y prime and wide double prime. So first I need to figure out my why Double prime. So again I have that y prime is equal to y squared minus four one. Take the derivative this using implicit differentiation. So I take the derivative of why Square and I get to why times y prime before goes away And then I'm going to plug in What? I know why Prime is so I get Teo. Why times why squared minus for she can Factoring that now you get why plus two why minus two. So now I still have my same zeros. Now you tin tube. But I'm going to add another zero here. Why was zero so gonna have thes different things going on? So it we already found the signs for Why Prime not wanna look up? Why double prime? So when Wydell crime is out of value less than negative too. I know this one's going to be greater than zero. However, this one is going to be negative. So this is going to be lesson zero when it's summer and let's say like a negative one. This is going to lessen zero. This one's going to lessons era but negative times negative is a positive sis immigrate er here, let's say I put a one in this was going to lessen zero two times one here between greater than zero positive times negative is a negative. So to lessen zero, and then this one's going to be greater than zero. So she was answered for part B in the now for part. See, we're asked to sketch several curves, several solution curves. So I make my graph here. So first off, when I'm doing this, I can go ahead and just put it in my you delivering value. So I have wanted to zero negative two doesn't know what kind of use my phase line here, Teo, make my curves. So first off, I know that she's in grade into Why prime? It's going dear. What still country and zero. And I know that is moving away from that equilibrium value. So that means I'm gonna have for anything greater than to you is going to kind of be close to it. But it's going to move away. Certainly I know for negative Teo anything blow it all my solution curves are actually going towards it. So I know if I want anything less than native to is going to start away, but then it's gonna get closer to that negative t value. So here again, then everything else. If we started somewhere in between here, it's going to start at this too. But it's going eventually. Come to me here again. As we saw here, everything between these two values goes towards the negative too, to these air at the other valley. So I promise. Listen zero my dull crimes lesson zero.

And this example, we're going to be plotting several solution curves to the autonomous differential equation we see displayed here. The first thing we need to do is solve that the first derivative is set equal to zero. And to do that, let's factor the left 10 side of this equation will have why Times y squared minus one equals zero. So, factoring further, we have y times. Why plus one times y minus one equals zero and the solutions are going to be. Why equals negative one y equals zero and why equals positive one? So these three are the equilibrium solutions. Next, let's start writing what the phase line is going to look like for these solutions. So we have a Y axis that's been written horizontally this way. Let's start by placing down the equilibrium solutions. We have one at negative one one at zero and 1/3 equilibrium solution at positive one. Next, let's look at what happens to the derivative when we pick values that are between these segments here, where we've come up, the real number line at negative one is ruined. Positive one say, for example, we pick a value here at two If we pluck to into this differential equation, we're going to get all together too times three times one which results in a positive quantity. So what this, that tells us, is that we have white prime is bigger than zero in this region, when wherever were greater than one. And because this factory's ation has no repeated factors, the sign will alternate. So why prime will be less than zero here? Why prime will be greater than zero here and blast why prime will be less than zero in this far region here. What this thing tells us is by the first derivative test, if we take of value for why that starts at a height less than negative one, the curve will then begin to decrease. If it's greater than one, it will increase and so on. So we have the solution curve either. Decreasing increasing will be decreasing in this interval and increasing in this interval. Now let's look next at a different way to look at this. If we take a solution curve that's less than negative. One will be decreasing this way, that increasing this way that tells us that negative one day clear broom solution is unstable since solution curves are repelled away from it. Next we have decreasing. So if we pick a solution, curve and go this way, we have zero is attracting the solution curves so at zero this is stable and for the less interval curves that start out at height, greater than one Onley get larger. And so at one solutions a repelled away and the solution curve is unstable or the equilibrium value is unstable at y equals one. Now let's extend this results further and look at the second derivative. So we have to start out why Prime is given to be White Cube minus y. So if we go to the second derivative a y, we will have three y squared, multiply by a white prime since we're differentiating implicitly minus the derivative of why but were denoting that by white prime. Let's factor next y prime from the right hand side so that the second derivative is now equal to why print prime times three y squared minus one. Then we know that why prime is this quantity here That's indicated in blue so we can write by substitution that the second derivative is equal to why Prime, which is by substitution. Why times y plus one times white minus one times the Group three y squared minus one. And as before, we're setting this equal to zero. And then we see that this has solutions. Why equals zero negative one positive one as before. And the next solution is found by solving three y squared minus one itself is equal to zero. If we isolate the Y squared, we'll have. Why squared equals 1/3. And this tells us that why is either positive or negative one over the square to three, which is approximately equal for the sake of sketching positive negative 0.6. So it's placed these two new solutions from the second derivative equation on our same Y axis that's provided here. I will have a value of negative one over Route three, which is about here and positive one over Route three, which is about here, then does partitions that real number line into more regions. Let's used black desh line to express those several regions, and this time we're going to check the sign of the second derivative in each one of these regions. So once again, if we pick, say a value to then go to the second derivative which is provided here Place to infer why and we're going to get a positive quantity altogether. Since mainly here we have four times three minus one here will have to minus one and the rest of the values are clearly positive. So if we pick our value to in this region will have that the second derivative is greater than zero. And then, as before noticed that there are no repeated factors in this factory ization which tells us that the sign is going to alternate for the second derivative. So here will have white double prime is less than zero in this region. The second derivative is positive. Then here it's negative, then positive again and then finally, the second derivative is negative in this region. This tells us information about con cavity. When we go to the last step of curve sketching where the second derivative is negative, the curve will be con cave down. All right, CD for conclave down. So that occurs here here. And let's put this here rather and here the remaining regions, it's con cave up. All right, See you for Khan gave up site curves here, here and here. So now we're ready to begin sketching the curve. First, let's place an ex and why Excess, where Recall this was the Y axis, but now we're going to write it vertically, as it normally is, and let's start writing down the equilibrium solutions to start out. So we had one at zero. I'll just put a heavy blue line here, overlapping the X axis. We have one and positive one. Let's write y equals positive one to denote it and another one at negative one. We also found interesting inflection points at these values here, so it's put them at two dashed lines. So first will have one at about 2.6 and another one at about negative 10.6, which will be roughly here. So we're now ready to begin sketching. I'll sketch the curve and purple and first note in this interval where we're less than negative. One were decreasing and con cave down. A typical curve then would look about like this. It's concave down, since all the tension lines will lie above this curve next in this interval, we're going to be increasing start out con cave up, then make it con cave down. So we increase towards zero, and then we have an inflection point. So we make it con Cave down after its con cave up for the next set, we're going to be decreasing start out con cave up, then concave down. So the curve is this time going to look about like this? Rick on cave up now past the green Dodd Linus con cave down. So that's a typical curve. So far, let's try this again. It should have been decreasing. So let's try first, decreasing where it's con cave up, then con cave down will give us the right sketch here and keep in mind Y equals zero is stable, which means the solution curves should have been attracted in this way towards the X axis. So now that we've made that correction, let's go to the last portion where the curve will be increasing as well as con cave up. So that typical solution curve might look about like this, and this completes our typical sketches for the solutions of this autonomous equation.


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