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The Fitzhugh-Nagumo model for the electrical impulse in a neuron states that, in the absence of relaxation effects, the electrical potential in a neuron $ v(t) $ obeys the differential equation

$ \frac {dv}{dt} = - v [v^2 - (1 + a) v + a] $

where $ a $ is a positive constant such that $ 0 < a < 1. $(a) For what values of $ v $ is $ v $ unchanging (that is, $ dv/dt = 0)?. $(b) For what values of $ v $ is $ v $ increasing?(c) For what values of $ v $ is $ v $ decreasing?

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(a) $\frac{d v}{d t}=0$ when $v=0,1, a$(b) $v$ is increasing when $v \in(-\infty, 0) \cup(a, 1)$(c) $v$ is decreasing when $v \in(0, a) \cup(1, \infty)$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 1

Modeling with Differential Equations

Matt S.

October 22, 2021

Thank you for the detailed solution

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Okay, so once again, we have a problem with us. This is the given the fits Fitz hug Nagumo model. And it happens to be a differential equation. Uh, with the falling expression, the view of the deity equals to negativity of the square minus one plus a in print. This is V plus A and then square bracket. So this is a differential equation, which means that it has, um, variables. Together with the derivatives, it's possible to have first order differential equations, a whole second order differential equations, for example. In our case, we have a fast order differential equation because this's just the fast river, too. On it's an electrical impulse in the absence ofsome kind of relaxation effect, and it obeys this type of differential equations. So another thing well given is the fact that the constant eh lies between zero and one. So that's a fraction. Anything between zero and one is a fractional decimal input. A. I was supposed to assume that devi over the T is zero, and this simply means that lousy um, venti is not changing. HVT is not changing. So this one would assume that vt is a constant to constant variable. Remember that home, vt. Just tons for the electrical impulse Electrical impulse off a new room. Sexual impulse of a neuron. So important, eh? I'm just gonna shift to the next page in part A they're telling us to find, but, hey, they're telling us to find Okay, My pen is so find V when Devi over DT equals to zero. So we're going to make the whole differential equation. We're going to make it equal to zero. So negativity, um V squared minus one plus a plus a V plus A that's equal to zero. So this simply means that either negative V equals zero in the whole expression. This squared minus one plus a V plus A equals to zero. If you have a product and you have two expressions that give you a product of zero, it means that one of them has to be equal to zero, or either both of them have to be equal to zero. So in this case, you divide both sides by negative one, and we already have one solution for the vehicles to zero. In the second case, we can distribute the V and also distribute the minus So we're distributing both of'em the miners, and that gives us a negative fee right there minus Evie plus A that gives us a zero. And then we follow a process called factoring by grouping where we group these two together and those two together. But remember, since there's a minus in between there, this plus is going to become a minus. When you introduce that parenthesis is so then we have V squared minus V on minus in parentheses minus a V plus. Hey equals to zero. We can factor out of the which is the juicy air from the fast sprint this's and we're left with V minus one. And then we can also factor out on a and we're left with another V minus one. Remember, we said that that is going to be a minor, so we don't have a plus anymore because we introduced apprentices. So this becomes a minus. So this is equal to zero, um, in the next page, going to simplify this new expression. So we have Z. We have V on DH, then the minus one minus. Hey, three minus one equals to zero. This is common for both three and a and therefore we can fact it out. We can pull it out and RV minus a. So combining this variable V and the variable, eh? And then putting the common factor there that becomes equal to zero. This simply means that either V minus a equals to zero. And so, Oh, if you are a on both sides, you get another solution. Vehicles too, eh? Or the minus one equals to zero. So you are dead, one in both sides, and so you get the equals to one. Okay, so we have three solutions in the previous page. You can see that the fast solution was when V equals zero. Now the second solution is when vehicles to a and then that that solution is when vehicles to one on DH that completes the process for solving Put a input B. The question is, for value is vey increasing. When a function is increasing, it simply means that it's derivative is obviously going to be a greater than zero. You know, that's what happens when a function is increasing. So Devi of Aditi eyes greater than zero. And so you still go back and take the same inequality. We had before. Oh, rather the same expression we had before. And we have negative v of this word minus one plus day V plus a one plus ain't Prentice's that has to be greater than zero will simplify the negative. We don't deal with the negative. So we divide both sides ofthe inequality by negative one. Uh, so we end up with V. But remember, from the previous step, this whole expression simplified too. V miners one and V minus V will be minus tae and vee minus one. That has to be less than zero. The reason why it's lasted zeroes when you divide by won't play by a negative number. The inequality fitze where did done becomes a less than on this To have a relationship, you can see that this squared minus one plus a V plus a simplified to the minus a V minus one. Um, so at this point, we can use the number line to test for locations where the expression is lessons here. So we're going to do that in the next page. So we have Harvey V minus e the minus one. If we check initially was greater than and then it shifted to less than so We have to take note of that. Then we're gonna have on inequality on specific test points. So the initial test point, if you can call, was a solution of zero. Then we had the second test, not test point. Sorry, but these were the critical point's not the test point. These are the critical points. So this is a and then we have another critical point, which is one amusing holes. Because of this inequality, if we had last under equal to the character off, the number line changed to solid points to reflect the equal sign. But since it's just a less than something happens to greater them, you only have holes right there. Then I'm gonna pick ah, specific test points. This is negative one how pig zero point one, which is very close to zero out click zero point nine, which is close to one just to make sure it's not close to a. And then I'll pick one point one to be on the safe side. So you plug in these numbers into the expression the V minus a V minus one. So when you plug in negative one, there's a certain outcome when you plug in negative one into the expression so you can see this is a negative negative one minus negative A. Regardless of what is a positive number anyway, because it lies between zero and one, it's going to give you a negative value and then a negative right here in the last parenthesis. The negative and the negative will give you another negative value. So you're simply multiplying three negative bodies. And that means thie initial part of the number line will all be negative. Uh, that's step one second step will take the zero point one and plug it into the expression. So this is zero point one minus one, and then it's positive is obviously greater than zero point one, because allies to the right with zero point one. So this part of the princesses of this part of the expression is going to be zero on, then zero point one minus one obviously has to be negative. So we have two negative numbers and, uh, rather too, too negative numbers will give your positive, so this part will be positive. The next thing is to pick the zero point nine. So we have zero point nine zero point nine, minus a and then zero point nine minus one. So zero point nine is positive, zero point nine minus a. That means we have zero point nine is to the right off a. So it's going to be giving us a positive number. In that practices, zero point nine is to the left of one, so it's going to be giving us a negative number, a positive times of positive times and negative. You can see that's going to give us a negative number. So in between and one, we have a negative number bond, then the last test point is one point one so one point one one point one minus a and then one point one minus one one point one is positive, one point one minus a obvious responsive because it's the right off A on one point one minus one is also going to be positive because it's to the right of one, so anything that lies to the right of one will be positive. So now we have a number line and we can use that to complete the solution and say that how inequality z the mine ISI. The minus one is less than zero. This is true when the inequality, when the solutions light to the left of zero and also when the solutions vibe between and one. So those are going to be our two solutions where the function happens to be increasing. So the function is increasing. When V is between negative infinity and zero and the union off stay in one. So those are the two location's way. It's increasing in the last part of the problem. Just going to go to the next page we're tasked to. They were asking us to find this is but not partly, but Patsy, because you already did party A and party. So this happens to be but see and in part see, you can see that we want to find where the function is decreasing where the function is decreasing. And that simply means that Devi over DT is less than zero. We're going to write out the expression again. So negativity, uh, v squared, This is square bracket. Make it easier. One plus que I think we had a V outside of there, so we just have to go back and verify that we have the the right expression. Yes, we have be outside. So, V, um and then plus C, that whole thing has to be less than zero again. We divide both sides by negative one so we can get rid of the negative on the left side of the inequality. So we end up with the A V squad minus one plus a V eyes greater than is greater than zero. If you can recall, one was simplified. The expression the squared minus Want less Evie Plus que we ended up with the V minus a V minus. One is greater than zero. And that means we're going to get a new number line. But it's no different from the previous number line since the test points are the same. So we have a test point here on that test point. Happened to be zero. We have another test point, not test One. Critical points are so we have a critical point there, which is zero. And then we have another critical point. A and another critical point one. I was still going to test the same points, a zero point one, and then we'LL test another zero point nine and then one point one going back to the previous slide. You can see that that when we tested those four test points, as in negative one zero point one zero point nine and one point one we were able to get negative solutions right there. Positive solutions between zero and a negative solutions between and one And then we also had positive solutions beyond one. And so, looking at our inequality, we want to find where the expression is greater than zero. And that happens happens between zero and a and also beyond one. So function is decreasing. Our function is decreasing. Ah, when a V or the solution lies between zero and a and also in our solution lies beyond one. So you know, though those of the steps on finding the solutions, just going back to clarify what happened Initially we had Devi of a TT being equal to zero. That means the the neural, the electric impulse of the neuron HVT is a constant and so we factored a couple of variables distribution process. So we use distributive law in second page and also used juicy a factory ization and that helps that helps us helped us to simplify the problem. You can call that we had a fast solution, which is vehicles to zero. And then by the time it got to the cut that page, we had two more solutions. Vehicles to eight and vehicles to one. And this is similar to what you do when you're solving quadratic equations. And then but be we had to find the function when it's increasing. And that means Devi over duty is greater than zero s o. We use that inequality we still fucked. Our problem will divide both sides by negative one because we wanna end up with a simplified expression that just changes thing. Equality coalesced inside with simplified. We use the critical values to build the number line. Remember, we have holds because we're dealing with inequalities on, then protest points negative one zero point one zero four nine and one point one. Simplify that we find this solution and we repeat the same process to find where the function is decreasing. I hope you enjoy the whole process of solving the problem, Phil. Free to send any comments or questions or anything that you need to go around to the problem. But remember increasing and decreasing functions. I used the large and apply problems in calculus. Thanks and bye and have a great day.

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