Lec 31: Market model with inventory, Phase diagrams, Higher order equations

Hello, and welcome to another lecture of this course, Mathematics for Economics. The module that we are covering right now is about Difference Equations. We have seen in the last few lectures how to solve a first-order difference equation, which is non-homogeneous, and right now we are talking about certain examples and applications where such difference equations are used.

Now, this is the model that I introduced in the last lecture, a market model with inventory. So here there is a demand side and there is a supply side, like in the Cobweb model. However, unlike the Cobweb model, we are not assuming that the market clears in every period. Here it is possible that there is some excess demand or extra supply in the market, and that affects the inventory of the producers. And because of this inventory and change of this inventory, the producers might change their prices in the next period; in particular, as you can see on your screen, the second point that I have written here: if the inventory has piled up, they may reduce the price so that the unintended inventory can be disposed of. Similarly, the price is revised upwards if the inventory has run down unexpectedly.

The idea is that in any period, suppose the inventory is piling up; then the producers will like to give some encouragement to the buyers to buy this unintentional rise in inventory, so in the next period they will reduce the price so as to give incentive to the buyers. And similarly, if the inventory is running down unexpectedly, in that case, they will like to stop that rundown, and they will raise the price so as to discourage the buyers. So this is how the model is working.

Mathematically, we represent this in terms of these three equations. So, like the Cobweb model, we are assuming the demand function and the supply function; they are linear. But, unlike the Cobweb model, the supply function is a function of the current price, and this third equation is the new thing. You do not have market clearance in each period, but what you have is adjustment of the price, and presumably this adjustment is done by the producers. So Pt+1 = Pt - σ(Qst - Qdt). The parameters are all positive: α, β, γ, δ, σ. Qst - Qdt is the excess supply in period t; this denotes the accumulated inventory—the part of the production that is not sold that will go to the inventory. So this is that accumulation of inventory, and the new parameter that we have introduced here is σ. It measures the response of accumulated inventory on the next period's price. So this is where the lag is coming. So if there is accumulated inventory, Qst - Qdt is positive. Then that is going to affect the price in the next period, and how much does it affect that response is measured by σ. In particular, Qst - Qdt affects the next year's price in a negative manner because it measures the excess supply—that accumulation of inventory and that reduces the price.

Now, these three equations, if we put them together, then that will give us the first-order difference equation, and this is how we are getting it. So we are basically substituting the demand function and the supply function here in the Pt+1 equation, and that will reduce everything to functions of Pt only. So it is a difference equation of first order in Pt: Pt+1 - (1 - σ)(β + δ)Pt = σ(α + γ). Now, I am not going to go into the steps of the solution; so we have to find out the particular integral and the complementary function. The particular integral turns out to be this, like in the Cobweb model: (α + γ)/(β + δ). And the complementary function is this—this is the complementary function, Ab^t, where if I know that in period 0 the price was P0, I use that information. So this becomes the complete solution: Pt = P0 - [(α + γ)/(β + δ)][1 - σ(β + δ)]^t + (α + γ)/(β + δ). And to make it look simple, I assume that this P* is the particular integral, P* = (α + γ)/(β + δ), I assume that, so I get rid of this cumbersome term. I only write P*, and, as you know, P* is the inter-temporal equilibrium value of price in this case.

Now, let us talk about the stability. As we know that in this solution the stability of the equilibrium depends on this term, β, which is 1 - σ(β + δ). Now, as we know, it can take seven possible different values or ranges of values. However, we can get rid of two such values because we know σ, β, δ, all are positive. Therefore, this possibility that this b is greater than or equal to 1 is ruled out—that is ruled out because all these parameters are positive, so 1 minus these things cannot be greater than 1. But, even if we have got rid of two possibilities, there are five possibilities that are left out. For each of these cases, one can find the condition in terms of the three parameters: σ, β, and δ. So five cases are left out, and let us try to see how they turn out to be. So these are the five cases that I am talking about: one, two, three, four, five. This B can be between one and zero. It can be exactly equal to zero. It can lie between zero and minus one. It can be equal to minus one, and it could be less than minus one, and in each of these cases, these conditions, if we play around with them, then they turn out to be like these. The first region or the first condition turns out to be that σ lies between 0 and 1/(β + δ), and this case, as we know, lying B between 0 and 1, it will give me a time path which is non-oscillatory and convergent. That comes from our initial results. Secondly, if this is exactly equal to 0, we know we are always in equilibrium because B is equal to 0, and this condition turns out to be this: that σ = 1/(β + δ). The third case is 1 - σ(β + δ) lies between -1 and 0, and this turns out to be this condition that σ lies between two limits: 1/(β + δ) and 2/(β + δ), and as we know, this gives me a damped oscillation case. The fourth case is 1 - σ(β + δ) is just equal to -1. We know this will be the case of uniform oscillation because B = -1, and this condition can be written as this: σ = 2/(β + δ). And finally, the fifth case is the B value that is 1 - σ(β + δ) is very low; it is less than -1, and in this case we know there is going to be explosive oscillation. And this condition once again turns out to be this: that σ is strictly greater than 2/(β + δ).

Now, out of these five cases, therefore, we have these first three as stable. In these cases, you have convergence, or you are always in equilibrium, whereas the last two are not stable. Now these five cases can, in fact, be shown in terms of a very convenient diagram. In this diagram, what we have done is that along the horizontal axis we have taken β + δ, and on the vertical axis I have taken σ. Now, why have I done so? Because if you look at the table, you will see that we are getting these two equations: one is σ = 1/(β + δ), this is the case, and here is the other case, where σ = 2/(β + δ). Now, if I am representing β + δ along the horizontal axis and σ along the vertical axis, these two relations can be shown as two rectangular hyperbolas. So this is the first one, and this is the second one: σ = 2/(β + δ) and σ = 1/(β + δ). Both of them are of the form xy = k, that k here is 2 and here is 1. And we know that is the typical equation of a rectangular hyperbola. So these two rectangular hyperbolas will help us demarcate these cases—these five cases. As we can see, the unstable cases are these: either σ = 2/(β + δ) or σ is strictly greater than 2/(β + δ). Now, which are these regions that we are talking about? These regions are actually these cases above the higher rectangular hyperbola or even on the rectangular hyperbola; these are the cases where there is no stability. So that is what I have written here: below the curve σ = 2/(β + δ), the price converges to the inter-temporal equilibrium value. What was the inter-temporal equilibrium value? It was that P*, the particular integral. So if we are picking up points from here—that is below the upper rectangular hyperbola—then we are actually in that region of parameters where there is stability. Either you are always in the equilibrium, or even if you are not on the equilibrium, you are going to converge to the equilibrium; and in other cases—that is, if you are on the northeast of this upper rectangular hyperbola or if you are on that rectangular hyperbola—there is no convergence.

Here is an example of an exercise: the following three equations characterize a model—that is a model of inventory market model. Find the inter-temporal equilibrium value of price and comment if there is convergence. So these three equations are given: the demand function, the supply function, and the price adjustment function. We have to answer two questions: we have to first find the inter-temporal equilibrium value of price. Now that is given by this formula right: P* = (α + γ)/(β + δ). So I just pick up the values from here and put it here, and if I do so, I will get 3; α is 21, γ is 3, and β = 2 and δ = 6. I put all of them here and I get 3: 21 + 3 is 24 divided by 2 + 6. That is 8; 24 divided by 8 is 3. Now then we have to answer the second question: comment if there is a convergence. Now we know that convergence depends on this particular expression: 1 - σ(β + δ), and again I put the values here. σ is 0.3, so I put that value here, and I know the rest of the values, and I get -1.4, and -1.4, as we know, it is less than -1, so there is going to be instability. And it is instability plus it is going to be oscillatory instability because this is negative. The B term is negative and it is less than -1. So there is going to be explosive oscillation.

Now we take up another topic of difference equations, which is non-linear difference equations and phase diagrams. So if you do not have linear difference equations, then how you deal with them. If there are non-linear first-order difference equations, a way to analyze them is to take the help of phase diagrams, so we are going to use phase diagrams.

So here is an example: yt+1 = 6yt² - 9yt + 4 is a first-order difference equation, but it is not linear; first order because the lag, as you can see, is of 1. But there is a power of yt here, which is 2. So it is not a linear difference equation; it is non-linear. Now the first thing to note is that in general, such equations can be expressed in this form: yt+1 = f(yt)—that is a function of yt. Note yt is the only independent variable in the above function. We plot this relation in a Cartesian plane with yt and yt+1 represented along the horizontal and vertical axes respectively, and we demarcate the 45-degree line. This line is of great service in phase diagram analysis. So let us see if we can find a diagram—here it is. Here is a Cartesian plane—that is a two-dimensional plane. On the horizontal axis I have taken yt, and on the vertical axis I have taken yt+1. So it is not xy, but the value of the variable that we are dealing with in two successive periods; the previous period is on the horizontal axis, and the subsequent period is on the vertical axis. Now, additionally, what we have done is that we have plotted this function. This function is what: yt+1 = f(yt)—this line. As you can see, it is not a straight line, and that is precisely the reason why it is a non-linear relation that we are dealing with, and plus we have demarcated the 45-degree line. This 45-degree line will be very critical in the analysis of phase diagrams.

Now, why is it going to be critical? Because the 45-degree line, along the 45-degree line, as we know, the value of the x and the y, the coordinates are equal. So that is the property that we are going to be using in this analysis. In the diagram, the difference equation yt+1 = f(yt) has been represented in this plane, Cartesian plane, yt+1, yt. It is, in this case, I have assumed to be an upward rising line. It is an upward rising line with slope greater than 0. It is not necessary that all f(yt)s will be like this. There could be downward sloping lines as well. We shall deal with them later on, but right now, to start with, I am assuming that this f(yt) is an increasing function. At the point where it cuts the 45-degree line, what we have is this: that yt+1 = f(yt) = yt; the y and the x, they are equal because you are talking about this point; in this point you are on the 45-degree line as well. So yt+1 and yt are the same at that point, and let us suppose that particular value of yt; it is denoted by y*. So what we essentially have is y* = f(y*). That is how y* is defined.

Now, start from an arbitrary y0 in t = 0. So what we are saying is suppose we are starting in the initial period where t = 0. Now, what is the value of y in that period? Let us suppose that is y0. That could be any value, an arbitrary value, so we are taking an y0 like here. It could be somewhere else also. Now, the question that we are asking is: does the time path of yt converge to y*? That is the question of dynamic stability or the lack of it. On the horizontal axis, we have taken y0 to be less than y*. Now y0 is here, then the question is: where is y1? Well, that can be easily found out by using this function. So from y0 we draw a vertical line and go to the point of at this curve, and that will give me y1 because that is the next period's y, which is determined by the function. So the value of y1 can be read off from the function yt+1 = f(yt). This is marked on the vertical axis. Since the function f(yt) is above the 45-degree line, therefore, y1 is strictly greater than y0. So at this point the function is above the 45-degree line. It is above this line, 45-degree line, so that obviously means that y1 will be more than y0. So this is y1. Now the question is: where is y2? We are interested to know the locations of y1, y2, y3, etc., because we want to find out whether we are approaching y* or not because y* is the point where if you go there, then you will stay there. If you are at y*, then in the next period also, you are going to be y* in the third period. Also, you are going to be at y*, etc., etc. So the question of dynamic stability is: when you are starting from an arbitrary y0, then the sequence of y's, do they go towards y* or not—to converge to y*? So that is why we are interested to find out where is y2? If we know about y1 to find this, we use the 45-degree line. The 45-degree line helps us project y1, which is located on the vertical axis on the horizontal axis, as shown. So here is y1. I am projecting this on the horizontal axis here. How am I doing that? I am taking this y1 and going to the 45-degree line and drawing a vertical straight line, and I am reading y1 from the horizontal axis. So this is y0, therefore, and on the same line on the same axis, I am able to locate where is y1, and if I know y1, then I will straight away know y2 because from here I draw a vertical line straight going to the f function, f(yt), and f(yt) will give me y2, which is here from the function. And y2 is here, but we wanted to find out what is y2 on the x-axis. So again, I project that on the x-axis, taking the help of the 45-degree line, so here is y2. So this is the method we are deploying here; we are using the 45-degree line to project the values on the y-axis on the x-axis. And simultaneously we are using the f function as well, so that is what we have written here: from y1 on the horizontal axis, we follow the same procedure as described above to locate y2 on the vertical axis, which is projected back on the horizontal axis with the help of the 45-degree line. Since the f(yt) curve is flatter than the 45-degree line, the increment falls. So this is very critical, the last statement: the f function. I have purposefully drawn it in a manner that it is flatter than the 45-degree line, at least in the neighborhood of y*, at least where I have located y0. Now, if it is flatter, then it basically means that subsequently, the gaps between ys will go on declining. So what I mean by that is that this gap, the gap between y2 - y1, is less than this gap, which is y1 - y0. You can see that with your naked eyes in the diagram, but that is coming because this y, the f function, is a flat function. So that is what I have written here: y1 - y0, which was the first increment, is strictly greater than y2 - y1. This was the second increment, so likewise you can go one step further. The next stage will be y3 - y2, and that y3 - y2 will be further lower than y2 - y1. We go repeating the steps described above at each iteration; the successive yts get closer. That means the gap between the two successive ys they keep on declining, as can be seen in the diagram. In the limit, as t goes to infinity, yt goes to y*—all, so this is how it is happening. First, you go here, then you go here, and like this, you ultimately reach y*. So these are the arrows one can draw to show how the path is demarcated. So in this case there is actually a dynamic stability. Thus, in this case, there is convergence to the inter-temporal equilibrium value. As t goes to infinity, yt actually tends towards y*. One can get the same result of dynamic stability if the initial y is greater than y*. So here we took y0, that is the initial value, to be less than y*, but if we had taken it somewhere here, then also the same result would have been applicable. So here this will be y1, and from here you can find y1. Here this will be y1, and from here you can find y2 here; this is y2, so you can see that actually you go and converge to the point of intersection which is y*, so it does not really matter where we are starting from; y0 could be above y*, it could be below y*; in either case we are able to converge to y*.

If the f(yt) has a shape as shown here—means in this diagram—then, however, there is no stability. So let us try to trace out one path. Here we are taking y0 here below y*, so y1 will be in this value, and again projecting back we get y1 here, so yt will be here, and as you can see, y0 was here; y1 is further away from y*, and y2 will be even further away, and so we are actually going away from the inter-temporal equilibrium value of y. Starting from y0, the time path takes one increasingly away from y*. The difference between this and the earlier case is: in this case, the shape of f(yt) is the slope. The slope of f(yt) is greater than 1, at least around the equilibrium point that you can see here. This line f(yt) should not be here; f(yt) is a steep line. In particular, it is intersecting the 45-degree line from below, which means that it has a slope which is greater than 1, which is the slope of the 45-degree line, and in this case, we are getting the case of dynamic instability. So in both these cases—one was stable, the other is unstable—there was one common point that the f function was an increasing function, but what happens if the function is not increasing? If it is downward sloping, then you have something like this right. I have drawn this f(yt) function. We can see from the diagram that price here fluctuates between below and above the inter-temporal y*. So here y* is at the intersection point like before, which is the long-run inter-temporal equilibrium value; but let us see how the time paths look like, so we are starting from y0 here. What is y1? We go to the f function; this is y1, and we are projecting it back right. So this is y1 on the horizontal axis. Now, what is y2? We go up right. This is y2, and so this should be y2. So this is y2, but as you can see that price first it was below y*. Then it goes y2, y1, which is above y*, and then it again comes back to y2, which is below y*, which means price is fluctuating in this case from below to above the inter-temporal equilibrium y*. That is the first thing to note, but the second thing to note is that the price over time is becoming closer and closer to y*. So thus, in this case, the time path is actually convergent. Unlike the case depicted here, if the absolute slope of f(yt) is greater than 1, then the time path will be divergent, although it will fluctuate as long as f(yt) has a negative though. So the reason why we are going to have fluctuation is because f(yt) is downward sloping—that is giving me fluctuation—but whether you are going to have stability or not that depends on the absolute value of the price. So what is being written here is this: suppose this is the 45-degree line, and you have a downward sloping line, but which is very steep like this. So in this case you are going to have instability. So let us start from—suppose this is P* or y*, and you are starting from here y0 going up to y1, so this is y1. This is y1; what is y2? y2 is this, so this is y2. What is y3? This is y3. So, as you can see, we are moving away from y*. First we were here quite close to y*. Then

We are going a little bit away from y star, y1, and then y2. Again, the gap is rising from y2 to y star and y3. The gap is rising even more so; actually, you are going away from the equilibrium. There is divergence, irrespective of positive or negative slope. If the absolute value of the slope is less than 1, there is dynamic stability.

So this is the conclusion: if we want to have dynamic stability, then the absolute value of the slope of f of yt that has to be less than 1. If it is positively sloped, you are going to have a monotonic convergence. If there is a negative slope, then there can be dynamic stability, but it will be a kind of fluctuation of price. If the slope of f of yt varies between less than 1, equal to 1, and greater than 1, and between positive and negative values, depending on the value of yt, one will see complicated patterns of the time path of price.

So this is a general case: it is possible that if you have yf of yt, but it is not always less than one and not always greater than one, but the slope goes on changing—it can be equal to 1, less than 1, greater than 1, and it can be positive and negative as well. Then we can have really very complex patterns of the time path of price; just think about it in terms of a diagram. Suppose you have something like—so you can see first, we are starting from a very steep line, but then it is flattening out, and again it is getting quite steep right at this range. So here, the slope is varying between less than 1, equal to 1, greater than 1, etc., etc. In this case, like s kind of function, you can have very complicated patterns of price over time.

I think we shall start this discussion of higher-order difference equations; maybe I will not be able to finish it today itself, but at least we can introduce this new topic. Now, so far we have talked about only first-order difference equations; there was only one lag. But if there are higher-order difference equations, then how we can deal with them? In particular, we are going to talk about second-order difference equations.

So here is the general form of a second-order difference equation with some additional properties: that this equation is a linear equation. It is non-homogeneous because you have a non-zero term on the right-hand side, and we have constant coefficients; a1, a2 are all constants, and on the right-hand side, I have a constant term c. So this is that equation, which satisfies all these properties, and we want to find out how to solve it.

Now the general pattern remains the same: first, we find out what is the particular integral and the complementary function. And given the initial conditions, we find out the value of the arbitrary constants, and we write the general solution. So that is the strategy of solving it. Let us first start with the particular integral, and we assume a simple solution, yp = k, k is a constant. And I put it here: yp = k, and so that means y is invariant with time. So I get this equation, and that boils down to k = c/(1 + a1 + a2). But I can write this if the denominator is non-zero, that is 1 + a1 + a2 ≠ 0. But what happens if 1 + a1 + a2 = 0? Then we have to try some alternate solution; I take yp, that is the particular integral, to be a function of t, k multiplied by t. So again, I put this in this equation, and if I do so and solve it, then I will get k = c/(t(1 + a1 + a2) + a1 + 2). However, this part drops out because it is equal to 0, so I am getting c/(a1 + 2). So therefore, the solution will be this multiplied by t. In this case, the particular integral will be k multiplied by t, that is ct/(a1 + 2), and this is called a moving equilibrium because it is a function of time—t is there—but this again is valid if the denominator is not 0, that is a1 + 2 ≠ 0. So I have to cover the case where a1 + 2 = 0.

This is the third case; now, if a1 + 2 = 0, then we try another solution, which is a little bit more complicated: yp = k multiplied by t². So earlier we took k multiplied by t. In the first case, we took only k, and now we are taking yp = kt², and again this can be substituted in the difference equation and solved for k, and we are going to get k = c/2. Therefore, yp = c/2 multiplied by t², and this is the case where this condition is satisfied and the earlier condition—this is also satisfied. So that these two conditions actually will give me these two values: a1 = -2 and a2 = 1. So this is the case of the particular integral; as you can see, there could be three possible cases.

Now we come to the complementary function, and yc = Ab^t. This is the complementary function that we use, and we substitute it to the homogeneous part of the difference equation, and this was the difference equation: yt+2 + a1yt+1 + a2yt = c, and this will give me this thing: Ab^t(b² + a1b + a2) = 0. So the first part, that is Ab^t, is not equal to zero. Therefore, only this part is equal to zero: b² + a1b + a2 = 0. This is a quadratic equation in b, and this is, in particular. This is called the characteristic equation, and its roots are called characteristic roots, and it is given this form: (-a1 ± √(a1² - 4a2))/2. These are two characteristic roots; one if we take the plus sign, minus can be taken, then that will give me the second root. If b1 and b2 are the two solutions, then the complementary function is written as this: A1b1^t + A2b2^t, where A1 and A2 are arbitrary constants.

So this is the general case, but there could be three possible sub-cases from here. One is where the roots are real and distinct, right, where within the square root this part, that is a1² - 4A2, is positive; that will give me real and distinct roots. So let us explore this possibility: a1² - 4a2 > 0. Now, in this case, let us suppose the roots that we are getting from here, right, are called b1 and b2. Then the solution, as we have just seen, is yc = A1b1^t + A2b2^t. So here is an example: suppose this is the difference equation that is given to us: yt+2 + yt+1 - 2yt = 1. Here, what is a1² - 4a2? It turns out to be 1 + 8, which is equal to 9 and which is, as we know, positive. So this is the thing that we are talking about here. The roots will be real and distinct, and, as it turns out, the roots are 1 and -2, so this 1 and -2 will be substituted here as b1 and b2. So this was the first case where the roots are real and distinct, and we have seen what is the complementary function in this case.

The second case is a case where there is repeated real roots; in this case, a1² = 4a2. So basically, we are talking about this part becoming 0. In this case, there will be no distinct roots; there will be a single root, and that is why we are calling it repeated real roots, and what are these two roots? Actually, they are the same root. It is equal to -a1/2, and this root is repeated twice. In this case, yc, that is the complementary function: A = A1b1t + A2b2^t, and since b1 and b2 are the same, this can be written simply as A3b^t, and in this case, the complementary function is actually written as this: yc = A3b^t + A4tb^t, where, like before, A3 and A4 are arbitrary constants, which can be found out if we are given the initial conditions.

And finally, is the case of complex roots, so this is the case where the square root term, a1² - 4a2, turns out to be negative. So this is the condition where it is negative. In this case, the characteristic roots are complex, in the form b1 and b2 = h ± vi. i is the imaginary number, where h = -a1/2 and v = √(4a2 - a1²)/2. So therefore, the complementary function is A1b1^t + A2b2^t, and I know b1, b2 are these h ± vi, so I can write them, and it turns out to be like this. Now, by De Moivre's Theorem, (h ± vi)^t can be written as R^t(cos θt ± i sin θt), where R = √(h² + v²). In this case, what is h² + v²? It is simply a2, and what about θ? We have to find out what is θ also. Well, θ is given by this relation: cos θ = h/r, and if we use the value of h and r, then it is coming out to be -a1/(2√a2), and similarly, sin θ = v/r, and that is coming out to be √(1 - a1²/4a2). So if we solve this, then we can find out θ. And if we solve this, I can find out r. So then we can put the value of r and θ here. Now, since this is known, then yc, I can just put in place of h + vi. I can put to the power t, h + vi to the power t, I can put R^t(cos θt + i sin θt). In place of h - vi to the power t, I can put R^t(cos θt - i sin θt), and this simplifies to be this: R^t(A3 cos θt + A4 sin θt), where A3 = A1 + A2 and A4 = i(A1 - A2). So this is my complementary function in this case. This was the case of complex roots, right, so we have discussed all the four cases of real, distinct roots, repeated real roots, and complex roots. These are basically the cases where the characteristic equation has to be solved, and the characteristic equations can have three separate cases of roots, so we have talked about that.

Now the next point is to take one particular equation, a second-order difference equation, and try to solve it where you can have complex roots. So here, yt+2 + (1/4)yt = 5. You can see that the lag is of two, so it is a second-order difference equation; there is no non-linearity here. So first we look at that term a1² - 4a2, that is crucial. So, in this case, it turns out to be -1, which basically means that the roots are going to be complex, and if the roots are complex, we know that the complementary function we have just found out is given by yc = R^t(A3 cos θt + A4 sin θt), where R = √a2. What is a2? a2 = 1/4, so √a2 will be 1/2. So remember we are taking the positive root here, and what is cos θt? We want to find out θ also, right, cos θt and sin θt, so cos θ is, as we know, given by -a1/(2√a2), and what is a1? a1 is 0 actually because there is no yt+1 term here. So the coefficient of yt+1 is zero. So therefore, a1 = 0, so cos θ = 0 means that θ = π/2, so that information is used to get the complementary function, which is (1/2)^t. This is R^t term multiplied by A3 cos(π/2)t + A4 sin(π/2)t. So this is the complementary function. We still do not know what is A3 and A4. But if we know the initial conditions, then we can find that out, but we have not found out the intertemporal equilibrium; the particular integral again we take yp = k, and if we substitute that here then we get k + k/4 = 5, that gives me k = 4, so both the yp and yc's are found out. Then the general solution will be the sum of these two; that is it. That is what we have written. If the initial conditions are known, the values of the arbitrary constants can be calculated. Note as t changes, the values of sin θt and cos θt go on changing in a cyclical pattern. So that we know from trigonometry that as this θ, that is π/2, is given to us, as t changes, so cos(π/2)t, t will take 0, then 1, then 2, etc., etc. So this angle is going to change accordingly, and the cos of that angle will change in a cyclical manner, and similarly, the sin of the angle will also change in a cyclical manner. That is what I have written: is going to change in a cyclical pattern. The cycles are not smooth because the variable t takes only discrete values. It is a step-like cyclical movement, right? So t will take only values like 0, 1, 2, 3, 4, like that discrete values, so corresponding to that the sins and the cos they will go on changing like that according to that, so the value of yt will change in a cyclical manner, but it is not going to be a smooth cycle. It is going to be something like this. As you can see, there are jumps between two time periods. So that is what I mean: it is a step-like cyclical movement, all right, so I will stop here, and in the next lecture, I am going to talk about the convergence. So we have talked about convergence and dynamic stability of first-order difference equations; we have solved the second-order difference equation, the general pattern, but what about the question of dynamic stability, under what conditions dynamic stability is going to be there? That we can discuss in the next lecture. Thank you.