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This is kind of a rambling question since I don't know how to ask it clearly.

I'm doing some reading on modeling Supply and Demand from economics using a second order differential equation (http://espin086.wordpress.com/2009/11/26/the-second-order-differential-equations-of-dynamic-market-equilibrium/). It's been a while since I've studied this (the DE part) and I have some questions.

This article describes their solution to a differential equation that models price as having 3 solutions:

  • A distinct real solution
  • A repeating real solution, and
  • A complex conjugate.

I presume that if one wanted to use the solutions to the differential equations and calculate expected prices, one would use the distinct real solution.

What's tripping me up though is the repeating real solution. Having plotted it (using graphsketch.com), I don't really see a repeating component in what they describe ($e^{-x}+x*e^{-x}$ is a generalization of the solution).

Logically I don't understand how this fits in to a supply and demand context. Is it something that can be outright ignored? It is a solution after all so that would lead me to day no. The complex conjugate I presume can be discarded entirely since you can't have imaginary money. I remember from my electrical engineering days that imaginary components were very important in calculating magnitudes and this is adding to the confusion.

Any help is appreciated. mj

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I really don't know what you are trying to ask. –  copper.hat Mar 6 '13 at 4:46
I also have no idea what's going on, but holy crap, do economists love using a whole lot of words to describe relatively simple mathematical concepts. –  Arkamis Mar 6 '13 at 5:08
@copper.hat, he might be asking about repeating component in the graph, I put that in grey. Arkamis, may be that is because Mathematics is the Language of Economics, and that happens if you are not good at your own language. (I am also Economist though, I am also in similar situation during my master's days). –  Ram Mar 6 '13 at 5:39
@Ram: Thanks! ${}{}$ –  copper.hat Mar 6 '13 at 5:41
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3 Answers

First let me take you through the generalized method of solving the nonhomogeneous second order ordinary differential equations. Then , I will try to explain the economics behind it.

As you recall when solving second order nonhonogeneus equations of the form $y''+p(t)y'+q(t)y=g(t)$ we need to find the solution of the coressponding homogeneous equation $y''+p(t)y'+q(t)y=0$ and the particular solution of the aformentioned second order nonhomogeneous ODE.

To find the solution of the corresponding homogenous equation we need to find the roots of the characteristic polynomial $r^2+p(t)r+q(t)=0$. This is an easy task. Let me remind you that the root(s) of this characteristic polynomial is(are) $r=\frac{-p(t)\pm \sqrt{p^2(t)-4q(t)}}{2}$. So now we have three possibilities:

  1. If the expression inside the square root sign is positive we will have two distinct roots $r_1=\frac{-p(t)+ \sqrt{p^2(t)-4q(t)}}{2}$ and $r_2=\frac{-p(t)- \sqrt{p^2(t)-4q(t)}}{2}$. And the general solution of the differential equation is going to be $y(t)=c_1e^{r_1t}+c_2e^{r_2t}$.

  2. If the expression inside the square root is zero then we will have only one root (or repeated root) $r_1=\frac{-p(t)}{2}$. And the general solution for the diff.eq. is going to be $y(t)=c_1e^{r_1t}+c_2te^{r_1t}$. Notice that there is extra $t$.

  3. If the expression inside the square root is negative then we will have to use imaginary numbers. I am not going to go through the whole process. You need to know just that the general solution is $y(t)=e^{\lambda t} (c_1cos\mu t + c_2sin\mu t$) where $\lambda$ and $\mu$ are real constants. As you see this solution does not have imaginary part because we can just drop the $i$ with the purpose of having the real part only.

To find the particular solution there are numerous methods (Undetermined coefficients, variation of parameters) that are irrelevant to your question. The bottom line is that the solution of the nonhomogeneous second order ordinary differential equation is $y(t)=y_h+y_p$ where $y_h$ is the solution of the corresponding homogeneous equation and $y_p$ is the particular solution.

Economics interpretation:

You dont disregard any of those solutions. Each solution of nonhomogenous second order ODE is valid and represents the equilibrium prices based on data gathered for the peoples' expectations of the prices and their rates of change (in the article coefficients $c_1,w_1,v_1,u_1,c_2,w_2,v_2,u_2$). If you plug in numbers in those solutions I doubt one will get real world price values, but it is gonna be easy to convert the numerical solutions to real world prices. So, by solving the differential equation in the article $P''+\frac{u}{v} P'+\frac{w}{v} P=\frac{-c}{v}$ you will find the current period equilibrium price P, its rate of change P', and the rate of the rate of change P" , which will allow you to predict equilibrium prices in the future provided your data about peoples' expectations of the prices and their rates of change is accurate.

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Dostre, thank you for your illuminating comment. So if I have made up coefficients and plug those into the three solutions, how does one know which solution to pick? Would I use the magnitude sqrt( sol1 ^ 2 + sol2 ^ 2 + sol3 ^ 2 )? Also, I see that the "t" variable has entered the picture. That's time, right? I just don't want to make any assumptions about anything. –  mj_ Mar 6 '13 at 15:17
@mj_ you are gonna have one solution to the nonhomogeneous differential equation. The way it looks is going to depend on the expression inside the square root. There are three ways the solution to the nonhomogeneous differential equation can look. I described them in my asnwers. So, you are gonna have one solution. The article also describes these three types of solution. First one with two real distinct roots. Second with one real repeated root. Third with two complex conjugates (imagenary numbers). t is time indeed, butnI am not sure what fraction of time (days, months, years). –  Koba Mar 7 '13 at 0:19
Aha, I got it now. Three pieces of the solution that all comprise one complete solution depending on whether or not (u/v)^2 is greater than, less than or equal to 4*w*v. Thank you for your patience with me. –  mj_ Mar 7 '13 at 15:24
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Aside: It is known mathematical fact that our Government runs on imaginary money everyday!

From his description, we have the following.

"We need to write the equations for supply and demand in terms of price (P), the rate of change of the price (P’), and the rate of change of the rate of change of the price (P’). The values given to w, u, and v depends on the peoples expectations about how prices are changing. If people think that prices are rising then the coefficient in front of the first derivative of price will be positive and if there is a belief that prices are falling then this coefficient will be negative. The magnitude and since of the v value reflects how fast people believe that prices are rising or falling. These values can be estimated using statistics and econometric methods, but the following solution is for the general case where these variable are arbitrary real numbers not equal to zero."

As mathematicians, what we like to do is to analyze our solutions using the general approach and to describe what the phases portraits will look like for these general cases. For this, we have several cases to consider. You can choose the variable representation to mean whatever you'd like and maybe in actuality you wouldn't get all those cases based on how you are representing reality (whatever that means).

If you are actually asking what the phase portraits will look like for the three cases, we can certainly draw them and then try and see if we can add actual meaning based on how the model and variables were chosen.

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I like the humor, and your answer, too +1 –  amWhy Apr 25 '13 at 0:22
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The "distinct real, repeated real, and complex conjugate" describe, not the solutions of the differential equation, but of an associated quadratic equation, called the characteristic equation. When that quadratic has distinct real solutions $r,s$, the differential equation has solutions $Ae^{rx}+Be^{sx}$. When the quadratic has a repeated real root $r$, the DiffEq has general solution $(Ax+B)e^{rx}$. When the quadratic has complex conjugate solutions, the DE has solutions that can be expressed entirely in real terms, using sines and cosines instead of exponentials.

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