[begin transmission 2/?]
As promised, the day has finally come. We're covering dynamics. God help us all.
I'd like to preface this explanation w/ the qualification that dynamics is a very general term.
Just like systems, dynamics can assume a mechanical, electrical, biological, financial, etc. flavor.
The entire point of dynamics is to describe how these types of systems change over time, using mathematical formalisms.
Sounds crazy, right? You're trying to capture the overwhelming complexity of the world in mere symbols.
With good reason that dynamics is a field of study in itself, w/ people dedicating their lives in the pursuit.
Myself, by no means am I an expert, but in my profession I have to have an acute understanding of them.
It's a piece of the greater picture, and a very important piece at that, but it is not the entire story.
Anyways, let's get to it.
Normal Mode: Dynamics. But First, Babby Mode: Algebra
Okay, so I'm kind of torn as to where to begin. I'm assuming most of my readers are literate in basic algebra, and understand the concept of a function. But for completeness sake, let me remind you that a function is merely an equation that establishes a relationship between an input variable and an output variable (ignore multivariate functions for now). You put a quarter into the machine and receive a gumball. Here's a simple function:\begin{equation} y(x) = x + 2 \tag{1}\end{equation}
You can take your input variable, x, and determine the value of y, the output variable. Suppose that x = 2, you can conclude that y = 4. If you conclude here that y is anything else other than four--and you're not a pure mathematician armed w/ a formal proof--please take your critical theory idiocy elsewhere. For the rest of you, onwards.
Functions can be expressed graphically. For a univariate function such as Eq. 1, this only takes two axes to describe:
 |
| Figure 1. Graph of y(x) = x + 2 |
I apologize if this is all so painfully rudimentary and unnecessary, but I read something like 40% of Americans can't read basic graphs. So here I hope it makes sense that, if you assume a value of x = 2, you can see that y = 4; if you assume a value of x = 3, y = 5, etc. String a bunch of these x and y value pairs together and you get the graphical representation of our function.
Normal Mode: Dynamics. No, no. Hold on. First Calculus.
But okay, you probably knew all of this. Let's graduate from algebra and tackle calculus. The central concept of calculus is that of the derivative: namely, that for every function (ignore discontinuous functions) there exists another function that describes its sensitivity to change in input variables. The common notation for a derivative is dy/dx. To illustrate the concept of the derivative, consider the gumball example above: suppose you put in two quarters--how would this affect the number of gumballs spat out? What if you insert three quarters? Four? A normal gumball machine will yield one gumball per quarter, so you get two, three, and four, respectively. Mathematically:\begin{equation} y(x) = x \tag{2}\end{equation}
Graphically:
 |
| Figure 2. Graph of y(x) = x |
Where y is the number of gumballs spat out, x is the number of quarters inserted. To graphically determine the derivative of this function, take any sample point for x, let's say 3, and increment it by one unit. Then observe how y changes accordingly. If you go from x = 3 to x = 4, then y behaves accordingly, going from y = 3 to y = 4. Expressing this change as a ratio defines the function's derivative:
\begin{equation} \frac{dy}{dx} = \frac{y(4) - y(3)}{4 - 3} = \frac{4 - 3}{4 - 3} = \frac{1}{1} = 1 \tag{3}\end{equation}
The derivative for our function is simply 1. Again, remember, a derivative describes a function's sensitivity to change in input. x, our input, is being changed, and as a result y changes. For every one unit change in x, y changes by one unit as well.
Now, suppose that the gumball manufacturing supply chain is disrupted by a global pandemic, lowering the supply of available gumballs. Also suppose that God awful Democratic policy has disincentivized gumball factory workers from working--by offering them more in unemployment than they'd otherwise earn by working--resulting in an additional decrease in the supply of available gumballs. Since gumballs are now scarce and demand hasn't proportionally decreased but remained constant or even increased, the price of gumballs has now increased. Now gumballs cost four quarters a piece. Mathematically:
\begin{equation} y(x) = 0.25x \tag{4}\end{equation}
And graphically:
 |
| Figure 3. Graph of y(x) = 0.25x |
Following the same procedure as before, we take x = 3, increment it by one unit, and observe the change in y. We then express that as a ratio to obtain the derivative:
\begin{equation} \frac{dy}{dx} = \frac{y(4) - y(3)}{4 - 3} = \frac{1 - 0.75}{4 - 3} = \frac{0.25}{1} = 0.25 \tag{5}\end{equation}
And yes, I botched the mathematical modeling here on the first cut, how embarrassing. I caught it on my own.
So far we've covered two basic examples of determining a function's derivative. These functions are linear, so determining them are child's play. We've obtained them w/ the help of graphs; this is cumbersome. A good student that paid attention in calculus class or an engineer worth their salt knows derivatives of several functions by heart; it's akin to memorizing one's times tables. Such functions are as follows:
\begin{equation} y(x) = cos(x), \frac{dy}{dx} = -sin(x)\tag{6}\end{equation}
\begin{equation} y(x) = sin(x), \frac{dy}{dx} = cos(x)\tag{7}\end{equation}
\begin{equation} y(x) = tan(x), \frac{dy}{dx} = sec^2(x)\tag{8}\end{equation}
\begin{equation} y(x) = ln(x), \frac{dy}{dx} = \frac{1}{x}\tag{9}\end{equation}
\begin{equation} y(x) = e^x, \frac{dy}{dx} = e^x\tag{10}\end{equation}
And if they're not memorized, there are several readily implementable heuristics available to obtain them, such as the power rule, chain rule, product rule, quotient rule...etc. We won't cover how these heuristics are derived because I don't remember how they are derived; that's left for mathematicians to fawn over because, much like how one doesn't need to understand the Carnot cycle to drive a car, an engineer doesn't need to understand the derivation of every single one of their mathematical tools to successfully employ them.
...Why on Earth did we use gumballs as an example? Let's illustrate an example that is easily comprehensible yet readily translatable to an engineering context. For your consideration, the scenario of going for a run. Why a run? Because we can describe a run in terms of spatial position, velocity, and acceleration. These three quantities are directly related to each other by, you guessed it, their derivatives. Specifically, by their time derivatives. What is a time derivative? It's the derivative of a function w/ respect to time. Want to know how you can tell an engineer from a mathematician? Ask a mathematician to write down the symbol for a derivative, and they'll ask "with respect to what?". Ask an engineer the very same thing and they'll instinctually write down something like dy/dt. This is because, in most dynamics problems, we're concerned w/ how systems change over time. An engineer is concerned w/ the sensitivity of a particular function as time changes.
Right, so the time derivative of a function that describes position yields velocity; take the time derivative of velocity and you get acceleration. This ought to make intuitive sense: you describe a point in space via length units, such as a meter; I am 5 meters ahead of you. You describe a velocity in terms of length units per time units, such as m/s; I am running at a speed of 4.5 m/s; for every increment of 1 time unit, my position changes by 4.5 length units. Accelerations are reported as...well, a variety of other units, but keeping w/ meters and seconds, as m/s^2; I am accelerating 2 m/s^2; for every increment of 1 time unit, my velocity changes by 2 velocity units. These three quantities are used frequently in classical mechanics types of problems.
Personally I like to go for a three mile run every other day, keeping up a pace of about six minutes per mile. In engineering units (meters and seconds), this translates to a total of 4827 meters, ran in 1080 seconds. Therefore, my velocity is, on average, 4.47 meters per second. Below are graphical representations of the run, the first describing position:
 |
| Figure 4. Position Graph |
So, at time 100 seconds (t = 100) I'll be 447 meters along the track, at time 101 seconds (t = 101) I'll be at 451.5 meters along the track, at 102 seconds (t = 102) 455.9 meters along the track, etc. Easy enough, right?
As mentioned before, my velocity, on average is 4.47. This is confirmed by the position graph, as moving 1 time unit increases my position by 4.47 meters. So if we take the derivative of the position at every time point, we obtain the following velocity graph:
 |
| Figure 5. Velocity Graph |
It's a straight, flat line. What the heck. Why? Well, that's because we're running at a constant velocity of 4.47 m/s. There are no changes in the velocity, hence it is flat. Why did I show you this, this is boring. It is because it helps one grasp what an acceleration, the second time derivative of position, is. Much like how velocity describes change in position per time unit, acceleration describes change in velocity per time unit. Here, because our velocity is constant (it doesn't change at all, but remains 4.47 m/s), acceleration is zero. Hence, our acceleration graph looks like this:
 |
| Figure 6. Acceleration Graph |
The second, equally important concept of calculus is that of the antiderivative. It is simply the inverse of the derivative, and is mathematically implemented via the integral operation. As such, taking the time integral of acceleration yields velocity; taking the time integral of velocity yields position. Thus, differentiating and integrating is commonplace in dynamics, as it yields several variables of interest. Much like derivatives, the antiderivatives of common functions are memorized like your times tables, and for those that aren't so easy to obtain, there are heuristics to follow to determine them.
Alright, I hope that I haven't lost too many of you. These things are really elementary to understand, but I'm sure my ham-fisted attempts at explaining them are making them seem much more complicated than they need to be. Part of that is because I've been dealing w/ these kinds of things for a thousand years, so tons of information is taken as for granted and lots of small, nuanced details are omitted in explanation. One of such nuances include notation; let's get square that circle here quickly.
Normal Mode: Dyna--No, Dot Notation
Let's take the following function as an example. Suppose it is our equation that describes our position at any given time (hence t):\begin{equation} y(t) = t^2 + 5t + 11 \tag{11}\end{equation}
Quick note on notation (heh): I like to represent functions w/ their arguments. Here, rather than just leaving it as y = blah blah blah, I like to write y(t) = blah blah blah. This reminds me that y is a function of time. One more note here, but notice that this function is non-linear. All equations that we've been dealing w/ before have been linear, making them trivial to differentiate or integrate. This one is a second degree polynomial (second because the 'highest' power contained by a term is 2). Although this equation is non-linear, it is still trivial to differentiate and obtain our velocity equation. It is obtained using the heuristic known as the power rule. For every t term, you take its power, multiply the term by it, and decrease the power by one. So, term-by-term:
\begin{equation} \frac{d}{dt}(t^2) = 2t \tag{12}\end{equation}
For the next term, t is implicitly raised to the power of 1, so:
\begin{equation} \frac{d}{dt}(5t) = 5 \tag{13}\end{equation}
And constants--terms that have no t--simply evaluate to zero and can be omitted:
\begin{equation} \frac{d}{dt}(11) = 0 \tag{14}\end{equation}
Putting all of our terms together, we obtain our velocity equation:
\begin{equation} \frac{dy}{dt} = 2t + 5 + 0 = 2t + 5\tag{15}\end{equation}
To obtain the second derivative, our acceleration equation (notice the difference in notation), apply power rule once more:
\begin{equation} \frac{d^2y}{dt^2}=\frac{d}{dt}(2t+5) = 2\tag{16}\end{equation}
I don't like this notation, I think it's too busy. So, allow me to introduce you to dot notation. Dot notation is much more compact and cuter. The first derivative:
\begin{equation} \dot y = 2t+5\tag{17}\end{equation}
And the second derivative:
\begin{equation} \ddot y = 2\tag{18}\end{equation}
...Okay. Armed w/ a fundamental understanding of functions, their time derivatives/integrals, and dot notation, we can now tackle dynamics in earnest.
...May I Say It Now? Normal Mode: Dynamics
Right. As stated before, dynamics seeks to capture the behavior of a system of interest and represent that behavior mathematically, to varying degrees of complexity. That representation is known as a model. Typically a model is expressed via a series of differential equations. What are differential equations? They are simply equations that have derivatives in them. This should make sense, since systems in the natural world evolve over time, so you must be able to express how that system's variables change w/ respect to time (the time derivative).Let's not get caught up on superfluous examples. Let's go straight to an engineering example. Let's step through the process a mathematical modeler would undertake to obtain dynamic equations. Consider the following spring-mass-damper mechanical system, commonly found in automobile shock absorbers:
 |
| Figure 7. (a) Schematic of spring-mass-damper system. (b) Free-body diagram of spring-mass-damper system. |
On the left is a simplified schematic representation of the system. Pertinent features are an object of mass M, walls that have a coefficient of friction b, and a spring with a spring constant k. Notice that these terms, M, b, and k are intrinsic properties of their corresponding entities. Thus, they are parameters, and not variables. They do not change with respect to time (though they can, and that would introduce more complexity to our model). On the right is what's called a free-body diagram of the same system; it is meant to demonstrate all of the forces acting on an object. Here we can see that the object of mass M is subject to three forces: a frictional force by, a spring force ky, and an applied force r(t). The fourth vector y (a quantity that specifies a magnitude and a direction, such as forces, typically designated as an arrow) refers to a displacement along the y-axis (the object moves strictly vertically, up and down).
\begin{equation} M\ddot y + b \dot y + ky = r(t) \tag{19}\end{equation}
This is known as a second order differential equation. Second order because, you guessed it, the highest derivative found here is 2. This particular equation--obtained by analyzing the pertinent forces involved in a system and applying the appropriate physical principles--describes the displacement y of the mass M, subject to the three aforementioned forces. And that's the core of what a modeler does, really. They examine the system under question, determine the pertinent forces at play, and use physics to formulate a series of dynamical equations that collectively form a model. This part is incredibly tricky as, once again, you're trying to capture the enormous amount of complexity in the world and represent it mathematically. Fidelity, that is, how accurately your model represents real-world phenomenon, increases as complexity increases.
This particular example is merely a toy; in reality that b parameter would probably be non-uniform--it's value changing depending on the position of mass M. Or worse yet, perhaps it does not vary just according to the position of mass M, but also according to the temperature of the system. Thus that parameter b is no longer a simple constant, but becomes a function. And not only is it a simple function, but a multivariate function, depending on position and temperature. Those two quantities, position and temperature are not simple constants either; they change w/ time. Hence, they need to be modeled w/ differential equations too. If you're following me so far, you can easily see how things can get hairy, as mathematical models of increasing fidelity incorporate more differential equations that are often times coupled. This is both the bane and beauty of dynamical systems: the quantities that differential equations describe are often times interdependent on each other.
Great, so now we have this mathematical model in the form of a second order differential equation. What now? The answer is, of course, to solve it. What the heck are we solving for? I like to conceptualize this part as analogous to solving for a variable in an algebraic equation. When you're solving an algebraic equation, such as x + 2 = 5, you apply some simple arithmetic to arrive at x = 3. You're solving for a scalar; that is, a simple number. When you're solving differential equations, you're not solving for a scalar, but rather a function; the very same thing that we defined at the start of this post. However, very much unlike algebraic equations, sometimes differential equations can't be solved w/ pen and paper, following a set procedure to solve them. This is known as an analytical method. No, almost all of the mathematical models that describe real-world complex systems must be solved via numerical methods--in short, by computer simulation.
Numerical approaches to solving differential equations is an ENTIRE discipline in it's own right. As far as I'm concerned, this is the fundamental mathematical basis for what will probably be referred to in the future (if not already) as "simulation theory". No, not the hypothesis that we're living in a simulation. I mean the formal discipline on how to construct simulations. But anyway, that's enough of that digression. Back to our example.
Lucky for us, our little toy model of a mass-spring-damper system is amenable to an analytical solution. There is a set, procedural, pen-and-paper way to work it out. We don't have to (but we very well could) simulate it in order to solve it. Though, now I have to wonder whether or not I should get into the nitty-gritty and introduce you all to the Laplace transform. I think I will, but this post has gone on long enough. We'll pick it up in the next one.
[end transmission 2/?]
