Math 555: Differential Equations
1.1 Intro to ODE and Direction Fields
1.1.1 Overview ¶
Start with this video to give you an overview of the subject of Differential Equations. This video will motivate the subjects we discuss in this course and hopefully motivate you to study more about differential equations.
1.1.2 Linear Algebra ¶
We need some material that will help us get started with Differential Equations. While these topics are not prerequisites, they will really help us to understand the mathematical tools we will use to study differential equations. These tools are used throughout the course. I will introduce these subjects when they come up in our course and refer you back to this page and these videos.
Before you view the lectures on Chapter 1 or attempt any homework assignments you should view and study the first five videos about linear algebra. Remember that they were mentioned in our overview video.
1.1.3 Calculus ¶
We are going to use differential and integral calculus to study, solve, and compute solutions to our differential equations. Keep in mind that we will extensively use your previous mathematics courses. In fact, I will need to use everything you were ever taught in high school and college about mathematics in this course. As engineers and scientists, we use the language of differential equations to describe the physics of our fields. Many of the topics you studied in pre-requisite courses, such as Taylor Series, were covered specifically to prepare you for solving differential equations.
1.1.4 How Do We Study Differential Equations? ¶
Much of the behavior of the physical world that we observe change over time. In fact, deep study of the physics of our natural world has taught us that measuring the precise state of a mixing solution or an orbiting moon is very difficult, however observing the changes that take place and measuring the rate at which things change is much easier. In mathematical terms rates and accelerations are derivatives and equations that express the derivatives of a function are called differential equations .
A differential equation that describes a physical or abstract process is called a mathematical model of that process. We will study many of those mathematical models and the differential equations that express those models and the processes they represent in this course. We will start with some simple differential equations and increase the complexity of the differential equations as the semester progresses.
We generally introduce a new type of differential equation, the physical system it describes, or the mathematical model the differential equation is used to represent by how they are solved or numerically computed . It may unsettle you at first to realize that we don't prioritize the subject according to the mathematical models we use, but by the differential equation used to express it. This is a skills class and you will be expected to use these mathematical tools in future courses to express the engineering and physics problems you encounter. Ultimately, you are expected to demonstrate understanding about the process you are modeling by providing a solution to the differential equation or computing a solution.
This means that early in the course we may use a simple differential equation to describe a very complex process, or later in the course require a difficult differential equation to describe a simple process.
We do this because we break down our differential equations by how hard they are to solve , not by how complex the process is to model .
With this in mind, let us start with a simple process and differential equation.
1.1.5 First Example - A Falling Object ¶
Let's start with
Newton's second law
which states that the mass of an object times its acceleration is equal to the net force acting on the object. We express this with the equation
$$ F = ma $$
where $m$ is the mass of the object measured in kilograms, $t$ is time measured in seconds, $v$ is the velocity measure in
$$ v = \dfrac{\text{meters}}{\text{second}}, $$
$a$ is the acceleration measured in
$$ a = \dfrac{\text{meters}/\text{second}}{\text{second}}, $$
and $F$ is the force exerted on the object measured in newtons. Since $a$ is related to the velocity by $a = \frac{dv}{dt}$ we can write Newton's second law as a differential equation
$$ F = m\dfrac{dv}{dt}. $$
In this model, falling downward is considered the positive direction so that no negative signs are hidden in our constants.
Forces Acting on the Falling Object
Examine the free body diagram. Another way to express the net forces acting on our object is to account for each force and add these vectors. Newton's second law states that gravity exerts a force on the object equal to its weight
$$ F_w = mg $$
where $g$ is the acceleration due to gravity. The value of $g$ has been determined experimentally to be $9.8\ m/s^2$ near the earth's surface. According to Newton's law of gravitation this this force acts toward the center of mass of the earth so the force acts downward and has the same sign as velocity.
Opposed to the force due to gravity $F_w$ the object encounters air resistance or drag. In subsequent physics and engineering courses we will find that drag can be difficult to model. For our first mathematical model we will express drag as proportional to the velocity and opposite in direction to the velocity of the object. Thus the force acting on the object due to drag is modeled by
$$F_d = -\gamma v.$$
Thus the net force acting on the falling object is given by
$$ F = F_w + F_d = mg - \gamma v.$$
Our First Differential Equation
Using the derivative form of the force equation $F = m\frac{dv}{dt}$ and the net force equation $F = mg - \gamma v$, we have our first differential equation:
$$ m\dfrac{dv}{dt} = mg - \gamma v. $$
This differential equation expresses a mathematical model for the velocity of a object falling in the atmosphere near sea level. Let's take a look at this model. The velocity of the falling object is the dependent variable and the independent variable is time. The equation contains both the function representing the changing velocity of the object $v$ and its first derivative $\frac{dv}{dt}$.
The symbols $m$, $g$ and $\gamma$ are constants as far as this model is concerned. However, they may have very different values. Different objects have differing masses. Different objects have different sizes, geometries and physical properties that affect drag. The acceleration due to gravity near mean sea level of earth is different from near the mean surface level of Mars. Objects falling at higher altitudes will use a different acceleration due to gravity according to Newton's Law of Gravitation and will encounter different drag forces due to differences in the atmosphere.
We may ask if indeed $m$, $g$ and $\gamma$ are variables instead of constants. We usually call these symbols parameters because we will not consider them to be an independent or dependent variable for each object or in each situation, however appropriate values must be chosen for each parameter to allow us to model the velocity of a specific falling object.
1.1.6 Second Example - A Mathematical Model ¶
In our second example let us consider the free body problem in example 1 with the parameters of our mathematical model to be
$$m = 10\ \text{kg,}\quad g = 9.8\ \text{m}/\text{s}^2,\quad \gamma = 2\ \text{kg}/{s}.$$
From above, we have
$$m\dfrac{dv}{dt} = mg - \gamma v$$
or
$$\dfrac{dv}{dt} = g - \dfrac{\gamma}{m}v.$$
Substituting the values of $g$, $m$ and $\gamma$ gives us
$$\dfrac{dv}{dt} = 9.8 - \dfrac{v}{5}.$$
In the next section we will find a function $v = v(t)$ so that when we substitute $v(t)$ and its derivative $v'(t) = \frac{dv}{dt}$ into the equation we have a true statement. We will call such a function $v(t)$ a solution to the differential equation. We will learn how to find such a solution in Section 1.2. A really good question to ask before we start pulling out all of our calculus, trigonometry and algebra tools is
"Is there anything we can find out about a differential equation before we go to all of the trouble to solve it? "
Indeed there is and we have a programs to help us do just that. As you saw in the video above, we can visualize the differential equation using a two-dimensional vector field called the
direction field
,
slope field
or simply
vector field
of the differential equation. If we write our differential equation a generally as possible what we are saying is that
$$\dfrac{dv}{dt} = f(t, v)$$
the derivative of the velocity is a function of two variables, time $t$ and velocity $v$. We have a function of two variables; more than that we have a vector-valued function of two variables because derivative (and acceleration) have both direction and magnitude.
We can draw our graph paper with variable $t$ along the horizontal axis and variable $v$ along the vertical axis. At every point we can draw an arrow whose whose tail sits at the point $(t,v)$ on the graph and its tip sits one unit from the tail with slope $v'= f(t,x)$.
Whenever we have a function like $f$ that assigns to every point on the plane a vector with its tail at that point, we call that function a vector field . The vector field formed by the function $f(t,v)$ in differential equation $\frac{dv}{dt} = f(t, v)$ is called a slope field or direction field .
1.1.7 Drawing a Direction Field with a Computer ¶
Drawing a Direction Field by hand is very time consuming and rough sketches can miss important details so we have tools for drawing the direction field of a first order differential equation. Here, I will be using one provided by Desmos .
Desmos provides several examples for use in its online graphing calculator , including this slope field generator . This will be the basis for the example covered in the following video.
1.1.8 Third Example - Investigating Solutions Graphically ¶
In this example we have the differential equation $$\dfrac{dp}{dt} = \dfrac{p}{2} - 450.$$ When $\frac{dp}{dt} = 0$ then $p(t) = 900$, a constant function. It is an equilibrium solution because the solution is a constant function. However notice that if a solution is at any time above the equilibrium solution then the solution grows without bound away from the equilibrium solution. Likewise if at any time the solution is below the equilibrium solution then solution grows without bound in the negative direction away from the equilibrium solution. We call this equilibrium solution unstable . Try this yourself with Desmos. No matter how close you start above or below the solution $p(t) = 900$, the solution will grow away from it.
1.1.9 Fourth Example - Another Direction Field ¶
Here is an interactive example, showing how the solution is related to the direction field. The point $(0,-1)$ is an
initial value
(this will be discussed later) for the solution of the differential equation
$$ \dfrac{dy}{dt} = y + x^2. $$
Examine the relationship between the solution and the slope field, and how it follows a path outlined by the direction vectors. In addition, in this example the initial value at $(0,-1)$ can be dragged around to see how changing the value affects the solution. Try it out!
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