A Time for Introductions: Lecture

Differential equations are a many-fangled thing and come in lots of varieties. Physicists and mathematicians make certain distinctions based mostly on what they model, and what their solutions look like. Computationally, there are yet more distinctions based on what kind of integration schemes can be applied to them. We will discuss a lot of these in due time, but really, whole courses are taught about differential equations (and some of them are often required for physics majors).

For now, it's worth focusing on just a few of the different kinds. An ordinary or total differential equation is one that relates various derivatives of some function \(x\) with only a single independent variable, \(t\), and can be written in the form \[ F \left( t; x, x', \dots, x^{(n)} \right) = 0 \] where \( F \) could, in general, be literally any function. A linear differential equation is one where \(F\) is a linear function of the form \[ y^{(n)} = \sum_{i=0}^{n-1} a_i(t) \, x^{(i)} + b(t) \] and these are a monumentally important topic in their own right; a non-linear differential equation is one that doesn't fit this scheme. Further, an autonomous differential equation is one where \(F\) doesn't depend on \(t\).

In class I claimed that Kepler's third law could be "derived" from Newton's law of universal gravitation, without solving a differential equation. This idea is important enough that it's well worth stepping through here.

Kepler's third law is what's known as a scaling law: it may not necessarily give an exact relation, but it does establish how one quantity changes with respect to another. In this case, we'd like to show that given a planet in an elliptical orbit around a star, its orbital period \( T \) should scale with its orbital semi-major axis \( a \) as a power law: \( T^2 \propto a^3 \).

To start with, let's write down the force acting on the planet in spherical coordinates: \[ \mathbf{F} = m_2 \, {d^2r \over dt^2} \, \hat{\mathbf{r}} = - {Gm_1m_2 \over r^2} \, \hat{\mathbf{r}} \] where \( m_1 \geq m_2 \) are the masses of the star and planet, respectively. Obviously the factors of \( m_2 \) cancel, and we're left with an \(r\)-component of \[ {d^2r \over dt^2} = - {Gm_1 \over r^2}. \] Now, the left- and right-hand sides of this equation both have overall units of acceleration. If we could isolate those and make this equation unitless, we'd be left with exactly one mathematical problem that could be re-scaled to model different masses.

We already know something about one family of solutions to this equation: they describe orbits, and orbits are inherently cyclical. Suppose that \( a \) and \( T \) are two constants describing the characteristic size and duration, respectively, of one cycle in a particular orbit. Then, write this equation in terms of the unitless variables \( \rho = r/a \) and \( \tau = t/T \): \[ {a \over T^2} \, {d^2\rho \over d\tau^2} = - {Gm_1 \over a^2} \, {1 \over \rho^2} \] We can now pull a sneaky little trick: since \( T \) and \( a \) are generic constants, we can lump all the constants into one (called \( k \), say), leaving us with a single differential equation \[ {d^2\rho \over d\tau^2} = -{k \over \rho^2} \] and the condition that \[ {Gm_1 T^2 \over a^3} = k. \] But this implies \( T^2 \propto a^3 \), exactly as we wanted to show.

Consider a culture of bacteria. Initially there is a single bacterium; after a while, through cell division, it splits into two. A while later, two split into four, and so on. As the culture grows, its growth rate speeds up in proportion.

Let \( N(t) \) represent the number of bacteria in the culture at time \( t \), with the initial condition \( N(0) = 1 \). Since the culture's growth rate is proportional to how many bacteria there currently are, we can model this system with the (very simple) differential equation \[ {dN \over dt} = \alpha N \] where \( \alpha \gt 0 \) is a proportionality constant. Isolating variables so that \( N \) is on the left-hand side and \( t \) is on the right, we can then integrate directly: \[ \int {dN \over N} = \alpha \int dt \] This leaves us with a logarithmic relationship: \[ \ln N = \alpha t + C \] where \( C \) is a constant of integration. Note, since \( N(0) = 1 \) we know that \( C = \ln[N(0)] = 0 \). Taking the exponent of both sides, we finally have \[ N(t) = \exp (\alpha t) \]

Suppose that, on your way to Pittsburgh from Meadville on I-79, you take great care to make sure the distance you have left to travel is always the same as the speed you're going, i.e. at 50 miles out you go 50 mph, and so on.

Let \( x(t) \) represent the distance (in miles) left to travel at time \( t \) (in arbitrary units). Since Meadville is about 86 miles from Pittsburgh on I-79, this sets your initial condition \( x(0) = x_0 \). For simplicity, let's assume your motion is one-dimensional with Pittsburgh at the origin, so you're moving in the \( -x \) direction on your way there. Your motion can be modeled as \[ {dx \over dt} = -{x \over H} \] where \( H \gt 0 \) is a (constant) unit conversion factor to get speed in miles per hour. Once again isolating variables and integrating, we get \[ \ln x = \int {dx \over x} = -{1 \over H} \int dt = -{t \over H} + C \] where \( C \) is a constant of integration. Taking the exponent of both sides and matching this integration constant to the initial condition, we get \[ x(t) = x_0 \exp (-t/H) \]