These notes are set so that you get to prove the main results
by solving smaller problems that when put together give the big result.
The answers to the problems are in the videos.
You will get the most out of these notes if you do (or try) the problems
before looking at the videos.
The basics of systems of differential equations.
A system of differential equations, as opposed to a single differential equation is a set of more than one equation for more than one unknown functions. Here we will mostly consider sets of two equations for two unknown functions. The most basic version of this is a system of the form \begin{align*} \frac{dx}{dt} &= f(t,x(t),y(t))\\ \frac{dy}{dt} &= g(t,x(t),y(t)) \end{align*} We first record the fact that such equations always have solutions
Theorem: (Existence and uniqueness for first order systems.) Assume the functions \(f(t,x,y)\) and \(t,x,y)\) have continuous first partial derivatives. Then the for any \(t_0\) and numbers \(x_0\) and \(y_0\) the initial value problem \begin{align*} \frac{dx}{dt} &= f(t,x(t),y(t)),&& x(t_0)=x_0\\ \frac{dy}{dt} &= g(t,x(t),y(t)),&& y(t_0)=y_0 \end{align*} has a unique solution.
(Note to any mathematician reading this: I am aware that this statement should be more precise about domains and that the solution is only necessarily defined on a neighborhood \(\{ t: |t-t_0|< \delta\}\) of \(t_0\).)
Those not interested in theory can skip ahead to the next section on Dealing with autonomous systems graphically.
To relate this back to equations we have looked at, consider the second order equations $$ x''(t) + p(t)x'(t) + q(t) x(t) = f(t). $$ If we introduce a variable for the derivative, that is let $$ y(t) = x'(x) $$ then $$ y'(t) = x''(x). $$ and we can rewrite the equation as $$ y'(t) + p(t)y(t) + q(t) x(t) = f(t) $$ Solving for \(y'(x)\) gives $$ y'= -q(t)x(t) - p(x) y(t) + f(t). $$ Putting this all together gives
Theorem: The second order initial value problem $$ x''(x)+p(t)x'(t) + q(t)x(t) = f(t) \qquad x(t_0)=x_0,\quad x'(t_0) = y_0 $$ is equivalent to the first order system \begin{align*} x'(t) &= y(y)&& x(t_0)=x_0\\ y'(t) &= -q(t)x(t) - p(x) y(t) + f(t)&&y(t_0)=y_0. \end{align*}
The point of this is that we can use the Existence and uniqueness for first order systems to see that we also have unique solutions for second order differential equations. (But this is probably mostly interesting to people, such as mathematicians, that wish to unify the theory and to some people writing computer programs for solving differential equations so that they can write programs that cover the widest number of types of equations.)
Dealing with autonomous systems graphically.
The system is autonomous if and only if the variable \(t\) is not explicitly in the equations. That is it is of form \begin{align*} \frac{dx}{dt} &= f(x(t),y(t)),&& x(t_0)=x_0\\ \frac{dy}{dt} &= g(x(t),y(t)),&& y(t_0)=y_0 \end{align*} Which we will usually write in the more compact form \begin{align*} \frac{dx}{dt} &= f(x,y),&& x(t_0)=x_0\\ \frac{dy}{dt} &= g(x,y),&& y(t_0)=y_0 \end{align*} or even \begin{align*} \dot x &= f(x,y),&& x(t_0)=x_0\\ \dot y &= g(x,y),&& y(t_0)=y_0 \end{align*} where the dot notation, which goes back to Issac Newton, is used to denote the derivative with respect to \(t\) (which is generally though of as time).
An autonomous system has a pleasant geometric interpolation. On the plane consider the vector field $$ \mathbf v(x,y) = f(x,y)\,\mathbf i + g(x,y)\,\mathbf j. $$ Then if \((x,y)\) = \((x(t),y(t))\) is a solution to the system \begin{align*} \dot x &= f(x,y)\\ \dot y &= g(x,y) \end{align*} the curve $$ \mathbf c(t) = (x(t),y(t)) $$ has $\mathbf v$ as its velocity vector at each point. That is $$ \dot{\mathbf c}(t)= \mathbf v(\mathbf c(t)) = \mathbf v(x(t),y(t)). $$ To see this in action let us consider the systemm \begin{align*} \dot x &= -2 x + y\\ \dot y &= x - 3y \end{align*} Go to the website
https://homepages.bluffton.edu/~nesterd/apps/slopefields.html
and click on the tab that which is labeled System. You should something that looks like:
This is for the default system on the site which is \begin{align*} \dot x &= x+y \\ \dot y &= xy-1 \end{align*} To change the system to the one we are after edit the formulas for \(dx/dt\) and \(dy/dt\) to our functions (\(-2x+y\) and \(x-3y\)). The picture now looks like
The arrows point in the direction that a solution to the equation will be moving. Now let us plot a couple of solutions. To start let us look at the solution with $$ x(0)=2, \qquad y(0)=-1 $$ Click on the PLOT NUMERICAL SOLUTION CURVES and where it says enter (x,y) enter \( (2,-1)\) and click on submit Things should now look like:
The red curve is the track of the solution starting at \((2,-1)\) and shows that if is ``pushed'' by the arrows to the origin. If we plot anther initial condition, say $$ x(0)= -2,\qquad y(0)=2.5 $$ we get both solutions (the second one in green) on the graph:
Here is a more interesting example. We look at the system \begin{align*} \frac{dx}{dt} &=y\\ \frac{dy}{d} &= -\sin(2x) \end{align*} and let us plot the solutions with initial conditions $$ (x(0),y(0)) = (1,0),\qquad (x(0),y(0))=(.2,-.3),\qquad (x(0),y(0)= (-3,2). $$ Before plotting click on the bottom that says Euler and change it to Improved Euler (Heun) The result should look like
If it does not and yu have some red and green spirals running off of the graph, then that most likely means you did not change the Euler to Improved Euler (Heun).
Problem: For the system \begin{align*} \frac{dx}{dt} &=y\\ \frac{dy}{dt} &= \sin(2x) - . 1 y \end{align*} with the four initial conditions $$ (x(0),y(0)) = (1,0),\qquad (x(0),y(0))=(-.2,-.3),\qquad (x(0),y(0)= (-3,2),\qquad (x(0),y(0)) = (1,-2) $$ using Improved Euler (Heun). You can save the image by clicking on the To save the image, right-click this thumbnail bottom. Your homework is to save the graph and e-mail me a copy midnight by Friday, November 13.