Third 3 week block, Part 1

Systems of linear differential equations

In this 3-week block we will be looking at the behavior of systems of linear differential equations.

A system of two linear differential equations is a pair of linked differential equations of the form

$\frac{dx}{dt}=ax+by$

$\frac{dy}{dt}=cx+dy$

where $a, b, c, d$ are constants.

There is a more general form for linear systems in which the $a, b, c, d$ vary with $t$ but in this block we will only look at the case in which $a, b, c, d$ are constant, and do not vary with $t$ .

The analysis of systems of linked linear differential equations revolves around the behavior of linear functions, so will first spend a little time looking at linear functions on the $x-y$ plane. Trying to get a picture of a linear function will lead us very naturally into systems of linear differential equations.

Linear transformations of vectors

The Euclidean plane consists of all ordered pairs $(x,y)$ where $x \textrm{ and } y$ are real numbers. We think of the point $(x,y)$ as the endpoint of a vector segment that starts at the origin $(0,0)$ :

Sometimes we will refer to “the vector $(x,y)$ ” meaning the directed vector segment from $(0,0) \textrm{ to }(x,y)$ .

A linear function on the Euclidean plane is a function of the form $T[(x,y)]=(ax+by, cx+dy)$ . A linear function is exactly a function that preserves vector addition and scalar multiplication: $T[(x_1,y_1)+(x_2,y_2)]=T[(x_1,y_1)]+T[(x_2,y_2)]$ and $T[\lambda(x,y)]=\lambda T[(x,y)]$ for all vectors $(x_1,y_1), (x_2,y_2), (x,y)$ and all real numbers $\lambda$ .

As a result, once we know what a linear function T does to the vectors $(1,0) \textrm{ and } (0,1)$ we know what T does to every vector. The reason is that $T[(x,y)]=T[x(1,0)+y(0,1)]=xT[(1,0)]+yT[(0,1)]$ .

As an example, we will look at how the linear function $T[(x,y)]:=(2x+y,x+y)$ transforms vectors. One way to begin to get a geometric picture of how T transforms vectors is to look at what T does to the unit square with corners $(0,0), (1,0), (1,1), (0,1)$ :

First, $T[(1,0)]= (2,1) \textrm{ and } T[(0,1)]=(1,1)$ so $T[(1,1)]=(3,2)$ .

The line segment joining $(0,0) \textrm{ to }(1,0)$ consists of vectors $(x,0) \textrm{ where } 0\leq x \leq 1$ and for these vectors we have $T[(x,0)]= (2x,x) = x(2,1)$ which, as $x \textrm{ varies from } 0 \textrm{ to } 1$ is a line segment running from $(0,0) \textrm{ to } (2,1)$ .

Similarly, T maps the line segment from $(0,0) \textrm{ to } (0,1)$ to the line segment from $(0,0) \textrm{ to } (1,1)$ . So, T maps the unit square to a parallelogram as follows:

Vector fields for linear transformations

In principle, we could just draw, for each vector $(x,y)$ a directed line segment from $(x,y)$ to $T[(x,y)]=(2x+y,x+y)$ . However, a moment’s thought tells us that all these directed line segments would overlap in a mess, and give us no sensible picture at all. One way around this is to draw a very short directed line segment from $(x,y)$ to $T[(x,y)]=(2x+y,x+y)$ , so that these small directed line segments do not overlap:

This was produced with the following MATLAB code:

>> [x,y]=meshgrid(-1:1/10:1,-1:1/10:1);
>> u=2*x+y;
>> v=x+y;
>> quiver(x,y,u,v,1)

We see a rather blurred line running from top left to bottom right of the picture, and the vectors near that line seem to curve over.

Generally, Mathematica® produces better graphics than MATLAB. Below is a graphic produced with the following Mathematica® code:

Needs["VectorFieldPlots`"];

VectorFieldPlot[{2*x + y, x + y}, {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20, ScaleFactor -> 0.2, Axes->True]

Here we can see a little more clearly the curving of the vectors near a line running from top left to bottom right.

Finding invariant lines

How could we find that line that seems to run from top left to bottom right of the picture?

The line seems to go through the origin, so if we could find any non-zero vector on that line, we would know what the line is, because all other vectors on the line would just be multiples of that vector.

The characteristic property that looks like it will help us find the line is that if $(x,y)$ is a vector on the line then $T[(x,y)]$ is also on the line – just look at how the vectors seem to point on the line. So, we are looking for a non-zero vector $(x,y)$ – that is, either $x\neq 0 \textrm{ or } y\neq 0$ – such that $T[(x,y)]=\lambda(x,y)$ for some real number $\lambda$ . In other words, $2x+y=\lambda x \textrm{ and } x+y = \lambda y$ for some $\lambda$ and for some $(x,y)$ with either $x \neq 0 \textrm{ or } y\neq 0$ (or both).

This gives us the two equations:

$(\lambda-2)x-y=0$

$-x + (\lambda-1)y=0$

Substituting $x=(\lambda-1)y$ gives us $((\lambda-2)(\lambda-1)-1)y=0$ , and substituting $y=(\lambda-2)x$ gives us $((\lambda-2)(\lambda-1)-1)x=0$ .

Since one of $x, y$ is non-zero this gives us $(\lambda-2)(\lambda-1)-1=0$ .

This is a quadratic for $\lambda$ with roots $\frac{1}{2}(3\pm\sqrt{5})$ .

Substituting $\lambda=\frac{1}{2}(3-\sqrt{5})$ in the equation $(\lambda-2)x-y=0$ gives us the equation $y=-\frac{1}{2}(1+\sqrt{5})x$ for the line we see running from the top left to the bottom right of the picture.

Of course, the other root, $\lambda =\frac{1}{2}(3+\sqrt{5})$ , gives us another invariant line with equation $y=\frac{1}{2}(-1+\sqrt{5})x$ . This line is a little harder to see from the vector field picture. Both lines are shown below:

This graphic was produced using the following Mathematica® code:

Needs["VectorFieldPlots`"];

A1 = VectorFieldPlot[{2*x + y, x + y}, {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20, ScaleFactor -> 0.2, Axes -> True];
A2 = Plot[-x*(1 + Sqrt[5])/2, {x, -1, 1}];
A3 = Plot[x*(-1 + Sqrt[5])/2, {x, -1.5, 1.5}];
Show[A1, A2, A3]

Connection with a system of linear differential equations

The curl of the directed line segments away from the line $y=-\frac{1}{2}(1+\sqrt{5})$ indicates a flow on the plane, such that if we placed a leaf in the flow it would follow the directed line segments as time progressed.

The introduction of time as a parameter means that we are thinking of the $x-y$ coordinates changing with time as the leaf follows the flow. The leaf’s path will be such that the instantaneous rate of change of the leaf’s coordinates will be given by the directed line segment at that point. This gives us a linked pair of differential equations for the coordinates, as functions of time, along a flow line:

$\frac{dx}{dt}=2x+y$

$\frac{dy}{dt}=x+y$

Some examples to try

1. (a) Plot the vector field for the linear function $T[(x,y)]=(x+y,x)$ .

(b) Find any invariant lines, and write down the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow.

2. Repeat this for the linear functions $T[(x,y)]=(ax+by, cx+dy)$ for several choices of integers $a, b, c, d$ chosen randomly in the range $-5 \leq a, b, c, d \leq 5$ .

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Third 3 week block, Part 1

Systems of linear differential equations

Linear transformations of vectors

Vector fields for linear transformations

Finding invariant lines

Connection with a system of linear differential equations

Some examples to try

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