Third 3 week block, Part 2

Common practice is to write a linear function $T[(x,y)]:=(ax+by, cx+dy)$ by picking out the coefficients $a, b, c, d$ in a square array:

$A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$

called a matrix.

This square matrix A encodes the essential information for the linear function T. The matrix operates on vectors to produce the exact same result as does T:

$\left( \begin{array}{cc} a & b \\ c & d \end{array} \right). \left( \begin{array}{c} x \\ y \end{array} \right)= \left( \begin{array}{c} ax + by \\ cx + dy \end{array} \right)$ .

The constants $\lambda$ associated with finding the invariant lines of T are the eigenvalues of $A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ , and they are the roots of the quadratic polynomial $(\lambda - a)(\lambda - d) - bc$ , which is known as the characteristic polynomial of A (if you have studied some linear algebra, this is the determinant of $\lambda I - A$ ).

The characteristic polynomial, being a quadratic in $\lambda$ can have two distinct real roots, one real root, and two complex roots which are complex conjugates.

The behavior of the associated differential equations

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

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

(which describe the smooth flow of the vector field corresponding to the linear function T)

is qualitatively different, according to the character of the solutions to the characteristic equation.

Case 1: Two distinct roots

The characteristic equation is $(\lambda-a)(\lambda-d)-bc = \lambda^2-(a+d)\lambda+(ad-bc)$ which has roots $\frac{1}{2}(a+d) \pm \frac{1}{2}\sqrt{(a-d)^2+4bc}$ so there are two distinct real roots exactly when $(a-d)^2+4bc >0$ .

This happens, for example, for the linear function with matrix $A=\left( \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right)$ , where the roots of the characteristic equation are $\frac{1}{2}(1 \pm \sqrt{5})$ . Let’s call these two roots $\lambda_1:=\frac{1}{2}(1-\sqrt{5}) \approx -0.618034$ and $\lambda_2:=\frac{1}{2}(1+\sqrt{5} \approx 1.618034)$ . There are two invariant lines for the linear function defined by the matrix A: the lines $y=(\lambda_1-1)x \approx -1.618034x$ and $y=(\lambda_2 -1)x\approx 0.618034x$ .

The corresponding differential equations (for the flow of the vector field) for this linear function are:

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

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

On the line $y=(\lambda_1-1)x \approx -1.618034x$ the differential equations reduce to a single differential equation:

$\frac{dx}{dt}=\lambda_1x \approx -0.618034x$ .

This simple linear differential equation has an exponential solution $x(t)=C_1e^{\lambda_1t} \approx C_1e^{-0.618034t}$ , where $C_1$ is an arbitrary constant, BUT ONLY ON THE LINE $y=(\lambda_1-1)x \approx -1.618034x$ .

On the other invariant line, $y=(\lambda_2-1)x \approx 0.618034x$ , the differential equations reduce to a single differential equation:

$\frac{dx}{dt}=\lambda_2x \approx 1.618034x$

which has the exponential solution $x(t)=C_2e^{\lambda_2t} \approx C_2e^{1.618034t}$ , where $C_2$ is an arbitrary constant, BUT ONLY ON THE LINE $y=(\lambda_2-1)x \approx 0.618034x$ .

What about solutions to the differential equations

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

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

that are not restricted to one or the other of these two invariant lines, but describe a flow through other parts of the plane?

The idea is to project such solutions onto both of the invariant lines, and add together the projected solutions we find on the invariant lines.

In this example we see that one of the eigenvalues is negative and the other is positive. This gives us a flow for which the origin $(0,0)$ is a saddle point, with the flow coming in toward the origin on one of the invariant lines, and away from the origin along the other invariant line.

A very nice picture of this can be obtained using TEMATH(Tools for Exploring Mathematics).

TEMATH is installed on the computers in the lab. Open it and run this example following the instructions here. You can also download TEMATH and run it on your own computer.

In the TEMATH instructions, an eigenvector is any non-zero vector on the invariant line for an eigenvalue.

Tasks for this 3-week block

Using TEMATH and/or MATLAB, go through all the cases for the roots of the characteristic equation, describing the geometry of the solutions of the differential equations

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

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

for the linear function corresponding to the matrix $A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ . Use examples different to those on the TEMATH page.

In each case, also describe the exact solutions to the differential equations obtained from MATLAB, and relate the form of those solutions to the geometry of the flow.

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Third 3 week block, Part 2

Case 1: Two distinct roots

Tasks for this 3-week block

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