Sem#4: Problem 7-02 (from 2020)

[Document copied from 2020 - hopefully I fixed the links!]

Just a clarification of ambition when it comes to phase diagrams. When I teach - and this applies now as well, since I'm responsible for the exam - I promise to never ask you to "draw" a phase diagram, only "sketch" it. 

The difference is that it does not need to be pretty as long as it is qualitatively correct: your null-isoclines should look about the way they actually do; paths across an x-isocline should be vertical or reasonably close; you should not reverse arrows or directions. 

I mean, look at the the front page of the Norwegian book. Links to an external site. Yes that is a phase diagram; the circle and the parabola are null-isoclines, and that curve that goes all over the picture from (-3,-1) to (4,3) is the convergent saddle path. Who wants to spend an hour of an exam making that thing pretty?
But I expect you to be able to draw a parabola and a circle, and please take note how the paths cross the nullclines; vertically or horizontally. 

System: 

$\begin{align}\dot x&= ax+2y+\alpha\\ \dot y&= 2x+ay+\beta\end{align}$

Because the question says "for all", we must expect that for some parameters, the matrix-based approach will get us into trouble (repeated roots, ...), so let us do it like 7-01: 

• Deduce a diff.eq. for x
• Solve that
• Find y. From the first equation, $y=\tfrac12\big(\dot x-ax-\alpha\big)$, so when x is found, we differentiate it and insert and that is it.

A differential equation for x

... will be of the form $\ddot x-\dot x\textsf{ tr }\mathbf A+x\textsf{ det }\mathbf A=f(t)$. (Here trace is 2a and determinant is a²-4.) You can look up in the book a formula for f(t), or you can just do the job: 

• Differentiate the first eq.: $\ddot x=a\dot x+2\dot y$
• Insert for $\dot y$: $\ddot x=a\dot x+2(2x+ay+\beta)$
• Eliminate the 2y using the first equation that says $2y=\dot x-ax-\alpha$, to get $\ddot x=a\dot x+4x+2\beta+a(\dot x-ax-\alpha)$
• Reorganize into $\ddot x-2a\dot x+(a^2-4)x=2\beta-a\alpha$

Solve the homogeneous equation for x:

Like 7-01, except with a parameter: 

• Characteristic roots: $r=a\pm\sqrt{a^2-(a^2-4)}=a\pm 2$, two distinct and real.
• Homogeneous equation has general solution $C_1e^{(a-2)t}+C_2e^{(a+2)t}$.
• For the inhomogeneous equation, the problem arises if $a=\pm 2$, for then the RHS solves the homogeneous equation. We therefore split up into those two cases (see next)

The inhomogeneous equation and y: case $|a|\neq 2.$

In this case, we have $x(t)=C_1e^{(a-2)t}+C_2e^{(a+2)t}+d$, with the constant d fit to the right-hand side: inserting into the equation, that gives $0-2a0+(a^2-4)d=2\beta-a\alpha$ so $x^*\equiv d=\frac{2\beta-a\alpha}{a^2-4}$and $\displaystyle x(t)=C_1e^{(a-2)t}+C_2e^{(a+2)t}+\frac{2\beta-a\alpha}{a^2-4}$.

To get y, we have $y=\tfrac12\big(\dot x-ax-\alpha\big)$. 
[Here you just do the rest.]

The inhomogeneous equation and y: case |a|=2

Here we must try some first-order ct for x*. (Why not ct+d? No problem, but we already have an arbitrary constant, so d is redundant.)
Inserting x* = ct into the equation, we get $-2ac = 2\beta-a\alpha$ and $c = \frac{a\alpha-2\beta}{2a}=\frac\alpha2-\frac\beta a$.
Then we have $\displaystyle x(t)=C_1e^{(a-2)t}+C_2e^{(a+2)t}+\big(\frac\alpha2-\frac\beta a\big)t$.

• [Here you can split into the two cases a=-2 or a=2, and insert into $y=\tfrac12\big(\dot x-ax-\alpha\big)$to get the answer on page 36.]
• [Alternatively, if you want to be a bit fancy: Either a=2 so a-2=0 and a+2=2a, or a=-2 so a-2=2a and a+2=0; in either case, one root is zero and the other is 2a, so you can write x as $\displaystyle x(t)=D+Ce^{2at}+\Big(\frac\alpha2-\frac\beta a\Big)t$ and take it from there.]

Part (c) - the phase diagram for the special case $\begin{align}\dot x&= -x+2y-4\\ \dot y&= 2x-y-1\end{align}$; 

More sketches for the phase diagram here. Download More sketches for the phase diagram here.

A bit of theory first: the associated matrix has determinant (-1)*(-1)-2*2=-3<0. Negative determinant => saddle point! 
So what is the stationary state [a.k.a. "equilibrium point"]? It satisfies 
-x+2y=4
2x-y=1
Scale the second by 2 and add up to get 3x=6 so x=2; then y=2x-1=4-1=3. So the system is equivalent to 
$\frac d{dt}\begin{pmatrix}x-2\\y-3\end{pmatrix}=\begin{pmatrix}-1&2\\2&-1\end{pmatrix}\begin{pmatrix}x-2\\y-3\end{pmatrix}$
which apart from moving the origin two steps horizontally and three steps vertically, behaves the same as $\begin{pmatrix}\dot x\\\dot y\end{pmatrix}=\begin{pmatrix}-1&2\\2&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

Null-isoclines:

• To get the null-isocline for x: $\dot x=0$ when 2y=x+4 i.e. y=2+x/2. Across that line, there is no left--right motion; the paths cross that line vertically.
• To get the null-isocline for y: $\dot y=0$ when y=2x-1. Across that line, there is no up--down motion; the paths cross that line horizontally.

Directions: Null-isoclines solve an equality. You are interested in inequalities:

• $\dot x>0$ (i.e. rightward motion - that does not mean "straight to the right", it can be NE or SE, but not W):
  When 2y-x-4>0. That is, when y>2+x/2, that is, above that null-isocline. So above that line, there is motion in the rightward direction. 
  Likewise, below that line, $\dot x=2y-x-4$ is < 0 and we have motion to the left.
• $\dot y>0$ (i.e. upwards motion - that means NE or NW or possibly straight N): 
  When 2x-y-1>0. That is, when y<2x-1, that is, below that null-isocline. So below that line, there is upward motion. 
  Likewise, above that line, $\dot y$ is < 0 and we have motion downwards. 

Saddle path: 

This is probably not needed for this question - but you could have been asked for the convergent non-constant path, so I refresh that theory. Above, we saw that the associated matrix has eigenvalues a plus/minus 2, and here, a=-1. A negative and a positive eigenvalue (that gives the saddle property), and the negative eigenvalue is -1. An associated eigenvector solves
(-1-(-3))x+2y=0 (exercise: why? And why only one equation?), that is, x+y=0 and the eigenvector has a slope of -1 (-45 degrees).