1

When you have two equations

$x' = 4x -3y$

$y' = 8x -6y$

The solution turns out to be

$x = c_1e^{-2t} + 3c_2$

$y = 2c_1e^{-2t} + 4c_2$

and I understand how the phase portrait is visualized.

However, what I don't understand is that whether or not type of this phase portrait is

considered to be a stable system. one of the solution seems to be approaching $0$

and other solution is a constant. However, we usually assume that determinant of a matrix

which contains these roots are not $0$ so that we can find out the critical points or

equilibrium solution of $x$ to see where the solutions are headed. Therefore, the assumption

of this is violated. Thus, I am not sure what this kind of phase portrait should be

classified as.

Amzoti
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EHMJ
  • 768

1 Answers1

3

For this system, we have the equilibrium points as the line:

$$y = \dfrac 43 x$$

You found that the solution to the system is given by:

$$X(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix} = c_1 e^{-2t} \begin{bmatrix}1\\2\end{bmatrix} + c_2 e^{0t}\begin{bmatrix}3\\4\end{bmatrix} $$

If we draw the phase portrait for this system (red line is the equilibrium line), we have:

enter image description here

Since the trace of the matrix is negative, we have a line of neutrally stable non-isolated fixed points. Sometimes this is called by other names like stable, but not asymptotically stable or marginally stable, etc.

It is also worth mentioning that in some settings (like Control), they consider this unstable.

Amzoti
  • 56,093