For these systems, you may have to pull out several tools to figure them out.
For the first one, you guessed at the energy function, but there is no known way to derive those (for some physical systems, you can, but no general way).
What other tools might we use?
- Find the critical points - for some pathological problems, this in itself can be very difficult - like some of your examples. For all of your examples, one of the critical points is $(0,0)$
- Find the Jacobian matrix and evaluate the eigenvalues at the critical points - sometimes this may not work.
- Plot the phase portrait and look at the qualitative behavior.
For the first problem, if we find the Jacobian matrix, evaluate it at the critical point $(0,0)$, we find the the eigenvalues are $-2$ and $0$, so what can you tell about the stability from that? You can try this approach for the others and see if it bears fruit. Note, for all of these, there is more than a single critical point and there are some strange behaviors going on, but that is to be expected given these systems.
Here are the four phase portraits for these systems (note, if you expand out the ranges, you can see wild behaviors).
Update
We are given:
$$ f(x, y) = x'= e^{x+2y}-\cos 3x\\ g(x,y) = y'= 2\sqrt{1+2x}-2e^y$$
It is clear that $(0,0)$ is a critical point.
Lets find the Jacobian matrix of this system, evaluate it at the critical point, determine the type of critical point and validate this behavior using the phase portrait. The Jacobian matrix is given by:
$$\displaystyle J(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix} e^{x+2y} + 3 \sin 3x & 2e^{x+2y} \\ \dfrac{2}{\sqrt{x^2+1}} & -2e^{y} \end{bmatrix}$$
Now, evaluated at the critical point $(0,0)$, we have:
$$J(0,0) = \begin{bmatrix} 1 & 2 \\ 2 & -2 \end{bmatrix}$$
The eigenvalues of this system are $2$ and $-3$, thus we have two real, distinct and opposite sign $\rightarrow$ a saddle point.
If you look at the coordinates $(0,0)$ of the third phase portrait, do you see the saddle point? What can you tell about the stability of a saddle point?
The phase portrait lets us see this qualitative behavior without having to do all that work.
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $\frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $\frac{dx}{dt} - \frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
Best Answer
I agree with your critical points.
Now, look at the signs of each of your eigenvalues. What can you tell about the type of curve from this linearization and these signs? Be careful with linearization for the marginally stable cases, but you are safe with the four critical points regarding stability.
Now, compare that to this phase portrait.
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If you write the system as $\dfrac{dy}{dx}$, you can then choose numbers to figure out the direction and magnitudes because you already know the type of critical point from the eigenvalues. You have:
$$\dfrac{dy}{dx} = \dfrac{y(2-y-0.75x)}{x(1.5-x-0.5y)}$$