Given is the system of differential equations:
$$\begin{cases}\dot x=4y \\ \dot y=-3x \end{cases}$$
(a) Write the first integral of the system. Is the system conservative? Explain.
(b) Sketch the phase portrait of the given system using level curves.
(c) Show that the fixed point of the system is stable but not asymptotically stable (according to the definition).
Attempt: (a) To find the first integral (a conserved quantity), we look for a function $H(x, y)$ such that its total derivative with respect to time $t$ is zero along the solutions of the system. This means:
$$
\frac{dH}{dt} = \frac{\partial H}{\partial x} \dot{x} + \frac{\partial H}{\partial y} \dot{y} = 0.
$$
Substituting $\dot{x} = 4y$ and $\dot{y} = -3x$, we get:
$$
\frac{\partial H}{\partial x} 4y + \frac{\partial H}{\partial y} (-3x) = 0.
$$
What function $H$ should I pick from here?
(b)
To sketch the phase portrait, we use the level curves of the conserved quantity $H(x, y)$:
$$
\frac{3x^2}{2} + \frac{4y^2}{2} = C,
$$
where $C$ is a constant. These are ellipses in the $xy$-plane. We can rewrite this as:
$$
3x^2 + 4y^2 = 2C.
$$
But what is this, is this even right? How from here?
(c) The fixed point of the system is at the origin $(0, 0)$. To determine its stability, we analyze the eigenvalues of the Jacobian matrix.
The Jacobian matrix of the system I think is:
$$
J = \begin{pmatrix}
\frac{\partial \dot{x}}{\partial x} & \frac{\partial \dot{x}}{\partial y} \\
\frac{\partial \dot{y}}{\partial x} & \frac{\partial \dot{y}}{\partial y}
\end{pmatrix} = \begin{pmatrix}
0 & 4 \\
-3 & 0
\end{pmatrix}.
$$
The eigenvalues $\lambda$ of $J$ are found by solving the characteristic equation:
$$
\det(J – \lambda I) = 0 \implies \det \begin{pmatrix}
-\lambda & 4 \\
-3 & -\lambda
\end{pmatrix} = \lambda^2 + 12 = 0 \implies \lambda = \pm 2i\sqrt{3}.
$$
Since the eigenvalues are purely imaginary (as I got), what can I conclude from here?
Best Answer
a) and b) $H=3x^2+4y^2$ is a solution. So the phase portraits are $ 3x^2+4y^2=C$ with different constant 'C'. c) You solve the eigenvalue of this system，and for a linear system, if all the eigenvalues have a negetive real part，then you can claim the system is asymptotically stable. But a pruely imaginary shows the system is not asymptotically stable.