Nonlinear Systems

Solution 1:

We are given:

$$f(x, y) = x'(t) = -2y + x^2$$ $$g(x, y) = y'(t) = 4x - 2y$$

We solve for the critical points by finding the simultaneous point where $x'$ and $y'$ both equal zero, so you found those points as: $(0,0)$ and $(4,8)$.

Next, we find the Jacobian matrix of the system as:

$$\displaystyle J(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix}2x & -2\\4 & -2\end{bmatrix}$$

Now, we evaluate the eigenvalues at the two critical points using the Jacobian matrix $J(x,y)$, so we have:

$$\displaystyle J(0,0) = \begin{bmatrix}0 & -2\\4 & -2\end{bmatrix}$$

The eigenvalues and eigenvectors for the CP $(0,0)$ yield:

• $\lambda_1 = ~~i (\sqrt 7+i), ~v_1 = \left(\dfrac{1}{4}(1+i \sqrt 7), 1\right)$
• $\lambda_2 = -i (\sqrt 7-i), ~v_2 = \left(\dfrac{1}{4} (1-i \sqrt 7), 1\right)$
• These are complex conjugate eigenvalues with negative real part $\rightarrow$ stable spiral (node).

The eigenvalues and eigenvectors for the CP $(4,8)$ yield:

$$\displaystyle J(4,8) = \begin{bmatrix}8 & -2\\4 & -2\end{bmatrix}$$

The eigenvalues and eigenvectors for the CP $(0,0)$ yield:

• $\lambda_1 = ~~3+ \sqrt {17}, ~v_1 = \left(\dfrac{1}{4} (5+ \sqrt {17}), 1\right)$
• $\lambda_2 = 3- \sqrt {17}, ~v_2 = \left(\dfrac{1}{4} (5- \sqrt {17}), 1\right)$
• These are real eigenvalues with positive and negative $\rightarrow$ unstable saddle point.

We can draw a phase portrait to show this behavior over time as:

Notice the analytical result at the two critical points.

We can also write the solution to these expressions as functions of time, but the phase portrait shows the local and global behavior over time.

Solution 2:

Steady state problems are time independent problems. So, in the case where we want a steady state solution, we are basically looking for solutions that won't change when the time progresses. In practice, to find the steady state, we require the derivative to be zero. So, in your case, you will have to solve the system

$$ x^2-2y=0 $$

$$4x-2y=0 . $$