# Drawing the phase portrait of the dynamical system?

I was thinking about drawing the phase portrait of the dynamical system given by

$$\dot{x} = e^{x^2 -y^2} - 1$$

$$\dot{y} = x^2y^{-1} + 2$$

Please find the image below describing my try to draw the phase portrait, any mistakes I have done, please help me draw it better with logic and understanding. Any graphing ideas which can help us in drawing it easily?

I also do not see why $$(0,0)$$ cannot be a fixed point of the 2D dynamical system? even though it is in the intersection of the nullclines, but when we substitute $$(0,0)$$ in the equation it does not give us 0.

• There is... sooo much here. Please try to find one answerable question to ask about drawing the phase portrait. As for why zero is not a fixed point; just plug in $x=y=0$ to the equations and observe the derivative is not zero. Commented Sep 30, 2018 at 14:02

We are given

$$x' = e^{x^2 -y^2} - 1 \\ y' = x^2y^{-1} + 2$$

Drawing the phase portrait (PP) is sometimes easy, but when the PP is complicated, we have to do a lot more analysis to figure out its true shape and that is the case here. Lets first do all the intermediate calculations. See the references below.

The critical points are given by

$$e^{\left(x^2-y^2\right)}-1=0,\dfrac{x^2}{y}+2=0 \implies (x, y) = (-2, -2), (2,-2)$$

The Jacobian matrix is given by

$$J_{(x,y)}(f)=\begin{pmatrix} \dfrac{\partial f_1}{\partial x} & \dfrac{\partial f_1}{\partial y} \\ \dfrac{\partial f_2}{\partial x} & \dfrac{\partial f_2}{\partial y} \end{pmatrix} = \begin{pmatrix} 2 e^{x^2-y^2} x & -2 e^{x^2-y^2} y \\ \dfrac{2 x}{y} & -\dfrac{x^2}{y^2} \\ \end{pmatrix}$$

The eigenvalues of the Jacobian at $$(x, y) = (-2,-2)$$ are

$$\lambda_{1,2} = \left\{\frac{1}{2} \left(-\sqrt{41}-5\right),\frac{1}{2} \left(\sqrt{41}-5\right)\right\} \implies \text{unstable saddle due to different signs}$$

The eigenvalues of the Jacobian at $$(x, y) = (2,-2)$$ are

$$\lambda_{1,2} = \left\{\frac{1}{2} \left(3+i \sqrt{7}\right),\frac{1}{2} \left(3-i \sqrt{7}\right)\right\} \implies \text{unstable spiral due to positive real part}$$

We now find the x-nullcline by solving for $$y$$ when $$x' = 0$$

$$x' = e^{x^2 -y^2} - 1 = 0 \implies y = \pm~ x$$

We now find the y-nullcline by solving for $$y$$ when $$y' = 0$$

$$x' = x^2y^{-1} + 2 = 0 \implies y = - \dfrac{x^2}{2}, ~ x \ne 0~\text{because we cannot divide by}~ y = 0$$

Along the nullclines the flow is either vertical $$(x' = 0)$$ or horizontal $$(y' = 0)$$. They divide the phase plane into regions where the flow mainly points towards one of the four directions:

$$x' <0, y' < 0 ~ \text{or} ~ x' >0, y' < 0 \implies \text{down}$$ $$x' <0, y' > 0 ~ \text{or} ~x' >0, y' > 0 \implies \text{up}$$

The intersection points of the nullclines give the fixed points of the flow. Since trajectories are not allowed to cross, the information given by the nullclines often allows to make a qualitative plot of the dynamics.

You now select various $$x$$ and $$y$$ values and determine the up/down as prescribed above. Note that I also like to calculate $$\dfrac{dy}{dx}$$ in addition to the typical $$x'$$ and $$y'$$ values because I do not do these by hand and I like to know the magnitudes of the slopes. For example, varying $$1 < x < 3$$ and fixing $$y = 0.5$$, we generate a table of $$\{x, y, \{y',x'\}, \dfrac{y'}{x'} = \dfrac{dy}{dx} \}$$ as (compare these values to the PP and make sure you can do it by hand, for example, notice what is happening to the slope as we move towards $$x = 3$$)

$$\left( \begin{array}{cccc} 1. & 0.5 & \{4.,1.117\} & 3.58102 \\ 1.1 & 0.5 & \{4.42,1.6117\} & 2.74245 \\ 1.2 & 0.5 & \{4.88,2.28708\} & 2.13372 \\ 1.3 & 0.5 & \{5.38,3.2207\} & 1.67045 \\ 1.4 & 0.5 & \{5.92,4.52896\} & 1.30714 \\ 1.5 & 0.5 & \{6.5,6.38906\} & 1.01736 \\ 1.6 & 0.5 & \{7.12,9.07442\} & 0.784623 \\ 1.7 & 0.5 & \{7.78,13.0132\} & 0.597854 \\ 1.8 & 0.5 & \{8.48,18.8857\} & 0.449017 \\ 1.9 & 0.5 & \{9.22,27.7892\} & 0.331784 \\ 2. & 0.5 & \{10.,41.5211\} & 0.240842 \\ 2.1 & 0.5 & \{10.82,63.0715\} & 0.171551 \\ 2.2 & 0.5 & \{11.68,97.4944\} & 0.119802 \\ 2.3 & 0.5 & \{12.58,153.47\} & 0.0819704 \\ 2.4 & 0.5 & \{13.52,246.151\} & 0.0549256 \\ 2.5 & 0.5 & \{14.5,402.429\} & 0.0360312 \\ 2.6 & 0.5 & \{15.52,670.826\} & 0.0231356 \\ 2.7 & 0.5 & \{16.58,1140.39\} & 0.0145389 \\ 2.8 & 0.5 & \{17.68,1977.31\} & 0.00894142 \\ 2.9 & 0.5 & \{18.82,3497.19\} & 0.00538147 \\ 3. & 0.5 & \{20.,6309.69\} & 0.00316973 \\ \end{array} \right)$$

Using all of this information (please fill in the blanks), we can start to draw the PP and here is a computer generated one (note the behavior for $$y = 0$$ (because it is a nullcline) of the phase portrait).

References: