I have a flow defined by $\dot{x} = x-x^4+1 :=f(x)$. I need to sketch its phase portrait. Firstly, I have to find its fixed points, these occur at $f(x)=x$. So, $x^4=1 \Rightarrow x=
\pm1$. Next, I find the Jacobian, in this case the derivative of $f(x)$, at these points. $f'(x)=1-4x^3 \Rightarrow f'(1)=-3<0, f'(-1)=5>0$. So the points; $x=1$ is stable and $x=-1$ is unstable. Is this the right way to go about obtaining the nature of the fixed points. Further, how do a sketch the phase portrait. I have only seen 2D cases where I solve the DE for $x_1$ and $x_2$ and graph them. Here I only have 1D so I am not sure whether I have to graph the solution of $\dot{x} = x-x^4+1$ and how would I graph the flow lines etc.
[Math] 1 dimensional flows and phase portraits
dynamical systems
Related Solutions
We are given:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
Recall, when we are doing polar coordinates, we have
$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$
When we differentiate this, we have:
$$2 x x' + 2 y y' = 2 r r'$$
This gives us:
$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$
To find the angle, we take:
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule gives us:
$$\theta' = \dfrac{x y' - y x'}{r^2} = \dfrac{x^2(x^2+y^2)+y^2(x^2+y^2)}{r^2}= \dfrac{r^4}{r^2} = r^2$$
Thus in polar coordinates, the original system becomes
- $r' = 0$
- $\theta' = r^2$
Can you now solve this system and draw the phase portrait?
The phase portrait should look like:
Phase portrait:
$\qquad\qquad\qquad\qquad\qquad$
Fixed points:
- At $(3,1)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-1&-3\end{pmatrix}$, trace $-4$ (negative), determinant $+2$ (positive), discriminant $(-4)^2-4\cdot(+2)=8$ (positive), hence two real negative eigenvalues: the point $(3,1)$ is a stable node
- At $(1,3)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-3&+1\end{pmatrix}$, trace $0$, determinant $-4$ (negative), hence two real eigenvalues of opposite signs: the point $(1,3)$ is a saddle point
...As explained there:
$\qquad$
Best Answer
The fixed points are the constant solutions $x(t)=k$ for some constant $k\in\mathbb R$. Thus we have $x'(t)=0$ or, by the differential equation, $x-x^4+1=0$. (Not $x-x^4+1=x$.) Now, there are two real numbers where $x-x^4+1=0$ and they are approximately at $x=-0.724492$ and $x=1.22074$. The graph is positive between these and negative outside. So the phase line consists of two negatively oriented arrows on the outside and one positively oriented arrow in the middle. You can easily draw it on the $x$-axis together with the graph of $x-x^4+1$.