Consider the planar system $$\dot{x} = 2x-y-\frac{2x^3+2xy^2-xy}{\sqrt{x^2+y^2}} \\ \dot{y} = 2y+x-\frac{2x^2y+2y^3+x^2}{\sqrt{x^2+y^2}}$$ which in polar coordinates is given by $$\ \dot{r}=2r(1-r) \\ \dot{\theta}=2sin^2(\theta/2)$$
Is the above system a gradient system? Can you explain? Also, 1. Find all equilibrium points of this system and determine their stability properties. Are equilibria asymptotically stable? 2. What are stable and unstable manifolds of the equilibria. From the polar coordinates, r=0 could be equilibrium but then x=0, y=0 will cause the system to blow up.
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$\begingroup$A little background: If $\dot{\mathbf{x}}=\nabla G(\mathbf{x})$, then $g(t):=G(\mathbf{x}(t))$ is non decreasing since $g'(t)=\|\nabla G(\mathbf{x}(t))\|^2$. Thus, the system cannot admit periodic solutions. If $\mathbf{x}_0$ is a steady solution ($\nabla G(\mathbf{x}_0)=\mathbf{0}$), then the matrix $D_{\mathbf{x}}\nabla G(\mathbf{x}_0)=\operatorname{Hessian}(G)(\mathbf{x_0})$, being symmetric, has eigenvalues that are real.
For a plannar system like yours, if $\nabla G((x_1,x_2))=(g_1(x,y),g_2(x,y))$, then $\partial_{x_2}g_1=\partial_{x_1}g_2$ when $G$ is twice differentiable.
All the above suggests a few direct strategies:
In $\dot{x}=f_1(x,y)$, $\dot{y}=f_2(x,y)$, check if $\partial_yf_1=\partial_xf_2$. If that is not the case, your system is not a gradient one.
See if there is a periodic orbit (try $r=1$). If this is indeed a period orbit, then your system is not a gradient system.
Look at stay solutions $(1,0)$ seems to be one of them an check whether all eigenvalues are real (if they are not, then this is not a gradient system)
Finally you may try, through integration, to match a function $G(x,y)$ such that $\nabla G$ is the field in your problem. The integrals are not terribly bad.
Wlog consider unit speed curve primed with respect to arc variable.
$$ \dfrac{dr}{ds}= \cos \psi = 2 r(1-r) \\ \dfrac{d\theta}{ds}= \dfrac{\sin \psi}{r} = (1-\cos \theta) $$
Eliminating arc length we obtain an $(r-\theta)$ relation as:
$$ (1-\cos^2\theta)= \dfrac{1}{r^4}- \dfrac{4 (1-r)^2}{r^2} $$
$$ \cos^2\theta=1+ \dfrac{4 (1-r)^2}{r^2} -\dfrac{1}{r^4} $$
which is clearly periodic and can be polar plotted to verify. . No orbit for $r<1$ as stable dynamic attractor, not a gradient system.
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