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My problem is, given a $V: \mathbb{R}^2 \rightarrow \mathbb{R}$ that is at least $C^2$, consider its gradient flow $$ \dot{x} = -\nabla V,\quad \text{ or }\quad \dot{x_i} = -\frac{\partial V}{\partial x_i}, x \in \{1,2\}.$$

Show that this system cannot have a homoclinic solution.

My progress so far is to use the identity $\dot{V} = \nabla V \cdot \dot{x} = \nabla V \cdot -\nabla V$, which means that $$\dot{V} = -\left(\frac{\partial V}{\partial x}\right)^2-\left(\frac{\partial V}{\partial y}\right)^2 \leq 0 \,\, , $$ and if $V$ is not constant with respect to $x$ and $y$, this is a strict inequality.

This would mean that any solutions to our system will have decreasing $V$ along their trajectory. Here is my reasoning. Assume that $\varphi_t(x_0)$ is a homoclinic solution to our system with respect to an equilibrium point $x^* \neq x_0$. Then we have $\varphi_t(x_0) \rightarrow x^*$ as $t \rightarrow \infty$, so by what we have shown above $V(x^*) < V(x_0)$. On the other hand we also have $\varphi_t(x_0) \rightarrow x^*$ as $t \rightarrow -\infty$, so by this it must hold that $V(x^*) > V(x_0)$. But this gives $V(x^*) > V(x_0) > V(x^*)$ which is a contradiction.

My problem is how to handle function $V$ which are partially constant. Any ideas? Does the above proof work?

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I'm supposing that your equilibrium $x^*$ is isolated. What does it mean for $x^*$ to be an equilibrium point? (i.e. look at the gradient of $V$). Now what is the connection between $\dot{V}$ and the gradient of $V$? Try to conclude that in a neighborhood of $x^*$, $\dot{V}$ is strictly negative (what I mean is that for any orbit passing through a neighborhood of $x^*$ experiences strictly decreasing $V$ value).

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