Differential equation I've working on the following problem:

Consider a solution $(x_1(t),x_2(t),\ldots,x_N(t))$ of the equation
$$\frac{dx_i}{dt}=-\sum_{j=1}^{N}\nabla W(x_i-x_j), i=1,\ldots,n $$
with $W\in C^1(\mathbb{R}^d)$ satisfying $$(\nabla W(x)-\nabla W(y))(x-y)\geq \lambda |x-y|^2, $$ for each $x,y\in \mathbb{R}^d,\lambda>0.$ If we define $$V(t):=\frac{1}{2}\sum_{i=1}^{N}|x_i(t)-\bar x(t)|^2,$$ where $\bar x(t)=\frac{1}{N}\sum_{i=1}^{N}x_i(t),$ then prove that there exists a constant $A>0$ such that:
$V(t)\leq V(0)e^{-At},~~t\geq0.$

I think that I have to use somehow the Grönwall's inequality using the given property of the function $W$, so I've tried to derive $V(t)$ and I've obtained this:
$\displaystyle \frac{dV(t)}{dt}=\sum_{i=1}^{N}\left[(x_i(t)-\bar x)\left(-\sum_{j=1}^{N}(\nabla W(x_i-x_j))+\frac{1}{N}\sum_{j=1}^{N}\sum_{k=1}^{N}\nabla W(x_j-x_k)\right) \right]$.
I don't know how can I continue from this or if I have chosen a good path to solve the problem.
 A: $\newcommand{\bu}{{\bf u}}\newcommand{\bv}{{\bf v}}\newcommand{\bx}{{\bf x}}\newcommand{\mM}{{\rm M}}$
In the case of scalar $x_i$ the expression for the derivative of $V$ has the structure $\dot V=2\bu^T\mM\bv$ with an anti-symmetric matrix $\mM=-\mM^T$,
$$
\mM_{ik}=-\frac{1}{2N}\sum_{j=1}^{N}\left(\nabla W(x_i-x_j) - \nabla W(x_k-x_j)\right),
$$
$\bv$ the vector of ones and $\bu$ the vector of $u_i=x_i-\bar x$.
Exploiting the anti-symmetry one can abstractly transform
$$
\dot V=\bu^T\mM\bv-\bu^T\mM^T\bv={\rm trace}((\bv\bu^T-\bu\bv^T)\mM)
=(\bu\bv^T-\bv\bu^T)\,{\bf :}\,\mM
$$
where $$(\bu\bv^T-\bv\bu^T)_{ik}=u_iv_k-u_kv_i=u_i-u_k=x_i-x_k$$
By the monotonicity condition
$$
(x_i-x_k)\left(\nabla W(x_i-x_j) - \nabla W(x_k-x_j)\right)\ge λ(x_i-x_k)^2.
$$
and thus for the terms inside the Euclidean/Frobenius matrix scalar product
$$
(\bu\bv^T-\bv\bu^T)_{ik}\mM_{ik}\le-\fracλ2(x_i-x_k)^2
$$
Over the $N^2$ terms of the scalar product this makes
\begin{align}
\frac{dV}{dt}&\le -\frac{λN^2}2\sum_{i=1}^N\sum_{k=1}^Nλ(x_i-x_k)^2\\
&=-\frac{λN^2}2\sum_{i=1}^N\sum_{k=1}^N(u_i-u_k)^2\\
&=-\frac{λN^2}2\left[4NV-2\sum_{i=1}^N\sum_{k=1}^Nu_iu_k\right]\\
&=-2λN^3·V
\end{align}
as $\sum_{i=1}^Nu_i=0$.

$\newcommand{\dotp}{\vcenter{\huge\cdot}}$
In the vector case for $\bx_i$, $\bu_i=\bx_i-\bar\bx$, the gradients in $M_{ik}$ are also vectors, still with $M_{ik}=-M_{ki}$. Then use the anti-symmetry directly without naming tensors of third order,
$$
2\sum_{i,k}\bu_i\dotp M_{ik}=\sum_{i,k}[\bu_i\dotp M_{ik}-\bu_i\dotp M_{ki}]
=\sum_{i,k}(\bu_i-\bu_k)\dotp M_{ik}
$$
the last step by switching the index labels in the second term. Then proceed as above.
