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\begin{align}
\delta E & = \int_{\Omega}\pars{\delta\verts{\nabla\mrm{u}}^{2}}\,\dd x =
\int_{\Omega}\pars{\verts{\nabla\mrm{u} + \nabla\delta\mrm{u}}^{2} -\verts{\nabla\mrm{u}}^{2}}\,\dd x =
\overbrace{\int_{\Omega}2\nabla\mrm{u}\cdot\nabla\delta\mrm{u}\,\dd x}
^{\ds{+\ \mbox{terms of order}\ \pars{\delta\mrm{u}}^{2}}}
\\[5mm] & =
\int_{\Omega}\bracks{2\nabla\cdot\pars{\delta\mrm{u}\nabla\mrm{u}} - 2\,\delta\mrm{u}\nabla^{2}\mrm{u}}\,\dd x
\\[5mm] & =
2\int_{\partial\Omega}\delta\mrm{u}\nabla\mrm{u}\cdot\dd\mathbf{S} -
2\int_{\Omega}\delta\mrm{u}\nabla^{2}\mrm{u}\,\dd x = 0
\end{align}
$\ds{\mrm{u}}$ is 'fixed' at $\ds{\partial\Omega}$ such that the first integral vanishes out and $\ds{\nabla^{2}\mrm{u} = 0}$
Poisson requires a constraint as $\ds{\Phi = \int_{\Omega}\rho\,\mrm{u}\,\dd x}$. It yields ( via Lagrange Multipliers ) to an additional term like $\ds{\mu\int_{\Omega}\delta\mrm{u}\,\rho\,\dd x}$ which leads to ( Poisson )
$\ds{\nabla^{2}\mrm{u} = -\mu\rho}$.
