Second order elliptic equation with nonlinearity depending on the gradient

Question

Let us consider the problem $$-\Delta u =f(x,u,\nabla u)\text{ in }\Omega$$ $$u=0 \text{ on }\partial\Omega,$$ where $\Omega\subset \mathbb{R}^n$ is a smooth and bounded domain. I have seen at many places, where it is written that since the nonlinearity depends on the gradient therefore it is not variational. I understand that it is not variational because we cannot find a functional of which critical points are the solutions of the given problem but I am not able to justify this statement.

Answer 1

Here is a suggestion for a proof. Some details are missing, but I am confident that these can be filled.

Let us assume that there is a functional $J : H_0^1(\Omega) \to \mathbb{R}$, such that $J$ is Fréchet-differentiable at $u \in H_0^1(\Omega)$ with $$J'(u)\,v = \int_\Omega \nabla u \cdot \nabla v - f(u,\nabla u)\,v\,\mathrm{d}x$$ for all $v \in H_0^1(\Omega)$. If $f$ is regular enough, we might be able to show that $J'$ is also Fréchet-differentiable with $$J''(u)(v,w) = \int_\Omega \nabla w \cdot \nabla v - \partial_1 f(u,\nabla u) \, v \, w - \partial_2 f(u,\nabla u) \, v \, \nabla w \, \mathrm{d}x$$ for $v,w \in H_0^1(\Omega)$. Here, $\partial_i f$ is the partial derivative of $f$ w.r.t. argument $i$. Hence, $J$ is twice Fréchet-differentiable, which implies that the second derivative is symmetric w.r.t. $v$ and $w$. But, in general, the last term might not be symmetric (try integration by parts).