Are there PDEs in which Hessian appears in the weak formulation Before stating the question, I would like to first use an example for the type of formulation that I'm interested in.
Suppose we consider the continuity equation $\partial_t \rho + \mathrm{div}( \rho v ) = 0$ with boundary conditions $\rho(0) = \rho_0$ and $\rho(1) = \rho_1$.
Now, if we were to test this equation with function $f$, we could arrive at the following equation (assuming zero boundary terms):
$$\int_\Omega f ( \rho_1 - \rho_0 ) \ \mathrm{d}x = \int_0^1 \int_\Omega \nabla f \cdot (\rho v) \ \mathrm{d}x \ \mathrm{d}t.
$$
On the left side of the equation, we have values of $f$, while on the right, its gradient.
By considering all $v$ and $\rho$ satisfying the continuity equation we could, in some sense, think that we are trying to "retrieve" $\nabla f$ back.

Question
I would like to ask whether there are PDEs which could be used for "retrieving" the hessian $\nabla^2 f$ back in the above sense. (Or just in which hessian and not, for example, laplacian, appears explicitly)
 A: Integration by parts of $\nabla^{2} f$ against a symmetric field $\sigma=(\sigma_{ij})$ yields eventually the formula,
$$
\int_{\Omega}(\nabla^{2}f,\sigma)dx=\int_{\Omega}f\,\nabla^{*}\nabla^{*}\sigma\,dx+\int_{\partial\Omega}\partial_nf\,(\sigma,n\otimes n)\,dS+\int_{\partial\Omega}f\,T\sigma\,dS
$$
where $(\nabla^{*}\sigma)_{j}=\sum_{i}\partial_{i}\sigma_{ij}$, $\nabla^{*}\nabla^{*}\sigma=\sum_{i,j}\partial_{i}\partial_{j}\sigma_{ij}$, and $T$ is a first order boundary operator mixing $\nabla^{*}\sigma$ and the derivatives of $\sigma\cdot n$ on the boundary. Hence, requiring zero boundary conditions, you could cook up a "second order" continuity equation of the form $\partial_{t}\rho-\nabla^{*}\nabla^{*}(\rho\sigma)=0$ in order to obtain,
$$
\int_{0}^{1}\int_{\Omega}(\nabla^{2}f,\rho\sigma)\,dx \,dt=\int_{0}^{1}\int_{\Omega}f\,\nabla^{*}\nabla^{*}(\rho\sigma)\,dx\,dt=\int_{0}^{1}\int_{\Omega}f\,\partial_{t}\rho\,dx\,dt=\int_{\Omega}f(\rho(1)-\rho(0))dt
$$
assuming that $f$ is either time independent or satisfies $f(0)=0$ and $f(1)=0$.
