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If $w$ is the vector of conservative variables, $f=f(w)$ the flux function, I think have read somewhere (I can not find it anymore) that the inflow boundary $\Sigma$_ is characterized by:

$\Sigma_{-} = \left\{x, \nabla_{w}f \cdot \mathbf{n} < 0 \right\}$.

What is the exact definition of $\nabla_{w}f$ and is it not $\nabla_{u}f$ instead, where $u$ is the velocity ?

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If $w=(w_1,w_2,\ldots,w_l)$ represents a vector of $l$ variables and $f$ is a scalar function, then $\nabla_w f$ should stand for the gradient with respect to variables $w$, i.e. it is the vector function $(\partial_{w_1} f,\partial_{w_2} f,\ldots,\partial_{w_l} f)$.

If $l>1$ then typically $f$ is a vector function of $l$ components and one prefers a notation $\bf{f}$ or $\vec{f}$ (me too). Then $\nabla_w \vec{f}$ is a Jacobian (matrix) given by entries $\partial_{w_l} f_k$ where $f_k$ is the $k$-th component of $\vec{f}$.

These are main tools when considering a system of conservation laws like $$ \partial_t w + \nabla \cdot \vec{f} = 0 $$ where now $\nabla$ stands for standard divergence operator with respect to space variables $x$, i.e. $w=w(x,t)=(w_1(x,t),w_2(x,t),\ldots,w_l(x,t))$. The system above can be rewritten (if derivatives exist) to the system $$ \partial_t w + \nabla_w \vec{f} \cdot \nabla w = 0 $$ that can be viewed as nonlinear advection equation, so $\nabla_w \vec{f}$ plays a role of (solution dependant) velocity. In this sense you might consider the definition of inflow boundary conditions.

The system of nonlinear conservation laws is a nontrivial topic, you can learn it e.g. from the book of LeVeque on FVM for Hyperbolic Problems.

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I have come across a similar expression in this paper by a group at INRIA. For more general discussion refer to section 1.2.4 in this notes from Uni Dortmund.

$\partial \Omega_{-} = \left\{x \in \partial d\Omega | \mathbf{n}\cdot\nabla_{u}f < 0 \right\}$.

In my understanding $\nabla_{u} f$ denotes the gradient. It's standard practice to use $u$ for velocity.

Edit: Peter Frolkovič's answer is accurate. Corrected my erroneous interpretation.

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  • $\begingroup$ Please, see the answer below for a correct answer. $\endgroup$ – Peter Frolkovič Aug 11 '16 at 11:26

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