Trying to prove an integration by parts formula Denote pde operator 
$$ Lu = - div \cdot (p \nabla u ) + qu$$
where $x \in D$ and $p=p(x) > 0$ and q=q(x) are continuous on $\bar{D}$ an p has continous first partial derivatives on $\bar{D}$. I want to prove that 
$$ \int\limits_D v Lu dx = \int\limits_D u Lv dx + \int\limits_{\partial D} p \left( u \frac{dv}{dn} - v \frac{du}{dn} \right) dA $$
What is the meaning of $n$ in this problem? This is why I am stuck as there is no indication as to what is $n$. IS it supposed to be x and it is a typo? or is it something else?
anyway, My approach is write 
$$ \int\limits_D (v Lu - u Lv ) $$
and simplify from there. Is this how we start this problem?
 A: The terms involving $n$ in question are normal derivatives, i.e.,
$$\frac{\partial u }{\partial n }(\mathbf{x}) = \nabla u(\mathbf{x}) \cdot \mathbf{n}(\mathbf{x}),$$
where $\mathbf{x} \mapsto \mathbf{n}(\mathbf{x})$ maps a point $\mathbf{x} \in \partial D$ to the outwardly directed unit normal vector at the surface $\partial D$.
Note that
$$\tag{1}vLu = - v\nabla \cdot(p \nabla u) +quv = - \nabla\cdot(pv\nabla u) + p\nabla u \cdot \nabla v + quv,$$
$$\tag{2}uLv = - u\nabla \cdot(p \nabla v) +quv = - \nabla\cdot(pu\nabla v) + p\nabla u \cdot \nabla v + quv,$$
Thus,
$$\tag{3}\int_D(v Lu - uLv) \, d\mathbf{x} = \int_D \left(\nabla \cdot (pu\nabla v) -\nabla \cdot (p v\nabla u) \right)\, d\mathbf{x} $$
Applying the divergence theorem, we have
$$\tag{4}\int_D \nabla \cdot (pu\nabla v)  \, d\mathbf{x} = \int_{\partial D}  pu\nabla v \cdot \mathbf{n}  \, dA  = \int_{\partial D}  pu\frac{\partial v}{\partial n}  \, dA,  \\ \int_D \nabla \cdot (pv\nabla u)  \, d\mathbf{x} = \int_{\partial D}  pv\nabla u \cdot \mathbf{n}  \, dA  = \int_{\partial D}  pv\frac{\partial u}{\partial n}  \, dA $$
Substitute (4) into (3) to finish.
