# 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?

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.

• For (1) and (2) we use the vector calculus identity $\nabla(\phi \mathbf{a}) = \phi \nabla \cdot \mathbf{a} + \nabla \phi \cdot \mathbf{a}$. – RRL Apr 22 at 23:42