2
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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?

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4
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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.

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

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