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I'm reading a textbook example on the finite element method:

$\nabla^T[D(x,y,z)\nabla u] - a(x,y,z)u + f = 0 $ in R

$\partial R= \partial R_1 \bigcup \partial R_2$, $\partial R_1 \bigcap \partial R_2 \neq \varnothing$

u=r(x,y,z) on $\partial R_1$

$D \nabla u \centerdot n = -p(x,y,z)u + q(x,y,z)$ on $\partial R_2$

where $n$ is a unit outward normal vector, and $D(x,y,z) > 0$ and $a(x,y,z),p(x,y,z),$ and $q(x,y,z)\ge0$.

I understand how to obtain the weak formulation using the test function $v$:

$\iiint\limits_R [-D\nabla u\centerdot \nabla v -auv + fv]dV + \iint\limits_{\partial R_2}[-puv+qv]dA$, where dV and dA are the volume and area differentials.

I'm looking for a bound on the Galerkin method error. The book I'm reading seems to use an energy functional of the form:

$F(u) = \iiint\limits_R \{ \frac{1}{2}D|\nabla u|^2 + \frac{1}{2}au^2 - fu \}dV + \iint\limits_{\partial R_2} \{ pu^2 - qu \} dA$.

I'm not sure exactly how this energy functional was derived. It almost looks like the weak formulation integrated once with respect to u, then integrated over the respective domains. But I don't think this is true for the $|\nabla u|^2$ term. How could I have come up with this appropriate energy functional on my own? What guidelines can I follow, in general, to obtain this functional for numerical error analysis?

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up vote 1 down vote accepted

To obtain the energy functional from the weak formulation, just put $v=u$. Generally, your weak formulation of the (linear) PDE is a bilinear form $B(u,v)$ and the corresponding energy functional is the quadratic form $B(u,u)$.

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Substituting v=u into the weak formulation accounts for most of the functional $F(u)$ except for the term $\frac{1}{2}au^2$. Where did the $\frac{1}{2}$ come from? –  Paul Aug 30 '12 at 15:51
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