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I am trying to understand a proof for critical points of certain energy functions being harmonic functions. It goes as follows: For a function $u(x_1,..,x_n)$, a functional E(u) is defined as

$E(u) = \frac{1}{2}\int |\nabla u|^2 $

For the proof, a curve is constructed in the space of functions by using $u+t\phi$. Restricting the energy functional to this curve gives:

$E(u+t\phi) = \frac{1}{2} \int | \nabla (u+t\phi)|^2 = \frac{1}{2} \int |\nabla u|^2 + t \int \langle u,\nabla \phi \rangle + \frac{t^2}{2} \int |\nabla \phi|^2 $

Then differentiating at $t=0$

$\frac{d}{dt_{t=0}}E(u+t\phi) = \int \langle\nabla u, \nabla \phi\rangle = - \int \phi \triangle u$

The proof then states that the last equality is due to divergence theorem which I know is

$\int \int_S F \cdot n d\sigma = \int \int \int \nabla \cdot F dV$

I do not understand the connection between divergence theorem and how it is used in this proof. Any help?

Link: (page 2)

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I also received this on mathoverflow from Spiro Karigiannis, I just put it here for addition "Try putting F=φ∇u into the divergence theorem. You either have no boundary or you are in the setting where the "bump" function φ can be taken to decay to zero by the time it reaches the boundary". I voted the article to be deleted from there, as it received -2 votes. I have no idea why. :) – BB_ML Apr 2 '11 at 13:10
It's a very basic and known method in the calculus of variations. MathOverflow is not suitable for this type of question, that's why it was downvoted. But it's certainly welcome here. :-) – Greg Graviton Apr 2 '11 at 13:21
ok, good to know. thanks. – BB_ML Apr 2 '11 at 13:38
up vote 3 down vote accepted

The identity

$$ \int_V d\vec x\, \langle \nabla \phi, \nabla u \rangle = -\int_V d\vec x\ \phi\Delta u $$

is a consequence of integration by parts and the divergence theorem. Namely, we have

$$ \langle \nabla \phi, \nabla u \rangle = \sum_i \partial_i\phi\,\partial_i u = \sum_i \left(\partial_i (\phi\,\partial_i u) - \phi\,\partial_i^2 u\right) = \nabla\cdot(\phi\nabla u) - \phi\Delta u$$

Hence, integrating over the volume gives

$$ \int_V d\vec x\,\langle \nabla \phi, \nabla u \rangle = \int_V d\vec x\,(\nabla\cdot(\phi\nabla u) - \phi\Delta u) = \int_{\partial V} d\vec n\cdot \phi\nabla u - \int_V d\vec x\ \phi\Delta u .$$

Since the variation $\phi$ is assumed to vanish on the boundary $\partial V$, the surface integral vanishes.

This trick of "shifting the derivatives of $\phi$ to $u$" is extremely common in the calculus of variations. I suggest that you absorb it by doing several exercises. For instance, I recommend to derive the Euler-Lagrange equations for a general functional

$$ S[u] := \int_V d\vec x\ \mathcal L(u,\partial_i u) .$$

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This is more often called Green's identity but is indeed a consequence of the divergence theorem as you know it.

To get an intuition of why this is true, one can go back to dimension $1$. Integration by parts gives $$ \int \phi' u'=[\phi u']-\int\phi u'', $$ and you recognize the LHS and the last term on the RHS. And the first term $[\phi u']$ on the RHS is zero because the functional $E:u\mapsto E(u)$ is defined on functions $u$ with a prescribed value on the boundary, and the tangent space of these is made of functions $\phi$ which are zero on the boundary. In other (simpler) words, for any $t\ne0$, both functions $u$ and $u+t\phi$ coincide with a prescribed function on the boundary if and only if $\phi=0$ on the boundary.

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