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I encountered a theorem in some online notes labeled the 'gradient theorem':

$$\int_\Omega \nabla u = \int_\Gamma u \mathbf n$$

I've never seen this theorem before, and searches for 'gradient theorem' yield the fundamental theorem of calculus for line integrals which appears to be a different theorem. I've also never seen an explicit integral of a vector - only integrals of scalar quantities derived from vector quantities, like in Stoke's theorem or the divergence theorem.

How do I prove this theorem, and how do I even interpret a vector integrand?

D G
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    This is the divergence theorem. I assume $u$ is a vector field, in which case the guy on the left is a volume integral and the guy on the right is a surface integral of $u\cdot n$, where $n$ is the surface normal – operatorerror Nov 28 '17 at 22:29
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    The divergence theorem relates the integral a divergence, which is a scalar quantity. It's not clear to me why this is equivalent to the theorem in the OP, since $u$ is a scalar field. – B. Mehta Nov 28 '17 at 22:31
  • This is one of the many manifestations of the fundamental theorem of calculus in the multidimensional case. Equivalent theorems are the divergence theorem, the Green formulas, the Stokes theorem... The general formula encompassing all of them is the differential form version of Stokes's theorem, but understanding that formalism is not strictly needed – Giuseppe Negro Nov 28 '17 at 22:34

2 Answers2

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Take a constant vector field $\mathbf a$. Then by Divergence Theorem $$ \mathbf a \cdot \int_\Omega \nabla u = \int_\Omega \mathbf a \cdot \nabla u = \int_\Omega \nabla \cdot (\mathbf a u) = \int_\Gamma \mathbf a u \cdot \mathbf n = \mathbf a \cdot \int_\Gamma u \mathbf n $$

Since this is valid for all $\mathbf a$ we have $$\int_\Omega \nabla u = \int_\Gamma u \mathbf n$$

md2perpe
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  • Wonderful! Alternatively, a colleague suggested breaking apart the equation into element-wise form, then applying a 1-dimensional divergence theorem to each element. – D G Nov 29 '17 at 16:10
  • And that is equivalent to taking $\mathbf a = \hat x, \hat y, \hat z, \ldots$ (the unit vectors along the axes). – md2perpe Nov 29 '17 at 16:22
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That's only a notation to write the formula $$\tag{1} \int_\Omega \frac{ \partial u}{\partial x_j}\, dV = \int_{\Gamma} u n_j\, dS,\quad j=1\ldots n.$$ Here $\Omega\subset \mathbb R^n$ is an open set with smooth enough boundary $\Gamma$ and $n_j$ are the components of the normal unit vector $\mathbf n$. The connection with the formula in the OP is given by $$ \nabla u = \sum_{j=1}^n \partial_{x_j} u\, \mathbf e_j, $$ (here $\mathbf e_j$ is an orthonormal basis of $\mathbb R^n$). This shows that the formula in this post is equivalent to the one in the OP by linearity of the integral.

EDIT I will add here some hints to prove (1).

Without loss of generality assume that $\Omega\subset \mathbb R^2$. Also, assume that $$ \Omega=\{ (x_1, x_2)\ : -\infty<x_1<\infty,\ \phi(x_1)\le x_2\},$$ for a smooth function $\phi\colon \mathbb R\to \mathbb R$. With this assumption, $\mathbf n(x_1, \phi(x_1))=\frac{\phi'(x_1)\mathbf e_1 -\mathbf e_2}{\sqrt{\phi'(x_1)^2+1}}$. And finally, assume that $u\colon \Omega\to\mathbb R$ is an integrable function such that $$u(x_1, x_2)=0\qquad \text{for all sufficiently big }x_1, x_2.$$ These assumptions are motivated by the machinery of the partition of unity, which is used to remove them.

The proof now goes as follows: $$\iint_\Omega \frac{\partial u}{\partial x_1}\, dx_1dx_2= \int_{-\infty}^\infty \left( \int_{\phi(x_1)}^\infty \frac{\partial u}{\partial x_1}\, dx_2\right)\, dx_1=\int_{-\infty}^\infty \left[\partial_{x_1}\left(\int_{\phi(x_1)}^\infty u(x_1, x_2)\, dx_2\right) +u(x_1, \phi(x_1))\phi'(x_1)\right]\, dx_1.$$ Here we have used Leibniz integral rule to take $\partial_{x_1}$ out of the integral. Now the first summand vanishes by the fundamental theorem of calculus, so we are left with $$\int_{-\infty}^\infty u(x_1, \phi(x_1))\phi'(x_1)\, dx_1$$ and you can check that this is equal to $\int_{\Gamma} u\, n_1\, dS.$

Giuseppe Negro
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