How to prove this gradient theorem in vector calculus

Question

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?

Answer 1

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

Answer 2

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 $\int_{\Gamma} u\, n_1\, dS.$