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I'm at a PDE class and my teacher gave a very generic definition of the divergence theorem. I can't find it anywhere. It's something like this:

Definition: let $k\in \{1,2,\cdots,\infty\}$, $N\ge 2$

An open $\Omega\in\mathbb{R}^N$ has border of class $C^k$ if:

$\Omega$ is bounded and given any $x_0\in\partial\Omega$, there is an open $U_0$ containing $x_0$ and a diffeomorphism $\psi:U_0\to Q$, where $Q = ]-1,1[^N$ such that

  1. $\psi(\Omega\cap U_0) = \{t\in Q: t_N>0\}$

  2. $\psi(x_0) = 0$

  3. $\psi(\partial \Omega\cap U_0) = \{t\in Q: t_N = 0\}$

Now he wrote these things:

$X(t) = \psi^{-1}(t_1,\cdots,t_{N-1}, 0)$

$X:\{t'\in\mathbb{R}^{N-1}:|t_j|<1, j=1,\cdots,N-1\}\to U_0\cap\partial\Omega$

$$\vec{\theta}(t') = \left(\frac{\partial(x_2,\cdots,x_N)}{\partial(t_1,\cdots,t_{N-1)}}(t'),\cdots,\frac{\partial(x_1,\cdots,x_N)}{\partial(t_1,\cdots,t_{N-1)}}(t')\right)$$

then $\vec{\theta}(t')$ is normal to $\partial\Omega\cap U_0$ at $X(t')$

What are those derivatives? I don't recognize this notation and I don't know how it can be normal to anything. And why he picked the inverse image of $\psi$?

The surface element $d\sigma$ of $\partial\Omega$ is represented in $U_0\cap \partial \Omega$ by

$$|\vec{\theta}(t')|dt_1\cdots dt_{N-1}$$

finally we can define the exterior unitary normal in $\partial \Omega$:

$$\vec{n}:\partial\Omega\to\mathbb{R}^N$$

$$|\vec{n}(p)| = 1, \forall p\in \partial \Omega$$

Divergence Theorem

Let $\Omega$ be an open with boundary class $C^k$ and $\vec{X}(x) = (X_1(x), \cdots, X_N(x))$ with $X_j\in C^1(\overline{\Omega}), j=1,\cdots, N$ then

$$\int_{\Omega}(\mbox{div} \vec{X})(x)dx = \int_{\partial\Omega}\vec{X}(y)\cdot\vec{n}(y)d\sigma(y)$$

This seems to be a very generic view of the divergence theorem. And it seems that I need to understand what is an open set with a border and what does it mean for it to have class $C^k$. I also need to understand the surface element $d\sigma$.

Is there a place where I can learn about those things but not too deep? They're just tools to solve PDEs

Also, what does $C^n(\Omega)$ means?

Also, the integral on $\Omega$ is supposed to be on an open of $\mathbb{R}^n$, but $div$ applied to $x$ is a real number, and $dx$ looks like a one dimensional thing.

UPDATE:

I'm downloading some books on vector calculus but I only find ones with a lot of wedge products and differential forms. I needed only a vector explanation of all this. Can someone point me a gook book?

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  • $\begingroup$ What do you mean by $Q=]-1, 1[$? $\endgroup$ – Jacky Chong Aug 11 '18 at 2:18
  • $\begingroup$ $C^n(\Omega)$ means $\vec X$ has continuous partial derivatives of order $n$. $\endgroup$ – manooooh Aug 11 '18 at 2:19
  • $\begingroup$ @manooooh it means that every partial derivative of order up to $n$ in any point $\Omega$ is continuous? $\endgroup$ – Guerlando OCs Aug 11 '18 at 2:22
  • $\begingroup$ @JackyChong I think the teacher meant open set $(1,1)$ $\endgroup$ – Guerlando OCs Aug 11 '18 at 2:28
  • $\begingroup$ Then how is it a diffeomorphism? $\endgroup$ – Jacky Chong Aug 11 '18 at 2:29
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This is just the usual divergence theorem written in an explicit coordinate notation. Any introductory book on multivariate calculus treats it.

First, $C^k(\Omega)$ is the space of $C^k$ functions on the set $\Omega$. The integral is an $N$-dimensional integral, where $dx$ is an $N$-dimensional volume element.

Here $\psi$ is a map that transforms a piece of the domain near its boundary into a domain with flat boundary (in this case the latter is just a half cube). Think of a bite out of an apple, and turning that small portion of the apple into one with flat boundary. This is called flattening out of the boundary. Boundary smoothness is defined by how smooth you can choose $\psi$ to be.

Then $X(t)$ is simply a parametrization of a boundary piece, where $t\in\mathbb{R}^{N-1}$. Since $\psi$ maps the boundary piece into a domain in $\mathbb{R}^{N-1}$, its inverse gives you a parametrization of that piece.

The derivative notation means the determinant of the matrix formed by the indicated partial derivatives. Let $\phi=\psi^{-1}$. Then $X$ is simply the restriction of $\phi$ on $\{t_N=0\}$. The Jacobian determinant of $\phi$ is $$ J = \frac{\partial(x_1,\ldots,x_N)}{\partial(t_1,\ldots,t_N)}=\begin{vmatrix}\frac{\partial x_1}{\partial t_1}&\frac{\partial x_1}{\partial t_2}&\cdots&\frac{\partial x_1}{\partial t_N}\\\frac{\partial x_2}{\partial t_1}&\frac{\partial x_2}{\partial t_2}&\cdots&\frac{\partial x_2}{\partial t_N}\\\vdots&\vdots&\ddots&\vdots\\\frac{\partial x_N}{\partial t_1}&\frac{\partial x_N}{\partial t_2}&\cdots&\frac{\partial x_N}{\partial t_N}\end{vmatrix}. $$ The vector $\theta$ is defined by the expansion coefficients of this determinant with respect to the last column (This is why I think there should be alternating signs in the formula of $\theta$. Just think of a surface in 3D space, where we use cross product). Note that the first $N-1$ columns of this matrix constitutes a basis of tangent vectors to the boundary. So taking an inner product of $\theta$ with any tangent vector $V$ would be equivalent to computing the above determinant with the last column replaced by $V$. Since $V$ is spanned by the first $N-1$ columns, the determinant will be $0$, meaning that $\theta$ is orthogonal to the tangent plane.

Let me elaborate on the tangent basis bit. Fix all $t_2,\ldots,t_N$ and look at $\phi(t_1,t_2,\ldots,t_N)$ as a function of $t_1$. It would draw a curve along the boundary of $\Omega$. If you take the derivative of this function, you would get a vector tangent to the boundary: $$ V_1 = \big(\frac{\partial x_1}{\partial t_1},\frac{\partial x_2}{\partial t_1},\ldots,\frac{\partial x_N}{\partial t_1}\big). $$ This is simply the first column of the Jacobian matrix of $\phi$. By the same reasoning, we see that the first $N-1$ columns of the Jacobian matrix of $\phi$ are vectors tangent to the boundary surface. Also, as $\phi$ is a diffeomorphism, its Jacobian matrix is invertible, and hence those $N-1$ vectors are linearly independent. This shows that they form a basis of the tangent space.

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  • $\begingroup$ Things are a LOT more clear now. But what do you mean by expansion coefficients? You mean that $\theta$ components are pieces of this jacobian determinant? I can't see how the first $N-1$ columns constitute a basis of tangent vectors to the boundary. Could you elaborate more on that? I can open a bounty if needed, I know this is a lot. $\endgroup$ – Guerlando OCs Aug 11 '18 at 2:51
  • $\begingroup$ You know you can expand a determinant in terms of its row or column? Cofactors or minors ring a bell? $\endgroup$ – timur Aug 11 '18 at 2:56
  • $\begingroup$ @GuerlandoOCs: Have a look at the update. $\endgroup$ – timur Aug 11 '18 at 3:06
  • $\begingroup$ If I start with $\frac{\partial x_1}{\partial t_1}$ then by eliminating its row and column I should get $\frac{\partial(x_2,\cdots,x_N}{t_2,\cdots,t_N}$ so it doesn't match the $\theta$'s first element $\endgroup$ – Guerlando OCs Aug 11 '18 at 3:08
  • $\begingroup$ Shuldn't it be $X(t_1,\cdots,t_N)$ instead of $\theta$? $\endgroup$ – Guerlando OCs Aug 11 '18 at 3:13

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