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2020-02-07, Friday

Today, we talk about Laplacian operator.

The formula

Take $\R^n$ to be our space. Let $x_1, \cdots, x_n$ be the standard coordinates. Let $u_1, \cdots, u_n$ be a general curvilinear coordinates. Then, we have vector basis transformation rule $$ \frac{\d}{\d u_i} = \sum_j \frac{\d x_j}{\d u_i}\frac{\d}{\d x_i}. $$

We can then compute the component of metric tensor in $u_i$ coordinates $$g_{ij} = g (\frac{\d}{\d u_i}, \frac{\d}{\d u_j}) = \sum_k \frac{\d x_k}{\d u_i} \frac{\d x_k}{\d u_j} $$

If we view $g_{ij}$ as entries of a matrix $[g_{ij}]$, then we denote $g^{ij}$ the entries of the inverse matrix. Let $|g| = \det( [g_{ij}] )$ be the determinant of the matrix. Then, our formula for the Laplacian operator is (where $\d_i = \d / \d u_i$)

$$ \Delta f = \sum_{i,j} \frac{1}{\sqrt{|g|}} \d_i (g^{ij} \sqrt{|g|} \d_j(f)) $$

Testing the formula

Let's consider spherical coordinates, we have $$ x = r \sin \theta \cos \phi, \quad y = r \sin \theta, \sin \phi, \quad z = r \cos \theta. $$

Denote $u_1 = r, u_2 = \theta, u_3 = \phi$. Then we have $$ g_{11} = 1 = \| \d_r \|^2, \quad g_{22} = r^2 = \| \d_\theta \|^2, \quad g_{33} = r^2 \sin^2 \theta = \| \d_\phi\|^2. $$ The $\sqrt{|g|} = r^2 \sin(\theta)$

This is an example of ortho-curvilinear coordinate, which means $\d_i \perp d_j$ , i.e. $g_{ij}=0$, for $i \neq j$, This simplifie the expression tremendorsly, because $g^{ii} = 1 / g_{ii}$.

$$\Delta f = \frac{1}{r^2 \sin \theta} \d_r (r^2 \sin \theta \d_r(f)) + \frac{1}{r^2 \sin \theta} \d_\theta ( \frac{r^2 \sin \theta}{ r^2} \d_\theta(f)) + \frac{1}{r^2 \sin \theta} \d_\phi (\frac{r^2 \sin \theta}{r^2 \sin^2 \theta} \d_\phi(f)) $$ $$ = \frac{1}{r^2 } \d_r (r^2 \d_r(f)) + \frac{1}{r^2 \sin \theta} \d_\theta ( \sin \theta \d_\theta(f)) + \frac{1}{r^2 \sin^2 \theta} \d^2_\phi f $$

Where does it come from?

Let $f$ be any smooth function on $\R^n$ and $\varphi$ be a “compactly supported” function, which means $\varphi(x)$ vanishes if $|x|$ is sufficently large. We define $\Delta f$ such that the following is true for all $\varphi$

$$ I = \int_{\R^n} (df(x), d \varphi(x))_g d\z{Vol}_g(x) = \int_{\R^n} (-\Delta f(x)) \varphi(x) d\z{Vol}_g(x)$$

(1) If we use Cartesian coordinate, then the above formula translates into

$$ I = \int \sum_i \d_{x_i} f \cdot \d_{x_i} \varphi dx_1 \cdots d x_n = \int (- \sum_i \d_{x_i}^2 f) \cdot \varphi dx_1 \cdots d x_n $$ where we used integration by part. Thus, we get $$ \Delta f(x) = \sum_i\d_{x_i}^2 f(x) $$ So this definition makes sense.

(2) If we use curvilinear cooridnate , then we claim that $$ d Vol_g(u) = \sqrt{|g|} du_1\cdots du_n \tag{1}$$ and that $$ (df, d\varphi)_g = \sum_{i,j} g^{ij} \d_{u_i} f \d_{u_j} \varphi \tag{2}$$.

Given these two claim, we get $$ I = \int \sum_{i,j} g^{ij} \d_{u_i} f \d_{u_j} \varphi \sqrt{|g|} du_1\cdots du_n = \int -\sum_{i,j} \d_{u_j}(g^{ij} \sqrt{|g|} \d_{u_i} f) \varphi du_1\cdots du_n$$ $$ = \int -\frac{1}{\sqrt{|g|}}\sum_{i,j} \d_{u_j}(g^{ij} \sqrt{|g|} \d_{u_i} f) \varphi \sqrt{|g|} du_1\cdots du_n $$ Hence, we have $$ \Delta f = \frac{1}{\sqrt{|g|}}\sum_{i,j} \d_{u_j}(g^{ij} \sqrt{|g|} \d_{u_i} f)$$

Tying up the loose ends

We will prove the two claims above. The first is about volume element.

Volume Element

I will motivate this using vector space. Suppose we are in $\R^n$, and we have a new basis $\gdef\t\tilde \t e_1, \cdots, \t e_n$. We are interested in getting the volume spanned by the skewed cube with sides given by these basis vectors.

You say, this is easy! Suppose we know the components of $\t e_i$ in the standard basis, then we can write $$ \t e_i = \sum_j a_{ij} e_j $$ and $| \det a_{ij} |$ is our desired volume.

The fancy way of saying this is that $$ \det a_{ij} = \frac{\t e_1 \wedge \cdots \wedge \t e_n }{e_1 \wedge \cdots \wedge e_n} $$ We take the volume of $e_1 \wedge \cdots \wedge e_n$ to be one. Then take the absolute number of $\det a_{ij}$, just in case it is a negative number.

However, we are not given $a_{ij}$! We are only given the $g_{ij}$, which are $$ g_{ij} = g(\t e_i, \t e_j) $$ Fortunately, we can obtain $g_{ij}$ from $a_{ij}$, and their determinants are also related. Concretely, we have $$ g_{ij} = \sum_k a_{ik} a_{jk} $$ Hence, if you view $G$ as the matrix with entry $g_{ij}$ and $A$ as a matrix with entries $a_{ij}$, we get $$G = A A^T \Rightarrow \det G = \det (A A^T) = \det A \cdot \det A^T = (\det A)^2$$ Thus, to get $|\det A|$, we can just do $$ |\det A| = \sqrt{\det G} $$ and don't worry, $G$ is positive definite, hence $\det G > 0$, so the square-root make sense.

Cotangent Vector

For $p \in \R^n$, we define $T_p \R^n$ to be the dual vector space of $T_p \R^n$, and an element of it is called a covector, or cotangnet vector.

If $f$ is a function, then $df(p)$ is an element of $T^*_p \R^n$. In Cartesian coordiante, we can write $$ df = \sum_i \frac{\d f}{\d x_i} d x_i $$

Actually, if we are given any coordinate system $(u_1, \cdots, u_n)$, then we have a basis of $T^*_p \R^n$, given by $du_1, \cdots, d u_n$. And any $df$ can be written as $$ df = \sum_i \frac{\d f}{\d u_i} d u_i. $$

Inner product of Covector

But, you wrote $ (df, d\varphi)_g$, what is that? It means, I am taking inner product of two co-vectors, using information of $g$.

Wait, $g$ is only inner product on $T_p \R^n$, how do you get inner product on $T^*_p \R^n$.

Well, in general, if $V$ has an inner product, we can equip $V^*$ with an inner product in the following way. Take $e_i$ an orthonormal basis of $V$, and $h^i$ the dual basis of $V^*$. We declare $h^i$ to be an orthonormal basis of $V^*$. This declaration defines an inner product on $V^*$, sometimes denoted as $g^*$.

That is still kind of abstract, let us use basis. Say, you pick a basis $e_i$ of $V$, and you know that $g_{ij} = g(e_i, e_j)$. You want to find out $g^*$ on $V^*$ using the dual basis $h^i$. We then know that $g^{ij} : = g^*(h^i, h^j)$ forms a matrix, and it is the inverse matrix for the one formed by $g_{ij}$. (Warning, we do not have $g^{ij} = \frac{1}{g_{ij}$, we should take the matrix inverse.)

That's it. Hence, we write $$ (df, d\varphi)_g = g^*( \sum_i \frac{\d f}{d u_i} d u_i, \sum_j \frac{\d \varphi}{d u_j} d u_j) = \sum_{i,j } \frac{\d f}{d u_i}\frac{\d \varphi}{d u_j} g^* (d u_i, d u_j) =\sum_{i,j } \frac{\d f}{d u_i}\frac{\d \varphi}{d u_j} g^{ij} $$

math121b/02-07.txt · Last modified: 2020/02/07 01:36 by pzhou