Today, we talk about Laplacian operator.

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 absolute value of the determinant of the matrix. Then, our formula for the Laplacian operator is (where $\d_i = \d / \d u_i$)

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

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 $g^{ii} = 1 / g_{ii}$.

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

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