User Tools

Site Tools


math121b:02-10

This is an old revision of the document!


2020-02-10, Monday

\gdef\div{\text{div}} \gdef\vol{\text{Vol}} We first finish up the derivation of Laplacian last time. See the lecture note 2020-02-07, Friday. Then, we introduce two vector operations, divergence and curl. Finally, we give a summary of the mathematical treatment of tensor analysis.

Then, we will follow Boas 10.9 and 10.10, to introduce the physic-engineer notations: the nabla operator \gdef\b{\mathbf} \b \nabla.

Differential of a function is a 1-form (covector field)

In Cartesian coordinate, the differential of a function ff is $$ df = \sum_i \frac{\df }{\d x_i} dx_i. $$

In general coordinate (u1,,un)(u_1, \cdots, u_n), the differential of a function ff is df=ifuidui. df = \sum_i \frac{\d f }{\d u_i} d u_i.

You can specify the differential of a function directly: dfdf at a point pRnp \in \R^n is a linear function on TpRnT_p \R^n, df(p)(TpRn)df(p) \in (T_p \R^n)^*. It does the following df(p):vpvp(f),vpTpRn df(p) : \b v_p \mapsto \b v_p(f), \quad \b v_p \in T_p \R^n where vp(f)\b v_p(f) is the directional derivative of ff along vp\b v_p.

Gradient of a function (is a vector field)

In Cartesian coordinate, the gradient of a function is  grad f=ifxixi. \gdef\grad{\text{ grad } } \grad f = \sum_i \frac{\d f}{\d x_i} \frac{\d }{\d x_i}.

In general coordinate, the gradient of a function is more complicated  grad f=i,jgijfuiuj, \grad f = \sum_{i,j} g^{ij} \frac{\d f}{\d u_i} \frac{\d }{\d u_j}, where gijg^{ij} is the entry of the inverse matrix of the matrix [gkl][g_{kl}]. And it just happens that, for Cartesian coordinate, gij=δijg^{ij} = \delta_{ij}.

Note that gijg_{ij} and gijg^{ij} depends on the coordinate system. gij=g(ui,uj),gij=g(dui,duj). g_{ij} = g(\frac{\d}{\d u_i}, \frac{\d}{\d u_j}), \quad g^{ij} = g^*(d u_i, d u_j).

Notation f= grad f. \nabla f = \grad f.

Divergence of a Vector field (is a function)

Let Rn\R^n be the flat space, with standard coordinates (x1,,xn)(x_1, \cdots, x_n).

Let V\b V be a vector field on Rn\R^n, that is, for each point pRnp \in \R^n, we specify a tangent vector V(p)=i=1nVi(p)iTpRn. \b V(p) = \sum_{i=1}^n V^i(p) \d_i \in T_p \R^n. We require that Vi(p)V^i(p) varies smoothly with respect to pp.

The divergence of V\b V is a function on Rn\R^n, div(V)=i=1ni(Vi) \div(\b V) = \sum_{i=1}^n \d_i( V^i ) recall that ViV^i is a function on Rn\R^n, and i\d_i is taking the partial derivative with respect to xix_i.

Notation V=div(V). \nabla \cdot \b V = \div (\b V).

What does divergence mean? Geometrically, it measure the relative change-rate of the volume of an infinitesimal cube situated at point pp. Suppose Φt:RnRn\Phi^t: \R^n \to \R^n is the flow generated by V\b V (every point moves as dictated by V\b V). And let C=C(p,ϵ)C= C(p, \epsilon) be a cube of side-length ϵ\epsilon, center at pp. Then, we have the geometrical interpretation as div(V)=limϵ01Vol(C)dVol(Φt(C))dtt=0. \div(\b V) = \lim_{\epsilon \to 0} \frac{1}{\vol(C)} \frac{d \vol(\Phi^t(C))}{dt} \vert_{t=0}. That is why, if SRnS \subset \R^n is an open domain, we can compute the change-rate of the volume of SS by dVol(Φt(S))dtt=0=Sdiv(V)(x)dVol(x). \frac{d \vol(\Phi^t(S)) } {dt}\vert_{t=0} = \int_{S} \div(\b V)(\b x) \, d \vol(\b x).

In curvilinear coordinate. The formula for computing the divergence is the following, suppose V=iViui\b V = \sum_i V^i \frac{\d }{\d u_i}, then div(V)=i=1n1g(gVi)ui \div (\b V) = \sum_{i=1}^n \frac{1}{\sqrt{|g|}} \frac{\d (\sqrt{|g|} V^i)}{\d u_i}

Exercise: prove that for any compactly supported function φ\varphi 1), we have Rn(V)(u)f(u)g(u)du1dun=Rng(V,f(u))g(u)du1dun \int_{\R^n} (\nabla \cdot \b V)(u) f(u) \sqrt{|g|(u)} du_1\cdots d u_n = \int_{\R^n} g(\b V, \nabla f(u)) \sqrt{|g|(u)} du_1\cdots d u_n

1)
a compactly supported function on Rn\R^n is a function that vanishes outside a sufficently large ball.
math121b/02-10.1581321315.txt.gz · Last modified: 2020/02/09 23:55 by pzhou