発散・勾配・ラプラシアン

ここから，各微分作用素を一般曲線座標系で表していく。まず，流束F ^{の発散は定義より，}

∇ ·F = lim

∆V→0

1

∆V

∫

∂∆V

F ·dS (A.8)

微小体積要素を図A.2のような平行六面体とすると，式(A.4)^{より面積分は，}

∫

∂∆V

F ·dS =

∑3 i=1

{

F ·(aj×ak)dξ^jdξ^k+− F ·(aj ×ak)dξ^jdξ^k−}

(A.9) であり，さらに式(A.7)^{を用いると，}

1

∆V

∫

∂∆V

F ·dS= 1 J

∑3 i=1

{F ·(aj×ak)dξ^jdξ^k+− F ·(aj ×ak)dξ^jdξ^k− dξⁱdξ^jdξ^k

}

= 1 J

∑3 i=1

{

F ·(aj×ak)|⁺−F ·(aj ×ak)|⁻ dξⁱ

}

(A.10) となる。極限∆V →0^{をとると，発散は，}

∇ ·F = 1 J

∑3 i=1

∂

∂ξⁱ {F ·(aj ×ak)} (A.11) で与えられる。これは発散の保存形表示である。勾配の導出のために発散の非保存形を導出しておく。流 束F が場所に依存せず，任意の定数cであると仮定する。このとき，

∇ ·F = 1 J

∑3 i=1

∂

∂ξⁱ{c·(aj ×ak)}= 0 (A.12) となるはずである。cは任意であるので，上式が成り立つためには，

∑3 i=1

∂

∂ξⁱ(aj×ak) = 0 (A.13)

でなければならない。これはメトリック恒等式と呼ばれる。この関係が成り立てば，式(A.11)^は，

∇ ·F = 1 J

∑3 i=1

(aj×ak)·∂F

∂ξⁱ (A.14)

と書くことができる。これが発散の非保存形である。

次に一般曲線座標系における勾配を導出する。流束F をデカルト座標系の単位基底ベクトルxˆ^，yˆ^，zˆ を使って，F =Fxxˆ+Fyyˆ+Fzzˆ^{と書き，式}(A.14)^{を展開すると，}

∂Fx

∂x +∂Fy

∂y + ∂Fz

∂z =∇ ·F = 1 J

∑3 i=1

(aj ×ak)· (∂Fx

∂ξⁱxˆ+∂Fy

∂ξⁱyˆ+ ∂Fz

∂ξⁱzˆ )

(A.15) となり，左辺と右辺の各項を比べると，

∂Fx

∂x =∇Fx·xˆ= 1 J

∑3 i=1

(aj×ak)·∂Fx

∂ξⁱxˆ (A.16a)

∂Fy

∂y =∇Fy·yˆ= 1 J

∑3 i=1

(aj ×ak)· ∂Fy

∂ξⁱyˆ (A.16b)

∂Fz

∂z =∇Fz·zˆ= 1 J

∑3 i=1

(aj ×ak)·∂Fz

∂ξⁱzˆ (A.16c)

となる。これらの関係からスカラー関数f^{の勾配は，}

∇f = 1 J

∑3 i=1

(aj×ak)∂f

∂ξⁱ (A.17)

と表せる。さらに式(A.13)の関係を使うと，勾配の保存形は次のように導ける。

∇f = 1 J

∑3 i=1

∂

∂ξⁱ{(aj×ak)f} (A.18)

式(A.11)^，(A.14)^，(A.17)^，(A.18)は，発散と勾配が共変基底ベクトルによって表されるということを 意味している。つまり，陽に与えられる変換関数x =X(ξ, η, ζ)の導関数によって，これらの微分作用 素を構成できるということである。

勾配を用いると，反変基底ベクトルを共変基底ベクトルで表すことができる。式(A.17) ^において f =ξ^{iとおくと，}

∇ξⁱ= 1 J

∑3 l=1

(aj×ak)∂ξⁱ

∂ξ^l (A.19)

であり，∂ξⁱ/∂ξ^l =δilと式(A.2)^より，

∇ξⁱ=aⁱ= 1

J(aj×ak) (A.20)

もしくは，

Jaⁱ=aj ×ak (A.21)

とも書ける。Neumann境界条件を与えるときに境界に垂直な単位法線ベクトルが必要になる。単位法線 ベクトルˆnⁱは反変基底ベクトルaⁱをそのノルムで割ればよく，

ˆ nⁱ= |J|

J

aj ×ak

∥aj ×ak∥ (A.22)

で求めることができる。式(A.21)を用いると，発散と勾配は反変基底ベクトルでも表すことができ，そ れぞれ式(A.11)^と(A.17)^より，

∇ ·F = 1 J

∑3 i=1

∂

∂ξⁱ(Jaⁱ·F) (A.23)

∇f = 1 J

∑3 i=1

Jaⁱ∂f

∂ξⁱ (A.24)

となる。

最後に一般曲線座標系におけるラプラシアンは上の二つの式より，

∇²f =∇ · ∇f = 1 J

∑3 i=1

∂

∂ξⁱ



aⁱ·

∑3 j=1

Ja^j ∂f

∂ξ^j



 (A.25)

と表すことができる。

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