Movatterモバイル変換

[0]ホーム
URL:

Jump to content
WikipediaThe Free Encyclopedia
Search

Divergence theorem

From Wikipedia, the free encyclopedia
(Redirected fromGauss's theorem)
Theorem in calculus which relates the flux of closed surfaces to divergence over their volume
"Gauss's theorem" redirects here. For the theorem concerning the electric field, seeGauss's law.
"Ostrogradsky's theorem" redirects here. For the theorem in mechanics, seeOstrogradsky instability.
Part of a series of articles about
Calculus
∫abf′(t)dt=f(b)−f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Invector calculus, thedivergence theorem, also known asGauss's theorem orOstrogradsky's theorem,[1] is atheorem relating theflux of avector field through a closedsurface to thedivergence of the field in the volume enclosed.

More precisely, the divergence theorem states that thesurface integral of a vector field over a closed surface, which is called the "flux" through the surface, is equal to thevolume integral of the divergence over the region enclosed by the surface. Intuitively, it states that "the sum of all sources of the field in a region (with sinks regarded as negative sources) gives the net flux out of the region".

The divergence theorem is an important result for the mathematics ofphysics andengineering, particularly inelectrostatics andfluid dynamics. In these fields, it is usually applied in three dimensions. However, itgeneralizes to any number of dimensions. In one dimension, it is equivalent to thefundamental theorem of calculus. In two dimensions, it is equivalent toGreen's theorem.

Explanation using liquid flow

[edit]
See also:Sources and sinks

Vector fields are often illustrated using the example of thevelocity field of afluid, such as a gas or liquid. A moving liquid has a velocity—a speed and a direction—at each point, which can be represented by avector, so that the velocity of the liquid at any moment forms a vector field. Consider an imaginary closed surfaceS inside a body of liquid, enclosing a volume of liquid. Theflux of liquid out of the volume at any time is equal to the volume rate of fluid crossing this surface, i.e., thesurface integral of the velocity over the surface.

Since liquids are incompressible, the amount of liquid inside a closed volume is constant; if there are nosources or sinks inside the volume then the flux of liquid out ofS is zero. If the liquid is moving, it may flow into the volume at some points on the surfaceS and out of the volume at other points, but the amounts flowing in and out at any moment are equal, so thenet flux of liquid out of the volume is zero.

However if asource of liquid is inside the closed surface, such as a pipe through which liquid is introduced, the additional liquid will exert pressure on the surrounding liquid, causing an outward flow in all directions. This will cause a net outward flow through the surfaceS. The flux outward throughS equals the volume rate of flow of fluid intoS from the pipe. Similarly if there is asink or drain insideS, such as a pipe which drains the liquid off, the external pressure of the liquid will cause a velocity throughout the liquid directed inward toward the location of the drain. The volume rate of flow of liquid inward through the surfaceS equals the rate of liquid removed by the sink.

If there are multiple sources and sinks of liquid insideS, the flux through the surface can be calculated by adding up the volume rate of liquid added by the sources and subtracting the rate of liquid drained off by the sinks. The volume rate of flow of liquid through a source or sink (with the flow through a sink given a negative sign) is equal to thedivergence of the velocity field at the pipe mouth, so adding up (integrating) the divergence of the liquid throughout the volume enclosed byS equals the volume rate of flux throughS. This is the divergence theorem.[2]

The divergence theorem is employed in anyconservation law which states that the total volume of all sinks and sources, that is the volume integral of the divergence, is equal to the net flow across the volume's boundary.[3]

Mathematical statement

[edit]
A regionV bounded by the surfaceS=∂V{\displaystyle S=\partial V} with the surface normaln

SupposeV is asubset ofRn{\displaystyle \mathbb {R} ^{n}} (in the case ofn = 3,V represents a volume inthree-dimensional space) which iscompact and has apiecewisesmooth boundaryS (also indicated with∂V=S{\displaystyle \partial V=S}). IfF is a continuously differentiable vector field defined on aneighborhood ofV, then:[4][5]

∭V(∇⋅F)dV={\displaystyle \iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(F⋅n^)dS.{\displaystyle (\mathbf {F} \cdot \mathbf {\hat {n}} )\,\mathrm {d} S.}

The left side is avolume integral over the volumeV, and the right side is thesurface integral over the boundary of the volumeV. The closed, measurable set∂V{\displaystyle \partial V} is oriented by outward-pointingnormals, andn^{\displaystyle \mathbf {\hat {n}} } is the outward pointing unit normal at almost each point on the boundary∂V{\displaystyle \partial V}. (dS{\displaystyle \mathrm {d} \mathbf {S} } may be used as a shorthand forndS{\displaystyle \mathbf {n} \mathrm {d} S}.) In terms of the intuitive description above, the left-hand side of the equation represents the total of the sources in the volumeV, and the right-hand side represents the total flow across the boundaryS.

Informal derivation

[edit]

The divergence theorem follows from the fact that if a volumeV is partitioned into separate parts, theflux out of the original volume is equal to the algebraic sum of the flux out of each component volume.[6][7] This is true despite the fact that the new subvolumes have surfaces that were not part of the original volume's surface, because these surfaces are just partitions between two of the subvolumes and the flux through them just passes from one volume to the other and so cancels out when the flux out of the subvolumes is summed.

A volume divided into two subvolumes. At right the two subvolumes are separated to show the flux out of the different surfaces.

See the diagram. A closed, bounded volumeV is divided into two volumesV1 andV2 by a surfaceS3(green). The fluxΦ(Vi) out of each component regionVi is equal to the sum of the flux through its two faces, so the sum of the flux out of the two parts is

Φ(V1)+Φ(V2)=Φ1+Φ31+Φ2+Φ32{\displaystyle \Phi (V_{\text{1}})+\Phi (V_{\text{2}})=\Phi _{\text{1}}+\Phi _{\text{31}}+\Phi _{\text{2}}+\Phi _{\text{32}}}

whereΦ1 andΦ2 are the flux out of surfacesS1 andS2,Φ31 is the flux throughS3 out of volume 1, andΦ32 is the flux throughS3 out of volume 2. The point is that surfaceS3 is part of the surface of both volumes. The "outward" direction of thenormal vectorn^{\displaystyle \mathbf {\hat {n}} } is opposite for each volume, so the flux out of one throughS3 is equal to the negative of the flux out of the other so these two fluxes cancel in the sum.

Φ31=∬S3F⋅n^dS=−∬S3F⋅(−n^)dS=−Φ32{\displaystyle \Phi _{\text{31}}=\iint _{S_{3}}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=-\iint _{S_{3}}\mathbf {F} \cdot (-\mathbf {\hat {n}} )\;\mathrm {d} S=-\Phi _{\text{32}}}

Therefore:

Φ(V1)+Φ(V2)=Φ1+Φ2{\displaystyle \Phi (V_{\text{1}})+\Phi (V_{\text{2}})=\Phi _{\text{1}}+\Phi _{\text{2}}}

Since the union of surfacesS1 andS2 isS

Φ(V1)+Φ(V2)=Φ(V){\displaystyle \Phi (V_{\text{1}})+\Phi (V_{\text{2}})=\Phi (V)}
The volume can be divided into any number of subvolumes and the flux out ofV is equal to the sum of the flux out of each subvolume, because the flux through thegreen surfaces cancels out in the sum. In (b) the volumes are shown separated slightly, illustrating that each green partition is part of the boundary of two adjacent volumes

This principle applies to a volume divided into any number of parts, as shown in the diagram.[7] Since the integral over each internal partition(green surfaces) appears with opposite signs in the flux of the two adjacent volumes they cancel out, and the only contribution to the flux is the integral over the external surfaces(grey). Since the external surfaces of all the component volumes equal the original surface.

Φ(V)=∑Vi⊂VΦ(Vi){\displaystyle \Phi (V)=\sum _{V_{\text{i}}\subset V}\Phi (V_{\text{i}})}
As the volume is subdivided into smaller parts, the ratio of the fluxΦ(Vi){\displaystyle \Phi (V_{\text{i}})} out of each volume to the volume|Vi|{\displaystyle |V_{\text{i}}|} approachesdiv⁡F{\displaystyle \operatorname {div} \mathbf {F} }

The fluxΦ out of each volume is the surface integral of the vector fieldF(x) over the surface

∬S(V)F⋅n^dS=∑Vi⊂V∬S(Vi)F⋅n^dS{\displaystyle \iint _{S(V)}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=\sum _{V_{\text{i}}\subset V}\iint _{S(V_{\text{i}})}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S}

The goal is to divide the original volume into infinitely many infinitesimal volumes. As the volume is divided into smaller and smaller parts, the surface integral on the right, the flux out of each subvolume, approaches zero because the surface areaS(Vi) approaches zero. However, from the definition ofdivergence, the ratio of flux to volume,Φ(Vi)|Vi|=1|Vi|∬S(Vi)F⋅n^dS{\displaystyle {\frac {\Phi (V_{\text{i}})}{|V_{\text{i}}|}}={\frac {1}{|V_{\text{i}}|}}\iint _{S(V_{\text{i}})}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S}, the part in parentheses below, does not in general vanish but approaches thedivergencedivF as the volume approaches zero.[7]

∬S(V)F⋅n^dS=∑Vi⊂V(1|Vi|∬S(Vi)F⋅n^dS)|Vi|{\displaystyle \iint _{S(V)}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=\sum _{V_{\text{i}}\subset V}\left({\frac {1}{|V_{\text{i}}|}}\iint _{S(V_{\text{i}})}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S\right)|V_{\text{i}}|}

As long as the vector fieldF(x) has continuous derivatives, the sum above holds even in thelimit when the volume is divided into infinitely small increments

∬S(V)F⋅n^dS=lim|Vi|→0∑Vi⊂V(1|Vi|∬S(Vi)F⋅n^dS)|Vi|{\displaystyle \iint _{S(V)}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=\lim _{|V_{\text{i}}|\to 0}\sum _{V_{\text{i}}\subset V}\left({\frac {1}{|V_{\text{i}}|}}\iint _{S(V_{\text{i}})}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S\right)|V_{\text{i}}|}

As|Vi|{\displaystyle |V_{\text{i}}|} approaches zero volume, it becomes the infinitesimaldV, the part in parentheses becomes the divergence, and the sum becomes avolume integral overV

∬S(V)F⋅n^dS=∭Vdiv⁡FdV{\displaystyle \;\iint _{S(V)}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=\iiint _{V}\operatorname {div} \mathbf {F} \;\mathrm {d} V\;}

Since this derivation is coordinate free, it shows that the divergence does not depend on the coordinates used.

Proofs

[edit]

For bounded open subsets of Euclidean space

[edit]

We are going to prove the following:[citation needed]

Theorem—LetΩ⊂Rn{\displaystyle \Omega \subset \mathbb {R} ^{n}} be open and bounded withC1{\displaystyle C^{1}} boundary. Ifu{\displaystyle u} isC1{\displaystyle C^{1}} on an open neighborhoodO{\displaystyle O} ofΩ¯{\displaystyle {\overline {\Omega }}}, that is,u∈C1(O){\displaystyle u\in C^{1}(O)}, then for eachi∈{1,…,n}{\displaystyle i\in \{1,\dots ,n\}},∫ΩuxidV=∫∂ΩuνidS,{\displaystyle \int _{\Omega }u_{x_{i}}\,dV=\int _{\partial \Omega }u\nu _{i}\,dS,}whereν:∂Ω→Rn{\displaystyle \nu :\partial \Omega \to \mathbb {R} ^{n}} is the outward pointing unit normal vector to∂Ω{\displaystyle \partial \Omega }.Equivalently,∫Ω∇udV=∫∂ΩuνdS.{\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\partial \Omega }u\nu \,dS.}

Proof of Theorem.[8]

  1. The first step is to reduce to the case whereu∈Cc1(Rn){\displaystyle u\in C_{c}^{1}(\mathbb {R} ^{n})}. Pickϕ∈Cc∞(O){\displaystyle \phi \in C_{c}^{\infty }(O)} such thatϕ=1{\displaystyle \phi =1} onΩ¯{\displaystyle {\overline {\Omega }}}. Note thatϕu∈Cc1(O)⊂Cc1(Rn){\displaystyle \phi u\in C_{c}^{1}(O)\subset C_{c}^{1}(\mathbb {R} ^{n})} andϕu=u{\displaystyle \phi u=u} onΩ¯{\displaystyle {\overline {\Omega }}}. Hence it suffices to prove the theorem forϕu{\displaystyle \phi u}. Hence we may assume thatu∈Cc1(Rn){\displaystyle u\in C_{c}^{1}(\mathbb {R} ^{n})}.
  2. Letx0∈∂Ω{\displaystyle x_{0}\in \partial \Omega } be arbitrary. The assumption thatΩ¯{\displaystyle {\overline {\Omega }}} hasC1{\displaystyle C^{1}} boundary means that there is an open neighborhoodU{\displaystyle U} ofx0{\displaystyle x_{0}} inRn{\displaystyle \mathbb {R} ^{n}} such that∂Ω∩U{\displaystyle \partial \Omega \cap U} is the graph of aC1{\displaystyle C^{1}} function withΩ∩U{\displaystyle \Omega \cap U} lying on one side of this graph. More precisely, this means that after a translation and rotation ofΩ{\displaystyle \Omega }, there arer>0{\displaystyle r>0} andh>0{\displaystyle h>0} and aC1{\displaystyle C^{1}} functiong:Rn−1→R{\displaystyle g:\mathbb {R} ^{n-1}\to \mathbb {R} }, such that with the notation

    x′=(x1,…,xn−1),{\displaystyle x'=(x_{1},\dots ,x_{n-1}),}it holds thatU={x∈Rn:|x′|<r and |xn−g(x′)|<h}{\displaystyle U=\{x\in \mathbb {R} ^{n}:|x'|<r{\text{ and }}|x_{n}-g(x')|<h\}}and forx∈U{\displaystyle x\in U},xn=g(x′)⟹x∈∂Ω,−h<xn−g(x′)<0⟹x∈Ω,0<xn−g(x′)<h⟹x∉Ω.{\displaystyle {\begin{aligned}x_{n}=g(x')&\implies x\in \partial \Omega ,\\-h<x_{n}-g(x')<0&\implies x\in \Omega ,\\0<x_{n}-g(x')<h&\implies x\notin \Omega .\\\end{aligned}}}

    Since∂Ω{\displaystyle \partial \Omega } is compact, we can cover∂Ω{\displaystyle \partial \Omega } with finitely many neighborhoodsU1,…,UN{\displaystyle U_{1},\dots ,U_{N}} of the above form. Note that{Ω,U1,…,UN}{\displaystyle \{\Omega ,U_{1},\dots ,U_{N}\}} is an open cover ofΩ¯=Ω∪∂Ω{\displaystyle {\overline {\Omega }}=\Omega \cup \partial \Omega }. By using aC∞{\displaystyle C^{\infty }}partition of unity subordinate to this cover, it suffices to prove the theorem in the case where eitheru{\displaystyle u} has compactsupport inΩ{\displaystyle \Omega } oru{\displaystyle u} has compact support in someUj{\displaystyle U_{j}}. Ifu{\displaystyle u} has compact support inΩ{\displaystyle \Omega }, then for alli∈{1,…,n}{\displaystyle i\in \{1,\dots ,n\}},∫ΩuxidV=∫RnuxidV=∫Rn−1∫−∞∞uxi(x)dxidx′=0{\displaystyle \int _{\Omega }u_{x_{i}}\,dV=\int _{\mathbb {R} ^{n}}u_{x_{i}}\,dV=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{\infty }u_{x_{i}}(x)\,dx_{i}\,dx'=0} by the fundamental theorem of calculus, and∫∂ΩuνidS=0{\displaystyle \int _{\partial \Omega }u\nu _{i}\,dS=0} sinceu{\displaystyle u} vanishes on a neighborhood of∂Ω{\displaystyle \partial \Omega }. Thus the theorem holds foru{\displaystyle u} with compact support inΩ{\displaystyle \Omega }. Thus we have reduced to the case whereu{\displaystyle u} has compact support in someUj{\displaystyle U_{j}}.
  3. So assumeu{\displaystyle u} has compact support in someUj{\displaystyle U_{j}}. The last step now is to show that the theorem is true by direct computation. Change notation toU=Uj{\displaystyle U=U_{j}}, and bring in the notation from (2) used to describeU{\displaystyle U}. Note that this means that we have rotated and translatedΩ{\displaystyle \Omega }. This is a valid reduction since the theorem is invariant under rotations and translations of coordinates. Sinceu(x)=0{\displaystyle u(x)=0} for|x′|≥r{\displaystyle |x'|\geq r} and for|xn−g(x′)|≥h{\displaystyle |x_{n}-g(x')|\geq h}, we have for eachi∈{1,…,n}{\displaystyle i\in \{1,\dots ,n\}} that∫ΩuxidV=∫|x′|<r∫g(x′)−hg(x′)uxi(x′,xn)dxndx′=∫Rn−1∫−∞g(x′)uxi(x′,xn)dxndx′.{\displaystyle {\begin{aligned}\int _{\Omega }u_{x_{i}}\,dV&=\int _{|x'|<r}\int _{g(x')-h}^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'\\&=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'.\end{aligned}}}Fori=n{\displaystyle i=n} we have by the fundamental theorem of calculus that∫Rn−1∫−∞g(x′)uxn(x′,xn)dxndx′=∫Rn−1u(x′,g(x′))dx′.{\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{n}}(x',x_{n})\,dx_{n}\,dx'=\int _{\mathbb {R} ^{n-1}}u(x',g(x'))\,dx'.}Now fixi∈{1,…,n−1}{\displaystyle i\in \{1,\dots ,n-1\}}. Note that∫Rn−1∫−∞g(x′)uxi(x′,xn)dxndx′=∫Rn−1∫−∞0uxi(x′,g(x′)+s)dsdx′{\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}u_{x_{i}}(x',g(x')+s)\,ds\,dx'}Definev:Rn→R{\displaystyle v:\mathbb {R} ^{n}\to \mathbb {R} } byv(x′,s)=u(x′,g(x′)+s){\displaystyle v(x',s)=u(x',g(x')+s)}. By the chain rule,vxi(x′,s)=uxi(x′,g(x′)+s)+uxn(x′,g(x′)+s)gxi(x′).{\displaystyle v_{x_{i}}(x',s)=u_{x_{i}}(x',g(x')+s)+u_{x_{n}}(x',g(x')+s)g_{x_{i}}(x').}But sincev{\displaystyle v} has compact support, we can integrate outdxi{\displaystyle dx_{i}} first to deduce that∫Rn−1∫−∞0vxi(x′,s)dsdx′=0.{\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}v_{x_{i}}(x',s)\,ds\,dx'=0.}Thus∫Rn−1∫−∞0uxi(x′,g(x′)+s)dsdx′=∫Rn−1∫−∞0−uxn(x′,g(x′)+s)gxi(x′)dsdx′=∫Rn−1−u(x′,g(x′))gxi(x′)dx′.{\displaystyle {\begin{aligned}\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}u_{x_{i}}(x',g(x')+s)\,ds\,dx'&=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}-u_{x_{n}}(x',g(x')+s)g_{x_{i}}(x')\,ds\,dx'\\&=\int _{\mathbb {R} ^{n-1}}-u(x',g(x'))g_{x_{i}}(x')\,dx'.\end{aligned}}}In summary, with∇u=(ux1,…,uxn){\displaystyle \nabla u=(u_{x_{1}},\dots ,u_{x_{n}})} we have∫Ω∇udV=∫Rn−1∫−∞g(x′)∇udV=∫Rn−1u(x′,g(x′))(−∇g(x′),1)dx′.{\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}\nabla u\,dV=\int _{\mathbb {R} ^{n-1}}u(x',g(x'))(-\nabla g(x'),1)\,dx'.}Recall that the outward unit normal to the graphΓ{\displaystyle \Gamma } ofg{\displaystyle g} at a point(x′,g(x′))∈Γ{\displaystyle (x',g(x'))\in \Gamma } isν(x′,g(x′))=11+|∇g(x′)|2(−∇g(x′),1){\displaystyle \nu (x',g(x'))={\frac {1}{\sqrt {1+|\nabla g(x')|^{2}}}}(-\nabla g(x'),1)} and that the surface elementdS{\displaystyle dS} is given bydS=1+|∇g(x′)|2dx′{\textstyle dS={\sqrt {1+|\nabla g(x')|^{2}}}\,dx'}. Thus∫Ω∇udV=∫∂ΩuνdS.{\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\partial \Omega }u\nu \,dS.}This completes the proof.

For compact Riemannian manifolds with boundary

[edit]

We are going to prove the following:[citation needed]

Theorem—LetΩ¯{\displaystyle {\overline {\Omega }}} be aC2{\displaystyle C^{2}} compact manifold with boundary withC1{\displaystyle C^{1}} metric tensorg{\displaystyle g}. LetΩ{\displaystyle \Omega } denote the manifold interior ofΩ¯{\displaystyle {\overline {\Omega }}} and let∂Ω{\displaystyle \partial \Omega } denote the manifold boundary ofΩ¯{\displaystyle {\overline {\Omega }}}. Let(⋅,⋅){\displaystyle (\cdot ,\cdot )} denoteL2(Ω¯){\displaystyle L^{2}({\overline {\Omega }})} inner products of functions and⟨⋅,⋅⟩{\displaystyle \langle \cdot ,\cdot \rangle } denote inner products of vectors. Supposeu∈C1(Ω¯,R){\displaystyle u\in C^{1}({\overline {\Omega }},\mathbb {R} )} andX{\displaystyle X} is aC1{\displaystyle C^{1}} vector field onΩ¯{\displaystyle {\overline {\Omega }}}. Then(grad⁡u,X)=−(u,div⁡X)+∫∂Ωu⟨X,N⟩dS,{\displaystyle (\operatorname {grad} u,X)=-(u,\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS,}whereN{\displaystyle N} is the outward-pointing unit normal vector to∂Ω{\displaystyle \partial \Omega }.

Proof of Theorem.[9]We use the Einstein summation convention. By using a partition of unity, we may assume thatu{\displaystyle u} andX{\displaystyle X} have compact support in a coordinate patchO⊂Ω¯{\displaystyle O\subset {\overline {\Omega }}}. First consider the case where the patch is disjoint from∂Ω{\displaystyle \partial \Omega }. ThenO{\displaystyle O} is identified with an open subset ofRn{\displaystyle \mathbb {R} ^{n}} and integration by parts produces no boundary terms:(grad⁡u,X)=∫O⟨grad⁡u,X⟩gdx=∫O∂juXjgdx=−∫Ou∂j(gXj)dx=−∫Ou1g∂j(gXj)gdx=(u,−1g∂j(gXj))=(u,−div⁡X).{\displaystyle {\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{O}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=-\int _{O}u\partial _{j}({\sqrt {g}}X^{j})\,dx\\&=-\int _{O}u{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}){\sqrt {g}}\,dx\\&=(u,-{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}))\\&=(u,-\operatorname {div} X).\end{aligned}}}In the last equality we used the Voss-Weyl coordinate formula for the divergence, although the preceding identity could be used to define−div{\displaystyle -\operatorname {div} } as the formal adjoint ofgrad{\displaystyle \operatorname {grad} }. Now supposeO{\displaystyle O} intersects∂Ω{\displaystyle \partial \Omega }. ThenO{\displaystyle O} is identified with an open set inR+n={x∈Rn:xn≥0}{\displaystyle \mathbb {R} _{+}^{n}=\{x\in \mathbb {R} ^{n}:x_{n}\geq 0\}}. We zero extendu{\displaystyle u} andX{\displaystyle X} toR+n{\displaystyle \mathbb {R} _{+}^{n}} and perform integration by parts to obtain(grad⁡u,X)=∫O⟨grad⁡u,X⟩gdx=∫R+n∂juXjgdx=(u,−div⁡X)−∫Rn−1u(x′,0)Xn(x′,0)g(x′,0)dx′,{\displaystyle {\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{\mathbb {R} _{+}^{n}}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=(u,-\operatorname {div} X)-\int _{\mathbb {R} ^{n-1}}u(x',0)X^{n}(x',0){\sqrt {g(x',0)}}\,dx',\end{aligned}}}wheredx′=dx1…dxn−1{\displaystyle dx'=dx_{1}\dots dx_{n-1}}.By a variant of thestraightening theorem for vector fields, we may chooseO{\displaystyle O} so that∂∂xn{\displaystyle {\frac {\partial }{\partial x_{n}}}} is the inward unit normal−N{\displaystyle -N} at∂Ω{\displaystyle \partial \Omega }. In this caseg(x′,0)dx′=g∂Ω(x′)dx′=dS{\displaystyle {\sqrt {g(x',0)}}\,dx'={\sqrt {g_{\partial \Omega }(x')}}\,dx'=dS} is the volume element on∂Ω{\displaystyle \partial \Omega } and the above formula reads(grad⁡u,X)=(u,−div⁡X)+∫∂Ωu⟨X,N⟩dS.{\displaystyle (\operatorname {grad} u,X)=(u,-\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS.}This completes the proof.

Corollaries

[edit]

By replacingF in the divergence theorem with specific forms, other useful identities can be derived (cf.vector identities).[10]

∭V[F⋅(∇g)+g(∇⋅F)]dV={\displaystyle \iiint _{V}\left[\mathbf {F} \cdot \left(\nabla g\right)+g\left(\nabla \cdot \mathbf {F} \right)\right]\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}gF⋅ndS.{\displaystyle g\mathbf {F} \cdot \mathbf {n} \mathrm {d} S.}
A special case of this isF=∇f{\displaystyle \mathbf {F} =\nabla f}, in which case the theorem is the basis forGreen's identities.
∭V∇⋅(F×G)dV=∭V[G⋅(∇×F)−F⋅(∇×G)]dV={\displaystyle \iiint _{V}\nabla \cdot \left(\mathbf {F} \times \mathbf {G} \right)\mathrm {d} V=\iiint _{V}\left[\mathbf {G} \cdot \left(\nabla \times \mathbf {F} \right)-\mathbf {F} \cdot \left(\nabla \times \mathbf {G} \right)\right]\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(F×G)⋅ndS.{\displaystyle (\mathbf {F} \times \mathbf {G} )\cdot \mathbf {n} \mathrm {d} S.}
∭V∇(F⋅G)dV=∭V[(∇G)⋅F+(∇F)⋅G]dV={\displaystyle \iiint _{V}\nabla \left(\mathbf {F} \cdot \mathbf {G} \right)\mathrm {d} V=\iiint _{V}\left[\left(\nabla \mathbf {G} \right)\cdot \mathbf {F} +\left(\nabla \mathbf {F} \right)\cdot \mathbf {G} \right]\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(F⋅G)ndS.{\displaystyle (\mathbf {F} \cdot \mathbf {G} )\mathbf {n} \mathrm {d} S.}
∭Vc⋅∇fdV={\displaystyle \iiint _{V}\mathbf {c} \cdot \nabla f\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(cf)⋅ndS−∭Vf(∇⋅c)dV.{\displaystyle (\mathbf {c} f)\cdot \mathbf {n} \mathrm {d} S-\iiint _{V}f(\nabla \cdot \mathbf {c} )\,\mathrm {d} V.}
The last term on the right vanishes for constantc{\displaystyle \mathbf {c} } or any divergence free (solenoidal) vector field, e.g. Incompressible flows without sources or sinks such as phase change or chemical reactions etc. In particular, takingc{\displaystyle \mathbf {c} } to be constant:
∭V∇fdV={\displaystyle \iiint _{V}\nabla f\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}fndS.{\displaystyle f\mathbf {n} \mathrm {d} S.}
∭Vc⋅(∇×F)dV={\displaystyle \iiint _{V}\mathbf {c} \cdot (\nabla \times \mathbf {F} )\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(F×c)⋅ndS.{\displaystyle (\mathbf {F} \times \mathbf {c} )\cdot \mathbf {n} \mathrm {d} S.}
By reordering thetriple product on the right hand side and taking out the constant vector of the integral,
∭V(∇×F)dV⋅c={\displaystyle \iiint _{V}(\nabla \times \mathbf {F} )\,\mathrm {d} V\cdot \mathbf {c} =}\oiintS{\displaystyle \scriptstyle S}(dS×F)⋅c.{\displaystyle (\mathrm {d} \mathbf {S} \times \mathbf {F} )\cdot \mathbf {c} .}
Hence,
∭V(∇×F)dV={\displaystyle \iiint _{V}(\nabla \times \mathbf {F} )\,\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}n×FdS.{\displaystyle \mathbf {n} \times \mathbf {F} \mathrm {d} S.}

Example

[edit]
The vector field corresponding to the example shown. Vectors may point into or out of the sphere.
The divergence theorem can be used to calculate a flux through aclosed surface that fully encloses a volume, like any of the surfaces on the left. It cannot directly be used to calculate the flux through surfaces with boundaries, like those on the right. (Surfaces are blue, boundaries are red.)

Suppose we wish to evaluate

\oiintS{\displaystyle \scriptstyle S}F⋅ndS,{\displaystyle \mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S,}

whereS is theunit sphere defined by

S={(x,y,z)∈R3 : x2+y2+z2=1},{\displaystyle S=\left\{(x,y,z)\in \mathbb {R} ^{3}\ :\ x^{2}+y^{2}+z^{2}=1\right\},}

andF is thevector field

F=2xi+y2j+z2k.{\displaystyle \mathbf {F} =2x\mathbf {i} +y^{2}\mathbf {j} +z^{2}\mathbf {k} .}

The direct computation of this integral is quite difficult, but we can simplify the derivation of the result using the divergence theorem, because the divergence theorem says that the integral is equal to:

∭W(∇⋅F)dV=2∭W(1+y+z)dV=2∭WdV+2∭WydV+2∭WzdV,{\displaystyle \iiint _{W}(\nabla \cdot \mathbf {F} )\,\mathrm {d} V=2\iiint _{W}(1+y+z)\,\mathrm {d} V=2\iiint _{W}\mathrm {d} V+2\iiint _{W}y\,\mathrm {d} V+2\iiint _{W}z\,\mathrm {d} V,}

whereW is theunit ball:

W={(x,y,z)∈R3 : x2+y2+z2≤1}.{\displaystyle W=\left\{(x,y,z)\in \mathbb {R} ^{3}\ :\ x^{2}+y^{2}+z^{2}\leq 1\right\}.}

Since the functiony is positive in one hemisphere ofW and negative in the other, in an equal and opposite way, its total integral overW is zero. The same is true forz:

∭WydV=∭WzdV=0.{\displaystyle \iiint _{W}y\,\mathrm {d} V=\iiint _{W}z\,\mathrm {d} V=0.}

Therefore,

\oiintS{\displaystyle \scriptstyle S}F⋅ndS=2∭WdV=8π3,{\displaystyle \mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S=2\iiint _{W}\,dV={\frac {8\pi }{3}},}

because the unit ballW hasvolume⁠4π/3⁠.

Applications

[edit]

Differential and integral forms of physical laws

[edit]

As a result of the divergence theorem, a host of physical laws can be written in both a differential form (where one quantity is the divergence of another) and an integral form (where the flux of one quantity through a closed surface is equal to another quantity). Three examples areGauss's law (inelectrostatics),Gauss's law for magnetism, andGauss's law for gravity.

Continuity equations

[edit]
Main article:continuity equation

Continuity equations offer more examples of laws with both differential and integral forms, related to each other by the divergence theorem. Influid dynamics,electromagnetism,quantum mechanics,relativity theory, and a number of other fields, there arecontinuity equations that describe the conservation of mass, momentum, energy, probability, or other quantities. Generically, these equations state that the divergence of the flow of the conserved quantity is equal to the distribution ofsources orsinks of that quantity. The divergence theorem states that any such continuity equation can be written in a differential form (in terms of a divergence) and an integral form (in terms of a flux).[12]

Inverse-square laws

[edit]

Anyinverse-square law can instead be written in aGauss's law-type form (with a differential and integral form, as described above). Two examples areGauss's law (in electrostatics), which follows from the inverse-squareCoulomb's law, andGauss's law for gravity, which follows from the inverse-squareNewton's law of universal gravitation. The derivation of the Gauss's law-type equation from the inverse-square formulation or vice versa is exactly the same in both cases; see either of those articles for details.[12]

History

[edit]

Joseph-Louis Lagrange introduced the notion of surface integrals in 1760 and again in more general terms in 1811, in the second edition of hisMécanique Analytique. Lagrange employed surface integrals in his work on fluid mechanics.[13] He discovered the divergence theorem in 1762.[14]

Carl Friedrich Gauss was also using surface integrals while working on the gravitational attraction of an elliptical spheroid in 1813, when he proved special cases of the divergence theorem.[15][13] He proved additional special cases in 1833 and 1839.[16] But it wasMikhail Ostrogradsky, who gave the first proof of the general theorem, in 1826, as part of his investigation of heat flow.[17] Special cases were proven byGeorge Green in 1828 inAn Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism,[18][16]Siméon Denis Poisson in 1824 in a paper on elasticity, andFrédéric Sarrus in 1828 in his work on floating bodies.[19][16]

Worked examples

[edit]

Example 1

[edit]

To verify the planar variant of the divergence theorem for a regionR{\displaystyle R}:

R={(x,y)∈R2 : x2+y2≤1},{\displaystyle R=\left\{(x,y)\in \mathbb {R} ^{2}\ :\ x^{2}+y^{2}\leq 1\right\},}

and the vector field:

F(x,y)=2yi+5xj.{\displaystyle \mathbf {F} (x,y)=2y\mathbf {i} +5x\mathbf {j} .}

The boundary ofR{\displaystyle R} is the unit circle,C{\displaystyle C}, that can be represented parametrically by:

x=cos⁡(s),y=sin⁡(s){\displaystyle x=\cos(s),\quad y=\sin(s)}

such that0≤s≤2π{\displaystyle 0\leq s\leq 2\pi } wheres{\displaystyle s} units is the length arc from the points=0{\displaystyle s=0} to the pointP{\displaystyle P} onC{\displaystyle C}. Then a vector equation ofC{\displaystyle C} is

C(s)=cos⁡(s)i+sin⁡(s)j.{\displaystyle C(s)=\cos(s)\mathbf {i} +\sin(s)\mathbf {j} .}

At a pointP{\displaystyle P} onC{\displaystyle C}:

P=(cos⁡(s),sin⁡(s))⇒F=2sin⁡(s)i+5cos⁡(s)j.{\displaystyle P=(\cos(s),\,\sin(s))\,\Rightarrow \,\mathbf {F} =2\sin(s)\mathbf {i} +5\cos(s)\mathbf {j} .}

Therefore,

∮CF⋅nds=∫02π(2sin⁡(s)i+5cos⁡(s)j)⋅(cos⁡(s)i+sin⁡(s)j)ds=∫02π(2sin⁡(s)cos⁡(s)+5sin⁡(s)cos⁡(s))ds=7∫02πsin⁡(s)cos⁡(s)ds=0.{\displaystyle {\begin{aligned}\oint _{C}\mathbf {F} \cdot \mathbf {n} \,\mathrm {d} s&=\int _{0}^{2\pi }(2\sin(s)\mathbf {i} +5\cos(s)\mathbf {j} )\cdot (\cos(s)\mathbf {i} +\sin(s)\mathbf {j} )\,\mathrm {d} s\\&=\int _{0}^{2\pi }(2\sin(s)\cos(s)+5\sin(s)\cos(s))\,\mathrm {d} s\\&=7\int _{0}^{2\pi }\sin(s)\cos(s)\,\mathrm {d} s\\&=0.\end{aligned}}}

BecauseM=Re(F)=2y{\displaystyle M={\mathfrak {Re}}(\mathbf {F} )=2y}, we can evaluate∂M∂x=0{\displaystyle {\frac {\partial M}{\partial x}}=0}, and becauseN=Im(F)=5x{\displaystyle N={\mathfrak {Im}}(\mathbf {F} )=5x},∂N∂y=0{\displaystyle {\frac {\partial N}{\partial y}}=0}. Thus

∬R∇⋅FdA=∬R(∂M∂x+∂N∂y)dA=0.{\displaystyle \iint _{R}\,\mathbf {\nabla } \cdot \mathbf {F} \,\mathrm {d} A=\iint _{R}\left({\frac {\partial M}{\partial x}}+{\frac {\partial N}{\partial y}}\right)\,\mathrm {d} A=0.}

Example 2

[edit]

Let's say we wanted to evaluate the flux of the followingvector field defined byF=2x2i+2y2j+2z2k{\displaystyle \mathbf {F} =2x^{2}{\textbf {i}}+2y^{2}{\textbf {j}}+2z^{2}{\textbf {k}}} bounded by the following inequalities:

{0≤x≤3},{−2≤y≤2},{0≤z≤2π}{\displaystyle \left\{0\leq x\leq 3\right\},\left\{-2\leq y\leq 2\right\},\left\{0\leq z\leq 2\pi \right\}}

By the divergence theorem,

∭V(∇⋅F)dV={\displaystyle \iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}(F⋅n)dS.{\displaystyle (\mathbf {F} \cdot \mathbf {n} )\,\mathrm {d} S.}

We now need to determine the divergence ofF{\displaystyle {\textbf {F}}}. IfF{\displaystyle \mathbf {F} } is a three-dimensional vector field, then the divergence ofF{\displaystyle {\textbf {F}}} is given by∇⋅F=(∂∂xi+∂∂yj+∂∂zk)⋅F{\textstyle \nabla \cdot {\textbf {F}}=\left({\frac {\partial }{\partial x}}{\textbf {i}}+{\frac {\partial }{\partial y}}{\textbf {j}}+{\frac {\partial }{\partial z}}{\textbf {k}}\right)\cdot {\textbf {F}}}.

Thus, we can set up the following flux integralI={\displaystyle I=}\oiintS{\displaystyle {\scriptstyle S}}F⋅ndS,{\displaystyle \mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S,}as follows:

I=∭V∇⋅FdV=∭V(∂Fx∂x+∂Fy∂y+∂Fz∂z)dV=∭V(4x+4y+4z)dV=∫03∫−22∫02π(4x+4y+4z)dV{\displaystyle {\begin{aligned}I&=\iiint _{V}\nabla \cdot \mathbf {F} \,\mathrm {d} V\\[6pt]&=\iiint _{V}\left({\frac {\partial \mathbf {F_{x}} }{\partial x}}+{\frac {\partial \mathbf {F_{y}} }{\partial y}}+{\frac {\partial \mathbf {F_{z}} }{\partial z}}\right)\mathrm {d} V\\[6pt]&=\iiint _{V}(4x+4y+4z)\,\mathrm {d} V\\[6pt]&=\int _{0}^{3}\int _{-2}^{2}\int _{0}^{2\pi }(4x+4y+4z)\,\mathrm {d} V\end{aligned}}}

Now that we have set up the integral, we can evaluate it.

∫03∫−22∫02π(4x+4y+4z)dV=∫−22∫02π(12y+12z+18)dydz=∫02π24(2z+3)dz=48π(2π+3){\displaystyle {\begin{aligned}\int _{0}^{3}\int _{-2}^{2}\int _{0}^{2\pi }(4x+4y+4z)\,\mathrm {d} V&=\int _{-2}^{2}\int _{0}^{2\pi }(12y+12z+18)\,\mathrm {d} y\,\mathrm {d} z\\[6pt]&=\int _{0}^{2\pi }24(2z+3)\,\mathrm {d} z\\[6pt]&=48\pi (2\pi +3)\end{aligned}}}

Generalizations

[edit]

Multiple dimensions

[edit]

One can use thegeneralised Stokes' theorem to equate then-dimensional volume integral of the divergence of a vector fieldF over a regionU to the(n − 1)-dimensional surface integral ofF over the boundary ofU:

∫⋯∫U⏟n∇⋅FdV=∮⋯∮∂U⏟n−1F⋅ndS{\displaystyle \underbrace {\int \cdots \int _{U}} _{n}\nabla \cdot \mathbf {F} \,\mathrm {d} V=\underbrace {\oint _{}\cdots \oint _{\partial U}} _{n-1}\mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S}

This equation is also known as the divergence theorem.

Whenn = 2, this is equivalent toGreen's theorem.

Whenn = 1, it reduces to thefundamental theorem of calculus, part 2.

Tensor fields

[edit]
Main article:Tensor field

Writing the theorem inEinstein notation:

∭V∂Fi∂xidV={\displaystyle \iiint _{V}{\dfrac {\partial \mathbf {F} _{i}}{\partial x_{i}}}\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}FinidS{\displaystyle \mathbf {F} _{i}n_{i}\,\mathrm {d} S}

suggestively, replacing the vector fieldF with a rank-n tensor fieldT, this can be generalized to:[20]

∭V∂Ti1i2⋯iq⋯in∂xiqdV={\displaystyle \iiint _{V}{\dfrac {\partial T_{i_{1}i_{2}\cdots i_{q}\cdots i_{n}}}{\partial x_{i_{q}}}}\mathrm {d} V=}\oiintS{\displaystyle \scriptstyle S}Ti1i2⋯iq⋯inniqdS.{\displaystyle T_{i_{1}i_{2}\cdots i_{q}\cdots i_{n}}n_{i_{q}}\,\mathrm {d} S.}

where on each side,tensor contraction occurs for at least one index. This form of the theorem is still in 3d, each index takes values 1, 2, and 3. It can be generalized further still to higher (or lower) dimensions (for example to 4dspacetime ingeneral relativity[21]).

See also

[edit]

References

[edit]
  1. ^Katz, Victor J. (1979). "The history of Stokes's theorem".Mathematics Magazine.52 (3):146–156.doi:10.2307/2690275.JSTOR 2690275. reprinted inAnderson, Marlow (2009).Who Gave You the Epsilon?: And Other Tales of Mathematical History. Mathematical Association of America. pp. 78–79.ISBN 978-0-88385-569-0.
  2. ^R. G. Lerner; G. L. Trigg (1994).Encyclopaedia of Physics (2nd ed.). VHC.ISBN 978-3-527-26954-9.
  3. ^Byron, Frederick;Fuller, Robert (1992),Mathematics of Classical and Quantum Physics, Dover Publications, p. 22,ISBN 978-0-486-67164-2
  4. ^Wiley, C. Ray Jr.Advanced Engineering Mathematics, 3rd Ed. McGraw-Hill. pp. 372–373.
  5. ^Kreyszig, Erwin; Kreyszig, Herbert; Norminton, Edward J. (2011).Advanced Engineering Mathematics (10 ed.). John Wiley and Sons. pp. 453–456.ISBN 978-0-470-45836-5.
  6. ^Benford, Frank A. (May 2007)."Notes on Vector Calculus"(PDF).Course materials for Math 105: Multivariable Calculus. Prof. Steven Miller's webpage, Williams College. Retrieved14 March 2022.
  7. ^abcPurcell, Edward M.; David J. Morin (2013).Electricity and Magnetism. Cambridge Univ. Press. pp. 56–58.ISBN 978-1-107-01402-2.
  8. ^Alt, Hans Wilhelm (2016). "Linear Functional Analysis".Universitext. London: Springer London. pp. 259–261,270–272.doi:10.1007/978-1-4471-7280-2.ISBN 978-1-4471-7279-6.ISSN 0172-5939.
  9. ^Taylor, Michael E. (2011). "Partial Differential Equations I".Applied Mathematical Sciences. Vol. 115. New York, NY: Springer New York. pp. 178–179.doi:10.1007/978-1-4419-7055-8.ISBN 978-1-4419-7054-1.ISSN 0066-5452.
  10. ^M. R. Spiegel; S. Lipschutz; D. Spellman (2009).Vector Analysis. Schaum's Outlines (2nd ed.). USA: McGraw Hill.ISBN 978-0-07-161545-7.
  11. ^abMathWorld
  12. ^abC.B. Parker (1994).McGraw Hill Encyclopaedia of Physics (2nd ed.). McGraw Hill.ISBN 978-0-07-051400-3.
  13. ^abKatz, Victor (2009). "Chapter 22: Vector Analysis".A History of Mathematics: An Introduction. Addison-Wesley. pp. 808–9.ISBN 978-0-321-38700-4.
  14. ^In his 1762 paper on sound, Lagrange treats a special case of the divergence theorem: Lagrange (1762) "Nouvelles recherches sur la nature et la propagation du son" (New researches on the nature and propagation of sound),Miscellanea Taurinensia (also known as:Mélanges de Turin ),2: 11 – 172. This article is reprinted as:"Nouvelles recherches sur la nature et la propagation du son" in: J.A. Serret, ed.,Oeuvres de Lagrange, (Paris, France: Gauthier-Villars, 1867), vol. 1, pages 151–316;on pages 263–265, Lagrange transforms triple integrals into double integrals using integration by parts.
  15. ^C. F. Gauss (1813)"Theoria attractionis corporum sphaeroidicorum ellipticorum homogeneorum methodo nova tractata,"Commentationes societatis regiae scientiarium Gottingensis recentiores,2: 355–378; Gauss considered a special case of the theorem; see the 4th, 5th, and 6th pages of his article.
  16. ^abcKatz, Victor (May 1979). "A History of Stokes' Theorem".Mathematics Magazine.52 (3):146–156.doi:10.1080/0025570X.1979.11976770.JSTOR 2690275.
  17. ^Mikhail Ostragradsky presented his proof of the divergence theorem to the Paris Academy in 1826; however, his work was not published by the Academy. He returned to St. Petersburg, Russia, where in 1828–1829 he read the work that he'd done in France, to the St. Petersburg Academy, which published his work in abbreviated form in 1831.
    • His proof of the divergence theorem – "Démonstration d'un théorème du calcul intégral" (Proof of a theorem in integral calculus) – which he had read to the Paris Academy on February 13, 1826, was translated, in 1965, into Russian together with another article by him. See: Юшкевич А.П. (Yushkevich A.P.) and Антропова В.И. (Antropov V.I.) (1965) "Неопубликованные работы М.В. Остроградского" (Unpublished works of MV Ostrogradskii),Историко-математические исследования (Istoriko-Matematicheskie Issledovaniya / Historical-Mathematical Studies),16: 49–96; see the section titled: "Остроградский М.В. Доказательство одной теоремы интегрального исчисления" (Ostrogradskii M. V. Dokazatelstvo odnoy teoremy integralnogo ischislenia / Ostragradsky M.V. Proof of a theorem in integral calculus).
    • M. Ostrogradsky (presented: November 5, 1828; published: 1831)"Première note sur la théorie de la chaleur" (First note on the theory of heat)Mémoires de l'Académie impériale des sciences de St. Pétersbourg, series 6,1: 129–133; for an abbreviated version of his proof of the divergence theorem, see pages 130–131.
    • Victor J. Katz (May1979)"The history of Stokes' theorem,"Archived April 2, 2015, at theWayback MachineMathematics Magazine,52(3): 146–156; for Ostragradsky's proof of the divergence theorem, see pages 147–148.
  18. ^George Green,An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism (Nottingham, England: T. Wheelhouse, 1838). A form of the "divergence theorem" appears onpages 10–12.
  19. ^Other early investigators who used some form of the divergence theorem include:
    • Poisson (presented: February 2, 1824; published: 1826)"Mémoire sur la théorie du magnétisme" (Memoir on the theory of magnetism),Mémoires de l'Académie des sciences de l'Institut de France,5: 247–338; on pages 294–296, Poisson transforms a volume integral (which is used to evaluate a quantity Q) into a surface integral. To make this transformation, Poisson follows the same procedure that is used to prove the divergence theorem.
    • Frédéric Sarrus (1828) "Mémoire sur les oscillations des corps flottans" (Memoir on the oscillations of floating bodies),Annales de mathématiques pures et appliquées (Nismes),19: 185–211.
  20. ^K.F. Riley; M.P. Hobson; S.J. Bence (2010).Mathematical methods for physics and engineering. Cambridge University Press.ISBN 978-0-521-86153-3.
  21. ^see for example:
    J.A. Wheeler; C. Misner; K.S. Thorne (1973).Gravitation. W.H. Freeman & Co. pp. 85–86, §3.5.ISBN 978-0-7167-0344-0., and
    R. Penrose (2007).The Road to Reality. Vintage books.ISBN 978-0-679-77631-4.

External links

[edit]
Precalculus
Limits
Differential calculus
Integral calculus
Vector calculus
Multivariable calculus
Sequences and series
Special functions
and numbers
History of calculus
Lists
Integrals
Miscellaneous topics
Retrieved from "https://en.wikipedia.org/w/index.php?title=Divergence_theorem&oldid=1280148436"
Category:
Hidden categories:
[8]ページ先頭
©2009-2025 Movatter.jp