# Satz von Green

Green's theorem This article is about the theorem in the plane relating double integrals and line integrals. For Green's theorems relating volume integrals involving the Laplacian to surface integrals, see Green's identities. Not to be confused with Green's law for waves approaching a shoreline. Part of a series of articles about Calculus Fundamental theorem Leibniz integral rule Limits of functionsContinuity Mean value theoremRolle's theorem show Differential show Integral show Series hide Vector GradientDivergenceCurlLaplacianDirectional derivativeIdentities Theorems GradientGreen'sStokes'Divergencegeneralized Stokes show Multivariable show Advanced show Specialized show Miscellaneous vte In vector calculus, Green's theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. It is the two-dimensional special case of Stokes' theorem.

Inhalt 1 Satz 2 Proof when D is a simple region 3 Proof for rectifiable Jordan curves 4 Validity under different hypotheses 5 Multiply-connected regions 6 Relationship to Stokes' theorem 7 Relationship to the divergence theorem 8 Area calculation 9 Geschichte 10 Siehe auch 11 Verweise 12 Weiterlesen 13 External links Theorem Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of (x, j) defined on an open region containing D and having continuous partial derivatives there, dann {Anzeigestil {scriptstyle C}} {Anzeigestil (L,dx+M,dy)=int _{D}links({frac {teilweise m}{teilweise x}}-{frac {partial L}{teilweise y}}Rechts)dx,dy} where the path of integration along C is anticlockwise.[1][2] In physics, Green's theorem finds many applications. One is solving two-dimensional flow integrals, stating that the sum of fluid outflowing from a volume is equal to the total outflow summed about an enclosing area. In plane geometry, und besonders, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

Proof when D is a simple region If D is a simple type of region with its boundary consisting of the curves C1, C2, C3, C4, half of Green's theorem can be demonstrated.

The following is a proof of half of the theorem for the simplified area D, a type I region where C1 and C3 are curves connected by vertical lines (possibly of zero length). A similar proof exists for the other half of the theorem when D is a type II region where C2 and C4 are curves connected by horizontal lines (wieder, possibly of zero length). Putting these two parts together, the theorem is thus proven for regions of type III (defined as regions which are both type I and type II). The general case can then be deduced from this special case by decomposing D into a set of type III regions.

If it can be shown that {Anzeigestil gesalbt_{C}L,dx=iint _{D}links(-{frac {partial L}{teilweise y}}Rechts)dA} (1) und {Anzeigestil gesalbt_{C} M,dy=iint _{D}links({frac {teilweise m}{teilweise x}}Rechts)dA} (2) are true, then Green's theorem follows immediately for the region D. We can prove (1) easily for regions of type I, und (2) for regions of type II. Green's theorem then follows for regions of type III.

Assume region D is a type I region and can thus be characterized, as pictured on the right, durch {displaystyle D={(x,j)mid aleq xleq b,g_{1}(x)leq yleq g_{2}(x)}} where g1 and g2 are continuous functions on [a, b]. Compute the double integral in (1): {Anzeigestil {Start{ausgerichtet}iint _{D}{frac {partial L}{teilweise y}},dA&=int _{a}^{b},int _{g_{1}(x)}^{g_{2}(x)}{frac {partial L}{teilweise y}}(x,j),dy,dx\&=int _{a}^{b}links[L(x,g_{2}(x))-L(x,g_{1}(x))Rechts],dx.end{ausgerichtet}}} (3) Now compute the line integral in (1). C can be rewritten as the union of four curves: C1, C2, C3, C4.

With C1, use the parametric equations: x = x, y = g1(x), a ≤ x ≤ b. Dann {Anzeigestil int _{C_{1}}L(x,j),dx=int _{a}^{b}L(x,g_{1}(x)),dx.} With C3, use the parametric equations: x = x, y = g2(x), a ≤ x ≤ b. Dann {Anzeigestil int _{C_{3}}L(x,j),dx=-int _{-C_{3}}L(x,j),dx=-int _{a}^{b}L(x,g_{2}(x)),dx.} The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (anticlockwise). On C2 and C4, x remains constant, meaning {Anzeigestil int _{C_{4}}L(x,j),dx=int _{C_{2}}L(x,j),dx=0.} Deswegen, {Anzeigestil {Start{ausgerichtet}int _{C}L,dx&=int _{C_{1}}L(x,j),dx+int _{C_{2}}L(x,j),dx+int _{C_{3}}L(x,j),dx+int _{C_{4}}L(x,j),dx\&=int _{a}^{b}L(x,g_{1}(x)),dx-int _{a}^{b}L(x,g_{2}(x)),dx.end{ausgerichtet}}} (4) Combining (3) mit (4), wir bekommen (1) for regions of type I. A similar treatment yields (2) for regions of type II. Putting the two together, we get the result for regions of type III.

Proof for rectifiable Jordan curves We are going to prove the following Theorem — Let {Anzeigestil Gamma } be a rectifiable, positively oriented Jordan curve in {Anzeigestil mathbb {R} ^{2}} und lass {Anzeigestil R} denote its inner region. Nehme an, dass {Anzeigestil A,B:{überstreichen {R}}zu mathbb {R} } are continuous functions with the property that {Anzeigestil A} has second partial derivative at every point of {Anzeigestil R} , {Anzeigestil B} has first partial derivative at every point of {Anzeigestil R} and that the functions {displaystyle D_{1}B,D_{2}EIN:Rto mathbb {R} } are Riemann-integrable over {Anzeigestil R} . Dann {Anzeigestil int _{Gamma }(EIN,dx+B,dy)=int _{R}links(D_{1}B(x,j)-D_{2}EIN(x,j)Rechts),d(x,j).} We need the following lemmas whose proofs can be found in:[3] Lemma 1 (Decomposition Lemma) — Assume {Anzeigestil Gamma } is a rectifiable, positively oriented Jordan curve in the plane and let {Anzeigestil R} be its inner region. For every positive real {Anzeigestil-Delta } , Lassen {Anzeigestil {mathematisch {F}}(Delta )} denote the collection of squares in the plane bounded by the lines {displaystyle x=mdelta ,y=mdelta } , wo {Anzeigestil m} runs through the set of integers. Dann, for this {Anzeigestil-Delta } , there exists a decomposition of {Anzeigestil {überstreichen {R}}} into a finite number of non-overlapping subregions in such a manner that Each one of the subregions contained in {Anzeigestil R} , sagen {Anzeigestil R_{1},R_{2},Punkte ,R_{k}} , is a square from {Anzeigestil {mathematisch {F}}(Delta )} . Each one of the remaining subregions, sagen {Anzeigestil R_{k+1},Punkte ,R_{s}} , has as boundary a rectifiable Jordan curve formed by a finite number of arcs of {Anzeigestil Gamma } and parts of the sides of some square from {Anzeigestil {mathematisch {F}}(Delta )} . Each one of the border regions {Anzeigestil R_{k+1},Punkte ,R_{s}} can be enclosed in a square of edge-length {displaystyle 2delta } . Wenn {Anzeigestil Gamma _{ich}} is the positively oriented boundary curve of {Anzeigestil R_{ich}} , dann {displaystyle Gamma =Gamma _{1}+Gamma _{2}+cdots +Gamma _{s}.} Die Nummer {displaystyle s-k} of border regions is no greater than {textstyle 4!links({frac {Lambda }{Delta }}+1Rechts)} , wo {Anzeigestil Lambda } is the length of {Anzeigestil Gamma } .

Lemma 2 — Let {Anzeigestil Gamma } be a rectifiable curve in the plane and let {Anzeigestil Delta _{Gamma }(h)} be the set of points in the plane whose distance from (the range of) {Anzeigestil Gamma } ist höchstens {Anzeigestil h} . The outer Jordan content of this set satisfies {Anzeigestil {überstreichen {c}},,Differenz _{Gamma }(h)leq 2hLambda +pi h^{2}} .

Lemma 3 — Let {Anzeigestil Gamma } be a rectifiable curve in {Anzeigestil mathbb {R} ^{2}} und lass {Anzeigestil f:{Text{range of }}Gamma to mathbb {R} } be a continuous function. Dann {displaystyle leftvert int _{Gamma }f(x,j),dyrightvert leq {frac {1}{2}}Lambda Omega _{f},} und {displaystyle leftvert int _{Gamma }f(x,j),dxrightvert leq {frac {1}{2}}Lambda Omega _{f},} wo {Anzeigestil Omega _{f}} is the oscillation of {Anzeigestil f} on the range of {Anzeigestil Gamma } .

Now we are in position to prove the theorem: Proof of Theorem. Lassen {displaystyle varepsilon } be an arbitrary positive real number. By continuity of {Anzeigestil A} , {Anzeigestil B} and compactness of {Anzeigestil {überstreichen {R}}} , given {displaystyle varepsilon >0} , es existiert {Anzeigestil 00.} We may as well choose {Anzeigestil-Delta } so that the RHS of the last inequality is {Anzeigestil 0} . Since this is true for every {displaystyle varepsilon >0} , wir sind fertig.

Validity under different hypotheses The hypothesis of the last theorem are not the only ones under which Green's formula is true. Another common set of conditions is the following: Die Funktionen {Anzeigestil A,B:{überstreichen {R}}zu mathbb {R} } are still assumed to be continuous. Jedoch, we now require them to be Fréchet-differentiable at every point of {Anzeigestil R} . This implies the existence of all directional derivatives, im Speziellen {displaystyle D_{e_{ich}}A=:D_{ich}EIN,D_{e_{ich}}B=:D_{ich}B,,i=1,2} , wo, as usual, {Anzeigestil (e_{1},e_{2})} is the canonical ordered basis of {Anzeigestil mathbb {R} ^{2}} . Zusätzlich, we require the function {displaystyle D_{1}B-D_{2}EIN} to be Riemann-integrable over {Anzeigestil R} .

As a corollary of this, we get the Cauchy Integral Theorem for rectifiable Jordan curves: Satz (Cauchy) — If {Anzeigestil Gamma } is a rectifiable Jordan curve in {Anzeigestil mathbb {C} } und wenn {Anzeigestil f:{Text{closure of inner region of }}Gamma to mathbb {C} } is a continuous mapping holomorphic throughout the inner region of {Anzeigestil Gamma } , dann {Anzeigestil int _{Gamma }f=0,} the integral being a complex contour integral.

Proof We regard the complex plane as {Anzeigestil mathbb {R} ^{2}} . Jetzt, definieren {Anzeigestil u,v:{überstreichen {R}}zu mathbb {R} } to be such that {Anzeigestil f(x+iy)= u(x,j)+iv(x,j).} These functions are clearly continuous. It is well known that {Anzeigestil u} und {Anzeigestil v} are Fréchet-differentiable and that they satisfy the Cauchy-Riemann equations: {displaystyle D_{1}v+D_{2}u=D_{1}u-D_{2}v={Text{zero function}}} .

Jetzt, analyzing the sums used to define the complex contour integral in question, it is easy to realize that {Anzeigestil int _{Gamma }f=int _{Gamma }u,dx-v,dyquad +iint _{Gamma }v,dx+u,dy,} the integrals on the RHS being usual line integrals. These remarks allow us to apply Green's Theorem to each one of these line integrals, finishing the proof.

Multiply-connected regions Theorem. Lassen {Anzeigestil Gamma _{0},Gamma _{1},Punkte ,Gamma _{n}} be positively oriented rectifiable Jordan curves in {Anzeigestil mathbb {R} ^{2}} befriedigend {Anzeigestil {Start{ausgerichtet}Gamma _{ich}subset R_{0},&&{Text{wenn }}1leq ileq n\Gamma _{ich}Teilmenge mathbb {R} ^{2}setminus {überstreichen {R}}_{j},&&{Text{wenn }}1leq i,jleq n{Text{ und }}ineq j,Ende{ausgerichtet}}} wo {Anzeigestil R_{ich}} is the inner region of {Anzeigestil Gamma _{ich}} . Lassen {displaystyle D=R_{0}setminus ({überstreichen {R}}_{1}Tasse {überstreichen {R}}_{2}cup cdots cup {überstreichen {R}}_{n}).} Vermuten {Anzeigestil p:{überstreichen {D}}zu mathbb {R} } und {Anzeigestil q:{überstreichen {D}}zu mathbb {R} } are continuous functions whose restriction to {Anzeigestil D} is Fréchet-differentiable. If the function {Anzeigestil (x,j)longmapsto {frac {partial q}{partial e_{1}}}(x,j)-{frac {partial p}{partial e_{2}}}(x,j)} is Riemann-integrable over {Anzeigestil D} , dann {Anzeigestil {Start{ausgerichtet}&int _{Gamma _{0}}p(x,j),dx+q(x,j),dy-sum _{i=1}^{n}int _{Gamma _{ich}}p(x,j),dx+q(x,j),dy\[5Punkt]={}&int _{D}links{{frac {partial q}{partial e_{1}}}(x,j)-{frac {partial p}{partial e_{2}}}(x,j)Rechts},d(x,j).Ende{ausgerichtet}}} Relationship to Stokes' theorem Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the {displaystyle xy} -plane.

We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for the vector-valued function {Anzeigestil mathbf {F} =(L,M,0)} . Start with the left side of Green's theorem: {Anzeigestil gesalbt_{C}(L,dx+M,dy)=Punkt _{C}(L,M,0)cdot (dx,dy,dz)=Punkt _{C}mathbf {F} cdot dmathbf {r} .} The Kelvin–Stokes theorem: {Anzeigestil gesalbt_{C}mathbf {F} cdot dmathbf {r} =int _{S}nabla mal mathbf {F} cdot mathbf {Hut {n}} ,dS.} The surface {Anzeigestil S} is just the region in the plane {Anzeigestil D} , with the unit normal {Anzeigestil mathbf {Hut {n}} } defined (by convention) to have a positive z component in order to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes {displaystyle nabla times mathbf {F} cdot mathbf {Hut {n}} =links[links({frac {teilweise 0}{teilweise y}}-{frac {teilweise m}{teilweise z}}Rechts)mathbf {ich} +links({frac {partial L}{teilweise z}}-{frac {teilweise 0}{teilweise x}}Rechts)mathbf {j} +links({frac {teilweise m}{teilweise x}}-{frac {partial L}{teilweise y}}Rechts)mathbf {k} Rechts]cdot mathbf {k} =links({frac {teilweise m}{teilweise x}}-{frac {partial L}{teilweise y}}Rechts).} Thus we get the right side of Green's theorem {displaystyle iint _{S}nabla mal mathbf {F} cdot mathbf {Hut {n}} ,dS=iint _{D}links({frac {teilweise m}{teilweise x}}-{frac {partial L}{teilweise y}}Rechts),dA.} Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives: {Anzeigestil gesalbt_{C}L,dx+M,dy=Punkt _{partial D}!omega = int _{D}domega =int _{D}{frac {partial L}{teilweise y}},dywedge ,dx+{frac {teilweise m}{teilweise x}},dxwedge ,dy=iint _{D}links({frac {teilweise m}{teilweise x}}-{frac {partial L}{teilweise y}}Rechts),dx,dy.} Relationship to the divergence theorem Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of the divergence theorem: {displaystyle iiint _{v}links(mathbf {nabla } cdot mathbf {F} Rechts),dV=} {displaystyle partial scriptstyle V} {Anzeigestil (mathbf {F} cdot mathbf {Hut {n}} ),dS.} wo {displaystyle nabla cdot mathbf {F} } is the divergence on the two-dimensional vector field {Anzeigestil mathbf {F} } , und {Anzeigestil mathbf {Hut {n}} } is the outward-pointing unit normal vector on the boundary.