Định lý Green

Trong toán học, định lý Green' đưa ra mối liên hệ giữa tích phân đường quanh một đường cong khép kín C vàa tích phân mặt trên một miền D bao quanh bởi C. Đây là trường hợp đặt biệt trong không gian 2 chiều của định lý Stokes, và được đặt tên theo nhà toán học người Anh tên George Green.

Định lý

Let C be a positively oriented, piecewise smooth, simple closed curve in the plane $\mathbb {R}$², and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then

$\oint _{C}(L\,\mathrm {d} x+M\,\mathrm {d} y)=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,\mathrm {d} x\,\mathrm {d} y.$

For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in the small circle in the integral symbol.

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area. In plane geometry, and in particular, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

Relationship to the Stokes theorem

Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the 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 $\mathbf {F} =(L,M,0)$. Start with the left side of Green's theorem:

$\oint _{C}(L\,dx+M\,dy)=\oint _{C}(L,M,0)\cdot (dx,dy,dz)=\oint _{C}\mathbf {F} \cdot d\mathbf {r} .$

Then by Kelvin–Stokes Theorem:

$\oint _{C}\mathbf {F} \cdot d\mathbf {r} =\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS.$

The surface $S$ is just the region in the plane $D$, with the unit normals $\mathbf {\hat {n}}$ pointing up (in the positive z direction) to match the "positive orientation" definitions for both theorems.

The expression inside the integral becomes

$\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} =\left[\left({\frac {\partial 0}{\partial y}}-{\frac {\partial M}{\partial z}}\right)\mathbf {i} +\left({\frac {\partial L}{\partial z}}-{\frac {\partial 0}{\partial x}}\right)\mathbf {j} +\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\mathbf {k} \right]\cdot \mathbf {k} =\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right).$

Thus we get the right side of Green's theorem

$\iint _{S}\nabla \times \mathbf {F} \cdot \mathbf {\hat {n}} \,dS=\iint _{D}\left({\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}\right)\,dA.$

Relationship to the divergence theorem

Considering only two-dimensional vector fields, Green's theorem is equivalent to the following two-dimensional version of the divergence theorem:

$\iint _{D}\left(\nabla \cdot \mathbf {F} \right)dA=\oint _{C}\mathbf {F} \cdot \mathbf {\hat {n}} \,ds,$

where $\mathbf {\hat {n}}$ is the outward-pointing unit normal vector on the boundary.

To see this, consider the unit normal $\mathbf {\hat {n}}$ in the right side of the equation. Since in Green's theorem $d\mathbf {r} =(dx,dy)$ is a vector pointing tangential along the curve, and the curve C is the positively-oriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be $(dy,-dx)$. The length of this vector is ${\sqrt {dx^{2}+dy^{2}}}=ds$. So $\mathbf {\hat {n}} \,ds=(dy,-dx).$

Now let the components of $\mathbf {F} =(P,Q)$. Then the right hand side becomes

$\oint _{C}\mathbf {F} \cdot \mathbf {\hat {n}} \,ds=\oint _{C}Pdy-Qdx$

which by Green's theorem becomes

$\oint _{C}-Qdx+Pdy=\iint _{D}\left({\frac {\partial P}{\partial x}}+{\frac {\partial Q}{\partial y}}\right)\,dA=\iint _{D}\left(\nabla \cdot \mathbf {F} \right)dA.$

The converse can also easily shown to be true.

Area Calculation

Green's theorem can be used to compute area by line integral. The area of D is given by:

$A=\iint _{D}\mathrm {d} A.$

Provided we choose L and M such that:

${\frac {\partial M}{\partial x}}-{\frac {\partial L}{\partial y}}=1.$

Then the area is given by:

$A=\oint _{C}(L\,\mathrm {d} x+M\,\mathrm {d} y).$

Possible formulas for the area of D include:

$A=\oint _{C}x\,\mathrm {d} y=-\oint _{C}y\,\mathrm {d} x={\tfrac {1}{2}}\oint _{C}(-y\,\mathrm {d} x+x\,\mathrm {d} y).$

See also

• Planimeter
• Method of image charges – A method used in electrostatics that takes advantage of the uniqueness theorem (derived from Green's theorem)

Further reading

• Calculus (5th edition), F. Ayres, E. Mendelson, Schuam's Outline Series, 2009, ISBN 978-0-07-150861-2.
• Advanced Calculus (3rd edition), R. Wrede, M.R. Spiegel, Schuam's Outline Series, 2010, ISBN 978-0-07-162366-7.

External links

• Green's Theorem on MathWorld