es

Green's theorems

Rectangular domain

Green's Theorems in Polar coordinate

Curvilinear coordinates

General case

Semi-infinite strip

Let U ⊂ ℝ² be an open set containing a compact, simply-connected region Ω whose boundary is parameterized by the piecewise smooth, simple closed curve C = ∂Ω. Take C to be positively oriented when region Ω is at its left. Given functions P and Q defined on U such that {P, Q} ∈ ℭ¹(U), we have the following identity
\begin{equation} \label{EqStrip.1} \oint_{\partial\Omega} P(x,y)\,{\text d}x + Q(x,y)\,{\text d}y = \iint_{\Omega} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) {\text d}A , \end{equation}
where dA = dxdy is the element of plane domain.

Let us consider the infinite half-strip RT = {(x, y) ∈ ℝ² : 0 ≤ x < ∞,    0 ≤ y < T < ∞}. This semi-infinite strip is a limit of enclosed bounded rectangles RT, n = {(x, y) ∈ ℝ² : 0 ≤ x < n,    0 ≤ y < T < ∞} whose boundary is parameterized by γ₁, γ₂, γ₃, and γ₄.

original = Polygon[{{0, 0}, {2, 0}, {2, 1}, {0, 1}}]; rec = Graphics[{EdgeForm[{Thick, Black}], Opacity[.3], Black, original, LightRed}, Axes -> False]; arx = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{-0.7, 0}, {2.7, 0}}]}]; ary = Graphics[{Black, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, -0.4}, {0, 1.6}}]}]; txt = Graphics[{Black, Text[Style["O", FontSize -> 15, Bold], {-0.15, -0.15}], Text[Style["x-axis", FontSize -> 18, Bold], {2.5, 0.15}], Text[Style["A", FontSize -> 15, Bold], {2, -0.15}], Text[Style["y-axis", FontSize -> 18, Bold], {-0.1, 1.7}], Text[Style["B", FontSize -> 15, Bold], {2, 1.15}], Text[Style["(0. T)", FontSize -> 15, Bold], {-0.3, 1.0}]}]; arT = Graphics[{Black, Thickness[0.01], Line[{{0, 1}, {2.7, 1}}]}]; ar2 = Graphics[{Black, Thick, Arrowheads[0.035], Arrow[{{2.1, 0.3}, {2.1, 0.7}}]}]; ar3 = Graphics[{Black, Thick, Arrowheads[0.035], Arrow[{{1.4, 1.1}, {0.6, 1.1}}]}]; ar4 = Graphics[{Black, Thick, Arrowheads[0.035], Arrow[{{-0.1, 0.7}, {-0.1, 0.3}}]}]; txt2 = Graphics[{Black, Text[Style[Subscript[R, T], FontSize -> 15, Bold], {1, 0.5}], Text[Style[Subscript[\[Gamma], 1], FontSize -> 15], {1.0, -0.15}], Text[Style[Subscript[\[Gamma], 2], FontSize -> 15], {2.25, 0.5}], Text[Style[Subscript[\[Gamma], 3], FontSize -> 15], {1.0, 1.25}], Text[Style[Subscript[\[Gamma], 4], FontSize -> 15], {-0.25, 0.5}]}]; Show[rec, arx, ary, txt, arT, ar2, ar3, ar4, txt2]
Figure 1: Semi-infinite strip RT.

We would like to apply Green’s theorem to this region. However, since the theorem concerns a compact domain of integration, we must examine a sequence of closed and bounded sub-domains RT, n whose limit is RT.

The rectangle in the depiction Figure 1 above of RT, n has width 𝑎n, which is the nth term in the arbitrary positive sequence {𝑎n} → ∞ as n → ∞. Without any loss of generality, we can set 𝑎n = n.

Theorem 1: Let R be an open, simply connected region with a boundary curve C = ∂R that is a piecewise smooth, simple closed curve oriented counterclockwise. Let 𝐅 =⟨𝑃, 𝑄⟩ be a vector field with component functions that have continuous partial derivatives on R. Then,

If P and Q satisfy the conditions for Green’s theorem and if

  1. {P(x, 0), P(x, T)} are Lebesque integrable on half-line, which we abbreviate as ℌ¹(ℝ≥0);
  2. Q(𝑎n, y) → 0 pointwise on [0, T];
  3. Q(x, y) is uniformly bounded on y ∈ [0, T] for sufficiently large x;
  4. (Qx − Py) ∈ ℌ¹(ℝT);
then we have the corresponding application of Green’s theorem on the half-strip with vanishing path integral on the side sent to infinity, \[ \int_0^{\infty} P(x,0)\,{\text d}x - \int_0^{\infty} P(x,T)\,{\text d}x - \int_0^T Q(0, y)\,{\text d}y = \iint_{R_T} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) {\text d}A . \]
Assume that P and Q possess the properties listed in the theorem. We first examine the limit of the sequence of closed contour integrals, \begin{align*} \lim_{n\to\infty} \oint_{C_n} P\,{\text d}x + Q\,{\text d}y &= \lim_{n\to\infty} \left( \int_{\gamma_1} + \int_{\gamma_2} + \int_{\gamma_3} + \int_{\gamma_4} \right) \left( P\,{\text d}x + Q\,{\text d}y \right) \\ &= \lim_{n\to\infty} \left( \int_{\gamma_1} P\,{\text d}x + \int_{\gamma_2} Q\,{\text d}y + \int_{\gamma_3} P\,{\text d}x + \int_{\gamma_4} Q\,{\text d}y \right) \\ &= \lim_{n\to\infty} \left( \int_0^{a_n} P(x, 0)\,{\text d} x + \int_0^T Q(a_n , y)\,{\text d}y \right. \\ & \quad \left. + \int_{a_n}^0 P(x, T)\,{\text d}x + \int_T^0 Q(0, y)\,{\text d} y \right) \end{align*} To show that the limit of the sum equals the sum of the limits, we will show that the limit of each term converges.

I) By condition I and the continuity of the absolute value, \begin{align*} \left\vert \lim_{n\to\infty} \int_0^{a_n} P(x, 0)\,{\text d} x \right\vert &= \lim_{n\to\infty} \left\vert \int_0^{a_n} P(x, 0)\,{\text d} x \right\vert \\ &\le \lim_{n\to\infty} \int_0^{a_n} \left\vert P(x, 0) \right\vert {\text d} x = \int_0^{\infty} \left\vert P(x, 0) \right\vert {\text d} x < \infty . \end{align*} Similarly, \begin{align*} \left\vert \lim_{n\to\infty} \int_{a_n}^0 P(x, T)\,{\text d} x \right\vert &= \lim_{n\to\infty} \left\vert \int_0^{a_n} P(x, T)\,{\text d} x \right\vert \\ &\le \lim_{n\to\infty} \int_0^{a_n} \left\vert P(x, T) \right\vert {\text d} x = \int_0^{\infty} \left\vert P(x, T) \right\vert {\text d} x < \infty . \end{align*} We also know that \[ \left\vert \lim_{n\to\infty} \int_T^0 Q(0, y)\,{\text d}y \right\vert = \left\vert \int_0^T Q(0, y)\,{\text d}y \right\vert < \infty \] since Q is continuous on RT, n and the interval [0, T] is compact.

II, III) In order for \[ \lim_{n\to\infty} \int_0^T Q(a_n , y)\,{\text d}y = 0 \] we make use of conditions II and III. Let fn(y) := Q(𝑎n, y). Then for arbitrary fixed y ∈ [0, T], we have \( \displaystyle \quad \lim_{n\to\infty} f_n (y) = 0 . \)

According to condition III, there exists some N ∈ ℕ and a corresponding c₀ > 0 such that whenever n ≥ N, Q(𝑎n, y) ≤ c₀ for all y ∈ [0, T]. Define hn(y) := fN+n(y), and let g(y) := c₀ be a constant function. The function g dominates hn(y) on [0, T], as for all n ∈ ℕ, \[ \left\vert h_n (y) \right\vert = \left\vert f_{N+n} (y) \right\vert = \left\vert Q(a_{N+n} , y) \right\vert \le c_0 = g(y) . \] Moreover, g is integrable over the compact set [0, T], and hn is measurable. By the dominated convergence theorem \[ \lim_{n\to\infty} \int_0^T h_n (y)\,{\text d}y = \int_0^T \lim_{n\to\infty} h_n (y)\,{\text d}y = 0 . \] Notice that this result implies that \begin{align*} \lim_{n\to\infty} \int_0^T Q( a_n , y)\,{\text d}y &= \lim_{n\to\infty} \int_0^T f_n (y) \,{\text d}y = \lim_{n\to\infty} \int_0^T f_{N+n} (y) \,{\text d}y \\ &= \lim_{n\to\infty} \int_0^T h_n (y) \,{\text d}y = 0 . \tag{T1.1} \end{align*} We have proved the limit of each term in the sum converges and can reexpress the left-hand side of Green’s formula, \begin{align*} \mbox{LHS} &= \lim_{n\to\infty} \oint_{C_n} P\,{\text d}x + Q\,{\text d}y = \lim_{n\to\infty} \left( \int_0^{a_n} P(x,0)\,{\text d}x + \int_0^T Q( a_n , y)\,{\text d}y \right. \\ &\quad \left. + \int_{a_n} P(x, T)\,{\text d}x + \int_Y^0 Q(0, y) \,{\text d}y \right) \\ &= \lim_{n\to\infty} \int_0^{a_n} P(x,0)\,{\text d}x + \lim_{n\to\infty} \int_0^T Q( a_n , y)\,{\text d}y \\ &\quad + \lim_{n\to\infty} \int_{a_n} P(x, T)\,{\text d}x +\lim_{n\to\infty} \int_T^0 Q(0, y)\,{\text d}y \\ &= \int_0^{\infty} P(x,0)\,{\text d}x - \int_0^{\infty} P(x,T)\,{\text d}x - \int_0^{\infty} Q(0, y)\,{\text d}y . \end{align*}

IV) To deal with the limit of the sequence of area integrals in the right-hand side of Green’s formula, we first see that \[ \lim_{n\to\infty} \iint_{R_{T,n}} \left( Q_x - P_y \right) {\text d}A = \lim_{n\to\infty} \iint_{R_{T,n}} \mathbb{I}_{R_{T,n}} (x, y) \cdot \left( Q_x - P_y \right) {\text d}A . \] Next, define \( \displaystyle \quad f_N (x, y) = \mathbb{I}_{R_{T,n}} \cdot \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \quad \) for all (x, y) ∈ RT. Then \begin{align*} \lim_{n\to\infty} f_n (x,y) &= \lim_{n\to\infty} \left[ \mathbb{I}_{R_{T,n}} \cdot \left( Q_x - P_y \right) \right] \\ &= \lim_{n\to\infty} \mathbb{I}_{R_{T,n}} \cdot \left( Q_x - P_y \right) \\ &= \mathbb{I}_{R_T} (x, y) \cdot \cdot \left( Q_x - P_y \right) . \end{align*} We also know that \begin{align*} \left\vert f_n (x, y) \right\vert &= \left\vert \mathbb{I}_{R_T} (x, y) \cdot \cdot \left( Q_x - P_y \right) \right\vert = \mathbb{I}_{R_T} (x, y) \cdot \cdot \left\vert Q_x - P_y \right\vert \\ & \le \left\vert Q_x - P_y \right\vert \in ℌ¹(R_T) \end{align*} because of condition IV. Set g(x, y) = |Qx − Py| to be the dominating function for sequence fn.

Given that Qx − Py is measurable, and hence fn(x, y) is measurable, we can pass the limit into the integral using the dominated convergence theorem: \begin{align*} \lim_{n\to\infty} \iint_{R_{T,n}} \left( Q_x - P_y \right) {\text d}A &= \lim_{n\to\infty} \iint_{R_{T,n}} \mathbb{I}_{R_{T,n}} (x, y) \cdot \cdot \left( Q_x - P_y \right) {\text d}A \\ &= \lim_{n\to\infty} \iint_{R_T} f_n (x,y)\,{\text d}A \tag{????} \\ &= \iint_{R_T} \lim_{n\to\infty} \,f_n (x,y)\,{\text d}A = \iint_{R_T} \left( Q_x - P_y \right) {\text d}A . \end{align*} Therefore, if L and M satisfy the conditions set in Theorem 1, the left-hand side in Eq.(1) is \begin{align*} \mbox{LHS} &= \int_0^{\infty} P(x,0)\,{\text d}x - \int_0^{\infty} P(x,T)\,{\text d}x - \int_0^T Q(0, y)\,{\text d}y \\ &= \lim_{n\to\infty} \oint_{C_n} P(x, y)\,{\text d}x + Q(x,y)\,{\text d}y \\ &= \lim_{n\to\infty} \iint_{R_{T,n}} \left( Q_x - P_y \right) {\text d}A = \iint_{R_T} \left( Q_x - P_y \right) {\text d}A . \end{align*}

   
Example 1:    ■
End of Example 1

 

Application to the generalized heat equation

We consider the generalized heat equation
\[ \frac{\partial u}{\partial t} = \alpha\, \frac{\partial^2 u}{\partial x^2} - \gamma\, u + f(x,t) \]
in the two-dimensional domain RT = {(x, t) : 0 < x < ∞,    0 < t ≤ T < ∞}. Here α is a positive constant and γ ≥ 0. Multiplying both sides of the heat equation above by \( \displaystyle \quad e^{-(\alpha k^2 + \gamma )t + {\bf j}\,kx} \ \) and integrating with respect to independent variables (x, t) ∈ RT, we obtain
\[ \iint_{R_T} \frac{\partial u}{\partial t}\,e^{-(\alpha k^2 + \gamma )t + {\bf j}\,kx} {\text d}A = \iint_{R_T} \left( \alpha\, \frac{\partial^2 u}{\partial x^2} - \gamma\, u \right) e^{-(\alpha k^2 + \gamma )t + {\bf j}\,kx} {\text d}A \]
\[ + \iint_{R_T} e^{-(\alpha k^2 + \gamma )t + {\bf j}\,kx}\, f(x,t)\,{\text d}A . \]

We would like to solve the following equation, which comes from analysis of the generalized heat equation,

\begin{align*} & \iint_{R_T} \left[ \left( e^{-(\alpha k^2 + \gamma )t + {\bf j}\,kx} \,u \right)_t - \left( e^{(\alpha k^2 + \gamma )t + {\bf j}\,kx} \, \alpha \left( u_x - {\bf j}\,ku \right) \right)_x \right. \\ & \left. - e^{(\alpha k^2 + \gamma )t + {\bf j}\,kx}\, f(x,y) \right] {\text d}A = 0 \end{align*}
or equivalently,
\begin{align*} & \iint_{R_T} \left[ \left( e^{(\alpha k^2 + \gamma )t + {\bf j}\,kx} \, \alpha \left( u_x - {\bf j}\,ku \right) \right)_x - \left( e^{(\alpha k^2 + \gamma )t + {\bf j}\,kx} \,u \right)_t \right. \tag{2} \\ & \left. + e^{(\alpha k^2 + \gamma )t + {\bf j}\,kx}\, f(x,y) \right] {\text d}A = 0 . \end{align*}
Here j is the unit imaginary vector on complex plane ℂ, so j² = −1; also, subscripts x and t in the formulas above mean partial derivatives with respect to x and t, respectively. We will determine the conditions such that the integral of the first two terms and the integral of the final term in equation (2) each converge. In particular, we wish to apply Theorem 1 to the integral of the first two terms.

First quadrant in in Polar coordinates

A two-dimensional vector field F can be written in either Cartesian or polar coordinates:
\[ \mathbf{F} = F_x \hat{\bf x} + F_y \hat{\bf y} = F_r \hat{\bf r} + F_{\theta} \hat{\bf \theta} \]
Since
\[ \begin{split} \hat{\bf r} &= \cos\theta\,\hat{\bf x} + \sin\theta \,\hat{\bf y} = \cos\theta\,\hat{\bf i} + \sin\theta \,\hat{\bf j} , \\ \hat{\bf \theta} &= -\sin\theta \,\hat{\bf i} + \cos\theta \,\hat{\bf j} , \end{split} \]
because it is a custom to use i and j as unit vectors in Cartesian coordinates. This yields
\[ F_x = F_r \cos\theta - \sin\theta\,F_{\theta} , \quad F_y = F_r \sin\theta + F_{\theta} \cos\theta \]
Therefore,
\[ F_r = F_x \cos\theta + F_y \sin\theta , \qquad F_{\theta} = F_y \cos\theta - F_x \sin\theta = F_y \,\frac{x}{r} - F_x \,\frac{y}{r} . \]
wedge = Append[ Table[{r Cos[t], r Sin[t]} /. r -> 1, {t, 0, Pi/2, 0.01}], {0, 0}]; ar1 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{-0., 0}, {0.7, 0}}]}, Axes -> True]; ar2 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{0, -0.1}, {0, 0.7}}]}, Axes -> True]; ar3 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{0.3, 0.57}, {0.01, 0.071}}]}, Axes -> True]; txt = Graphics[{Black, Text[Style["O", FontSize -> 18, Bold], {0, -0.1}], Text[Style["A", FontSize -> 18, Bold], {0.9, 0.5}], Text[Style["B", FontSize -> 18, Bold], {0.5, 0.9}]}]; ar4 = Graphics[{Black, Thick, Circle[{0, 0}, 0.77, {Pi/6, Pi/3}]}, Axes -> False]; ar5 = Graphics[{Black, Thick, Arrowheads[0.03], Arrow[{{0.4, 0.66}, {0.38, 0.669}}]}, Axes -> False]; Show[Graphics[{Green, Polygon[wedge]}], ar1, ar2, ar3, txt, ar4, ar5]
Figure 3: Finite piece of the Wedge.

Recall that Green's theorem in polar coordinates reads as
\begin{equation} \label{EqGreen.2} %\oint_{\partial R} \mathbf{F} \bullet {\text d}\mathbf{r} = \oint_{\partial R} P\,{\text d}r + Q\,r\,{\text d} \theta = \iint_R \left( \frac{\partial \left( r\,Q \right)}{\partial r} - \frac{\partial P}{\partial \theta} \right) {\text d}r\,{\text d}\theta , \end{equation}
subjext that the corner condition holds:
\begin{equation} \label{EqGreen.3} \lim_{r\to 0} r\,F_{\theta} (r, \theta ) = \lim_{(x,y) \to 0} \left( x\,Q - y\,P \right) = 0 , \end{equation}