Abstract
As the foundation of double integral, we propose a triangular integral, which is an antisymmetric double integral by single limit of double dependent sums of triangularly divided areas. Extending integrand from scalar function to tensor one, we derive the curl theorem based on this triangular double integral. It is derived by substituting the total differentials in the transformation lemma, which is based on this triangular double integral. We may thus infer that this triangular integral is the inverse operation of the total differential.
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1 Introduction
The variational principle of the 2D theory is conventionally given as
where the integrand ℒ = ℒ(X, Y, X x , X y , Y x , Y y ) is a scalar functional and D is a domain. Here, X = X(x, y), Y = Y(x, y), The double integral in (1.1) is conventionally defined as
where Δx i ≡ x i - xi-1and Δy j ≡ y j - yj-1. According to the conventional method of the perpendicularly combined form of the Riemann and the Lebesgue integrals [1, 2], the area of double integral demands double limits at infinities, k → ∞ and n → ∞, of double independent sums, j = 1,2, . . . , k and i = 1,2, . . . , n, of rectangularly divided areas as shown in (1.2). Based on this definition of the conventional rectangular double integral (1.2), the curl theorem on the 2D plane is formulated as
where ∂D is an integral path.
Meanwhile, the total differential is widely used even in the exterior derivative [3]. However, it is not known how to derive the curl theorem (1.3) by substituting the total differentials in an integral formula based on the conventional rectangular double integral method. Extending integrand from scalar function to tensor one, we may derive the curl theorem by substituting the total differentials in an integral formula. It depends on how to define a new kind of double integral. We extend the variational principle (1.1) to
where the integrand L αβ = L αβ (X μ , X μ,v ) is a tensor functional and indices are summed over α, β = 1, 2. Here, X μ = X μ (xλ ) and A new type of double integral in (1.4) is defined as
where Δ(xα ) k ≡ (xα ) k - (xα )k-1, Δ(xβ ) j ≡ (xβ ) j - (xβ )j-1and indices are summed over α,β = 1,2. It makes possible to introduce a new kind of triangular double integral by the following two properties:
-
1.
replacing rectangular area by triangular one;
-
2.
replacing double limits of independent double sums by single limit of dependent double sums.
We propose an antisymmetric triangular double integral. It demands only single limit at infinity n → ∞ of double dependent sums, j = 1, 2, . . . , k and k = 1, 2, . . . , n, of triangularly divided areas as shown in Definition 1. We succeed to define a new kind of triangular integral method, which may derive the curl theorem by substituting the total differentials in an integral formula. In this article, we formulate the curl theorem based on a new kind of integral formula (1.5). We name it as the curl theorem of a triangular integral on the 2D plane as shown in the main theorem (4.6). In detail, we derive (4.1) by substituting the total differentials (4.3) and (4.4) in (4.2). The curl theorem of a triangular integral on the 2D plane (4.6) is finally derived from (4.1) and (4.5) under the condition of a closed curve (4.8). We may thus infer that this triangular integral is the inverse operation of the total differential.
There are three advantages of this theory. One is the conceptual coherence despite of its complicated procedure of calculation in the derivation of the curl theorem (see Section 4.1). Another one is that this theory is applicable for finite element method in the case of 1 < n < ∞ (see (2.12) and (2.18)). The other one is applicability to the integral in the variational principle of multiple variables in the case that the integrand is extended to tensor (see (1.4) and (1.5)).
This article is structured as follows. In Section 2, a triangularly divided area is introduced. This triangular double integral is defined by single limit of double sums of triangularly divided areas. In Section 3, the combination and the transformation lemmata are derived. In Section 4, the curl theorem on the 2D plane is derived by substituting the total differentials in the transformation lemma. In Section 5, the curl theorems of a triangular integral in the 3D space and the 4D hyper-space are presented.
2 Single limit of double sums
A triangular integral as the foundation of double integral is proposed in Section 2.1 and its example is shown in Section 2.2.
2.1 Sum of triangular areas
The sum of triangular areas is introduced as follows.
First of all, the respective triangular area is introduced. Let f(x, y) = 0 be a piecewise smooth curve of equation on the xy-plane, expressed in terms of the Cartesian coordinates (x, y) ∈ ℝ2. Assume there are two fixed points of A(xA, yA) and B(xB, yB). Suppose there is a sequence of points {P k (x k , y k ) | k = 0, 1, 2, . . . , n} on f(x, y) = 0, where the initial and the terminal points are respectively P0 (x0, y0) = A(xA, yA) and P n (x n , y n ) = B(xB, yB). The respective triangular area ΔS k (k = 1, 2, . . . , n), which is surrounded by three vertices of O(0, 0), Pk-1(xk-1, yk-1) and P k (x k , y k ), is introduced as in Figure 1 by
The increments of x k and y k , i.e., Δx k and Δy k , are respectively denoted as
Substituting Δx k and Δy k in ΔS k of (2.1), it is modified to be
Next, we introduce triangular sum S n (n = 1, 2, 3, . . .) as a sum of n triangular areas ΔS k , i.e.,
Substituting (2.4) in (2.5), it is modified to be
Furthermore, the sum of triangular areas is modified to be the sum of a triangular area and double sums as follows. Using the notations of Δx k in (2.2) and Δy k in (2.3), x k and y k are respectively expressed as
Substituting (2.7) and (2.8) in (2.6), it is finally modified to be
The cases of n = 1, 2, 3 and n → ∞ of
in (2.9) are shown in the following. Let the coordinates of point Q be (x n , y0).
-
1.
In the case of n = 1, i.e., for two points of P 0(x 0, y 0) and P 1(x 1, y 1) = P n (x n , y n ), it holds
(2.11) -
2.
In the case of n = 2, i.e., for three points of P 0(x 0, y 0), P 1(x 1, y 1) and P 2(x 2, y 2) = P n (x n , y n ), as shown in Figure 2, the area of P 0 P 1 P 2 is
(2.12)
Introducing R2.0 as the area of triangle P0QP n , we obtain
Introducing R2.1 as the area of triangle P0P1Q, we obtain
Introducing R2.2 as the area of triangle P1P2Q, we obtain
We introduce σ2 as the area of triangle P0P1P2 as
Substituting (2.13), (2.14) and (2.15) in (2.16), it is modified to be
We thus see the coincidence of (2.12) and (2.17).
-
3.
In the case of n = 3, i.e., for four points of P 0(x 0, y 0), P 1(x 1, y 1), P 2(x 2, y 2) and P 3(x 3, y 3) = P n (x n , y n ), as shown in Figure 3, the area of P 0 P 1 P 2 P 3 is
(2.18)
Introducing R3.0 as the area of triangle P0QP n , we obtain
Introducing R3.1 as the area of triangle P0P1Q, we obtain
Introducing R3.2 as the area of triangle P1P2Q, we obtain
Introducing R3.3 as the area of triangle P2P3Q, we obtain
We introduce σ3 as the area of P0P1P2P3 as
Substituting (2.19), (2.20), (2.21) and (2.22) in (2.23), it is modified to be
We thus see the coincidence of (2.18) and (2.24).
-
4.
In the case of n → ∞, i.e., for a segment of and a piecewise smooth curve of equation f(x, y) = 0, as shown in Figure 4, the area σ surrounded by the segment and the curve is
(2.25)
Definition 1. (Definition of a triangular integral) The triangular double integral of is defined in the case of a piecewise smooth curve of equation f(x, y) = 0 by the formula,
where dx' and dy' respectively correspond to increments of Δx j and Δy j , while dx and dy respectively correspond to those of Δx k and Δy k .
Suppose S is the limit of S n at infinity n → ∞, i.e.,
Theorem 1. The area S of OAB surrounded by, and the graph of a piecewise smooth curve of equation f (x, y) = 0 is expressed as
where O(0,0), A(xA, yA), B(xB, yB), f(xA, yA) = 0 and f(xB, yB) = 0.
Proof. As n → ∞ in (2.9), we obtain S as
as shown in Figure 4. Q.E.D.
Remark. The antisymmetric double integral here introduced demands only single limit of double sums of triangularly divided areas.
2.2 Two kinds of solutions of an example of a parabola
An example that the integrand is constant is shown in Example 1. An example that the integrand is not constant is shown in Section 4.2. Two kinds of solutions of the problem of a parabola are shown in the following. The first and the second solutions are respectively given by the arithmetic and the geometric sequences.
Example 1. An area surrounded by a parabola of
and a segment of , where A(xA, yA) = (1,8) and B(xB, yB) = (2,5).
-
1.
Integration by the arithmetic sequence (The first kind of solution)
The arithmetic sequences x j and x k are respectively
Using (2.2), the increments of the arithmetic sequences x j and x k , i.e., Δx = Δx j = Δx k , are
Substituting (2.30) in (2.3), the increments of the arithmetic sequences y j and y k , i.e., Δy j and Δy k , are respectively
Thus, the antisymmetric double increment of the arithmetic sequence (2.31) and (2.32) is
The double dependent sums of (2.36) for j = 1,2, . . . , k and k = 1,2,. . . , n is
As n → ∞ in (2.37), we obtain
-
2.
Integration by the geometric sequence (The second kind of solution)
The geometric sequences x j and x k are respectively
The increments of the geometric sequences x j , x k , y j and y k , i.e., Δx j , Δx k , Δy j and Δy k , are respectively
Thus, the antisymmetric double increment of the geometric sequence (2.39) and (2.40) is
The double dependent sums of (2.45) for j = 1,2, . . . , k and k = 1, 2, . . . , n is
As n → ∞ in (2.46), we obtain
3 The combination lemma and the transformation lemma
The combination lemma is shown in Section 3.1 and the transformation lemma is shown in Section 3.2.
Let y = f(x) be a differentiable function on the xy-plane. Assume there are two fixed points of A = (xA, yA) and B = (xB, yB) on y = f(x), where yA = f(xA) and yB = f(xB). An ordinary integral is denoted in accordance with the notation of double integral denoted in Definition 1. New notation for an ordinary integral of y = f(x) in x ∈ [xA, xB] is
3.1 The combination lemma on the 2D plane
The combination lemma shows that the sum of an integral along x-axis and that along y-axis between [A, B] is equal to the subtraction of two rectangles, x B y B and x A y A .
Lemma 1. (The combination lemma) Assume y = f(x) is a differentiable function on the xy-plane, and x = f-1(y) is also a differentiable one, where f-1is the inverse function of f. Between A(xA, yA) = (x0, y0) and B(xB, yB) = (x n , y n ), it holds
where yA = f(xA) and yB = f(xB).
We name (3.2) as the combination lemma.
Proof. It is divided into the following cases.
-
1.
For a monotonically increasing function y = f(x) in x ∈ [x A, x B] or x = f -1(y) in y ∈ [y A, y B], it is indicated that the first term of the left-hand side of (3.2) is an integral of y = f(x) along x-axis i.e.,
(3.3)
and the second term of the left-hand side of (3.2) is an integral of x = f-1(y) along y-axis, i.e.,
as shown in Figure 5.
-
2.
For a monotonically decreasing function y = f(x) in x ∈ [x A, x B] or x = f -1(y) in y ∈ [y B, y A], it holds
(3.5)
as shown in Figure 6. We thus obtain (3.2). Q.E.D.
3.2 The transformation lemma on the 2D plane
The transformation lemma transforms an integrand to the second integral variable. It transforms a single integral to an antisymmetric double integral of which integrand is constant.
Proposition 1. (Integral along x-axis) For a piecewise smooth curve of equation f(x, y) = 0, it holds
Proof. Substituting (2.8) into (3.3), it is modified to be
Q.E.D.
Proposition 2. (Integral along y-axis) For a piecewise smooth curve of equation f(x, y) = 0, it holds
Proof. Substituting (2.7) into (3.4), it is modified to be
Q.E.D.
Lemma 2. (The transformation lemma) For a piecewise smooth curve of equation f(x, y) = 0, it holds
Proof. (Combining integral along x-axis with that along y-one) Combining (3.6) with (3.8), we obtain
Substituting (3.2) into (3.11), we obtain the lemma. Q.E.D.
4 The curl theorem of a triangular integral on the 2D plane
The curl theorem of the conventional rectangular integral (1.3) is modified to be that of a new kind of triangular integral (4.6) in Section 4.1 and its example is shown in Section 4.2.
4.1 Proof of the curl theorem on the 2D plane
We present two lemmata in the following before the proof of Theorem 2.
Lemma 3. Let X = X(x, y) be a partially differentiable function with respect to x and y. It holds
for a piecewise smooth curve of equation f(x, y) = 0 between A(xA, yA) and B(xB, yB), where XA = X(xA, yB) and XB = X(xB, yB).
Proof. Rewriting the transformation lemma (3.10) for X = X(x, y) and f(x, y) = 0, it is expressed as
Substituting two kinds of the total differentials of X, i.e., dX and dX',
in (4.2), it is modified to be (4.1). Q.E.D.
Remark. Precise modifications in the proof of Lemma 3 are shown in Appendix 1.
Lemma 4. Let Y = Y(x, y) be a partially differentiable function with respect to x and y. It holds
for a piecewise smooth curve of equation f(x, y) = 0 between A(xA, yA) and B(xB, yB), where YA = Y(xA, yA) and YB = Y(xB, yB).
Proof. In the similar procedure as Lemma 3, we obtain (4.5). Q.E.D.
Theorem 2. (The curl theorem of a triangular integral on the 2D plane) Let ∂D be a piecewise smooth curve of equation f(x, y) = 0 on the xy-plane, which is expressed in terms of the Cartesian coordinates (x, y) ∈ ℝ2. Let D be the region inside and on ∂D. Let X1 = X = X(x, y) and X2 = Y = Y(x, y) be partially differentiable functions with respect to x1 = x and x2 = y in D. It holds
where indices are summed over α, β = 1,2.
Proof. Combining (4.1) with (4.5), we obtain
For an integral on a closed curve, the initial point A(xA, yA) coincides with the terminal one B(xB, yB), i.e., A(xA, yA) = B(xB, yB). It then holds
regardless of values of XA, XB, YA and YB. We thus obtain (4.6). Q.E.D.
Remark. and must be rigorously distinguished as integrands. See (5.20) and (5.19) in detail. An inequality holds for and in an integral form, i.e.,
Here, ΔX[Δx k ] is denoted as ΔX[Δx k ] ≡ X(x k , y k )-X(xk-1, y k ). See an example in (5.23). However, they coincide as explicit forms of derivatives. An equality holds for and in arbitrary derivative form, i.e.,
Similar formulae hold for and , for and and for and
Corollary 1. In the case of which it holds
where indices are summed over α, β = 1, 2, a sufficient condition to hold (4.11) is
Proof. Using (4.6), (4.11) is modified to be
where indices are summed over α, β = 1, 2. A sufficient condition to hold (4.13) is (4.12). Q.E.D.
Corollary 2. In the case of (4.11), we obtain an antisymmetric property of E αβ as
Proof. Interchanging α and β in (4.12), it also holds
Comparing (4. 12) with (4. 15), we obtain (4. 14). Q.E.D.
4.2 An example of the curl theorem on the 2D plane
An example of Theorem 2 is shown in Example 2. The integral path of this problem is an ellipse of
Definition 2. (Definition of elliptical sequence) Using (4.16) and (4.17), we may respectively introduce elliptical sequences x j , y j , x k and y k as
The increments of x j and y j in (4.18), i.e., Δx j and Δy j , of the angular arithmetic sequence θ j are respectively
The increments of x k and y k in (4.19), i.e., Δx k and Δy k , of the angular arithmetic sequence θ k are respectively
Example 2. We calculate both of the line integral and the area integral in the counterclockwise direction. The example is the curl theorem of a triangular integral on the 2D plane in the case of X = X(x, y) = -x2y and Y = Y(x, y) = xy2, where the closed curve is an ellipse of (4.16) and (4.17).
The arithmetic sequences of θ j and θ k are respectively
The increments of θ j and θ k , i.e., Δθ = Δθ j = Δθ k , are
-
1.
Calculation by line integral
The line integral is calculated to be
The integral in (4.27) is executed in the formula of
Substituting (4.28) into (4.27), we finally obtain
-
2.
Calculation by area integral
The area double integral in the region of is calculated to be
Calculations of the respective terms of (4.30) are shown in Appendix 2. We hence conclude that Theorem 2 holds for this problem.
Remark. It is very complicated to execute this triangular integral only by human's brain. Computer and formula manipulation software are recommended to verify these calculations in the case of n → ∞. This theory is also applicable for numerical calculation in the case of 1 < n < ∞.
5 The curl theorem of a triangular integral in the higher dimensions
The curl theorem of a triangular integral in the 3D space is derived in Section 5.1 and that in the 4D hyper-space is derived in Section 5.2.
5.1 The curl theorem in the 3D space
We extend the curl theorem of a triangular integral in the 3D space. We present three lemmata in the following before the proof of Theorem 3.
Lemma 5. Let X = X(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where XA = X(xA, yA, zA) and XB = X(xB, yB, zB).
Proof. Rewriting the transformation lemma (3.10) for X = X(x, y, z) and g(x, y, z) = 0, it is expressed as
Substituting two kinds of the total differentials of X, i.e., dX and dX',
in (5.2), it is modified to be (5.1). Q.E.D.
Lemma 6. Let Y = Y(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where YA = Y(xA, yA, zA) and YB = Y(xB, yB, zB).
Proof. In the similar procedure as Lemma 5, we obtain (5.5). Q.E.D.
Lemma 7. Let Z = Z(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where ZA = Z(xA, yA, zA) and ZB = Z(xB, yB, zB).
Proof. In the similar procedure as Lemma 5, we obtain (5.6). Q.E.D.
Theorem 3. (The curl theorem of a triangular integral in the 3D space) Let D be a piecewise smooth surface in the xyz-space, which is expressed in terms of the Cartesian coordinates (x, y, z) ∈ ℝ3. Let ∂D be the boundary of D. Let X1 = X = X(x, y, z), X2 = Y = Y(x, y, z) and X3 = Z = Z(x, y, z) be partially differentiable functions with respect to x1 = x, x2 = y and x3 = z in D. It holds
where indices are summed over α, β = 1,2,3.
Proof. Combining (5.1) with (5.5) and (5.6), we obtain
for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where XA = X(xA, yA, zA), XB = X(xB, yB, zB), YA = Y(xA, yA, zA), YB = Y(xB, yB, zB), ZA = Z(xA, yA, zA) and ZB = Z(xB, yB, zB). For an integral on a closed curve, the initial point A(xA, yA, zA) coincides with the terminal one B(xB, yB, zB), i.e., A(xA, yA, zA) = B(xB, yB, zB). It then holds
regardless of the values of XA, XB, YA, YB, ZA and ZB. We thus obtain (5.7). Q.E.D.
Corollaries 1 and 2 also hold for α, β = 1,2,3.
5.2 The curl theorem in the 4D hyper-space
We extend the curl theorem of a triangular integral in the 4D hyper-space.
Theorem 4. (The curl theorem of a triangular integral in the 4D hyper-space) Let D be a piecewise smooth surface in the txyz-hyper-space, which is expressed in terms of the Cartesian coordinates (t, x, y, z) ∈ ℝ4. Let ∂D be the boundary of D. Let X0 = T = T(t, x, y, z), X1 = X = X(t, x, y, z), X2 = Y = Y(t, x, y, z) and X3 = Z = Z(t, x, y, z) be partially differentiable functions with respect to x0 = t, x1 = x, x2 = y and x3 = z in D. It holds
where indices are summed over α, β = 0,1,2,3.
Proof. In the 4D, it holds
for a piecewise smooth curve of equation h(t, x, y, z) = 0 between A(tA, xA, yA, zA) and B(tB, xB, yB, zB), where TA = T(tA, xA, yA, zA), TB = T(tB, xB, yB, zB), XA = X(tA, xA, yA, zA), XB = X(tB, xB, yB, zB), YA = Y(tA, xA, yA, zA), YB = Y(tB, xB, yB, zB), ZA = Z(tA, xA, yA, zA) and ZB = Z(tB, xB, yB, zB). For an integral on a closed curve, the initial point A(tA, xA, yA, zA) coincides with the terminal one B(tB, xB, yB, zB), i.e., A(tA, xA, yA, zA) = B(tB, xB, yB, zB). It then holds
regardless of the values of TA, TB, XA, XB, YA, YB, ZA and ZB. We thus obtain (5.10). Q.E.D.
Corollaries 1 and 2 also hold for α, β = 0 1,2,3.
Appendix 1
Using Definition 1, the first term of the right-hand side of (4.2) in the proof of Lemma 3 is expressed as
The total increments ΔX j and ΔX k in the right-hand side of (5.13) are respectively expressed as
Substituting (5.14) and (5.15) in the right-hand side of (5.13), it is modified to be
Each term of (5.16) is respectively expressed as
We thus obtain (4.1) in Lemma 3.
Appendix 2
Substituting (4.24) in (4.20) and (4.21), substituting (4.25) in (4.22) and (4.23), the respective terms of (4.30) are calculated as follows.
-
1.
The first term of the left-hand side of (4.30) is
(5.21) -
2.
The second term of the lefthand side of (4.30) is
(5.22) -
3.
The third term of the left-hand side of (4.30) is
(5.23) -
4.
The fourth term of the left-hand side of (4.30) is
(5.24)
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Acknowledgements
The author thanks to Professor Susumu Ishikawa at Fukuoka Institute of Technology for refining this theory. In the process of reviewing, the author also thanks to the referees for their instructive comments.
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Tokunaga, K. The curl theorem of a triangular integral. Adv Differ Equ 2012, 23 (2012). https://doi.org/10.1186/1687-1847-2012-23
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DOI: https://doi.org/10.1186/1687-1847-2012-23