The Gravity Anomaly of a 2D Polygonal Body Having Density Contrast Given by Polynomial Functions

Abstract

An analytical solution is presented for the gravity anomaly produced by a 2D body whose geometrical shape is arbitrary and where the density contrast is a polynomial function in both the horizontal and vertical directions. Approximating the real shape of the body by a polygon, the solution is expressed as sum of algebraic quantities that depend only upon the coordinates of the vertices of the polygon and upon the polynomial density function. The solution presented in the paper, which refers to a third-order polynomial function as a maximum, exhibits an intrinsic symmetry that naturally suggests its extension to the case of higher-order polynomials describing the density contrast. Furthermore, the gravity anomaly is evaluated at an arbitrary point that does not necessarily coincide with the origin of the reference frame in which the density function is assigned. Invoking recent results of potential theory, the solution derived in the paper is shown to be singularity-free and numerically robust. The accuracy and effectiveness of the proposed approach is witnessed by the numerical comparisons with examples derived from the existing literature.

Appendices

Appendix 1: Some Useful Differential Identities

We prove hereafter some differential identities that are useful for the derivations illustrated in the main body of the paper; they are reported in the same order in which they are required.

Let us begin with the component expression of the divergence of the rank-three tensor

$$\begin{aligned} \mathrm{div}[\psi (\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})]_{ij}=\psi (\mathbf{a}_i\mathbf{b}_j\mathbf{c}_k)_{/k} \end{aligned}$$

(117)

where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ ($\psi$) are vector (scalar) differentiable fields and $(\cdot )_{/k}$ means derivation with respect to the $k$th variable. Applying the chain rule to (117), Tang (2006), one obtains

$$\begin{aligned} \psi (\mathbf{a}_i\mathbf{b}_j\mathbf{c}_k)_{/k}&= \psi _{/k}\mathbf{a}_i\mathbf{b}_j\mathbf{c}_k + \psi \mathbf{a}_{i/k}\mathbf{b}_j\mathbf{c}_k +\psi \mathbf{a}_i\mathbf{b}_{j/k}\mathbf{c}_k +\psi \mathbf{a}_i\mathbf{b}_j\mathbf{c}_{k/k}\nonumber \\&= (\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})_{ijk}(\mathrm{grad}\psi )_{k} + \psi [(\mathrm{grad}\,\mathbf{a})\mathbf{c}]_{i}\mathbf{b}_j\nonumber \\&\quad +\,\psi \mathbf{a}_i[(\mathrm{grad}\,\mathbf{b})\mathbf{c}]_{j}+\psi (\mathbf{a}\otimes \mathbf{b})_{ij}\mathrm{div}\,\mathbf{c}\end{aligned}$$

(118)

Thus, combining (117) and (118), one has

$$\begin{aligned} \mathrm{div}[\psi (\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})]&= (\mathbf{a}\otimes \mathbf{b}\otimes \mathbf{c})\mathrm{grad}\,\psi + \psi [(\mathrm{grad}\,\mathbf{a})\mathbf{c}]\otimes \mathbf{b}\nonumber \\&\quad +\,\psi \mathbf{a}\otimes [(\mathrm{grad}\,\mathbf{b})\mathbf{c}]+\psi (\mathbf{a}\otimes \mathbf{b})\mathrm{div}\,\mathbf{c}\end{aligned}$$

(119)

A further useful identity concerns the gradient of a scalar field expressed as scalar product of two vector fields

$$\begin{aligned} \mathrm{grad}\, (\mathbf{a}\cdot \mathbf{b})= [\mathrm{grad}\,\mathbf{a}]^\mathrm{t}\mathbf{b}+[\mathrm{grad}\,\mathbf{b}]^\mathrm{t} \mathbf{a} \end{aligned}$$

(120)

where $(\cdot )^\mathrm{t}$ stands for transpose. It stems from the relation

$$\begin{aligned}{}[\mathrm{grad}\, (\mathbf{a}\cdot \mathbf{b})]_{i}=(\mathbf{a}_j\mathbf{b}_j)_{/i} \end{aligned}$$

(121)

Actually, carrying out the derivations in the previous expression yields

$$\begin{aligned} (\mathbf{a}_j\mathbf{b}_j)_{/i} = \mathbf{a}_{j/i}\mathbf{b}_j + \mathbf{a}_j\mathbf{b}_{j/i} = [(\mathrm{grad}\,\mathbf{a})^\mathrm{t}]_{ij}\mathbf{b}_j + [(\mathrm{grad}\,\mathbf{b})^\mathrm{t}]_{ij}\mathbf{a}_j, \end{aligned}$$

(122)

which represents the component form of (120).

Appendix 2: Recursive Computation of Integrals

Application of formula (56) requires the analytical computation of integrals of the kind

$$\begin{aligned} I_{k}=\int \limits _0^1\frac{x^k}{p x^2 + 2 q x + u}\hbox {d}x \end{aligned}$$

(123)

where $p>0$ and the discriminant of the denominator, i.e., $\varDelta =q^2-pu$, is assumed to be negative. Hence, the quadratic function $p x^2 + 2 q x + u$ is always positive on the real interval [0, 1]. The case of a null discriminant $\varDelta$ will be directly addressed in Sect. 4 where the evaluation of $I_{ki}$, which makes use of the formulas derived hereafter for $\varDelta < 0$, will be detailed.

As previously shown by Zhou (2010), the generic integral (123) can be computed recursively as a function of two integrals, namely

$$\begin{aligned} I_{0}=\int \limits _0^1\frac{1}{p x^2 + 2 q x + u}\hbox {d}x = \frac{1}{\sqrt{-\varDelta }}\left[ \hbox { arctan}\frac{p+q}{\sqrt{-\varDelta }} - \hbox { arctan}\frac{q}{\sqrt{-\varDelta }}\right] \end{aligned}$$

(124)

and

$$\begin{aligned} I_{1}=\int \limits _0^1\frac{x}{p x^2 + 2 q x + u}\hbox {d}x = \frac{1}{2p}\hbox {log}\frac{p+2q+u}{u} - \frac{q}{p} I_0 \end{aligned}$$

(125)

Both results can be obtained, after some manipulation, by setting $t=x+q/p$ in the integrand functions above. To make the paper self-contained, we rephrase the result given in Zhou (2010)

$$\begin{aligned} J_{k}&= \int \frac{x^k}{p x^2 + 2 q x + u}\hbox {d}x = \frac{x^{k-1}}{p(k-1)} -\frac{2q}{p}\int \frac{x^{k-1}}{p x^2 + 2 q x + u}\hbox {d}x\nonumber \\&\quad -\, \frac{u}{p} \int \frac{x^{k-2}}{p x^2 + 2 q x + u}dx \end{aligned}$$

(126)

where $k>1$ and the terminology of this paper has been adopted.

For instance, if $k=2$, one has

$$\begin{aligned} J_{2}&= \int \frac{x^2}{p x^2 + 2 q x + u}\hbox {d}x = \frac{1}{p}\int \frac{px^2+(2qx+u-2qx-u)}{p x^2 + 2 q x + u}\hbox {d}x \nonumber \\&= \frac{1}{p}\left[ \int \hbox {d}x - \int \frac{2qx+u}{p x^2 + 2 q x + u}\hbox {d}x \right] \nonumber \\&= \frac{1}{p}\left[ \int \hbox {d}x - 2q\int \frac{x}{p x^2 + 2 q x + u}\hbox {d}x - u\int \frac{\hbox {d}x}{p x^2 + 2 q x + u} \right] \nonumber \\&= \frac{1}{p}[x-2qJ_1-uJ_0] \end{aligned}$$

(127)

Analogously,

$$J_{3}= \frac{1}{p}\left[ \frac{x^2}{2}-2qJ_2-uJ_1\right] = \frac{1}{p}\left[ \frac{x^2}{2}-\frac{2qx}{p}+\frac{4q^2-pu}{p}J_1 + \frac{2qu}{p}J_0 \right]$$

(128)

and

$$\begin{aligned} J_{4}= \frac{x^3}{3p}-\frac{q}{p^2}x^2 +\frac{4q^2-pu}{p^3}x - \frac{2q(4q^2-2pu)}{p^3}J_1 - \frac{u(4q^2-pu)}{p^3}J_0 \end{aligned}$$

(129)

Hence,

$$\begin{aligned} I_{2}&= \frac{1}{p}[1-2qI_1-uI_0] \end{aligned}$$

(130)

$$\begin{aligned} I_{3}&= \frac{1}{p}\left[ \frac{1}{2}-\frac{2q}{p}+\frac{4q^2-pu}{p}I_1 + \frac{2qu}{p}I_0 \right] \end{aligned}$$

(131)

and

$$\begin{aligned} I_{4}= \frac{1}{p} \left[ \frac{1}{3}-\frac{q}{p} +\frac{4q^2-pu}{p^2} - \frac{2q(4q^2-2pu)}{p^2}I_1 - \frac{u(4q^2-pu)}{p^2}I_0 \right] \end{aligned}$$

(132)