On Backus Average for Generally Anisotropic Layers

Abstract

In this paper, following the Backus (in J. Geophys. Res. 67(11):4427–4440, 1962) approach, we examine expressions for elasticity parameters of a homogeneous generally anisotropic medium that is long-wave-equivalent to a stack of thin generally anisotropic layers. These expressions reduce to the results of Backus (1962) for the case of isotropic and transversely isotropic layers.

In the over half-a-century since the publications of Backus (1962) there have been numerous publications applying and extending that formulation. However, neither George Backus nor the authors of the present paper are aware of further examinations of the mathematical underpinnings of the original formulation; hence this paper.

We prove that—within the long-wave approximation—if the thin layers obey stability conditions, then so does the equivalent medium. We examine—within the Backus-average context—the approximation of the average of a product as the product of averages, which underlies the averaging process.

In the presented examination we use the expression of Hooke’s law as a tensor equation; in other words, we use Kelvin’s—as opposed to Voigt’s—notation. In general, the tensorial notation allows us to conveniently examine effects due to rotations of coordinate systems.

Appendices

Appendix A: Average of Derivatives (Lemma 1)

Proof

We begin with the definition of averaging,

$$ \overline{f}(x_{3}):= \int _{-\infty}^{\infty}w(\xi-x_{3})f( \xi)\mathrm{d} \xi. $$

(12)

We make the nonrestrictive assumption that $f(\xi)$ is sufficiently regular to be able to perform the standard calculus operations that we make below. The derivatives with respect to $x_{1}$ and $x_{2}$ can be written as

$$\begin{aligned} \begin{aligned} \frac{\partial\overline{f}}{\partial x_{i}} &=\frac{\partial}{ \partial x_{i}} \int _{-\infty}^{\infty}w(\xi-x_{3})f(x_{1},x _{2},\xi)\mathrm{d}\xi \\ &= \int _{-\infty}^{\infty}w(\xi-x_{3})\frac{\partial f(x _{1},x_{2},\xi)}{\partial x_{i}} \mathrm{d}\xi=:\overline{\frac{ \partial f}{\partial x_{i}}} , \quad i=1,2 , \end{aligned} \end{aligned}$$

where the last equality is the statement of definition (12), as required. For the derivatives with respect to $x_{3}$, we need to verify that

$$ \frac{\partial}{\partial x_{3}} \int _{-\infty}^{\infty}w( \xi-x_{3})f(x_{1},x_{2}, \xi)\mathrm{d}\xi= \int _{-\infty} ^{\infty}w(\xi-x_{3})\frac{\partial f(x_{1},x_{2},\xi )}{\partial \xi} \mathrm{d}\xi. $$

(13)

Applying integration by parts, we write the right-hand side as

$$ w(\xi-x_{3})f(x_{1},x_{2},\xi)|_{-\infty}^{\infty}- \int _{-\infty}^{\infty}w'(\xi-x_{3}) f(x_{1},x_{2},\xi) \mathrm{d}\xi, $$

where $w$ is a function of a single variable. Since

$$ \lim_{x_{3}\rightarrow\pm\infty}w(x_{3})=0 , $$

the product of $w$ and $f$ vanishes at $\pm\infty$, and we are left with

$$ - \int _{-\infty}^{\infty}w'(\xi-x_{3}) f(x_{1},x_{2},\xi) \mathrm{d}\xi. $$

Let us consider the left-hand side of expression (13). Since only $w$ is a function of $x_{3}$, we can interchange the operations of integration and differentiation to write

$$ - \int _{-\infty}^{\infty}w'(\xi-x_{3}) f(x_{1},x_{2},\xi) \mathrm{d}\xi; $$

the negative sign arises from the chain rule,

$$ \frac{\partial w(\xi-x_{3})}{\partial x_{3}} =w'(\xi-x_{3})\frac{ \partial(\xi-x_{3})}{\partial x_{3}} =-w'(\xi-x_{3}) . $$

Thus, both sides of expression (13) are equal to one another, as required. In other words,

$$ \frac{\partial\overline{f}}{\partial x_{3}} = \overline{\frac{ \partial f}{\partial x_{3}}} , $$

which completes the proof. □

Appendix B: Stability of Equivalent Medium (Lemma 2)

Proof

The stability of layers means that their deformation requires work. Mathematically, it means that, for each layer,

$$ W=\frac{1}{2} \sigma\cdot\varepsilon>0 , $$

where $W$ stands for work, and $\sigma$ and $\varepsilon$ denote the stress and strain tensors, respectively, which are expressed as columns in (4): $\sigma=C\varepsilon$. As an aside, we can say that, herein, $W>0$ is equivalent to the positive definiteness of $C$, for each layer.

Performing the average of $W$ over all layers and using—in the scalar product—the fact that the average of a sum is the sum of averages, we write

$$ \overline{W}=\frac{1}{2} \overline{\sigma\cdot\varepsilon} >0 . $$

Thus, $W>0\implies\overline{W}>0$.

Let us proceed to show that this implication—in turn—entails the stability of the equivalent medium, which is tantamount to the positive definiteness of $\langle C\rangle$.

Following Lemma 3—if one of two functions is nearly constant—we can approximate the average of their product by the product of their averages,

$$ \overline{W}=\frac{1}{2} \overline{\sigma}\cdot \overline{\varepsilon} >0 . $$

(14)

Herein, we use the property stated in Sect. 2.4 that $\sigma_{i3}$, where $i\in\{1,2,3\}$, are constant, and $\varepsilon_{11}$, $\varepsilon_{12}$ and $\varepsilon_{22}$ vary slowly, along the $x_{3}$-axis, together with Lemma 3, which can be invoked due to the fact that each product in expression (14) is such that one function is nearly constant and the other possibly varies more rapidly.

By definition of Hooke’s law, $\overline{\sigma}:=\langle C\rangle \overline{\varepsilon}$, expression (14) can be written as

$$ \frac{1}{2} \bigl( \langle C\rangle\overline{\varepsilon} \bigr) \cdot\overline{\varepsilon} >0 , \quad\forall \overline{ \varepsilon}\neq0 , $$

which means that $\langle C\rangle$ is positive-definite, and which—in view of this derivation—proves that the equivalent medium inherits the stability of individual layers. □

Appendix C: Approximation of Product (Lemma 3)

For a fixed $x_{3}$, we may set $W(\zeta):=w(\zeta-x_{3})$. Then, $W\geq0$ and $\int_{-\infty}^{\infty}W(\zeta)\mathrm{d}\zeta =1$. With this notation, (3) becomes

$$\overline{f}:= \int _{-\infty}^{\infty}f(x)W(x)\mathrm{d}x . $$

Similarly,

$$ \overline{g}:= \int _{-\infty}^{\infty}g(x)W(x)\mathrm{d}x \quad\mbox{and} \quad\overline{fg}:= \int _{-\infty}^{\infty}f(x)g(x)W(x) \mathrm{d}x. $$

Proposition 1

Suppose that the first derivatives of $f$ and $g$ are uniformly bounded; that is, both

$$ \bigl\| f'\bigr\| _{\infty}:=\sup_{-\infty< x< \infty}\bigl|f'(x)\bigr| \quad \textit{and} \quad\bigl\| g'\bigr\| _{\infty}:=\sup_{-\infty< x< \infty}\bigl|g'(x)\bigr| $$

are finite. Then, we have

$$|{\overline{fg}}-{\overline{f}} {\overline{g}}|\leq2\bigl(\ell '\bigr)^{2} \bigl\| f'\bigr\| _{\infty} \bigl\| g'\bigr\| _{\infty}. $$

Proof

We may calculate

$$\begin{aligned} \begin{aligned} &\int _{-\infty}^{\infty} \bigl( f(x)-\overline{f} \bigr) \bigl( g(x)- \overline{g} \bigr) W(x)\mathrm{d}x \\ &\quad= \int _{-\infty}^{\infty}f(x)g(x)W(x)\mathrm{d}x- \overline{f} \int _{-\infty}^{\infty}g(x)W(x)\mathrm{d}x \\ &\qquad{}- \overline{g} \int _{-\infty}^{\infty}f(x)W(x)\mathrm{d}x+ \int _{-\infty}^{\infty}\overline{f}\overline{g}W(x) \mathrm{d}x \\ &\quad= \overline{fg}-\overline{f} \int _{-\infty}^{\infty}g(x) W(x)\mathrm{d}x -\overline{g} \int _{-\infty}^{\infty}f(x)W(x) \mathrm{d}x+\overline{f} \overline{g} \\ &\quad=\overline{fg}-\overline{f} \overline{g}-\overline{g} \overline{f}+ \overline{f} \overline{g}= \overline{fg}-\overline{f} \overline{g} ; \end{aligned} \end{aligned}$$

that is,

$$ \overline{fg}-\overline{f} \overline{g} = \int _{-\infty} ^{\infty} \bigl( f(x)-\overline{f} \bigr) \bigl( g(x)-\overline{g} \bigr) W(x) \mathrm{d}x. $$

(15)

Now,

$$f(x)-\overline{f}=f(x)- \int _{-\infty}^{\infty}f(y)W(y) \mathrm{d}y = \int _{-\infty}^{\infty} \bigl( f(x)-f(y) \bigr) W(y) \mathrm{d}y , $$

so that

$$ \bigl|f(x)-\overline{f}\bigr|\leq\bigl\| f'\bigr\| _{\infty} \int _{-\infty} ^{\infty}|x-y|W(y)\mathrm{d}y , $$

(16)

and hence, by the Cauchy-Schwartz inequality,

$$\bigl|f(x)-\overline{f}\bigr|^{2}\leq\bigl\| f'\bigr\| _{\infty}^{2} \int _{-\infty}^{\infty}|x-y|^{2}W(y)\mathrm{d}y \int _{-\infty}^{\infty}1^{2}W(y)\mathrm{d}y= \bigl\| f'\bigr\| _{\infty} ^{2} \int _{-\infty}^{\infty}|x-y|^{2}W(y)\mathrm{d}y. $$

Thus,

$$\begin{aligned} \int _{-\infty}^{\infty}\bigl|f(x)-\overline{f}\bigr|^{2}W(x) \mathrm{d}x & \leq\bigl\| f'\bigr\| _{\infty}^{2} \int _{-\infty}^{\infty} \int _{-\infty}^{\infty}|x-y|^{2}W(x)W(y)\mathrm{d}x \mathrm{d}y \\ &= \bigl\| f'\bigr\| _{\infty}^{2} \int _{-\infty}^{\infty} \int _{-\infty}^{\infty}\bigl(x^{2}-2xy+y^{2} \bigr)W(x)W(y)\mathrm{d}x \mathrm{d}y \\ &= \bigl\| f'\bigr\| _{\infty}^{2} \biggl( 2 \int _{-\infty}^{\infty}x ^{2}W(x)\mathrm{d}x-2 \biggl( \int _{-\infty}^{\infty}xW(x) \mathrm{d}x \biggr) ^{2} \biggr) . \end{aligned}$$

It follows, by the Cauchy-Schwartz inequality applied to (15), that

$$\begin{aligned} |\overline{fg}-\overline{f} \overline{g}|^{2} &\leq \int _{-\infty}^{\infty}\bigl|f(x)-\overline{f}\bigr|^{2}W(x) \mathrm{d}x \int _{-\infty}^{\infty}\bigl|g(x)-\overline{g}\bigr|^{2}W(x) \mathrm{d}x \\ &\leq\bigl\| f'\bigr\| _{\infty}^{2} \bigl\| g'\bigr\| _{\infty}^{2} \biggl( 2 \int _{-\infty}^{\infty}x^{2}W(x)\mathrm{d}x-2 \biggl( \int _{-\infty}^{\infty}xW(x)\mathrm{d}x \biggr) ^{2} \biggr) ^{2} . \end{aligned}$$

Note that

$$\int _{-\infty}^{\infty}xW(x)\mathrm{d}x= \int _{-\infty}^{ \infty}xw(x-x_{3})\mathrm{d}x = \int _{-\infty}^{\infty}(x+x _{3})w(x) \mathrm{d}x=x_{3}, $$

using the defining properties of $w(\zeta)$. Similarly

$$\int _{-\infty}^{\infty}x^{2}W(x)\mathrm{d}x= \int _{-\infty}^{\infty}x^{2}w(x-x_{3}) \mathrm{d}x= \int _{-\infty}^{\infty}(x+x_{3})^{2}w(x) \mathrm{d}x=\bigl(\ell'\bigr)^{2}+x _{3}^{2} . $$

Consequently,

$$2 \int _{-\infty}^{\infty}x^{2}W(x)\mathrm{d}x-2 \biggl( \int _{-\infty}^{\infty}xW(x)\mathrm{d}x \biggr) ^{2}=2\bigl(\bigl( \ell'\bigr)^{2}+x_{3}^{2}-x_{3}^{2} \bigr)=2\bigl(\ell'\bigr)^{2} $$

and we have

$$|\overline{fg}-\overline{f} \overline{g}|\le2\bigl(\ell' \bigr)^{2}\bigl\| f'\bigr\| _{\infty}\bigl\| g' \bigr\| _{\infty}, $$

as claimed. □

Hence, if $f$ and $g$ are nearly constant, which means that $\|f'\|_{\infty}$ and $\|g'\|_{\infty}$ are small, then $\overline{fg}\approx\overline{f} \overline{g}$.

Corollary 1

Since the error estimate involves the product of the norms of the derivatives, it follows that if one of them is small enough and the other is not excessively large, then their product can be small enough for the approximation, $\overline{fg}\approx\overline{f} \overline{g}$, to hold.

The exact accuracy of this property will be examined further by numerical methods in a future publication.

If $g(x)\geq0$, we can say more, even if $g(x)$ is wildly varying. If $f$ is continuous and $g(x)\geq0$, then, by the Mean-value Theorem for Integrals,

$$\overline{fg}= \int _{-\infty}^{\infty}f(x)g(x)W(x)\mathrm{d}x=f(c) \int _{-\infty}^{\infty}g(x)W(x)\mathrm{d}x=f(c) \overline{g} , $$

for some $c$. Hence,

$$\overline{fg}-\overline{f}\overline{g}=f(c)\overline{g}- \overline {f} \overline{g}= \bigl(f(c)-\overline{f}\bigr)\overline{g}. $$

This implies that

$$|\overline{fg}-\overline{f}\overline{g}|\leq\bigl|f(c)- \overline {f}\bigr| \overline{g} \leq\bigl\| f'\bigr\| _{\infty} \biggl( \int _{-\infty}^{\infty}|x-y|W(y)\mathrm{d}y \biggr) \overline{g} . $$

Hence, even for $g$ wildly varying—as long as $\overline{g}$ is not too big in relation to $\|f'\|_{\infty}$—it is still the case that the average of the product is close to the product of the averages. A bound on ${\int_{-\infty}^{\infty}|x-y|W(y)\mathrm{d}y}$ would depend on the weight function, $w$, used.

An alternative estimate is provided by the following proposition.

Proposition 2

Suppose that $m:=\inf f(x)>-\infty$ and $M:=\sup_{-\infty < x<\infty}f(x)<\infty$ and ${\sup_{-\infty< x<\infty }|g(x)|<\infty}$. Then,

$$ |\overline{fg}-\overline{f}\overline{g}|\leq\Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) (M-m). $$

Proof

$$ \overline{fg}= \int _{-\infty}^{\infty}f(x)g(x)W(x)\mathrm{d}x = \int _{-\infty}^{\infty}\bigl(f(x)-m\bigr)g(x)W(x) \mathrm{d}x+m \int _{-\infty}^{\infty}g(x)W(x)\mathrm{d}x, $$

which—by the definition of the average—is

$$ \overline{fg} = \int _{-\infty}^{\infty}\bigl(f(x)-m\bigr)g(x)W(x) \mathrm{d}x+m\overline{g} . $$

Hence,

$$\begin{aligned} |\overline{fg}-\overline{f}\overline{g}| &= \biggl\vert \int _{-\infty}^{\infty}\bigl(f(x)-m\bigr)g(x)W(x)\mathrm{d}x+ ( m- \overline{f} ) \overline{g}\biggr\vert \\ &\leq \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) \int _{-\infty}^{\infty}\bigl(f(x)-m\bigr)W(x)\mathrm{d}x+ \vert m- \overline{f}\vert |\overline{g}| \\ &= \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) ( \overline{f}-m ) + ( \overline{f}-m ) |\overline{g}| \\ &\leq2 \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) ( \overline{f}-m ) . \end{aligned}$$

(17)

Similarly,

$$ |\overline{fg}-\overline{f}\overline{g}|\leq 2 \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) ( M-\overline{f} ) . $$

(18)

Taking the average of expressions (17) and (18), we obtain

$$\begin{aligned} |\overline{fg}-\overline{f}\overline{g}|&\leq 2 \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) \frac{ ( \overline{f}-m ) + ( M-\overline{f} ) }{2} \\ &= \Bigl( \sup _{-\infty< x< \infty}\bigl|g(x)\bigr| \Bigr) (M-m) , \end{aligned}$$

as required. □

Consequently, if $f(x)$ is almost constant—which means that $m\approx M$—then $\overline{fg}\approx\overline{f}\overline{g}$.