P-Impedance and V_p/V_s prediction based on AVO inversion scheme with deep feedforward neural network: a case study from tight sandstone reservoir

Abstract

The low-frequency component of seismic data is an inevitable part to obtain absolute P-impedance ($I_{p}$) and $V_{p} /V_{s}$ ratio of the subsurface, especially for the reservoir sweet spot. In this work, we train the deep feedforward neural network (DFNN) with band-pass seismic data and well log data to obtain favorable low-frequency components. Specifically, the Bayesian inference strategy is first applied to the pre-stack constrained sparse spike inversion process, obtaining an “initial” inverted band-pass parameters, which are subsequently used as input when applying the DFNN algorithm to predict low- and band-pass parameters. Moreover, the high linear correlation coefficient between the DFNN-based inversion results and the realistic well logging curves of the blind wells demonstrates that the DFNN-based inversion scheme exhibits strong robustness and good generalization ability. Ultimately, we apply the proposed DFNN-based inversion strategy to a tight sandstone reservoir located at the Sichuan basin field from onshore China. Both low- and band-pass $I_{p}$ and $V_{p} /V_{s}$ inverted for the clastic formation of the Sichuan basin show a strong correlation with the corresponding $I_{p}$ and $V_{p} /V_{s}$ logs.

Appendix A

This appendix differentiates the regularization term $R(m)$ in Eq. 6. $R(m)$ can be written as:

$$\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial {\mathbf{m}}}} = \left[ {\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{1} }} \cdot \cdot \cdot \frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{k} }} \cdot \cdot \cdot \frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{3N} }}} \right]^{T}.$$

(17)

Taking the derivative of $R(m)$ with respect to $m_{k}$, then we can get:

$$\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{k} }} = 2\sum\limits_{i = 1}^{N} {\left( {\frac{1}{{1 + {\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}}}} \right)} \frac{\partial }{{\partial m_{k} }}{\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}.$$

(18)

If expanding the second term of Eq. (18), then we have:

$$\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{k} }} = 2\sum\limits_{i = 1}^{N} {\left( {\frac{1}{{1 + {\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}}}} \right)} \frac{\partial }{{\partial m_{k} }}\left( {\sum\nolimits_{l = 1}^{3N} {\sum\nolimits_{N = 1}^{3N} {m_{l} m_{n} {{\varvec{\Phi}}}_{\ln }^{i} } } } \right).$$

(19)

By using the fact $\frac{{\partial m_{p} }}{{\partial m_{q} }} = \delta_{pq} \left\{ {\begin{array}{*{20}l} { = 1,} \hfill & {{\text{if}}\,p = q} \hfill \\ { = 0,} \hfill & {{\text{if}}\,p \ne q} \hfill \\ \end{array} } \right.$, Eq. (19) can be written as:

$$\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{k} }} = 2\sum\nolimits_{i = 1}^{N} {\left( {\frac{{2\sum\nolimits_{l = 1}^{3N} {m_{n} {{\varvec{\Phi}}}_{kn}^{i} } }}{{1 + {\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}}}} \right)}.$$

(20)

After changing the order of summation of Eq. (20), then:

$$\frac{{\partial R\left( {\mathbf{m}} \right)}}{{\partial m_{k} }} = 2\sum\nolimits_{i = 1}^{3N} {\left( {\sum\limits_{l = 1}^{N} {\frac{{2{{\varvec{\Phi}}}_{kn}^{i} }}{{1 + {\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}}}} } \right)} m_{n} { = }2\sum\nolimits_{i = 1}^{3N} {\Phi_{kn} } m_{n} ,$$

(21)

where $Q_{kn} = \sum\nolimits_{i = 1}^{N} {\frac{{2{{\varvec{\Phi}}}_{kn}^{i} }}{{1 + {\mathbf{m}}^{T} {{\varvec{\Phi}}}^{i} {\mathbf{m}}}}, \, k,n = 1,2,3, \ldots ,3N}.$