Block Hankel matrix formulation for 5D seismic data

Consider a block of 5D data $\mathbf{D}_{time}(t,hx,hy,x,y)(t=1\cdots N_t)$, where $hx$, $hy$, $x$, $y$ denote $x$ and $y$ offsets and $x$ and $y$ midpoints. The rank-reduction method operates on the 5D seismic data in the following way: first, rank-reduction method transforms $\mathbf{D}_{time}(t,hx,hy,x,y)$ into $\mathbf{D}_{freq}(w,hx,hy,x,y)(w=1\cdots N_w)$ of complex values in the frequency domain. In the condition that the spatial variables are regularly sampled, each observation at a given frequency slice can be represented via a 4D spatial hypercube $\mathbf{D}_{k1,k2,k3,k4}$ with $k_i=1\cdots X_i$, $i=1,2,3,4$. In order to make our target matrices close to square matrices, parameters $Y_i$ are defined as $\lfloor \frac{X_i}{2} \rfloor +1$, $i=1,2,3,4$. Gao et al. (2015a) outlined the usefulness of square Hankel matrices in rank reduction based regularization and described why square matrices are preferred. The symbol $\lfloor\cdot\rfloor$ denotes the integer part of its argument. The 4D spatial hypercube $\mathbf{D}_{k1,k2,k3,k4}$ can be embedded in a level-four block Hankel matrix as follows. We have to address in advance that each frequency slice of data is operated the exactly same way. We first embed the seismic data in a level-one Hankel matrix using all data components in the first dimension of the tensor $\mathbf{D}_{k1,k2,k3,k4}$ at a given frequency $w_0$. This generates the following Hankel matrices,

$$\mathbf{M}_{k2,k3,k4}^{(1)}=\left(\begin{array}{cccc} D_{1,k2,k3,... ..._{Y_1,k2,k3,k4}&D_{Y_1+1,k2,k3,k4} &\cdots&D_{X_1,k2,k3,k4} \end{array}\right).$$ (1)

Matrices in Equation 1 are of size $Y_1 \times (X_1-Y_1+1)$. These matrices are embedded in a level-two block Hankel matrix:

$$\mathbf{M}_{k3,k4}^{(2)}=\left(\begin{array}{cccc} M_{1,k3,k4}^{(... ...k4}^{(1)}&M_{Y_2+1,k3,k4}^{(1)} &\cdots&M_{X_2,k3,k4}^{(1)} \end{array}\right).$$ (2)

Matrices in Equation 2 are of size $(Y_1 Y_2) \times (X_1-Y_1+1)(X_2-Y_2+1)$. These matrices are now embedded in a level-three block Hankel matrix:

$$\mathbf{M}_{k4}^{(3)}=\left(\begin{array}{cccc} M_{1,k4}^{(2)} & ... ..._{Y_3,k4}^{(2)}&M_{Y_3+1,k4}^{(2)} &\cdots&M_{X_3,k4}^{(2)} \end{array}\right).$$ (3)

Matrices in Equation 3 are of size $(Y_1 Y_2 Y_3) \times (X_1-Y_1+1)(X_2-Y_2+1)(X_3-Y_3+1)$. At last, these matrices are embedded in our final target level-four block Hankel matrix:

$$\mathbf{M}^{(4)}=\left(\begin{array}{cccc} M_{1}^{(3)} & M_{2}^{(... ...ots \\ M_{Y_4}^{(3)}&M_{Y_4+1}^{(3)} &\cdots&M_{X_4}^{(3)} \end{array}\right).$$ (4)

Equation 4 is the target matrix of our problem. The size of level-four block Hankel matrix $\mathbf{M}^{(4)}$ is $(Y_1 Y_2 Y_3 Y_4) \times (X_1-Y_1+1)(X_2-Y_2+1)(X_3-Y_3+1)(X_4-Y_4+1)$. If $X_i$ is an odd integer, the size of $\mathbf{M}^{(4)}$ is $(Y_1 Y_2 Y_3 Y_4) \times(Y_1 Y_2 Y_3 Y_4)$.