Demonstration of equivalence

Although shot-profile migration and source-receiver migration look totally different, they obtain both the same image and CIG. In this section, we prove that the mono-frequency image $I(x,z,\omega)$ and CIG $I(x,h,z,\omega)$ in the shot-profile migration are exactly the same mono-frequency image $P(x,h=0,z,\omega)$ and CIG $P(x,h,z,\omega)$ in the source-receiver migration, respectively.

We define a new wavefield $Q_s(x_U,x_D,z,\omega)$, which is the cross-correlation between the source wavefield $D(x_D,z,\omega,s)$ and the receiver wavefield $U(x_U,z,\omega,s)$ in the shot-profile migration for shot s, that is

$$Q_s(x_U,x_D,z,\omega)=U(x_U,z,\omega,s)\bar{D}(x_D,z,\omega,s),$$ (8)

and the wavefield $Q(x_U,x_D,z,\omega)$ is the stack of $Q_s(x_U,x_D,z,\omega)$ along all the shots,

$$Q(x_U,x_D,z,\omega)=\sum_s Q_s(x_U,x_D,z,\omega).$$ (9)

Obviously, from equation (7), $Q_s(x_U,x_D,z=0,\omega)$ is the surface data in source-receiver migration. We will demonstrate that the wavefield $Q_s(x_U,x_D,z,\omega)$ satisfies the DSR equation,  

$$\frac{\partial }{\partial z}Q_s=\left( \frac{i\omega}{v(x_U... ..._D,z)} {\omega^2}\frac{\partial^2}{\partial x_D^2}}\right)Q_s,$$ (10)

where $Q_s=Q_s(x_U,x_D,z,\omega)$. By extension, $Q(x_U,x_D,z,\omega)$ also satisfies the DSR equation. Thus shot-profile migration and source-receiver migration are two different ways to obtain wavefield Q at the subsurface. In shot-profile migration, source and receiver wavefields are downward continued into the subsurface with the one-way wave equation, and the wavefield $Q(x_U,x_D,z,\omega)$ is formed by cross-correlating the source wavefields and receiver wavefields and stacking over all shots at all depths. But in source-receiver migration, the wavefield $Q(x_U,x_D,z,\omega)$ at the surface $Q(x_U,x_D,z=0,\omega)$ is obtained by cross-correlating the source wavefield and the receiver wavefield at the surface, and $Q(x_U,x_D,z,\omega)$ is formed by extrapolating $Q(x_U,x_D,z=0,\omega)$ to all depths with the DSR equation.

From the Leibniz rule, we have  

$$\frac{\partial Q_s}{\partial z}=\frac{\partial U}{\partial z}\bar{D}+ U\frac{\partial \bar{D}}{\partial z},$$ (11)

where $U=U(x_U,z,\omega,s)$ and $D=D(x_D,z,\omega,s)$.Since $U(x_U,z,\omega,s)$ is an up-going wavefield, it satisfies the up-going wave equation(1), so we have  

$$\frac{\partial U}{\partial z}\bar{D}= \left( \frac{i\omega}... ...ega^2} \frac{\partial^2}{\partial x_U^2}}U\right)\cdot\bar{D}.$$ (12)

$\bar{D}(x_D,z,\omega,x)$ is not dependent on x_U, so it is constant with respect to the operator $\sqrt{1+\frac{v^2(x_U,z)}{\omega^2}\frac{\partial^2}{\partial x_U^2}}$, and we have  

$$\left(\sqrt{1+\frac{v^2(x_U,z)}{\omega^2}\frac{\partial^2}{\... ...frac{v^2(x_U,z)}{\omega^2}\frac{\partial^2}{\partial x_U^2}}Q_s$$ (13)

Summarizing equation (12) and (13), we have  

$$\frac{\partial U}{\partial z}\bar{D}= \frac{i\omega}{v(x_U,z... ...rac{v^2(x_U,z)}{\omega^2}\frac{\partial^2}{\partial x_U^2}}Q_s.$$ (14)

It is easy to prove that  

$$\frac{\partial \bar{D}}{\partial z}=\overline{\frac{\partial D}{\partial z}}.$$ (15)

So the second term of equation (11) changes to

$$U\frac{\partial \bar{D}}{\partial z}= U\overline{\frac{\partial D}{\partial z}}.$$ (16)

Since $D(x_D,z,\omega,s)$ is a down-going wavefield, it satisfies the down-going wave equation (2), and we have

$$\begin{eqnarray} U\cdot\overline{\frac{\partial D}{\partial z}}&=& U\cdot\overl... ..._D,z)}{\omega^2}\frac{\partial^2}{\partial x_D^2}}\bar{D} \right).\end{eqnarray}$$ (17) (18)

Again, $U(x_U,z,\omega,s)$ does not depend on x_D, so we have

$$\begin{eqnarray} U\left( \frac{i\omega}{v(x_D,z)} \sqrt{1+\frac{v^2(x_D,z)}{\om... ...\frac{v^2(x_D,z)}{\omega^2}\frac{\partial^2}{\partial x_D^2}} Q_s\end{eqnarray}$$ (19) (20)

Summarizing equations (15-20), we have  

$$U\frac{\partial \bar{D}}{\partial z}=\frac{i\omega}{v(x_D,z)... ...frac{v^2(x_D,z)}{\omega^2}\frac{\partial^2}{\partial x_D^2}}Q_s$$ (21)

Finally, from equation (11), equation (14) and equation (21), we know Q_s satisfies the DSR equation (10). Q is the stack of Q_s over all shots, so by extension Q satisfies the DSR equation also.

It is obvious that the image of shot-profile migration in equation (3) is

$$I(x,z,\omega)=Q(x_U=x,x_D=x,z,\omega),$$ (22)

and the corresponding CIG in equation (5) is

$$I(x,h,z,\omega)=Q(x_U=x+h,x_D=x-h,z,\omega).$$ (23)

In traditional source-receiver migration, $Q(x_U=x+h,x_D=x-h,z=0,\omega)$ is the stack of the cross-correlation between the impulse source and the recorded data along shots, which is the CMP gather $P(x,h,z=0,\omega)$ of the recorded data at the surface. Since both $Q(x_U=x+h,x_D=x-h,z,\omega)$ and $P(x,h,z,\omega)$ are obtained by propagating $Q(x_U=x+h,x_D=x-h,z=0,\omega)$ to the subsurface with the DSR equation (10), they are equivalent. If the source in source-receiver migration is not an impulse function, $Q(x_U=x+h,x_D=x-h,z=0,\omega)$ is the stack of the cross-correlation between the source wavefield and the receiver wavefield, and the same conclusion is reached. Thus we have

$$I(x,z,\omega)=P(x,h=0,z,\omega)=Q(x_U=x,x_D=x,z,\omega),$$ (24)

and

$$I(x,h,z,\omega)=P(x,h,z,\omega)=Q(x_U=x+h,x_D=x-h,z,\omega).$$ (25)