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A natural extension of dot products in (t,x)-space and $(\tau,m)$-space in a continuous domain is
(\bold d_1,\bold d_2)&=&\int\!\! \int d_1(t,x)d_2(t,x)\, dt \, ...
 ..._1,\bold u_2)&=&\int\!\! \int u_1(\tau,m)u_2(\tau,m)\, d\tau\, dm.\end{eqnarray} (22)
The definition of an adjoint operator in a continuous domain cannot be done straightforwardly by transposing a corresponding matrix (we simply do not have the matrix). Instead another equivalent property is used for definition:
(\bold L \bold u,\bold d)=(\bold u,\bold L^{\bold T}\bold d).\end{displaymath} (24)
This is the equation that is used for the dot product tests (Claerbout, 1985).

The definition of a velocity stacking operator $\bold H$ in (t,x)-space is
u(\tau,m)=\int d(\sqrt{\tau^2+x^2m},x) \, dx.\end{displaymath} (25)
Similarly we can define a stacking operator $\bold P$ in $(\tau,m)$-space:
d(t,x)=\int u(\sqrt{t^2-x^2m},m)\, dm.\end{displaymath} (26)
Now the question is what is the relationship between the two operators. It can be shown (Thorson, 1984) that they are adjoint for dot products defined by equations
(\bold d_1,\bold d_2)&=&\int\!\! \int t \, d_1(t,x) d_2(t,x) \,...
 ...2)&=&\int\!\! \int \tau \, u_1(\tau,m) u_2(\tau,m) \, d\tau \, dm.\end{eqnarray} (27)
They are not, however, adjoint for the dot products defined by equations (23). If we want to apply an algorithm using Euclidean norms, then our operators should be adjoint with respect to the Euclidean norms.

Let us derive the adjoint operator $\bold H^{\bold T}$:
(\bold u,{\bf Hd})&=&\int\!\! \int dm \, d\tau \, u(m,\tau) \in...
 ...\int\!\! \int dm \,d\tau \,dx \, u(m,\tau) d(\sqrt{\tau^2+x^2m},x)\end{eqnarray} (29)
After substitution $t=\sqrt{\tau^2+x^2m}$, $d\tau=t/\tau \,dt$ we obtain
(\bold u,{\bf Hd})=\int\!\! \int dt \, dx \,d(t,x) \int dm \,u(\sqrt{t^2-x^2m},m){t \over \sqrt{t^2-x^2m}}.\end{displaymath} (31)
Finally we have
{\bf H^T u}=\int u(\sqrt{t^2-x^2m},m){t \over \sqrt{t^2-x^2m}} \, dm.\end{displaymath} (32)
Similarly the adjoint operator ${\bf P^T}$ may be derived:
{\bf P^T d}=\int d(\sqrt{\tau^2+x^2m},x){\tau \over \sqrt{\tau^2+x^2m}}\, dx.\end{displaymath} (33)
For short reference we will write the above equations symbolically
{\bf H^T} = {t \over \tau} \bold P \\ {\bf P^T} = {\tau \over t} \bold H\end{eqnarray} (34)
We may weight both operators in the offset and velocity space. Here an infinite number of weighting factors can be obtained and a problem is which pair of them to choose. Here are some of them:

offset 1 $\tau/t$ $\sqrt{\tau/t}$ 1/t $\tau$
sloth $t/\tau$ 1 $\sqrt{t/\tau}$ $1/\tau$ t

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