Global Analysis of Stationary Points in Hyperbolic Moveout Approximation

Until further notice regard F etc. as depending on RMS square slowness u rather than on interval velocity v. Dependence on v, through the relatively easily analyzed map $v \mapsto u$, will be reintroduced at the end.

A short calculation shows that

$$p(t,x)= \frac{x}{t}u(T_0(t,x)).$$

Introduce the quantity $\Gamma$, with units of time:

$$\Gamma(t_0,x)=T_0(T^*(t_0,x),x)$$

That is, $\Gamma(t_0,x)$ is the zero offset time for which the time at offset x is the same in the slowness u as the time one obtains for t₀,x in slowness u^*.

Then introducing the expression for p, and changing variables from t to t₀ in the integral above, yields

$$J_0[u]=\int\int \,dt_0\,dx\,B_0(t_0,x)(u(\Gamma(t_0,x))-u^*(t_0))^2 (r^*(t_0))^2 + O(\lambda)$$

where

$$B_0(t_0,x)= \phi(T^*(t_0,x),x)B^*(T^*(t_0,x),x)\left(\frac{x}{T^*(t_0,x)}\right)^2$$

depends only on u^* and $\cal{A}$.

It is now straightforward to compute the first order perturbation $\delta J_0$ of J₀ with respect to u. First,

$$\delta T(t_0,x) = \frac{\frac{x^2}{2}\delta u(t_0)}{T(t_0,x)}$$

$$0 = \delta (T(T_0(t,x),x)) = \delta T(T_0(t,x),x) + \frac{\partial T}{\partial t_0}(T_0(t,x),x)\delta T_0(t,x)$$

$$=\frac{\frac{x^2}{2}\delta u(T_0(t,x))}{t}+\frac{\delta T_0(t,x)}{s(t,x)}$$

so

$$\delta T_0(t,x)=-\frac{\frac{x^2}{2}s(t,x)\delta u(T_0(t,x))}{t}$$

whence

$$\delta \Gamma(t_0,x) = \delta T_0(T^*(t_0,x),x) =-\frac{\frac{x^2}{2}s(T^*(t_0,x),x)\delta u(\Gamma(t_0,x))}{T^*(t_0,x)}$$

and

$$\delta (u(\Gamma(t_0,x))) = \delta u(\Gamma(t_0,x)) + \frac{du}{dt_0}(\Gamma(t_0,x))\delta \Gamma(t_0,x))$$

$$=\delta u(\Gamma(t_0,x))\left(1-\frac{\frac{x^2}{2}s(T^*(t_0,x),x)\frac{du}{dt}(\Gamma(t_0,x))}{T^*(t_0,x)}\right)$$

Recall that

$$s(t,x)=\frac{t}{T_0(t,x)+\frac{x^2}{2} \frac{\partial u}{\partial t_0}(T_0(t,x))}$$

so that

$$s(T^*(t_0,x),x)= \frac{T^*(t_0,x)}{\Gamma(t_0,x)+\frac{x^2}{2} \frac{\partial u}{\partial t_0}(\Gamma(t_0,x))}$$

so

$$\delta (u(\Gamma(t_0,x))) = \delta u(\Gamma(t_0,x))\left(1-... ...{x^2}{2} \frac{\partial u}{\partial t_0}(\Gamma(t_0,x))}\right)$$

$$=\frac{\delta u(\Gamma(t_0,x))}{1+\frac{x^2}{2\Gamma(t_0,x)}... ...Gamma(t_0,x)}{T^*(t_0,x)}s(T^*(t_0,x),x)\delta u(\Gamma(t_0,x))$$

Putting this all together,

$$\delta J_0[u]= \int\int \,dt_0\,dx\,B_1(t_0,x)(u(\Gamma(t_0,x))-u^*(t_0)) (r^*(t_0))^2\delta u(\Gamma(t_0,x)) + O(\lambda)$$

where

$$B_1(t_0,x)=\frac{B_0(t_0,x)}{1+\frac{x^2}{2\Gamma(t_0,x)} \f... ...x))}=\frac{\Gamma(t_0,x)}{T^*(t_0,x)} B_0(t_0,x)s(T^*(t_0,x),x)$$

depends on u, u^*, and $\cal{A}$.To compute the gradient, change variables again to $t_0'=\Gamma(t_0,x)$ for each x. Since

$$\frac{\partial \Gamma}{\partial t_0}(t_0,x) =\frac{\partial}{\partial t_0}(T^*_0(T(t_0,x),x))$$

$$=\frac{\partial T_0^*}{\partial t}(T(t_0,x),x)\frac{\partial T}{\partial t_0} (t_0,x) =\frac{s^*(T(t_0,x),x)}{s(T(t_0,x),x)}$$

and so

$$\frac{\partial \Gamma^{-1}}{\partial t_0}(t_0,x) =\frac{s(T^*(t_0,x),x)}{s^*(T^*(t_0,x),x)}$$

you get

$$\delta J_0[u]= \int\int \,dt_0\,dx\,B_1^*(t_0,x)(u(t_0)-u^*(... ...t_0,x))) (r^*(\Gamma^{-1}(t_0,x)))^2\delta u(t_0)) + O(\lambda)$$

with

$$B_1^*(t_0,x)=B_1(\Gamma^{-1}(t_0,x),x)\frac{s(T^*(t_0,x),x)}{s^*(T^*(t_0,x),x)}.$$

Thus the L² gradient of J₀ is

$$\nabla J_0[u](t_0) = \int \,dx\,B_1^*(t_0,x)(u(t_0)-u^*(\Gamma^{-1}(t_0,x))) (r^*(\Gamma^{-1}(t_0,x)))^2 + O(\lambda)$$

Both expressions for J₀ and its gradient suggest that these quantities are comparing the trial square slowness u and the target square slowness u^* at different points (eg. t₀ vs. $\Gamma^{-1}(t_0,x)$), and this in turn makes understanding of the implications for determination of u difficult. Fortunately this is not really the case:

Key Lemma: There exists a function h(t₀,x), depending on velocity v (or slowness u) and also on u^* and $\cal{A}$, having the following properties:

• h(t₀,x)>0 over the mute zone, and $\log h(t_0,x)$ is uniformly bounded for t₀,x in the mute zone and $v \in \cal{A}$;
• $u(t_0)-u^*(\Gamma^{-1}(t_0,x)) = h(t_0,x)(u(t_0)-u^*(t_0))$

Proof of Key Lemma: Note first that since

T^*(T^*₀(t,x),x)=t

$$\frac{\partial T^*}{\partial x}(T_0^*(t,x),x)+\frac{\partial... ...l t_0} (T^*_0(t,x),x)\frac{\partial T^*_0}{\partial x}(t,x) = 0$$

$$=\frac{xu^*(T_0^*(t,x))}{t}+\frac{1}{s^*(t,x)}\frac{\partial T^*_0}{\partial x}(t,x)$$

so

$$\frac{\partial T^*_0}{\partial x}(t,x)=-s^*(t,x)\frac{xu^*(T_0^*(t,x))}{t}.$$

It follows that since

$$\Gamma^{-1}(t_0,x)=T_0^*(T(t_0,x),x),$$

$$\frac{\partial \Gamma^{-1}}{\partial x}(t_0,x)= \frac{\parti... ...partial x}(t_0,x)+\frac{\partial T_0^*}{\partial x}(T(t_0,x),x)$$

$$=s^*(T(t_0,x),x)\frac{xu(t_0)}{T(t_0,x)}-s^*(T(t_0,x),x)\frac{xu^*(T_0^*(T(t_0,x),x))}{T(t_0,x)}$$

$$=\frac{xs^*(T(t_0,x),x)}{T(t_0,x)}(u(t_0)-u^*(\Gamma^{-1}(t_0,x)))$$

Thus

$$u(t_0)-u^*(\Gamma^{-1}(t_0,x)) = u(t_0)-u^*(t_0) + \int_0^x\... ...\,\frac{\partial}{\partial x'}(u(t_0)-u^*(\Gamma^{-1}(t_0,x')))$$

$$=u(t_0)-u^*(t_0) - \int_0^x\,dx'\,\frac{\partial u^*}{\parti... ...a^{-1}(t_0,x'))\frac{\partial \Gamma^{-1}}{\partial x'}(t_0,x')$$

$$=u(t_0)-u^*(t_0) - \int_0^x\,dx'\,\frac{\partial u^*}{\parti... ...'s^*(T(t_0,x'),x')}{T(t_0,x')}(u(t_0)-u^*(\Gamma^{-1}(t_0,x')))$$

$$=u(t_0)-u^*(t_0) + \int_0^x\,dx'g(t_0,x')(u(t_0)-u^*(\Gamma^{-1}(t_0,x')))$$

where

$$g(t_0,x)=-\frac{\partial u^*}{\partial t_0} (\Gamma^{-1}(t_0,x'))\frac{x's^*(T(t_0,x'),x')}{T(t_0,x')}$$

This simple integral equation has the solution

$$u(t_0)-u^*(\Gamma^{-1}(t_0,x')) = h(t_0,x)(u(t_0)-u^*(t_0))$$

where

$$h(t_0,x)=\exp \left(\int_0^x\,dx'g(t_0,x')\right)$$

has the properties claimed for it in the statement of the lemma. Q.E.D.

Now changing variables in the asymptotic formula for J₀, and applying the above relation to both this and the formula for $\nabla J_0$, you obtain

$$J_0[u]=\int\int \,dt_0(u(t_0)-u^*(t_0))^2 \,dx\,B^*_0(t_0,x)h(t_0,x)(r^*(\Gamma^{-1}(t_0,x)))^2 + O(\lambda)$$

$$\nabla J_0[u](t_0) = (u(t_0)-u^*(t_0)) \int \,dx\,B_1^*(t_0,x)h(t_0,x)(r^*(\Gamma^{-1}(t_0,x)))^2 + O(\lambda)$$

where

$$B_0^*(t_0,x)=B_0(\Gamma^{-1}(t_0,x),x)\frac{s(T^*(t_0,x),x)}{s^*(T^*(t_0,x),x)}.$$

Now B₀^* and B₁^* differ at each point in the mute zone by factors or divisors of s, s^*, T, and the like, and these are bounded over the mute zone uniformly in $v \in \cal{A}$. Therefore there exists a constant C>0 depending only on $\cal{A}$ for which

$$J_0[u] \le C \int\, dt_0\, (u(t_0)-u^*(t_0))\nabla J_0[u](t_0) + O(\lambda)$$

and we have proved the

Theorem: If u, the RMS square slowness for $v \in \cal{A}$,is a stationary point of J₀[u], then $J_0[u] = O(\lambda)$.

That is, for noise free data, any stationary point of J₀ is a global minimizer, up to an asymptotically vanishing error.