PDE to reject two dips

You can reject two dips with the operator  

$$(\partial_x-p_1\partial_t)(\partial_x-p_2\partial_t)$$ (1)

Finding the two values p₁ and p₂ is a nonlinear problem that is easy. Let u be the input signal and v be the output signal. Consider  

$$\left( {\partial^2 \over \partial x^2} + a {\partial^2 \o... ...\partial t^2} \right) \, u(t,x) {=}v(t,x) \quad\approx\quad 0$$ (2)

Now recognize that (2) (which is a separate equation at each point in the (t,x) space of v(t,x)) is an overdetermined set of linear equations for the two unknowns a and b. It is easy to find a and b which by comparison with equation (1) gives p₁ and p₂ by the nonlinear but easy equations a=p₁+p₂ and b=p₁p₂. The minimum signal required is three seismograms (on which to express $\partial_{xx}$).

To recapitulate, first a simple procedure gives us the required coefficients for a filter that fits two waves to the dataset u(t,x). Second, the same dip filter coefficients can be applied on a mesh in which t and x are interleaved, thus introducing new data locations on which we need data. Third, we interpolate t by any method. Fourth, we find missing traces by minimizing the power in v(t,x).