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Second space derivatives

The defining equation for a second difference operator is  
 \begin{displaymath}
{ \delta^2 \ \over \delta x^2 } \ P \eq 
{P(x\ +\ \Delta x)\ -\ 2P(x)\ +\ P(x\ -\ \Delta x) \over ( \Delta x )^2 }\end{displaymath} (30)
The second derivative operator is defined by taking the limit  
 \begin{displaymath}
{\partial^2 \over \partial x^2} \ P \ \ 
\matrix{\ \ \cr = \cr \lim \ \Delta x \to 0}\ \ {\delta^2\over \delta x^2}\ P\end{displaymath} (31)
Many different definitions can all go to the same limit as $\Delta x$ goes to zero. The problem is to find an expression that is accurate when $\Delta x$ is larger than zero and, on a practical level, is not too complicated. Our first objective is to see how the accuracy of equation (30) can be evaluated quantitatively. Second, we will look at an expression that is slightly more complicated than (30) but much more accurate.

The basic method of analysis we will use is Fourier transformation. Take the derivatives of the complex exponential $ P = P_0 \ \exp \ ( i k x ) $ and look at any errors as functions of the spatial frequency k. For the second derivative,  
 \begin{displaymath}
{ \partial^2 \ \over \partial x^2 } \ P
 \eq -\, k^2 \,P\end{displaymath} (32)
Define $ \hat k $ by an expression analogous to the difference operator:  
 \begin{displaymath}
{ \delta^2 \ \over \delta x^2 }
\ P
 \eq -\, \hat k^{2\,} P\end{displaymath} (33)
Ideally $ \hat k $ would equal k. Inserting the complex exponential $ P = P_0 \ \exp \ ( i k x ) $ into (30), we see that the definition (33) gives an expression for $ \hat k $ in terms of k:  
 \begin{displaymath}
- \hat k^2 P \eq {P_0 \over \Delta x^2 } \ 
\left[ \ e^{ik(x+\Delta x)}\ -\ 2\,e^{ikx} + e^{ik( x - \Delta x )} \right]\end{displaymath} (34)
 
 \begin{displaymath}
-\ { \delta^2 \ \over \delta x^2 }
 \eq 
\hat k^2 \eq {2 \over \Delta x^2 } \ 
[ 1 \ -\ \cos (k \Delta x ) ]\end{displaymath} (35)
It is a straightforward matter to make plots of $ \hat k \Delta x $versus $ k \Delta x $ from (35). The half-angle trig formula allows us to take an analytic square root of (35), which is  
 \begin{displaymath}
{\hat k \, \Delta x \over 2 } \eq \sin \ { k \, \Delta x \over 2 }\end{displaymath} (36)
Series expansion shows that for low frequencies $ \hat k $ is a good approximation to k. At the Nyquist frequency, defined by $ k \Delta x = \pi $,the approximation $ \hat k \Delta x = 2 $is a poor approximation to $ \pi $.


previous up next print clean
Next: The 1/6 trick Up: FREQUENCY DISPERSION IN WAVE-MIGRATION Previous: Spatial aliasing various space
Stanford Exploration Project
10/31/1997