# Potential functions used to solve wave equations

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.9a

Show that equation (2.9a) relating the potential functions ${\displaystyle {\phi }}$ and ${\displaystyle {\chi }}$ to the vector displacement ${\displaystyle {\zeta }}$ requires that ${\displaystyle {\Delta }}$ and ${\displaystyle {\theta _{z}}}$ [see equations (2.1e) and (2.1g)] be solutions of the P- and S-wave equations, that is, of equation (2.5a) with ${\displaystyle {\psi }}$ replaced by ${\displaystyle {\Delta }}$ and ${\displaystyle {\theta _{z}}}$, respectively.

 ${\displaystyle {\mathbf {\zeta =\nabla \left(\mathrm {\phi } +{\frac {\partial \chi }{\partial z}}\right)-\nabla ^{2}\mathrm {\chi } k} },}$ (2.9a)

${\displaystyle {\phi }}$ and ${\displaystyle {\chi }}$ being solutions of the P- and S-wave equations, respectively.

### Background

The dilatation ${\displaystyle \mathrm {\Delta } }$ and component of rotation ${\displaystyle \mathrm {\theta } _{x}}$ are defined in equations (2.1e,g).

While solutions of the wave equation (see problem 2.5) furnish values of ${\displaystyle \mathrm {\Delta } }$ or a component of rotation ${\displaystyle \mathrm {\theta } _{i}}$, we often need to know the displacements ${\displaystyle \left(u,\;v,\;w\right)}$ (defined in problem 2.1) which are not easily found given ${\displaystyle \mathrm {\Delta } }$ or ${\displaystyle \mathrm {\theta } _{i}}$. This difficulty can be avoided by using potential functions that are solutions of the wave equations and from which ${\displaystyle \left(u,\;v,\;w\right)}$, hence ${\displaystyle \mathrm {\theta } _{i}}$ also, can be found by differentiation.

Note that derivatives of a solution of a differential equation are also solutions.

The vector operator ${\displaystyle \nabla }$ (called “del”) and its properties are discussed in Sheriff and Geldart, 1995, Section 15.1.2c.

### Solution

From equation (2.1e) and the definition of ${\displaystyle \nabla }$, we get for the dilatation ${\displaystyle \Delta }$

 {\displaystyle {\begin{aligned}\Delta &={\nabla }{\cdot }{\zeta }=\nabla ^{2}\left(\mathrm {\phi } +{\frac {\partial \mathrm {\chi } }{\partial z}}\right)-\left[{\nabla }{\cdot }{k}\right]\left(\nabla ^{2}\mathrm {\chi } \right)\\&=\nabla ^{2}\mathrm {\phi } +\nabla ^{2}\left({\frac {\partial \mathrm {\chi } }{\partial z}}\right)-\left({\frac {\partial }{\partial z}}\right)\left(\nabla ^{2}\mathrm {\chi } \right)\\&=\nabla ^{2}\mathrm {\phi } ,\end{aligned}}} (2.9b)

since ${\displaystyle \nabla ^{2}\left({\frac {\partial \mathrm {\chi } }{\partial z}}\right)={\frac {\partial }{\partial z}}\left(\nabla ^{2}\mathrm {\chi } \right)}$ and ${\displaystyle \nabla \cdot k=\partial /\partial z}$. Because ${\displaystyle \mathrm {\phi } }$ is a solution of the P-wave equation, ${\displaystyle \mathrm {\Delta } }$ must also be a solution.

We have

{\displaystyle {\begin{aligned}{\theta }&={\frac {1}{2}}\nabla \times {\zeta }={\frac {1}{2}}{\nabla }\times {\nabla }\left(\mathrm {\phi } +{\frac {\partial \mathrm {\chi } }{\partial z}}\right)-{\frac {1}{2}}{\nabla }\times \left[\left(\nabla ^{2}\mathrm {\chi } \right)k\right]\\&=0-{\frac {1}{2}}{\nabla }\times \left[\left(\nabla ^{2}\mathrm {\chi } \right)k\right]={\frac {1}{2}}\nabla ^{2}\left(\chi _{x}-\chi _{y}\right)\left(\chi _{y}i-\chi _{x}j\right)\end{aligned}}}

[see Sheriff and Geldart, 1995, equations (15.13) and (15.9)]. Since ${\displaystyle \mathrm {\chi } }$ is a solution of the S-wave equation, ${\displaystyle {\theta }}$ is also a solution.

## Problem 2.9b

In two dimensions, the potential function ${\displaystyle \zeta }$ can be defined as

 {\displaystyle {\begin{aligned}{\zeta ={\mathbf {\nabla \phi } }+{\nabla }\times {\chi },\quad {\chi }=-\left|{\chi }\right|j}.\end{aligned}}} (2.9c)

Show how to obtain the displacements ${\displaystyle {\mathbf {\left(u,\;v,\;w\right)} }}$, the dilatation ${\displaystyle {\Delta }}$, and rotation ${\displaystyle {\theta }}$ from this equation (see Sheriff and Geldart, 1995, Section 15.1.2c and problem 15.5c).

### Solution

From equation (2.1d) we see that ${\displaystyle u}$ is the ${\displaystyle x}$-component of ${\displaystyle {\zeta }}$, that is, of ${\displaystyle {\nabla }\mathrm {\phi } }$ and ${\displaystyle {\nabla }\times {\chi }}$, so ${\displaystyle u={\frac {\partial \mathrm {\phi } }{\partial x}}+x}$-component of ${\displaystyle {\nabla }\times {\chi }}$. From Sheriff and Geldart, 1995, equation (15.13) we have

{\displaystyle {\begin{aligned}{\nabla }\times {\chi }=\left|{\begin{array}{ccc}i&j&k\\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\{0}&{-\mathrm {\chi } }&{0}\end{array}}\right|={\frac {\partial \mathrm {\chi } }{\partial z}}i-{\frac {\partial \mathrm {\chi } }{\partial x}}k=\mathrm {\chi } _{z}i-\mathrm {\chi } _{x}k.\end{aligned}}}

Thus,

 {\displaystyle {\begin{aligned}u=\mathrm {\phi } _{x}+\mathrm {\chi } _{z}\end{aligned}}} (2.9d)

and

 {\displaystyle {\begin{aligned}w=\mathrm {\phi } _{z}-\mathrm {\chi } _{x}.\end{aligned}}} (2.9e)

To get the dilatation ${\displaystyle \Delta }$, we use equation (2.1e) and Sheriff and Geldart, 1995, problem 15.5c and obtain

 {\displaystyle {\begin{aligned}\Delta ={\nabla }{\cdot }{\zeta }=\nabla ^{2}\mathrm {\phi } .\end{aligned}}} (2.9f)

The rotation ${\displaystyle {\theta }}$ can be obtained by taking the curl of equation (2.9c) but an easier method is to substitute equations (2.9d) and (2.9e) in equation (2.1g). This gives

{\displaystyle {\begin{aligned}{\theta }={\frac {1}{2}}\left|{\begin{array}{ccc}i&j&k\\{\frac {\partial }{\partial x}}&0&{\frac {\partial }{\partial z}}\\u&0&w\end{array}}\right|={\frac {1}{2}}\left({\frac {\partial u}{\partial z}}-{\frac {\partial w}{\partial x}}\right)j\end{aligned}}}

Thus ${\displaystyle {\theta }}$ has only a ${\displaystyle y}$-component given by

{\displaystyle {\begin{aligned}\mathrm {\theta } _{y}={\frac {1}{2}}\left[{\frac {\partial }{\partial z}}\left(\mathrm {\phi } _{x}+\mathrm {\chi } _{z}\right)-{\frac {\partial }{\partial x}}\left(\mathrm {\phi } _{z}-\mathrm {\chi } _{x}\right)\right]={\frac {1}{2}}\nabla ^{2}\mathrm {\chi } .\end{aligned}}}