# General solutions of the wave equation

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 2.5a

Verify that ${\displaystyle \psi =f\left(x-Vt\right)}$ and ${\displaystyle \psi =g\left(x+Vt\right)}$ are solutions of the wave equation (2.5b).

### Background

When unbalanced stresses act upon a medium, the strains are propagated throughout the medium according to the general wave equation

 {\displaystyle {\begin{aligned}{\nabla }^{2}\psi ={\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }y^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }z^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}},\end{aligned}}} (2.5a)

${\displaystyle \psi }$ being a disturbance such as a compression or rotation. ${\displaystyle \psi }$ is propagated with velocity ${\displaystyle V}$ (see Sheriff and Geldart, 1995, Section 2.2). The disturbance is the result of unbalanced normal stresses, shearing stresses, or a combination of both. When normal stresses create the wave, the result is a volume change and ${\displaystyle \psi }$ is the dilitation [see equation (2.1e)], and we get the P-wave equation, ${\displaystyle V}$ becoming the P-wave velocity ${\displaystyle {\mathrm {\alpha } }}$. Shearing stresses create rotation in the medium and ${\displaystyle \psi }$ is one of the components of the rotation given by equation (2.lg) ; the result is an S-wave traveling with velocity ${\displaystyle {\mathrm {\beta } }}$. Various expressions for ${\displaystyle {\mathrm {\alpha } }}$ and ${\displaystyle {\mathrm {\beta } }}$ in isotropic media are given in Table 2.2a.

In one dimension the wave equation (2.5a) reduces to

 {\displaystyle {\begin{aligned}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}}.\end{aligned}}} (2.5b)

### Solution

We use subscripts to denote partial derivatives and primes to denote derivatives with respect to the argument of the function. Then, writing ${\displaystyle \zeta =\left(x-Vt\right)}$, we have

{\displaystyle {\begin{aligned}{{{\mathrm {\psi } }_{x}}={\frac {\partial {\mathrm {\psi } }}{\partial x}}=\left({\frac {df}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial x}}\right)={\frac {df}{d{\mathrm {\zeta } }}}=f',}\\{{{\mathrm {\psi } }_{xx}}=\left({\frac {df'}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial x}}\right)=f'',}\\{{{\mathrm {\psi } }_{t}}=\left({\frac {df}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial t}}\right)=-Vf',}\\{{{\mathrm {\psi } }_{tt}}=-V\left({\frac {df'}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial t}}\right)={V^{2}}f''}\end{aligned}}}

Substituting in equation (2.5b), we get the identity ${\displaystyle f''=f''}$ so ${\displaystyle f\left(x-Vt\right)}$ is a solution. We get the same result when ${\displaystyle \psi =g\left(x+Vt\right)}$. A sum of solutions is also a solution, so ${\displaystyle f\left(x-Vt\right)+g\left(x+Vt\right)}$ is a solution.

## Problem 2.5b

Verify that ${\displaystyle {\psi =f\left(\ell x+my+nz-Vt\right)+g\left(\ell x+my+nz+Vt\right)}}$ is a solution of equation (2.5a), where ${\displaystyle {\left(\ell ,{\rm {\;}}m,\;n\right)}}$ are direction cosines.

### Solution

Let ${\displaystyle \left(\ell x\;+my+nz-Vt\right)=\zeta ,\left(\ell x\;+my+nz+Vt\right)=\xi }$. We now must show that ${\displaystyle \psi =f\left(\zeta \right)+g\left(\xi \right)}$ is a solution of equation (2.5a). Proceeding as before, we have

{\displaystyle {\begin{aligned}{{\mathrm {\psi } }_{x}=\left({\frac {df}{d\zeta }}\right)\left({\frac {{\mathrm {\partial } }\zeta }{{\mathrm {\partial } }x}}\right)+\left({\frac {dg}{d\xi }}\right)\left({\frac {{\mathrm {\partial } }\xi }{{\mathrm {\partial } }x}}\right)=\ell (f'+g'),}\\{{\mathrm {\psi } }_{xx}=\ell ^{2}(f''+g'').}\end{aligned}}}

In the same way we get

{\displaystyle {\begin{aligned}\psi _{yy}=m^{2}(f''+g''),\;{\quad }\psi _{zz}=n^{2}(f''+g'').\end{aligned}}}

But ${\displaystyle \left(\ell ^{2}+m^{2}+n^{2}\right)=1}$ (see Sheriff and Geldart, 1995, problem 15.9a), so ${\displaystyle \left(\psi _{xx}+\psi _{yy}+\psi _{zz}\right)=(f''+g'')}$.

Following the same procedure we find that ${\displaystyle \left(1/V^{2}\right)\psi _{tt}=(f''+g'')}$ thus verifying that ${\displaystyle \psi =f\left(\ell x+my+nz-Vt\right)+g\left(\ell x+my+nz+Vt\right)}$ is a solution of equation (2.5a).

## Problem 2.5c

Show that

{\displaystyle {\begin{aligned}\psi \left(r,\;t\right)=\left(1/r\right)f\left(r-Vt\right)+\left(1/r\right)g\left(r+Vt\right)\end{aligned}}}

is a solution of the wave equation in spherical coordinates (see problem 2.6b) when the wave motion is independent of ${\displaystyle {\theta }}$ and ${\displaystyle \phi }$:

 {\displaystyle {\begin{aligned}{\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }t^{2}}}={\frac {1}{r^{2}}}\left[{\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\frac {{\mathrm {\partial } }{\mathrm {\psi } }}{{\mathrm {\partial } }r}}\right)\right].\end{aligned}}} (2.5c)

### Solution

The wave equation in spherical coordinates is given in problem 2.6b. When we drop the derivatives with respect to ${\displaystyle {\mathrm {\theta } }}$ and ${\displaystyle {\mathrm {\phi } }}$, the equation reduces to equation (2.5c). Writing ${\displaystyle {\mathrm {\zeta } }=(r-Vt)}$, we proceed as in part (a). Starting with the right-hand side, we ignore ${\displaystyle g\left(r+Vt\right)}$ for the time being and obtain

{\displaystyle {\begin{aligned}{\mathrm {\psi } }_{r}=-\left(1/r^{2}\right)f\left({\mathrm {\zeta } }\right)+\left(1/r\right)f'\left({\mathrm {\zeta } }\right),\\r^{2}{\mathrm {\psi } }_{r}=-f\left({\mathrm {\zeta } }\right)+rf'\left({\mathrm {\zeta } }\right),\\{\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\mathrm {\psi } }_{r}\right)=-f'\left({\mathrm {\zeta } }\right)+f'\left({\mathrm {\zeta } }\right)+rf''\left({\mathrm {\zeta } }\right)=rf''\left({\mathrm {\zeta } }\right),\\\left(1/r^{2}\right){\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\mathrm {\psi } }_{r}\right)=\left(1/r\right)f''\left({\mathrm {\zeta } }\right),\\\psi _{t}=-\left(V/r\right)f';{\quad }{\mathrm {\phi } }_{tt}=\left(V^{2}/r\right)f''.\end{aligned}}}

Substitution in equation (2.5c) shows that ${\displaystyle \left(1/r\right)f\left({\mathrm {\zeta } }\right)}$ is a solution. In the same way we can show that ${\displaystyle \left(1/r\right)g\left({\mathrm {\xi } }\right)}$ is also a solution, hence the sum is a solution.

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