General solutions of the wave equation

Problems in Exploration Seismology and their Solutions

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

Problem 2.5a

Verify that $\psi =f\left(x-Vt\right)$ and $\psi =g\left(x+Vt\right)$ 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

${\begin{aligned}{\nabla }^{2}\psi ={\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }y^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }z^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}},\end{aligned}}$

(2.5a)

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

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

${\begin{aligned}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}}.\end{aligned}}$

(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 $\zeta =\left(x-Vt\right)$, we have

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

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

Problem 2.5b

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

Solution

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

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

In the same way we get

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

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

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

Problem 2.5c

Show that

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

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

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

(2.5c)

Solution

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

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

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

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Table of Contents (book)

Also in this chapter

• The basic elastic constants
• Interrelationships among elastic constants
• Magnitude of disturbance from a seismic source
• Magnitudes of elastic constants
• Wave equation in cylindrical and spherical coordinates
• Sum of waves of different frequencies and group velocity
• Magnitudes of seismic wave parameters
• Potential functions used to solve wave equations
• Boundary conditions at different types of interfaces
• Boundary conditions in terms of potential functions
• Disturbance produced by a point source
• Far- and near-field effects for a point source
• Rayleigh-wave relationships
• Directional geophone responses to different waves
• Tube-wave relationships
• Relation between nepers and decibels
• Attenuation calculations
• Diffraction from a half-plane

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