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Next: Transformation of Laplace Tidal Up: Terrestrial Ocean Tides Previous: Non-Global Ocean Tides

Useful Lemma

Let $ X(\theta,\phi,t)$ , $ Y(\theta,\phi,t)$ , $ P(\theta,\phi,t)$ , and $ Q(\theta,\phi,t)$ be well-behaved functions of $ \theta $ and $ \phi$ . Suppose that

$\displaystyle D\,P = -\frac{1}{\sin\theta}\left[\frac{\partial}{\partial\theta}\,(\sin\theta\,X) +\frac{\partial Y}{\partial\phi}\right],$ (12.186)

where

$\displaystyle \frac{1}{\sin\theta}\,\frac{\partial P(\theta,\phi_\pm,t)}{\partial\phi}=-Y(\theta,\phi_\pm,t),$ (12.187)

Suppose, further, that

$\displaystyle D\,Q =\frac{1}{\sin\theta}\left[\frac{\partial}{\partial\theta}\,(\sin\theta\,Y) - \frac{\partial X}{\partial\phi}\right],$ (12.188)

where

$\displaystyle Q(\theta,\phi_\pm,t) = 0.$ (12.189)

We wish to demonstrate that

$\displaystyle X$ $\displaystyle = -\frac{\partial P}{\partial\theta} - \frac{1}{\sin\theta}\,\frac{\partial Q}{\partial\phi},$ (12.190)
$\displaystyle Y$ $\displaystyle =-\frac{1}{\sin\theta}\,\frac{\partial P}{\partial\phi} + \frac{\partial Q}{\partial\theta}.$ (12.191)

Let

$\displaystyle X'$ $\displaystyle =X + \frac{\partial P}{\partial\theta} + \frac{1}{\sin\theta}\,\frac{\partial Q}{\partial\phi},$ (12.192)
$\displaystyle Y'$ $\displaystyle = Y + \frac{1}{\sin\theta}\,\frac{\partial P}{\partial\phi} - \frac{\partial Q}{\partial\theta}.$ (12.193)

Note, from Equations (12.192) and (12.194), that

$\displaystyle Y'(\theta,\phi_\pm,t) = 0.$ (12.194)

It follows from Equations (12.191), (12.193), (12.197), and (12.198) that

$\displaystyle \frac{\partial}{\partial\theta}\,(\sin\theta\,X') +\frac{\partial Y'}{\partial\phi}$ $\displaystyle = 0,$ (12.195)
$\displaystyle \frac{\partial}{\partial\theta}\,(\sin\theta\,Y')-\frac{\partial X'}{\partial\phi}$ $\displaystyle =0.$ (12.196)

Equations (12.199) and (12.201) imply that

$\displaystyle X'$ $\displaystyle =\frac{\partial P'}{\partial\theta},$ (12.197)
$\displaystyle Y'$ $\displaystyle =\frac{1}{\sin\theta}\,\frac{\partial P'}{\partial\phi},$ (12.198)

and

$\displaystyle \frac{1}{\sin\theta}\,\frac{\partial P'(\theta,\phi_\pm,t)}{\partial\phi} = 0.$ (12.199)

Furthermore, Equation (12.200) yields

$\displaystyle D\,P'=0.$ (12.200)

Multiplying by $ P'$ , integrating over the ocean, and making use of the boundary condition (12.204), we obtain

$\displaystyle \int_{\mit\Omega}\left[\left(\frac{\partial P'}{\partial\theta}\r...
...}{\sin\theta}\,\frac{\partial P'}{\partial\phi}\right)^2\right]d{\mit\Omega}=0.$ (12.201)

Here, $ {\mit\Omega}$ is the surface of the Earth that is covered by ocean, and $ d{\mit\Omega}=\sin\theta\,d\theta\,d\phi$ . It follows that $ P'$ is a constant. Thus, Equations (12.202) and (12.203) imply that $ X'=Y'=0$ , and, hence, that Equations (12.195) and (12.196) are valid.


next up previous
Next: Transformation of Laplace Tidal Up: Terrestrial Ocean Tides Previous: Non-Global Ocean Tides
Richard Fitzpatrick 2016-01-22