Page:EB1911 - Volume 26.djvu/989

Solving (20), we have

$\left.{\begin{aligned}&(\Sigma b_{i}x_{i})(f^{2}-\cos ^{2}\theta )={\frac {I}{4m}}\left[{\frac {d}{d\theta }}\Sigma b_{i}u_{i}+{\frac {s\cos \theta }{f\sin \theta }}\Sigma b_{i}u_{i}\right]\\&(\Sigma b_{i}y_{i})\sin ^{2}\theta (f^{2}-\cos ^{2}\theta )=-{\frac {I}{4m}}\left[{\frac {\cos \theta }{f}}{\frac {d}{d\theta }}\Sigma b_{i}u_{i}+{\frac {s}{\sin \theta }}\Sigma b_{i}u_{i}\right]\end{aligned}}\right\}{\text{(21)}}$

Then substituting from (21) in (19) we have

(22)

${\begin{aligned}{\frac {I}{\sin \theta }}&{\frac {d}{d\theta }}\left[{\frac {\gamma (\sin \theta {\frac {d}{d\theta }}\Sigma b_{i}u_{i}+{\frac {s}{f}}\cos \theta \Sigma b_{i}u_{i})}{f^{2}-\cos ^{2}\theta }}\right]\\&-{\frac {s\gamma \left[{\frac {\cos \theta }{f}}{\frac {d}{d\theta }}\Sigma b_{i}u_{i}+{\frac {s}{\sin \theta }}\Sigma b_{i}u_{i}\right]}{\sin \theta (f^{2}-\cos ^{2}\theta )}}+4ma\Sigma (u_{i}+e_{i})=o.\end{aligned}}$

This is almost the same as Laplace's equation for tidal oscillations in an ocean whose depth is only a function of latitude. If indeed we treat b_i as unity (thereby neglecting the mutual attraction of the water) and replace Σu_i and Σe_i by u and e, we obtain Laplace's equation.

When u_i is found from this equation, its value substituted in (21) will give x_i and y_i.

§ 17. Zonal Oscillations.—We might treat the general harmonic oscillations first, and proceed to the zonal oscillations by putting s=0. These waves are, however, comparatively simple, and it is well to begin with them. The zonal tides are those which Laplace describes as of the first species, and are now more usually called the tides of long period. As we shall only consider the case of an ocean of uniform depth, γ the depth of the sea is constant. Then since in this case s=0, our equation (22), to be satisfied by u_i; or h_i-e_i, becomes

${\frac {d}{d\theta }}\left[{\frac {\sin \theta {\frac {d}{d\theta }}\Sigma b_{i}u_{i}}{f^{2}-\cos ^{2}\theta }}\right]+{\frac {4ma}{\gamma }}\sin \theta \Sigma h_{i}=o.$

This may be written

(23)

${\frac {d}{d\theta }}\Sigma b_{i}u_{i}+{\frac {4ma}{\gamma }}{\frac {f^{2}-\cos ^{2}\theta }{\sin \theta }}\int ^{\theta }\Sigma h_{i}\sin \theta d\theta +A=o,$

where A is a constant.

Let us assume

h_i=C_iP_i, e_i=E_iP_i

where P_i denotes the ith zonal harmonic of cos θ. The coefficients C_i are unknown, but the E_i are known because the system oscillates under the action of known forces.

If the term involving the integral in this equation were expressed in terms of differentials of harmonics, we should be able to equate to zero the coefficient of each dP_i/dθ in the equation, and thus find the conditions for determining the C's.

The task then is to express P; sin θdθ in differentials of zonal harmonics.

It is well known that P_i satisfies the differential equation

(24)

${\frac {d}{d\theta }}\left(\sin \theta {\frac {dP_{i}}{d\theta }}\right)+i(i+I)P_{i}\sin \theta =o.$

Therefore $\int P_{i}\sin \theta d\theta =-{\frac {I}{i(i+I)}}\sin \theta {\frac {dP_{i}}{d\theta }}$, and

${\frac {f^{2}-\cos ^{2}\theta }{\sin \theta }}\int P_{i}\sin \theta d\theta =-{\frac {I}{i(i+I)}}(f^{2}-\cos ^{2}\theta ){\frac {dP_{i}}{d\theta }}=-{\frac {I}{i(i+I)}}(f^{2}-I){\frac {dP_{i}}{d\theta }}--{\frac {I}{i(i+I)}}\sin ^{2}\theta {\frac {dP_{i}}{d\theta }}$

Another well-known property of zonal harmonics is that

(25)

$\sin \theta {\frac {dP_{i}}{d\theta _{i}}}={\frac {i(i+I)}{2i+I}}(P_{i+I}-P_{i-I})$.

If we differentiate (25) and use (24) we have

(26)

${\frac {i(i+I)}{2i+I}}\left({\frac {dP_{i+I}}{d\theta }}-{\frac {dP_{i-I}}{d\theta }}\right)i(i+I)P_{i}\sin \theta =o.$

Multiplying (25) by sin θ, and using (26) twice over,

$\sin ^{2}\theta {\frac {dP_{i}}{d\theta }}={\frac {i(i+I)}{2i+I}}\left\{-{\frac {I}{2i+3}}\left({\frac {dP_{i+2}}{d\theta }}-{\frac {dP_{i}}{d\theta }}\right)+{\frac {I}{2i-I}}\left({\frac {dP_{i}}{d\theta }}-{\frac {dP_{i-2}}{d\theta }}\right)\right\}$

Therefore

${\begin{aligned}{\frac {f^{2}-\cos ^{2}\theta }{\sin \theta }}\int P_{i}\sin \theta d\theta ={\frac {I}{(2i-I)(2i+I)}}{\frac {dP_{i-2}}{d\theta }}\\-\left\{{\frac {f^{2}-I}{i(i+I)}}+{\frac {2}{(2i-I)(2i+3)}}\right\}{\frac {dP_{i}}{d\theta }}+{\frac {I}{(2i+I)(2i+3)}}{\frac {dP_{i+2}}{d\theta }}\end{aligned}}$

This expression, when multiplied by 4ma/γ and by C_i and summed, is the second term of our equation.

The first term is

$\Sigma b_{i}(C_{i}-E_{i}){\frac {dP_{i}}{d\theta }}$

In order that the equation may be satisfied, the coefficient of each dP_i/dθ must vanish identically. Accordingly we multiply the whole by γ/4ma and equate to zero the coefficient in question, and obtain

(27)

${\frac {b_{i}\gamma }{4ma}}(C_{i}-E_{i})+{\frac {C_{i-2}}{(2i-I)(2i-3)}}-\left\{{\frac {f^{2}-I}{i(i+I)}}+{\frac {2}{(2i-1)(2i+3)}}\right\}C_{i}+{\frac {C_{i+2}}{(2i+3)(2i+5)}}=o.$

This equation (27) is applicable for all values of i from 1 to infinity, provided that we take C₀, E_O, C_-1, E_-1 as being zero.

We shall only consider in detail the case of greatest interest, namely that of the most important of the tides generated by the attraction of the sun and moon. We know that in this case the equilibrium tide is expressed by a zonal harmonic of the second order; and therefore all the E_i, excepting E₂, are zero. Thus the equation (27) will not involve E_i in any case excepting when i=2.

If we write for brevity

$L_{i}={\frac {f^{2}-I}{i(i+I)}}+{\frac {2}{(2i-I)(2i+3)}}-{\frac {b_{i}\gamma }{4ma}}$

the equation (27) is

(28)

${\frac {C_{i+2}}{(2i+3)(2i+5)}}-L_{i}C_{i}+{\frac {C_{i-2}}{(2i-3)(2i-1)}}=o.$

Save that when i=2, the right-hand side is b₂γE₂/4ma, a known quantity ex hypothesi.

The equations naturally separate themselves into two groups in one of which all the suffixes are even and the other odd. Since our task is to evaluate all the C's in terms of E₂, it is obvious that all the C's with odd suffixes must be zero, and we are left to consider only the cases where i=2, 4, 6, &c.

We have, said that C₀ must be regarded as being zero; if however we take

$C_{0}=-3b_{2}\gamma E_{2}/4ma,$

so that C₀ is essentially a known quantity, the equation (28) has complete applicability for all even values of i from 2 upwards.

The equations are

${\begin{aligned}&{\frac {C_{0}}{1.3}}-L_{2}C_{2}+{\frac {C_{4}}{7.9}}=o\\&{\frac {C_{2}}{5.7}}-L_{4}C_{4}+{\frac {C_{6}}{11.13}}=o.\end{aligned}}$

It would seem at first sight as if these equations would suffice to determine all the C's in terms of C₂, and that C₂ would remain indeterminate; but we shall show that this is not the case.

For very large values of i the general equation of condition (28) tends to assume the form

${\frac {C_{i+2}+C_{i-2}}{C_{i}}}+{\frac {2^{3}\gamma }{ma}}=o.$

By writing successively i+2, i+4; i+6 for i in this equation, and taking the differences, we obtain an equation from which we see that, unless C_i/C_i+2 tends to became infinitely small, the equations are satisfied by C_i=C_i+1 in the limit for very large values of i.

Hence, if C_i does not tend to zero, the later portion of the series for h tends to assume the form C_i(P_i+P_i+2+P_i+4...). All the P's are equal to unity at the pole; hence the hypothesis that C, does not tend to zero leads to the conclusion that the tide is of infinite height at the pole. The expansion of the height of tide is essentially convergent, and therefore the hypothesis is negatived. Thus we are entitled to assume that C_i tends to zero for large values of i.

Now writing for brevity

a_i=1/(2i+1)(2i+3)²(2i+5),

we may put (28) into the form

${\frac {C_{i-2}/C_{i}}{(2i-3)(2i-1)}}=L_{i}-{\frac {\frac {\alpha _{i}}{C_{i}/C_{i+2}}}{(2i+1)(2i+3)}}$

By successive applications of this formula we may write the right hand side in the form of a continued fraction.

Let

$K_{i}={\frac {\alpha _{i}-2}{L_{i}-}}{\frac {\alpha _{i}}{L_{i+2}-}}{\frac {\alpha _{i}+2}{L_{i+4}-}}\dots$

Then we have

${\frac {C_{i-2}/C_{i}}{(2i-3)(2i-1)}}={\frac {\alpha _{i-2}}{K_{i}}}$,

or

$C_{i}/C_{i-2}=(2i-1)(2i+1)K_{i}$.

Thus

${\begin{aligned}C_{2}&=3.5K_{2}C_{0};\\C_{4}&=3.5.7.9K_{2}K_{4}C_{0};\\C_{6}&=3.5.7.9.11.13K_{2}K_{4}K_{6}C_{0},\end{aligned}}$&c.

If we assume that any of the higher C's, such as C₁₄ or C₁₆, is of negligible smallness, all the continued fractions K₂, K₄, K₆, &c., may be computed; and thus we find all the C's in terms of C₀, which is equal to -3b₂γE₂/4ma. The height of the tide is therefore given by

${\begin{aligned}h&=\Sigma h_{i}\cos(2nft+\alpha )\\&=-{\frac {3b_{2}\gamma }{4ma}}E_{2}\left\{3.5K_{2}P_{2}+3.5.7.9K_{2}K_{4}P_{4}+\dots \right\}\cos(2nft+\alpha )\end{aligned}}$.

It is however more instructive to express h as a multiple of the