Page:ThomsonMagnetic1889.djvu/7

Rh Equating the coefficients of $$\epsilon^{ipt}$$ to zero in equations (2) and (3) we have

$-\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}_{1}=-\dfrac{d\mathrm{F}_{1}}{dt}-\dfrac{d\psi_{1}}{dx}$ inside the sphere, |undefined

with similar equations for $$\mathrm{G}_{1}$$ and $$\mathrm{H}_{1}$$; outside the sphere we have

with similar equations for $$\mathrm{G}_{1}$$ and $$\mathrm{H}_{1}$$.

The form of equation (4) suggests that we should put

$\psi_{1}=B\dfrac{d}{dz}\dfrac{1}{r}$

A particular integral of (4) is then

$\mathrm{F}_{1}=\dfrac{c\mathrm{B}}{p}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}-\dfrac{\omega e}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}.$|undefined

The complementary function is that solution of the differential equation

$\nabla^{2}\mathrm{F}_{1}+\mu\mathrm{K}p^{2}\mathrm{F}_{1}=0$

which, when considered as a function of the angular coordinates of a point, varies as $$r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}$$; this (see Proc. Math. Soc. vol. xv. p. 212) is

$\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}$|undefined

where

$\mathrm{E}_{2}(i\lambda r)=\dfrac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{3}}+\dfrac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{2}}+\dfrac{\epsilon^{-i\lambda r}}{i\lambda r}$|undefined

and

$\lambda^{2}=\mu\mathrm{K}p^{2}$

Thus, outside the sphere,

$\begin{array}{l} \mathrm{F}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}+\dfrac{i\mathrm{B}\epsilon^{pt}}{p}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dx\ dz}\dfrac{1}{r},\\ \\ \mathrm{G}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r}+\dfrac{i\mathrm{B}\epsilon^{pt}}{p}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dy\ dz}\dfrac{1}{r},\\ \\ \mathrm{H}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}-2\mathrm{C}\mathrm{E}_{2}(i\lambda r)+\dfrac{i\mathrm{B}}{p}\epsilon^{pt}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}-\dfrac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{r}, \end{array}$|undefined

where $$\mathrm{E}_{0}(i\lambda r)=\epsilon^{-i\lambda r}/i\lambda r$$, and is introduced into the expression