Page:BatemanConformal.djvu/11

Now

${\frac {\partial P}{\partial \xi }}={\frac {b-c}{\xi }}\theta H'(\theta )={\frac {\theta }{z-a}}H'(\theta ),$ ${\begin{array}{cl}{\frac {dP}{dz}}&={\frac {\partial P}{\partial \xi }}{\frac {\partial \xi }{\partial z}}+{\frac {\partial P}{\partial \eta }}{\frac {\partial \eta }{\partial z}}+{\frac {\partial P}{\partial \zeta }}{\frac {\partial \zeta }{\partial z}}=(b-c){\frac {\partial P}{\partial \xi }}+\left(c-a\right){\frac {\partial P}{\partial \eta }}+(a-b){\frac {\partial P}{\partial \zeta }}\\\\&=\left({\frac {b-c}{z-a}}+{\frac {c-a}{z-b}}+{\frac {a-b}{z-c}}\right)\theta H'(\theta )\\\\&=-{\frac {(b-c)(c-a)(a-b)}{(z-a)(z-b)(z-c)}}\theta H'(\theta ).\end{array}}$

Hence

${\frac {\partial P}{\partial \xi }}={\frac {(z-b)(z-c)}{(a-b)(a-c)}}{\frac {1}{b-c}}{\frac {dP}{dz}}.$

Also, since

$\xi {\frac {\partial P}{\partial \xi }}=(b-c)\theta H'(\theta )=\phi (\theta ),$

we have

$\xi {\frac {\partial ^{2}P}{\partial \xi ^{2}}}+{\frac {\partial P}{\partial \xi }}={\frac {\partial }{\partial \xi }}\phi (\theta )={\frac {(z-b)(z-c)}{(a-b)(a-c)}}{\frac {1}{b-c}}{\frac {d}{dz}}\phi (\theta ),$

this relation being obtained in the same way as the one above.

The last expression may be written

${\frac {(z-b)(z-c)}{(a-b)(a-c)}}{\frac {d}{dz}}\theta H'(\theta ),$

i.e.,

$-{\frac {(z-b)(z-c)}{(a-b)(a-c)}}{\frac {d}{dz}}{\frac {(z-a)(z-b)(z-c)}{(b-c)(c-a)(a-b)}}{\frac {dP}{dz}},$

or

${\frac {(z-a)(z-b)^{2}(z-c)^{2}}{(b-c)(c-a)^{2}(a-b)^{2}}}\left[{\frac {d^{2}P}{dz^{2}}}+\left({\frac {1}{z-a}}+{\frac {1}{z-b}}+{\frac {1}{z-c}}\right){\frac {dP}{dz}}\right].$

Now

${\frac {(z-b)(z-c)}{(c-a)(a-b)}}+{\frac {(z-c)(z-a)}{(a-b)(b-c)}}+{\frac {(z-a)(z-b)}{(b-c)(c-a)}}=-1,$

and

${\begin{array}{cl}{\frac {\partial P}{\partial \xi }}+{\frac {\partial P}{\partial \eta }}+{\frac {\partial P}{\partial \zeta }}&=\left({\frac {1}{z-a}}+{\frac {1}{z-b}}+{\frac {1}{z-c}}\right)\theta H'(\theta )\\\\&=-{\frac {(z-a)(z-b)(z-c)}{(b-c)(c-a)(a-b)}}\left({\frac {1}{z-a}}+{\frac {1}{z-b}}+{\frac {1}{z-c}}\right){\frac {dP}{dz}}\end{array}}$;

consequently,

$\xi {\frac {\partial ^{2}P}{\partial \xi ^{2}}}+\eta {\frac {\partial ^{2}P}{\partial \eta ^{2}}}+\zeta {\frac {\partial ^{2}P}{\partial \zeta ^{2}}}=-{\frac {(z-a)(z-b)(z-c)}{(b-c)(c-a)(a-b)}}{\frac {d^{2}P}{dz^{2}}}.$