Page:BatemanConformal.djvu/5

is a solution of

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}=0$

when expressed in terms of x, y, z, w.

Next, let U be a homogeneous function of degree -1 in (l, m, n, λ, μ, ν), then we can show in a similar way that

${\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}+{\frac {\partial ^{2}U}{\partial w^{2}}}\equiv 4\left[{\frac {\partial ^{2}U}{\partial l\ \partial \lambda }}+{\frac {\partial ^{2}U}{\partial m\ \partial \mu }}+{\frac {\partial ^{2}U}{\partial n\ \partial \nu }}\right]\equiv \sum \limits _{k=1}^{6}{\frac {\partial ^{2}U}{\partial a^{2}}}.$

Hence, if U is a solution of

${\frac {\partial ^{2}U}{\partial l\ \partial \lambda }}+{\frac {\partial ^{2}U}{\partial m\ \partial \mu }}+{\frac {\partial ^{2}U}{\partial n\ \partial \nu }}=0,$

i.e., of

$\sum \limits _{k=1}^{6}{\frac {\partial ^{2}U}{\partial a_{k}^{2}}}=0,$

it is a solution of

${\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}+{\frac {\partial ^{2}U}{\partial w^{2}}}=0$

when expressed in terms of (x, y, z, w).

When ($a_{1},\ a_{2},\ \dots ,\ a_{6}$) are interpreted as the coordinates of a point in a space of six dimensions, the expressions

$\sum \limits _{k=1}^{6}a_{k}^{2},\ \sum \limits _{k=1}^{6}\left({\frac {\partial U}{\partial a_{k}}}\right)^{2},\ \sum \limits _{k=1}^{6}{\frac {\partial ^{2}U}{\partial a_{k}^{2}}}$

remain unaltered in form after any change of rectangular axes in which the origin remains the same. Any change of this kind corresponds to a transformation in the (x, y, z, w) space, enabling us to pass from one solution of the equation

${\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}+{\frac {\partial ^{2}U}{\partial w^{2}}}=0$

to another, and a similar remark applies to the equation

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}=0.$

To illustrate the method of formation of the transformation, we may consider the effect of simply interchanging n and -ν. The functions V and U of formulæ (8) and (4) then transform into

$F\left({\frac {l+\lambda }{2n}},\ {\frac {l-\lambda }{2in}},\ {\frac {m+\mu }{2n}},\ {\frac {m-\mu }{2in}}\right)=F\left({\frac {x}{r^{2}}},\ {\frac {y}{r^{2}}},\ {\frac {z}{r^{2}}},\ {\frac {w}{r^{2}}}\right)$

and

${\frac {1}{n}}f\left({\frac {l+\lambda }{2n}},\ {\frac {l-\lambda }{2in}},\ {\frac {m+\mu }{2n}},\ {\frac {m-\mu }{2in}}\right)={\frac {1}{r^{2}}}f\left({\frac {x}{r^{2}}},\ {\frac {y}{r^{2}}},\ {\frac {z}{r^{2}}},\ {\frac {w}{r^{2}}}\right)$

respectively, r² being written in place of x² + y² + z² + w².