186.] THEOREM III. If
V
{\displaystyle V}
x
′
{\displaystyle x'}
y
′
{\displaystyle y'}
x
′
{\displaystyle x'}
y
′
{\displaystyle y'}
x
{\displaystyle x}
y
{\displaystyle y}
∬
(
d
2
V
d
x
2
+
d
2
V
d
y
2
)
d
x
d
y
=
∬
(
d
2
V
d
x
′
2
+
d
2
V
d
y
′
2
)
d
x
′
d
y
′
{\displaystyle \iint \left({\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}\right)dx\ dy=\iint \left({\frac {d^{2}V}{dx'^{2}}}+{\frac {d^{2}V}{dy'^{2}}}\right)dx'\ dy'}
integration being between the same limits.
For
d
V
d
x
=
d
V
d
x
′
d
x
′
d
x
+
d
V
d
y
′
d
y
′
d
x
,
d
2
V
d
x
2
=
d
2
V
d
x
′
2
(
d
x
′
d
x
)
2
+
2
d
2
V
d
x
′
d
y
′
d
x
′
d
x
d
y
′
d
x
+
d
2
V
d
y
′
2
(
d
y
′
d
x
)
2
+
d
V
d
x
′
d
2
x
′
d
x
2
+
d
V
d
y
′
d
2
y
′
d
x
2
;
{\displaystyle {\begin{array}{c}{\frac {dV}{dx}}={\frac {dV}{dx'}}{\frac {dx'}{dx}}+{\frac {dV}{dy'}}{\frac {dy'}{dx}},\\\\{\frac {d^{2}V}{dx^{2}}}={\frac {d^{2}V}{dx'^{2}}}\left({\frac {dx'}{dx}}\right)^{2}+2{\frac {d^{2}V}{dx'dy'}}{\frac {dx'}{dx}}{\frac {dy'}{dx}}+{\frac {d^{2}V}{dy'^{2}}}\left({\frac {dy'}{dx}}\right)^{2}+{\frac {dV}{dx'}}{\frac {d^{2}x'}{dx^{2}}}+{\frac {dV}{dy'}}{\frac {d^{2}y'}{dx^{2}}};\end{array}}}
and
d
2
V
d
y
2
=
d
2
V
d
x
′
2
(
d
x
′
d
y
)
2
+
2
d
2
V
d
x
′
d
y
′
d
x
′
d
y
d
y
′
d
y
+
d
2
V
d
y
′
2
(
d
y
′
d
y
)
2
+
d
V
d
x
′
d
2
x
′
d
y
2
+
d
V
d
y
′
d
2
y
′
d
y
2
{\displaystyle {\frac {d^{2}V}{dy^{2}}}={\frac {d^{2}V}{dx'^{2}}}\left({\frac {dx'}{dy}}\right)^{2}+2{\frac {d^{2}V}{dx'dy'}}{\frac {dx'}{dy}}{\frac {dy'}{dy}}+{\frac {d^{2}V}{dy'^{2}}}\left({\frac {dy'}{dy}}\right)^{2}+{\frac {dV}{dx'}}{\frac {d^{2}x'}{dy^{2}}}+{\frac {dV}{dy'}}{\frac {d^{2}y'}{dy^{2}}}}
Adding the last two equations, and remembering the conditions of conjugate functions (1), we find
d
2
V
d
x
2
+
d
2
V
d
y
2
=
d
2
V
d
x
′
2
(
(
d
x
′
d
x
)
2
+
(
d
x
′
d
y
)
2
)
+
d
V
d
y
′
2
(
(
d
y
′
d
x
)
2
+
(
d
y
′
d
y
)
2
)
=
(
d
2
V
d
x
′
2
+
d
2
V
d
y
′
2
)
(
d
x
′
d
x
d
y
′
d
y
−
d
x
′
d
y
d
y
′
d
x
)
{\displaystyle {\begin{array}{ll}{\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}&={\frac {d^{2}V}{dx'^{2}}}\left(\left({\frac {dx'}{dx}}\right)^{2}+\left({\frac {dx'}{dy}}\right)^{2}\right)+{\frac {dV}{dy'^{2}}}\left(\left({\frac {dy'}{dx}}\right)^{2}+\left({\frac {dy'}{dy}}\right)^{2}\right)\\\\&=\left({\frac {d^{2}V}{dx'^{2}}}+{\frac {d^{2}V}{dy'^{2}}}\right)\left({\frac {dx'}{dx}}{\frac {dy'}{dy}}-{\frac {dx'}{dy}}{\frac {dy'}{dx}}\right)\end{array}}}
Hence
∬
(
d
2
V
d
x
2
+
d
2
V
d
y
2
)
d
x
d
y
=
∬
(
d
2
V
d
x
′
2
+
d
2
V
d
y
′
2
)
(
d
x
′
d
x
d
y
′
d
y
−
d
x
′
d
y
d
y
′
d
x
)
d
x
′
d
y
′
=
∬
(
d
2
V
d
x
′
2
+
d
2
V
d
y
′
2
)
d
x
′
d
y
′
{\displaystyle {\begin{array}{ll}\iint \left({\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}\right)dx\ dy&=\iint \left({\frac {d^{2}V}{dx'^{2}}}+{\frac {d^{2}V}{dy'^{2}}}\right)\left({\frac {dx'}{dx}}{\frac {dy'}{dy}}-{\frac {dx'}{dy}}{\frac {dy'}{dx}}\right)dx'\ dy'\\\\&=\iint \left({\frac {d^{2}V}{dx'^{2}}}+{\frac {d^{2}V}{dy'^{2}}}\right)dx'\ dy'\end{array}}}
If
V
{\displaystyle V}
d
2
V
d
x
2
+
d
2
V
d
y
2
+
4
π
ρ
=
0
{\displaystyle {\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}+4\pi \rho =0}
and we may write the result
∬
ρ
d
x
d
y
=
∬
ρ
′
d
x
′
d
y
′
{\displaystyle \iint \rho \ dx\ dy=\iint \rho '\ dx'\ dy'}
or the quantity of electricity in corresponding portions of two systems is the same if the coordinates of one system are conjugate functions of those of the other.
