tour les différentielles partielles de toutes, par rapport aux paramètres indépendans
et
et à leurs fonctions
on se trouvera avoir
équations, entre lesquelles on n’aura à éliminer que
quantités, savoir, les
paramètres et les
rapports
Que par exemple l’équation proposée soit
![{\displaystyle \operatorname {F} (x,y,z,a,b,c,a',b',c')=V=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4fdb8d8cc07d7e0f79fbd3678c5c4ff873a16e0)
ces six paramètres étant liés par les quatre relations
![{\displaystyle \operatorname {f} (a,b,c)=S=0,\qquad \operatorname {f} '(a',b',c')=S'=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3df223368397cdf0fcfb3127444c512c28c6bfda)
![{\displaystyle \varphi (a,b,c,a',b',c')=P=0,\qquad \psi (a,b,c,a',b',c')=Q=0\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f5d5c3f7a0eb56dff9b9e44b7d877f888e3fff5)
en considérant
comme des fonctions des deux paramètres indépendans
il faudra éliminer ces six paramètres et les huit rapports
![{\displaystyle {\begin{array}{llll}{\frac {\operatorname {d} a}{\operatorname {d} a'}},&{\frac {\operatorname {d} b}{\operatorname {d} a'}},&{\frac {\operatorname {d} c}{\operatorname {d} a'}},&{\frac {\operatorname {d} c'}{\operatorname {d} a'}},\\\\{\frac {\operatorname {d} a}{\operatorname {d} b'}},&{\frac {\operatorname {d} b}{\operatorname {d} b'}},&{\frac {\operatorname {d} c}{\operatorname {d} b'}},&{\frac {\operatorname {d} c'}{\operatorname {d} b'}},\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a6ef37be21a7f8d9bda0bed5dc0d11b0ff8885b)
entre les quinze équations
![{\displaystyle V=0,\qquad S=0,\qquad S'=0,\qquad P=0,\qquad Q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42f0912a7b485ec8836799616f2fde58de26b81c)
![{\displaystyle \left({\frac {\operatorname {d} V}{\operatorname {d} a}}\right){\frac {\operatorname {d} a}{\operatorname {d} a'}}+\left({\frac {\operatorname {d} V}{\operatorname {d} b}}\right){\frac {\operatorname {d} b}{\operatorname {d} a'}}+\left({\frac {\operatorname {d} V}{\operatorname {d} c}}\right){\frac {\operatorname {d} c}{\operatorname {d} a'}}+\left({\frac {\operatorname {d} V}{\operatorname {d} a'}}\right)+\left({\frac {\operatorname {d} V}{\operatorname {d} c'}}\right){\frac {\operatorname {d} c'}{\operatorname {d} a'}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/725054f3193c0ec495a03dc5bcd98302ce3b0ff2)
![{\displaystyle \left({\frac {\operatorname {d} V}{\operatorname {d} b}}\right){\frac {\operatorname {d} a}{\operatorname {d} b'}}\,+\left({\frac {\operatorname {d} V}{\operatorname {d} b}}\right){\frac {\operatorname {d} b}{\operatorname {d} b'}}\,+\left({\frac {\operatorname {d} V}{\operatorname {d} c}}\right){\frac {\operatorname {d} c}{\operatorname {d} b'}}\,+\left({\frac {\operatorname {d} V}{\operatorname {d} b'}}\right)+\left({\frac {\operatorname {d} V}{\operatorname {d} c'}}\right){\frac {\operatorname {d} c'}{\operatorname {d} b'}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4ae31a6bfcdb0f90e69e938d428a195c957456b)