## 7-1-18. Calculs

 $\displaystyle V=S_{0}RMN\quad\text{intér}\qquad V=R_{0}SMN\quad\text{extér.}$ $\displaystyle V_{1}=S^{0}_{1}R_{1}M_{1}N_{1};\quad\int V\frac{dV_{1}}{dn}d% \omega=0;\int RMNR^{\prime}_{1}M_{1}N_{1}\frac{d\rho}{dn}d\omega=0$ $\displaystyle\frac{d\rho}{dn}=\frac{1}{\alpha}=\frac{OP}{\rho}=\frac{\sqrt{\mu% ^{2}-\nu^{2}}A}{Q};\quad OP=\frac{\sqrt{\mu^{2}-\nu^{2}}\rho A}{Q}=\frac{\sqrt% {(\rho^{2}-a^{2})(\rho^{2}-b^{2})(\rho^{2}-c^{2})}}{\sqrt{(\rho^{2}-\mu^{2})(% \rho^{2}-\nu^{2})}}$ $\displaystyle\ell=\frac{1}{\sqrt{(\rho^{2}-\mu^{2})(\rho^{2}-\nu^{2})}}=\frac{% OP}{\sqrt{(\rho^{2}-a^{2})(\rho^{2}-b^{2})(\rho^{2}-c^{2})}};\quad\frac{d\rho}% {dn}=\ell iA$ $\displaystyle\int\ell MNM_{1}N_{1}d\omega=0;\quad F=\sum K_{i}M_{i}N_{i};\quad% \int\ell M_{j}N_{j}Fd\omega$ $\displaystyle=\sum K_{i}\int\ell M_{i}N_{i}M_{j}N_{j}d\omega=K_{j}\int\ell M^{% 2}_{j}N^{2}_{j}d\omega$

Problème de Dirichlet.11endnote: 1 Ces calculs sont datés par leur support : une lettre de G. Cres à Poincaré, 25.01.1912.

 $F=\sum KMN;\quad\text{int.}\;V=\sum\frac{KRMN}{R_{0}};\quad\text{ext.}\;V=\sum% \frac{KSMN}{S_{0}}$

Simple couche.

 $\displaystyle\frac{\zeta}{\ell}=\sum C\cdot MN;\quad\text{int.}\;V=\sum\frac{% KRMN}{R_{0}};\quad\text{ext.}\;V=\sum\frac{KSMN}{S_{0}}$ $\displaystyle\frac{dV}{dn}=\sum\ell iA\frac{KR^{\prime}\cdot MN}{R}\quad\text{% int.}\qquad\text{ext.}\quad\sum\frac{\ell iAKS^{\prime}\cdot MN}{S}$ $\displaystyle\ell iAK\left(\frac{R^{\prime}}{R}-{S^{\prime}}{S}\right)=4\pi\ell C$ $\displaystyle A(R^{\prime}S-S^{\prime}R)=-i(2n+1);\quad\frac{+K(2n+1)}{RS}=4\pi C$ $\displaystyle K=+\frac{4\pi CRS}{2n+1};\quad V=+\sum\frac{4\pi CS_{0}RMN}{2n+1}$

Couche ellipsoïdale;

 $\displaystyle\text{rapport}\;\varepsilon;\quad\zeta=PP_{1}=\varepsilon OP=$ $\displaystyle\frac{\varepsilon\ell}{\sqrt{(\rho^{2}-a^{2})(\rho^{2}-b^{2})(% \rho^{2}-c^{2})}}=\frac{4}{3}\pi\frac{\varepsilon\ell}{T};\;T\;\text{vol. % ellips.}$ $\displaystyle V=\frac{(4\pi)^{2}}{3}\frac{\varepsilon}{T}S\quad\text{ou}\quad R% =1.$

Ellipsoïde plein.

 $\varepsilon\frac{dV}{dx}$
 $\displaystyle MM_{1}=\varepsilon$ $\displaystyle MP=\zeta\qquad\text{angle}\;PMM_{1}=\theta.$ $\displaystyle\zeta=\varepsilon\cos\theta.$ $\displaystyle M\;(\rho,\mu,\nu,)\;Q\;\rho+d\rho,\mu,\nu$ $\displaystyle x,y,z\qquad x+dx\;\text{etc.}\quad R=\sqrt{\rho^{2}-a^{2}}$
 $\displaystyle\cos\theta=\frac{dx}{d\rho}\frac{d\rho}{dn}=\frac{x\rho}{\sqrt{% \rho^{2}-a^{2}}}\frac{d\rho}{dn};\quad RMN=x\sqrt{(a^{2}-b^{2})(a^{2}-c^{2})}$ $\displaystyle\cos\theta=\frac{x\rho}{\sqrt{\rho^{2}-a^{2}}}\frac{d\rho}{dn}=% \ell iA\frac{x\rho}{\sqrt{\rho^{2}-a^{2}}}=\ell x\sqrt{(\rho^{2}-b^{2})(\rho^{% 2}-c^{2})}$ $\displaystyle\zeta=\ell\varepsilon RMN\sqrt{\frac{(\rho^{2}-b^{2})(\rho^{2}-c^% {2})}{(a^{2}-b^{2})(a^{2}-c^{2})}}=\frac{\ell\varepsilon T}{\left(\frac{4}{3}% \pi\right)}\frac{MN}{\sqrt{(a^{2}-b^{2})(a^{2}-c^{2})}}$ $\displaystyle\frac{dV}{dx}=\sqrt{(\rho^{2}-b^{2})(\rho^{2}-c^{2})}\frac{4}{3}% \pi\frac{RMNS_{0}}{\sqrt{(a^{2}-b^{2})(a^{2}-c^{2})}}$
 $\displaystyle\triangle u=0,\quad S:u=Q;\quad M-Q=\varpi;\quad S:\varpi=0,\quad% \triangle\varpi=-\triangle Q=f$ $\displaystyle\int\left[\sum\left(\frac{d\varpi}{dx}\right)^{2}+2\varpi f\right% ]d\tau=J;\quad\int\left[\sum\frac{d\varpi}{dx}\frac{d\delta\varpi}{dx}+f\delta% \varpi\right]d\tau=0$ $\displaystyle\int\delta\varpi\frac{d\varpi}{dn}d\omega-\int\delta\varpi(% \triangle\varpi-f)d\tau=0;\quad\varpi=\sum A_{i}\psi_{i}$

$\psi$ continue sur $R$ et sur son bord ainsi que ses dérivées principales. $m,n,p=0,1,2$
2° Sur le bord $\psi=0$; $\zeta$
$\cdots$ $\zeta$ représ. par séries, toutes les fois que $\zeta$ continue ainsi que ses dérivées princip.
$\zeta_{m}=0$ entraîne $A_{1}=A_{2}=\cdots=A_{m}=0$ si $\zeta_{m}=\sum A_{i}\psi_{i}$

On fera $\psi_{i}=F\begin{smallmatrix}\sin\\ \cos\end{smallmatrix}mx\begin{smallmatrix}\sin\\ \cos\end{smallmatrix}ny\begin{smallmatrix}\sin\\ \cos\end{smallmatrix}pz$; $F=0$.

 $\displaystyle\int\left(\varpi_{m}\triangle\zeta_{m}-f\zeta_{m}\right)d\tau=0;% \quad\int\left[\sum\frac{d\varpi_{m}}{dx}\frac{d\zeta_{m}}{dx}+f\zeta_{m}% \right]d\tau=0$ $\displaystyle\int\varpi_{m}\frac{d\zeta_{m}}{dn}d\omega-\int\left(\varpi_{m}% \triangle\zeta_{m}-f\zeta_{m}\right)d\tau=0$
 $\displaystyle J_{p}$ $\displaystyle=\int\left[\sum\left(\frac{d\varpi_{p}}{dx}\right)^{2}+2f\varpi_{% p}\right]d\tau\qquad(1)\qquad pq$ $\displaystyle J_{q}$ $\displaystyle=\int\left[\sum\left(\frac{d\varpi_{q}}{dx}\right)^{2}+2f\varpi_{% q}\right]d\tau\qquad(-1)$ $\displaystyle Q$ $\displaystyle=\int\left[\sum\frac{d\varpi_{q}}{dx}\frac{d(\varpi_{q}-\varpi_{p% }}{dx}+f(\varpi_{q}-\varpi_{p})\right]d\tau\qquad(2)$
 $J_{p}-J_{q}=\int\sum\left[\frac{d(\varpi_{q}-\varpi_{p}}{dx}\right]^{2}d\tau$

AD 3p. Collection particulière, Paris 75017.

Time-stamp: " 5.05.2019 01:04"

### Notes

• 1 Ces calculs sont datés par leur support : une lettre de G. Cres à Poincaré, 25.01.1912.