Page:Mémoires de l’Académie des sciences, Tome 22.djvu/252

(91)

$f(x)=\mathbf {S} _{0}^{\infty }\left(\mathrm {A} \sin {\frac {n\pi x}{a}}\right)\qquad \qquad$

$=\mathrm {A} _{0}+\mathrm {A} _{1}\sin {\frac {\pi x}{a}}+\mathrm {A} _{2}\sin {\frac {2\pi x}{a}}+{\text{etc}}\ldots ,$

pour toutes les valeurs de $x$ comprises entre $x=0$ et $x=a.$

Troisième exemple. Intégrer l’équation

(92)

${\frac {d^{2}u}{dx^{2}}}+{\frac {d^{2}u}{dy^{2}}}+{\frac {d^{2}u}{dz^{2}}}=0,$

de manière que l’on ait

${\begin{array}{lll}{\text{pour }}z=h,&&{\cfrac {du}{dz}}=0,\\{\text{pour }}z=0,&&g{\cfrac {du}{dz}}-{\cfrac {d^{2}u}{dt^{2}}}=0,\\{\text{pour }}z=0&{\text{et}}\quad t=0,\qquad \qquad &u=0\quad {\text{et}}\quad {\cfrac {du}{dt}}=f(x,y).\end{array}}$

Solution. On trouvera, en mettant $x,y,z$ en évidence,

(93)

$u=e^{z\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{0}(x,y)+e^{-z\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{1}(x,y),$

(94)

$0={\sqrt {\alpha ^{2}+\beta ^{2}}}\left[e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{0}(x,y)-e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{1}(x,y)\right].$

Soient

(95)

$\left\{{\begin{aligned}e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{0}(x,y)+e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{1}(x,y)=&\varpi _{0}(x,y),\\e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{0}(x,y)-e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\psi _{1}(x,y)=&\varpi _{1}(x,y)\,;\end{aligned}}\right.$

l’équation (94) deviendra

(96)

${\sqrt {\alpha ^{2}+\beta ^{2}}}.\varpi _{1}(x,y)=0,$

et l’on aura

${\begin{aligned}\psi _{0}(x,y)=&e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\left[\varpi _{0}(x,y)+\varpi _{1}(x,y)\right],\\\psi _{1}(x,y)=&e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}\ \ \left[\varpi _{0}(x,y)-\varpi _{1}(x,y)\right].\end{aligned}}$

Or, on tire de l’équation (96), $\alpha ^{2}+\beta ^{2}$ étant égal au pro-