Page:Mémoires de l’Académie des sciences, Tome 22.djvu/253

duit $(\alpha -\beta \mathrm {i} )(\alpha +\beta \mathrm {i} ),$

${\begin{aligned}\varpi _{1}(x,y)=&e^{\beta x\mathrm {i} }{\text{Ȣ}}_{0}(y)+e^{-\beta x\mathrm {i} }{\text{Ȣ}}_{1}(y)\\=&{\text{Ȣ}}_{0}(y+x\mathrm {i} )+{\text{Ȣ}}_{1}(y-x\mathrm {i} ).\end{aligned}}$

Cela posé, la valeur de $u$ deviendra

(97)

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

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

et la seconde condition donnera

(98)

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

Si l’on fait, pour abréger ${\text{Ȣ}}_{0}=0,$ ${\text{Ȣ}}_{1}=0,$ et de plus

${\sqrt {\alpha ^{2}+\beta ^{2}}}=\lambda ,\qquad {\frac {d}{dt}}=\theta ,\qquad \varpi _{0}(x,y)=\upsilon ,$

$\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}{\frac {e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}-e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}}{e^{h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}+e^{-h\mathrm {i} {\sqrt {\alpha ^{2}+\beta ^{2}}}}}}=\mu ^{2},$

l’équation (98) deviendra

(99)

$\left[\theta ^{2}\left(e^{h\lambda \mathrm {i} }+e^{-h\lambda \mathrm {i} }\right)+g\lambda \mathrm {i} \left(e^{h\lambda \mathrm {i} }-e^{-h\lambda \mathrm {i} }\right)\right]\upsilon =0,$

et l’on en tirera, en mettant $t$ en évidence,

$\upsilon =e^{+\mu t\mathrm {i} }\chi _{0}(x,y)+e^{-\mu t\mathrm {i} }\chi _{1}(x,y),$

(100)

$u=\left[e^{(h-z)\lambda \mathrm {i} }+e^{-(h-z)\lambda \mathrm {i} }\right]\left[e^{\mu t\mathrm {i} }\chi _{0}(x,y)+e^{-\mu t\mathrm {i} }\chi _{1}(x,y)\right].$

Par suite, la dernière condition donnera

(101)

$\left\{{\begin{aligned}&\left[e^{h\lambda \mathrm {i} }+e^{-h\lambda \mathrm {i} }\right]\left[\chi _{0}(x,y)+\chi _{1}(x,y)\right]=0,\quad {\text{et}}\\\mu \mathrm {i} &\left[e^{h\lambda \mathrm {i} }+e^{-h\lambda \mathrm {i} }\right]\left[\chi _{0}(x,y)-\chi _{1}(x,y)\right]=f(x,y).\end{aligned}}\right.$

Or, on vérifie les équations (101) en prenant

$\chi _{0}(x,y)=-\chi _{1}(x,y)={\frac {f(x,y)}{2\mu \mathrm {i} \left(e^{h\lambda \mathrm {i} }+e^{-h\lambda \mathrm {i} }\right)}}.$