正在加载图片...
V+V(x, z)v(x,y, z)=Ey(x, y, z) 设:V(x,y,)=V1(x)+V2(y)+V3(z) 令:ψ(x,y,z)=X(x)Y(y)z(z) x(x)()2(2)+k(x)+(0)+(y(x,y,)=Ev(x,yz) 2pldx' dydz u dr2 x +V(x)y+xz-h'd2 h2 d h- d 2 2u dy r+v(ww +XY 2udZ +V,(a)y=Ey(x, J,z) 尊式两边除以(x,yz=X(x)Y(y)Z(z) h- d X 2u dx 计1计 n- d 2u dx2+l(o)x(x)=E,X(x) 其中 1-2d2+20)()=E,(y) E=E +e +E n- d 1-2ah2x+p(2)2(z)=E2(x)( , , ) ( , , ) ( , , ) 2 2 2 V x y z  x y z E x y z   =  −  +  ( ) ( , , ) 2 ( ) 2 ( ) 2 2 3 2 2 2 2 2 2 2 1 2 2 Z V z E x y z dzd Y V y XY dyd X V x XZ dxd YZ         + =    + + −    + + −  −    ( ) ( ) ( )  ( ) ( ) ( ) ( , , ) ( , , ) 2 2 1 2 3 2 22 2 2 2 X x Y y Z z V x V y V z x y z E x y z dzd dyd dxd     + + + =   − + +  Z V z E dzd Z Y V y dyd Y X V x dxd X =    +  + −    +  + −    + − ( ) 2 1 ( ) 2 1 ( ) 2 1 2 3 2 2 2 2 2 2 2 1 2 2       ( , , ) ( ) ( ) ( ) 1 2 3 设:V x y z =V x +V y +V z 等式两边除以 ( x, y,z) = X(x)Y( y)Z(z) ( )] ( ) ( ) 2 [ ( )] ( ) ( ) 2 [ ( )] ( ) ( ) 2 [ 2 3 2 22 2 2 22 1 2 2 V z Z z E Z z dzd V y Y y E Y y dyd V x X x E X x dxd z yx − + = − + = − + =     其中 E = E x + E y + E z 令:  ( x, y,z ) = X ( x ) Y ( y ) Z ( z ) 返回
<<向上翻页向下翻页>>
©2008-现在 cucdc.com 高等教育资讯网 版权所有