X=Xn(x)=x2m+I=sin (2m+1) 对应每一个本征值,解方程(1),可得 Do (=4 (1)= t+d sin m7 Tml(t=Emcs- (2m+1)m (2m+1)m t2=λ (2m+1)丌 Fns u(x, t)=Co!+Do+2(Cm cos"/t+Dm sin-1)cos 7 +∑(En0"+ 1) F 2/1)sin(2m+1)z (2m+1)m 代入初始条件可得, a=D0+ (2m+1) cos -x+ 27 0=c0+>Dc3 IIx+>F(2n 1)m(2n+1)x sIn 21 所以系数分别为 arax aX COs dx=o E sin (m+1) I'-Exsin (2m+)7r (2m+1)2z C0=0,Dn=0,Fn=0,因此 (x,D) (2m+1)mt(2m+1)z cos (2m+1) 2 10.9求二维无限深势阱的 Schrodinger方程( ) x l m X Xn x X m 2 2 1 ( ) 2 1 sin + π = = + = 对应每一个本征值，解方程（1），可得 0 0 0 T (t) = C t + D ( 0) λ = λ0 = t l m a t D l m a T m t Cm m π π 2 ( ) = cos + sin               = = 2 2 l m m π λ λ t l m a t F l m a T m t Em m 2 (2 1) sin 2 (2 1) ( ) cos 2 1 π + π + + + =               + = + = 2 2 1 2 (2 1) l m m π λ λ x l m t l m a t F l m a E x l m t l m a t D l m a u x t C t D C m m m m m m 2 (2 1) )sin 2 (2 1) sin 2 (2 1) ( cos ( , ) ( cos sin ) cos 0 1 0 0 π π π π π π + + + + + = + + + ∑ ∑ ∞ = ∞ = 代入初始条件可得，        + + = + + + − = + + ∑ ∑ ∑ ∑ ∞ = ∞ = ∞ = ∞ = 1 1 0 1 0 0 2 (2 1) sin 2 (2 1) 0 cos 2 (2 1) cos sin m m m m m m m m x l n l n a x F l m l m a C D x l m x E l m x D C π π π π π π ε 所以系数分别为 d 0 d 1 0 = − = ∫ ∫ − − l l l l x x x D ε cos d 0 cos d 1 2 − =       = ∫ ∫ − − l l l l m x l m x x x l m x C π ε π 2 2 1 2 (2 1) 8 d ( 1) 2 (2 1) sin d 2 (2 1) sin 1 π π ε ε π + = − + −       + = + − − ∫ ∫ m l x l m x x x l m x E m l l l l m C0 = 0 ， Dm = 0 ， Fm = 0，因此， ∑ ∞ = + + + + = − 0 2 1 2 2 (2 1) sin 2 (2 1) cos (2 1) 1 ( 1) 8 ( , ) m m l m x l m at m l u x t π π π ε 10.9 求二维无限深势阱的 Schrodinger 方程