X()=√2A2cos√l-=0 所以本征值为2=2「(2n+1)z (n=0,1,2,3…) 相应的本征函数为X=Xn(x)=sin (2n+1)丌 (2n+1) + ]2T=0 T(o=T,(0=Ccos (2n+1)m +1) t+D sin 27 27 n+1)m l(x,1)=∑(C D 27 27 C (2n+1)丌 代入初始条件可得, Dn (2n+1)m.smn(2n+1)z 27 2 系数C,D分别为 (2n+1)mx 8al (2n+1)mx B=0 (x,)=8a/ ∑(-1)-1 (2n+1) (2n+1)mx (2n+1)2 按定解问题,此解仅适用于0≤x≤l。根据对称性上式在-1≤x≤l 也成立。 [解二] 将坐标原点取在杆的左端,则解定解问题为 =0 =0. 其解为 (x,)= ∑ COs cos (2n+1X '(l) λ A cos λl = 2 =0 ⇒ cos λl =0 所以 本征值为 2 2 (2 1)       + = = l n n π λ λ (n=0,1,2,3…) 相应的本征函数为 x l n X X x n 2 (2 1) ( ) sin + π = = (6) ] 0 2 (2 1) '' [ 2 = + + T l n a T π t l n a t D l n a T t Tn t Cn n 2 (2 1) sin 2 (2 1) ( ) ( ) cos π + π + + = = x l n t l n a t D l n a u x t C n n n 2 (2 1) ) sin 2 (2 1) sin 2 (2 1) ( , ) ( cos 0 π π + π ⋅ + + + = ∑ ∞ = 代入初始条件可得，        + ⋅ + = + − = ∑ ∑ ∞ = ∞ = x l n l n a D x l n x C n n n n 2 (2 1) sin 2 (2 1) 0 2 (2 1) sin 0 0 π π π ε 系数Cn Dn , 分别为 2 2 1 0 0 2 (2 1) 8 ( 1) 2 (2 1) sin 2 (2 1) sin 1 π π ε ε π + = − + −       + = + ∫ ∫ n l dx l n x x dx l n x C n l l n Bn = 0 l n x l n at n l u x t n n 2 (2 1) sin 2 (2 1) cos (2 1) 1 ( 1) 8 ( , ) 2 1 0 2 π π π ε + + + = − + ∞ = ∑ 按定解问题，此解仅适用于0 ≤ x ≤ l 。根据对称性上式在 − l ≤ x ≤ l 也成立。 [解二] 将坐标原点取在杆的左端，则解定解问题为 ( )      = = = − = = ≤ ≤ = = = = 0, 0 ( ); 0; 0 2 0 2 0 0 2 x x x x l t t t tt xx u u u l x u u a u x l ε 其解为 ∑ ∞ = + + + = 0 2 2 2 (2 1) cos 2 (2 1) cos (2 1) 8 1 ( , ) n l n x l n at n l u x t π π π ε