不同边界条件的通解第一类边界条件u - a2u= 0xx，u(0, t)= u (/, t )= 0u第二类边界条件ttux-a 2u = 0(0,t)= u (l,t)= 0第三类边界条件Iu - a2u = 0u (0,t)= u(l,t)= 0第三类边界条件IIu - a2u = 0u(0,t)= u (l,t)= 0Solution:方程 u - a2u = 0设 u (x, t) = X (x)T (t)1 T X =人 a 2 T XX (x)= Aexx + Be-LX (x)= A/XeXx -B、iKeW第一类边界条件X(0)= 0A + B = 0 A = -BX (l)= 0e xi + e-x兀=in l.兀本征值 X = n2 = -n2 k2 k l 7x (x )=K本征函数-ein lx = 2i sin n x lT = - n 2 k 2 a 2T = - n 20 2TT (t)= Ceinot + De-inot = A cos nOt + B sin notka =2k2k-ZTT-=2lT时间频率u (x,t)=(A cos not + B sin not )sin nkx nnx = 0的情况： 单独讨论X (x ) = a + P xT (t) = y +51u(0,t)= 0X (0)= 0u (l, t ) = 0 X (l ) = 0p -1 = 0 P = 0X 0 (x )= 0(x/) = 0第一类边界条件的通解u (x，D=尤(Ann=1cos not + B sin not )sin nkx 苴中 k =，第二类边界条件X (0)= 0X(l)= 0e 入 1 exi = 0 e2 入 1 = 1 = 02兀-JX = in 1本征值x = n2=n 2 k 2X (x )= einkx + einkx =cosnkxk = 2 - 空间频率T = - n 2 k 2 a 2T = n 20 2TT (t)= Ceinot + De-inot = A cos not + B sin notka =222厂=t = o时间频率u 顷)=(4 cos not + B sin not )cos nkxx = 0的情况：单独讨论X (x ) = a + g xT(t ) = y+81u(0,t)= 0 X (0)= 0 p = 0u(1, t)= 0 X (1 )= 0X 0 (x ) = au (x,t) = a(y+8t) = 1+at其中k=1，第二类边界条件的通解 u (x, t)= 1 + 以t + (A cos not + B sin not)cosnkxn =1第三类边界条件IX0)= 0A-B = 0 A = BX (/)= 0e 山 + e z = 0 e2项i = 1 = e沅+以皿 (X兀/l=i n+1/(1 .本征值X = n + - k2V 2)X (x) = e(n+2奴 + e,n+2奴=cos n + kx本征函数V 2 J讨论n的取值范围n零和正整数X0 (x) = a + Bx，X 0)= 0p = 0 ； X (l)= 0 a= 0， 故X 0)(x)= 0第三类边界条件I的通解n=0.r 1 cA cos n + t + BV2)coskx其中*=1， 第三类边界条件IIX (0)= 0A + B = 0A = BX1 )= 0ei + e-、xi = 0e2 X1 = 1 = ei兀+i2n兀=i本征函数(x, t)=A cosr 1)n + 皿+ Bsinr 1)n + -皿sinr 1)n + n=0nV 2)nV 2)V2)X (0)= 0a = 0，故 X0 (x)= 0第三类边界条件I的通解kxr 1本征值X = n + - k2V2)r q,.r，皿 r 1 )kxX (x ) = Mn+2火eTn+2)kx = sin n + 1V2J讨论n的取值范围n零和正整数X0 (x) = a + Px，X 1) = 0 p = 0 ；其中k = i写的啰嗦，可精简