11=?[cos(x?t)?cos(x?t)]???[cos(x?(t??))?cos(x?(t??))]d?
220tt =sinxsint?sinx?sin(t??)d?
0? =sinxsint?sinx[?cos(t??)?sin(t??)]0 =tsinx 即 u(x,t)?tsinx 为所求的解。 9．求解波动方程的初值问题。
ttx?2u?au?xx?tt(1?x2)2? ??u|?0,u|?1t?0tt?0?1?x2?11??d??解： u(x,t)??1??2??(1??2)2d?d? 2ax?at0x?a(t??)x?atx?attx?a(t??)x?at1???12d??arctg(x?at)?arctg(x?at)
???1x?a(t??)d?d???[??(1??2)2?2(1??2)]x?a(t??)d?
0x?a(t??)01???]d? =?[22201?(x?a(t??)1?(x?a(t??))1x?at?u1x?at?u?du?=?a2(1?u2)?a2(1?u2)du 2x?2atx?attdutdudu??=??1?u22a?1?u2 zax?2a2x?at1?u2atx?at?1x?atxxttx?a(t??)tx?uxxx11?(x?at)2(arctg(x?at)?arctg(x?at))?2ln= 222a4a1?(x?at)t[2arctgx?arctg(x?at)?arctg(x?at)] 2a11=(x?at)arctg(x?at)?(x?at)arctg(x?at) 2a22a2+
t11?(x?at)2ln+arctgx? 22a4a1?(x?at)
所以
u(x,t)?14a22{(x?at?2a)arctg(x?at)?(x?at?2a)?311?(x?at)2?arctg(x?at)?2atarctgx?ln}221?(x?at)
§3混合问题的分离变量法 1. 用分离变量法求下列问题的解：
(1)
2??2u2?u?2?a2?t?x?3?x?u?,?ut?0?sinl?t??u(0,t)?u(l,t)?0??
t?o?x(1?x)(0?x?l)
解：边界条件齐次的且是第一类的，令
u(x,t)?X(x)T(t)
得固有函数Xn(x)?sinn?x，且 lan?an?Tn(t)?Ancost?Bnsint，(n?1,2?)
ll 于是 u(x,t)??(Ancosn?1?an?an?n?t?Bnsint)sinx lll今由始值确定常数An及Bn，由始值得
3?x?n???Ansinx sinlln?1x(l?x)??an?n?Bnsinx lln?1?所以 A3?1,An?0,当n?3
2 Bn?an?2?an?l?x(l?x)sin0n?xdx l??l2n?x??xcosx ??l??n???ln?l2n??xcosx?sin?l??lln2?2??n?
2l2xn?2l3n??22sinx?33cosxlln?n?因此所求解为
??l04l3?44(1?(?1)n) an?3a?3?4l3 u(x,t)?costsinx?lla?4
2??2u2?u?0?2?a2?x??t?(2) ?u(0,t)?0??u(x,0)?hx,?l?1?(?1)nan?n?sintsinx ?4llnn?1??u(l,t)?0 ?t?u(x,0)?0?t解：边界条件齐次的，令 u(x,t)?X(x)T(t)
?X????X?0得：??X(0)?0,2X?(l)?0 (1)
及 T???a?X?0(2)。
求问题(1)的非平凡解，分以下三种情形讨论。
1? ??0时，方程的通解为
X(x)?C1e??x?C2e???x
由X(0)?0得c1?c2?0 由X?(l)?0得C1??e??l?C2??e???l?0
解以上方程组，得C1?0，C2?0，故??0时得不到非零解。
2? ??0时，方程的通解为X(x)?c1?c2x
由边值X(0)?0得c1?0，再由X?(l)?0得c2?0，仍得不到非零解。
3???0时，方程的通解为
X(x)?c1cos?x?c2sin?x
由X(0)?0得c1?0，再由X?(l)?0得
c2?cos?l?0
为了使c2?0，必须 cos?l?0，于是
?2n?1????n???? (n?0,1,2?)
?2l?且相应地得到Xn(x)?sin22n?1?x (n?0,1,2?) 2l将?代入方程(2)，解得
Tn(t)?Ancos?2n?12n?1a?t?Bnsina?t (n?0,1,2?) 2l2l2n?12n?12n?1a?t?Bnsina?t)sin?x 2l2l2l于是 u(x,t)?再由始值得
n?0?(Ancos?2n?1?hx?Asin?x?n??l2ln?0 ??2n?12n?1?0??a?Bnsin?x?2l2ln?0?容易验证?sinl??2n?1??x?(n?0,1,2?)构成区间[0,l]上的正交函数系： 2l??2m?12n?1?0当m?n sin?xsin?xdx??l?当m?n2l2l?0?2利用?sin??2n?1??x?正交性，得 2l?2h2n?1?xdx An??xsinl0l2ll?2h?2l2n?12l? ?2??xcos?x???(2n?1)?(2n?1)?2ll??? ?????22n?1??sin?x?
2l??0l8h(2n?1)2?2(?1)n
Bn?0
(?1)n2n?12n?1cosa?tsin?x 所以 u(x,t)?2?22l2l?n?0(2n?1)8h?
2。设弹簧一端固定，一端在外力作用下作周期振动，此时定解问题归结为
2??2u2?u?2?a?x2??tu(l,t)?Asin?t 求解此问题。 ?u(0,t)?0,??uu(x,0)?(x,0)?0??t?解：边值条件是非齐次的，首先将边值条件齐次化，取U(x,t)?足
U(0,t)?0，U(l,t)?Asin?t
令u(x,t)?U(x,t)?v(x,t)代入原定解问题，则v(x,t)满足
2??2vA?22?v?xsin?t?2?a2l?x??t?v(0,t)?0,v(l,t)?0??vA?v(x,0)?0(x,0)??x??tl?Axsin?t，则U(x,t)满l(1)
v(x,t)满足第一类齐次边界条件，其相应固有函数为Xn(x)?sin故设 v(x,t)?n?x，(n?0,1,2?) l?Tn(t)sinn?1?n?xl(2)
?x?展成级数，得 ?A?2A??n?xsin?t及初始条件中?将方程中非齐次项x按?sinlll??A?2n?xsin?t??fn(t)sinx lln?12A?2n?xsin?tsinxdx 其中 fn(t)??l0ll?l2A?2n?l2n???sin?t?xcosx?sinx? ?22n?lll2n???0ll2A?2A??(?1)n?1sin?t?x
n?l???nsinn?1?n?x l