?2A?2n?2A?nxsinxdx?(?1)其中 ?n? ?l0lln?l??an???T(t)????n?l将(2)代入问题(1)，得Tn(t)满足??T(0)?0,n??2A??(?1)n?1sin?t?Tn(t)?n??2A?Tn?(0)?(?1)nn?22
an?an?2A?2sin?t解方程，得通解Tn(t)?Ancos t?Bnsint?(?1)n?1?an?2lln?()??2l由始值，得An?0
1(?1)n?12A?3l2(?1)n2A?aln2A?{(?1)?}? Bn? 222222an?n?n?((an?)??l)(an?)??l(?1)n2A?alan?sint 所以 v(x,t)??{22l(an?)?(?l)n?1?
(?1)n?12A?2l21n???sin?t}sinx 22l(an?)?(?l)n??(?1)2an??ln?{asint?sin?t}sinx ?2A?l?22ln?ln?1(an?)?(?l)因此所求解为
?A(?1)2 u(x,t)?xsin?t?2A?l? 22ln?1(an?)?(?l)
?{asinan??ln?t?sin?t}sinx lntl3．用分离变量法求下面问题的解
2??2u2?u?bshx?2?a2?x??t?u?|t?0?0 ?u|t?0? ?t??u|x?0?u|x?l?0??解：边界条件是齐次的，相应的固有函数为 Xn(x)?sinn?xl(n?1,2,?)
设 u(x,t)??T(t)sinnn?1?n?x l将非次项bshx按{sinn?x}展开级数，得 l?bshx??fn(t)sinn?1n?x l2bn?(?1)n?1shxsinxdx?222bn?shl 其中 fn(t)??2l0ln??l将 u(x,t)?ln?T(t)sinx代入原定解问题，得Tn(t)满足 ?nln?1?an?22bn??n?1??T(t)?()T(t)?(?1)shl?nn222 ln??l???Tn(0)?0,Tn?(0)?0方程的通解为
Tn(t)?Ancosan?an?l22bn?t?Bnsint?()?222(?1)n?1shl llan?n??l由Tn(0)?0，得：An??(由Tn?(0)?0，得Bn?0 所以 Tn(t)?(所求解为
l22bn?)222(?1)n?1shl an?n??l122bn?an?)222(?1)n?1shl(1?cost) an?n??ll?2bl2(?1)n?1an?n?(1?cost)sinx u(x,t)?2shl?222lla?n?1n(n??l)4．用分离变量法求下面问题的解：
2??2u?u2?u?a(b?0)?2?2b2?t?t?x?? ?u|x?0?u|x?l?0
?h?u?u|t?0?x,|t?0?0?l?t?解：方程和边界条件都是齐次的。令 u(x,t)?X(x)T(t) 代入方程及边界条件，得
T\?2bT'X\ ???? 2XaT X(0)?X(l)?0
由此得边值问题
?X\??X?0 ?X(0)?X(l)?0??n??因此得固有值???n???，相应的固有函数为
l?? Xn(x)?sin又T(t)满足方程
T?2bT?a?T?0
将???n代入，相应的T(t)记作Tn(t)，得Tn(t)满足 T\2n?x,n?1,2,? l\'2n?an??'?2bTn???T?0
l??22一般言之，b很小，即阻尼很小，故通常有
?an?? b???,n?1,2,?
?l?2故得通解 Tn(t)?e?bt(Ancos?nt?Bnsin?nt)
2?an??2其中 ?n????b
?l?所以
?u(x,t)?e?bt?(Ancos?nt?Bnsin?nt)sinn?1n?x l??hn?x?Asinx??nll?n?1再由始值，得 ?
??0?(?bA?B?)sinn?x?nnn?ln?1?所以
An?2hl2l?xsin0n?2hxdx?(?1)n?1 ln? Bn?所求解为
b?nAn?2bh(?1)n?1 n??nu(x,t)?2h?e?bt(?1)n?1bn?(cos?t?sin?t)sinx. ?nnn?lnn?1?
§4 高维波动方程的柯西问题
1． 利用泊松公式求解波动方程 utt?a(uxx?uyy?uzz)
32??ut?0?x?yz的柯西问题 ?
??utt?0?02解：泊松公式
????1??1?ds?ds u????????t?4?aMr?4?aMrSatSat??现 ??0,??x?yz
32?且 ??ds????(r,?,?)rsin?d?d?|r?at
r00sMat?2?其中 ?(r,?,?)??(x?rsin?cos?,y?rsin?sin?,z?rcos?) ?(x?rsin?cos?)?(y?sin?sin?)(z?rcos?)
?x?yz?3xrsin?cos??3xrsin?cos??rsin?cos? ?2yzrsin?sin??rzsin?sin??yrcos? ?2yrsin?cos?sin??rsin?sin?cos?
?2?32222223332222232计算
???(r,?,?)rsin?d?d?
00
?2???(x003?y2z)rsin?d?d??r(x3?y2z)?2?(?cos?)?0
?4?r(x3?y2z)?2???3x002rsin?cos??rsin?d?d??3xr22??sin0322??d??cos?d??0
032?0?2??00?3xrsin?cos??rsin?d?d??3xr222?2sin?d?cos???d? 01?12? ?3xr3[cos3??cos?]??[?sin2?]00324?4xr3???r?2??2?00?4r3sin?cos3??rsin?d?d?
2?4??0sin?d??0cos3?d??4?xr3
?222?
??2yzrsin?sin??rsin?d?d??2yzr?sin000?d??sin?d??0
0?2??22232??002rzsin?sin??rsin?d?d??rzsin?d?sin????d?001?1432??r3z[cos3??cos?]??[?sin2?]??rz003243?2??22?
??y00rcos??rsin?d?d??yr22?cos?sin?d??d??0
00?2???2yr002sin?coc?sin???rsin?d?d??2?
?2yr3?sin2?cos?d??sin?d??000?2?322rsin?sin?cos??rsin?d?d???00?2?
?r4?sin3?cos?d???sin2?d??000所以