loading迭加原理和傅立叶级数求解系数试探解
??n?n??(r)???n(r)??Ansinxshyaan?1n?1??Y=b处B.C.
?n?n??(r)|y?b??Ansinxshb?VS(x)aan?1?29
延拓为周期函数（0 ????x???U?x?定义域中的周期函数。 VS?x??2a?aa2ax30 傅氏系数?U(x)??0?V(x?2ka)?的S为U(x) ??VS(2ka?x)?(2k?1)a?x?2ka?(x?ka)?2ka?x?(2k?1)a?k?0,?1,?2,?3,???傅氏系数??n?n??(r)|y?b??Ansinxshb?VS(x)aan?1n???Ansh?an?1??n??b?sinx?U(x)a?31求解n?1Anshb?aan?U(x)sinxdx?a?aa2An?n?bashan?VS(x)sinxdx?a0a??(r)??(n?1?2n?n?n?V(x)sinxdx)sinxshy(V)Sn?b?aaa0ashaa0?x?a,0?y?b32