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数学物理方法习题集解答(完整编辑版)

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由?hu?ku????a?f????????qsin??0?????,得

?????2????0?cosm??Bm?sinm???f???, hC0???ha?km?am?1?Amm?1上式相当于在?0,2??区间上将f???展成傅里叶级数,由展开系数公式得:

C0?12h??2?0f???d??q2h???0sin?d??q, h?1??Am?ha?km?am?1??02?f???cosm?d?

?1?ha?km?am?1??0qsin?cosm?d?

????cos?m?1??cos?m?1???q?? m?1?2?m?1???ha?km?a??2?m?1?0m?1???1?m?1?1??1?m?1?1?1???1?qq??? ???m?1m?12?ha?km?a??2?m?1?2?m?1???ha?km?a?1?m???0?当m?1为奇数,m?2n?1,n?1,2,?????2q1??, ?当m为偶数,m?2n,n?1,2,???????ha?2nk?a2n?1?1?4n2???A1??1?ha?k???0qsin?cos?d? ?0??q ?2?ha?k????Bmsin2?d????1?cos2?0?0。

4?ha?k??1?ha?km?am?1??0qsin?sinm?d?

???sin?m?1??sin?m?1???q???0,当m?1, m?1?2?m?1???ha?km?a??2?m?1?0qB1???ha?k???0?qsin2?d??q。

2?ha?k?-*

qq2q??2ncos2n?。 ?u??,?????sin???2n?1h?2?ha?k??n?1a?ha?2nk??1?4n2?43、用傅里叶变换求解定解问题

??2u?2u??x2??y2?0,???uy?0???x?,??uy???0,??????x??,y?0?????x???????x???。

解:由于x在???,??内变化,对x进行傅里叶变换

2?%%??u?uyy?0?,(*) ?%%??%u?0???,u?y???y?0%%其中u?u??,y?,

%(*)的通解为u?C1???e?y?C2???e??y,?????,??。

%由uy??%??C2???e??y?当??0时,C1????0,u?0,得?, ?y%?C1???e??当??0时,C2????0,u??y??C2???e,??0????y%???C?e所以总的可写为u, ????y?C1???e,??0???其中C???????C2???,??0。

??C1???,??0%%%由u。 ?????,得C??y?0%%?u?????e??y。

进行傅里叶反变换,得

1u?x,y??2?1%u?,yed???????2??i?x??????%????e??yei?xd?

1?2??12??1?????2??????????????e?i??d??e??y?i?xd?

????y?i??x-???????ed??d?, ??????-*

而?e?????y?i??x-??d???e??0?y?i??x-??d???e0???y?i?x-???d?

?ey?i?x????y?i??x-??0???e11 ???y?i?x???y?i?x????y?i?x???0??y?i??x-????2y?x???2?y2,

代入上、下限时应注意到y?0,

?u?x,y?????x?????1?y????d?2?y2。

第七章习题解答

44、试用平面极坐标系把二维波动方程分离变数。 解:二维波动方程在极坐标系中可表为 utt?a2?u???u????1??。 (1) u2?????1令u??,?,t??T?t?R???????,代入(1),得:

??11T??R??a2?TR????TR???2TR????。 (2)

????(2)的两边除以a2TR?,得

T??R??R?1???????。 (3) a2TR?R?2?(3)的左边仅是t的函数,而右边却是?,?的函数,

?(3)的两边只能等于同一常数,记为?k2,从而

T???t??k2a2T?t??0?T?t??Acoskat?Bsinkat。

R??R?1?????2???k2, (4) R?R??(4)的两边乘以?2,并移项得

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?2R??R????。 ???k2?2??RR?同理上式两边只能等于同一常数,记为?。于是

???????0,?????????2??,???????????2??

??m2,?m?0,1,2,????,?????Ccosm??Dsinm?。

?2R??R????k2?2???m2, RR?2R????R???k2?2?m2?R?0,

令x?k?,R????y?x?,得

x2y???x??xy??x???x2?m2?y?x??0, m阶Bessel eq.

45、用平面极坐标系把二维输运方程分离变数。 解:在平面极坐标系中,二维输运方程为

ut?a2?u???u??2u??? (1)

?????11?令u??,?,t??T?t?R???????,代入(1),得:

T?R??a2?TR????TR???2TR????。 (2)

?????11?(2)的两边除以a2TR?,得

T?R??R?1???????. a2TR?R?2?上式左边仅是t的函数,而右边却是?,?的函数,

?上式两边只能等于同一常数,记为?k2,从而

T??t??k2a2T?t??0?T?t??Ae?kat。

22R及?与上一题的相同。

46、求证Pl?x??Pl??1(x)?2xPl??x??Pl??1?x?,l?1。

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证:勒让德多项式的生成函数为

11?2xr?r2??rlPl?x?,r?1。(1)

l?0?两边对x求导,得

r(1?2xr?r)322??rlPl??x?。

l?0?两边乘以(1?2xr?r2),得

r1?2xr?r2?(1?2xr?r)?rlPl??x?,(2)

2l?0?(1)代入(2),得

?rl?0?l?1Pl?x???rPl??x??2x?rPl??x???rl?2Pl??x?,

ll?1l?0l?0l?0???比较两边rl?1项的系数,得

?Pl?x??Pl??1?x??2xPl?x??Pl??1?x?。

47、利用上题和?l?1?Pl?1(x)??2l?1?xPl(x)?lPl?1(x)?0,l?1,

求证?2l?1?Pl?x??Pl??1?x??Pl??1?x?,l?1。 证:对勒让德多项式的递推公式

?l?1?Pl?1?x???2l?1?xPl?x??lPl?1?x??0 (1)

两边对x求导,得

?l?1?Pl??1?x???2l?1?Pl?x???2l?1?xPl??x??lPl??1?x??0。 (2)

?又由上题,得:Pl?x??Pl??1?x??2xPl?x??Pl??1?x?, (3)

?2??l??3?,得Pl??1?x??xPl??x???l?1?Pl?x?。 (4)

从(3)及(4)中消去Pl??1?x?,得

?xPl?x??Pl??1?x??lPl?x?。 (5)

数学物理方法习题集解答(完整编辑版)

-*由?hu?ku????a?f????????qsin??0?????,得?????2????0?cosm??Bm?sinm???f???,hC0???ha?km?am?1?Amm?1上式相当于在?0,2??区间上将f???展成傅里叶级数,由展开系数公式得:C0?12h??2?0f???d??q2h???0sin?d??q,h?1??Am
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