数学物理方程与特殊函数第三版课后答案王元明.pdf

22222222 2 2 2 2 0,0;0,0;0,0;0,0; 22222222 uuuuuuuu axltaxltaxltaxlt txtxtxtx = , , , , 2 22 22 22 2 llllllll l l l l 0 0 0 0 例例 设弦的两端固定于 设弦的两端固定于x=x=x=x=0 0 0 0 和 和x=lx=lx=lx=l，弦的初始位移 如下图，初速度为零，求弦满足的定解条件。 ，弦的初始位移 如下图，初速度为零，求弦满足的定解条件。 解：解： ,0,0,0,0 2 2 2 2 ,0,0,0,0 0 0 0 0 0 0 0 0 , , , , 2 2 2 2 l l l l xxxxxxxx u u u u u u u u t t t t ltltltlt t t t t lxxllxxllxxllxxl = = = = = = = = = 0; 0; 0; 0; 0 0 0 0 uuuuuuuu xxlxxlxxlxxl = = 练习练习： 求下列定解问题的解求下列定解问题的解 ( ( ( () ) ) ) ( ( ( () ) ) )( ( ( () ) ) ) ( ( ( () ) ) )( ( ( ( ) ) ) ) 2 2 2 2 ,0,0,0,0,0,0,0,0 0,0,00,0,00,0,00,0,0 ,0,0,0,0,0,0,0,0 txxtxxtxxtxx xxxxxxxx ua uxltua uxltua uxltua uxlt utul ttutul ttutul ttutul tt uxxxluxxxluxxxluxxxl = = ( ( ( () ) ) ) 2 2 2 2 1 1 1 1 0 0 0 0 :,cos:,cos:,cos:,cos 1 1 1 1 2 2 2 2 nananana t t t t l l l l n n n n n n n n a a a a n n n n u x ta exu x ta exu x ta exu x ta ex l l l l = = = = = = = =+ + + + 解为解为 ( ( ( ( ) ) ) ) 0 0 0 0 2 2 2 2 cos0,1,2,.cos0,1,2,.cos0,1,2,.cos0,1,2,. l l l l n n n n n n n n axxdxnaxxdxnaxxdxnaxxdxn llllllll = ， 其中其中 0 0 0 0 0 0 0 0 0(0, 0)0(0, 0)0(0, 0)0(0, 0) 0,( );0,( );0,( );0,( ); 0,0.0,0.0,0.0,0. xxyyxxyyxxyyxxyy xx axx axx axx a yyyyyyyy yy byy byy byy b uuxaybuuxaybuuxaybuuxayb uuf yuuf yuuf yuuf y uuuuuuuu = = +=+=+=+= = = (0, 0)(0, 0)(0, 0)(0, 0) : : : : xaybxaybxaybxayb：在矩形中求解拉普拉斯 方程的 ：在矩形中求解拉普拉斯 方程的 练练 定解问题定解问题 习习 提示：将第三式看作边界条件。提示：将第三式看作边界条件。 ( , )( , )( , )( , ) P54 (1P54 (1P54 (1P54 (1 (- ),(- ),(- ),(- ), 3)3)3)3) 0 0 0 0 ( , ).( , ).( , ).( , ). a a a a u aTu aTu aTu aT u u u u = = = = 一半径为的半圆形平板,其圆周边界上的温度 保持而直径边界上的温度保持 为 度,板的侧面绝缘,试求稳恒状态下的温度分布 规律 一半径为的半圆形平板,其圆周边界上的温度 保持而直径边界上的温度保持 为 度,板的侧面绝缘,试求稳恒状态下的温度分布 规律 例例 : : : :提示提示 定解问题定解问题 ( ( ( () ) ) ) 22222222 222222222222 11111111 0,0,0,0,0,0,0,0, ( , )(),( , )(),( , )(),( , )(), ( ,0)( , )( ,0)( , )( ,0)( , )( ,0)( , ) 0 0 0 0 , , , , , , , , 0 0 0 0 0,00,00,00,0 uuuuuuuuuuuu a a a a u aTu aTu aTu aT uuauuauuauua +=+=+=+= = = | (0, )| (0, )| (0, )| (0, )|. . . ., , , ,0 0 0 0u u u u + + + + ( , )( ) ( ),( , )( ) ( ),( , )( ) ( ),( , )( ) ( ),uRuRuRuR =令经过分离变量后得令经过分离变量后得 ( ( ( ( ) ) ) )( ( ( ( ) ) ) ) 0,0,0,0,0,0,0,0, 00,00,00,00, + = + = + = + = = = = = = ( ( ( ( ) ) ) ) 2 2 2 2 0, 0, 0, 0, 0.0.0.0. RRRRRRRRRRRR R R R R +=+=+=+= + + + + 特征值特征值 2 2 2 2 ,1,2,.,1,2,.,1,2,.,1,2,. n n n n nnnnnnnn = 特征函数特征函数( ( ( ( ) ) ) ) sin,1,2,.sin,1,2,.sin,1,2,.sin,1,2,. nnnnnnnn bnnbnnbnnbnn= ( ),1,2,.( ),1,2,.( ),1,2,.( ),1,2,. n n n n nnnnnnnn RcnRcnRcnRcn= ( ( ( () ) ) ) 1 1 1 1 ,sin,sin,sin,sin, n n n n n n n n n n n n udnudnudnudn = = = = = = = = ( , )(),( , )(),( , )(),( , )(),u aTu aTu aTu aT =将代入上式将代入上式 1 1 1 1 sin()sin()sin()sin() n n n n n n n n n n n n d anTd anTd anTd anT = = = = = 2 2 2 2 00000000 sind()sindsind()sindsind()sindsind()sind n n n n n n n n d anTnd anTnd anTnd anTn = 3 3 3 3 4 4 4 4 (1cos),(1cos),(1cos),(1cos), n n n nn n n n T T T T dndndndn a na na na n = ( ( ( () ) ) ) 3 3 3 3 1 1 1 1 4 4 4 4 ,1( 1) sin.,1( 1) sin.,1( 1) sin.,1( 1) sin. n n n n n n n n n n n n n n n n T T T T unununun a na na na n = = = = = = = = sinsinsinsin cos ,cos ,cos ,cos , xxxxxxxx = coscoscoscos sin ,sin ,sin ,sin , yyyyyyyy = 同理同理 coscoscoscos sinsinsinsin x x x x y y y y = = = = = = = = 1cossin1cossin1cossin1cossin 0sincos0sincos0sincos0sincos xxxxxxxx xxxxxxxx = =+=+=+=+ ( , )( , )( , )( , )u u u u 22222222 22222222 0 0 0 0 uuuuuuuu xyxyxyxy +=+=+=+= 已知已知 sinsinsinsin coscoscoscos x x x x uuuuuuuuuuuuuuuu x x x xx x x x u u u u =+=+=+=+= 极坐标系下的拉普拉斯方程的表达式极坐标系下的拉普拉斯方程的表达式 222222222222 22222222 22222222 2 2 2 2 2 2 2 2 cos( sin )cos( sin )cos( sin )cos( sin ) sinsinsinsin 1sin1sin1sin1sin coscoscoscos uuuuuuuuuuuuuuuu x x x x u u u u xxxxxxxxxxxx xxxxxxxx xxxxxxxx u u u u u u u u =+ =+ =+ =+ + 222222222222 222222222222 .,0.,0.,0.,0 uuuuuuuuuuuu yxyyxyyxyyxy =+=+=+=+= 同理代入同理代入 22222222 222222222222 11111111 0 0 0 0 uuuuuuuuuuuu +=+=+=+= sinsinsinsin coscoscoscos uuuuuuuu 练习练习 求解定解问题 求解定解问题 22222222 2 2 2 2 22222222 0 0 0 0 00000000 22222222 sincos,0,0;sincos,0,0;sincos,0,0;sincos,0,0; |3,|6;0|3,|6;0|3,|6;0|3,|6;0 4 4 4 4 |3 1,|sin,0.|3 1,|sin,0.|3 1,|sin,0.|3 1,|sin,0. xx lxx lxx lxx l tttttttt uuuuuuuu axxxl taxxxl taxxxl taxxxl t txlltxlltxlltxll uutuutuutuut xuxuxuxu uxxluxxluxxluxxl ltlltlltlltl = = =+=+=+=+= =+=+=+=+= ( ( ( () ) ) )( ( ( () ) ) )( ( ( ( ) ) ) ), , , , , , , , , ,u x tv x tu x tv x tu x tv x tu x tv x tw xw xw xw x+ + + += = = =提示：提示： 其中其中 w w w w( ( ( (x x x x) ) ) ) 满足满足 ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )( ( ( ( ) ) ) ) 2 2 2 2 22222222 sincos0sincos0sincos0sincos0 03,603,603,603,6 a wxxxa wxxxa wxxxa wxxx llllllll ww lww lww lww l +=+=+=+= = 22222222 22222222 ( , ),0,0,( , ),0,0,( , ),0,0,( , ),0,0, uuuuuuuu f x yxaybf x yxaybf x yxaybf x yxayb xyxyxyxy +=+=+=+= ( ( ( () ) ) )( ( ( () ) ) ) 12121212 ,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.u xxu x bxxau xxu x bxxau xxu x bxxau xxu x bxxa= ( ( ( () ) ) )( ( ( ( ) ) ) )( ( ( () ) ) )( ( ( ( ) ) ) ) 12121212 0,0,0,0,0,0,0,0,uyyu a yyybuyyu a yyybuyyu a yyybuyyu a yyyb= 例例 在矩形域内求下面定解问题在矩形域内求下面定解问题 P54 (16)P54 (16)P54 (16)P54 (16) 提示提示: : : : 先把一组边界条件化成齐次的。比如要把先把一组边界条件化成齐次的。比如要把 x=x=x=x=0 0 0 0 及及 x=a x=a x=a x=a 上的边界条件化成齐次的，可以令上的边界条件化成齐次的，可以令 ( ( ( () ) ) )( ( ( () ) ) )( ( ( () ) ) ),u x tv x tu x tv x tu x tv x tu x tv x tw x yw x yw x yw x y+ + + += = = = 其中其中 21212121 1 1 1 1 ( )( )( )( )( )( )( )( ) ( , )( ),( , )( ),( , )( ),( , )( ), yyyyyyyy w x yyxw x yyxw x yyxw x yyx a a a a =+=+=+=+ 通过代换后得到关于通过代换后得到关于 v v v v 的定解问题的定解问题 22222222 1 1 1 1 22222222 ( , ),0,0,( , ),0,0,( , ),0,0,( , ),0,0, vvvvvvvv fx yxaybfx yxaybfx yxaybfx yxayb xyxyxyxy +=+=+=+= ( ( ( () ) ) )( ( ( () ) ) ) 12121212 ,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.v xxv x bxxav xxv x bxxav xxv x bxxav xxv x bxxa= ( ( ( () ) ) )( ( ( () ) ) )0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,vyv a yybvyv a yybvyv a yybvyv a yyb= 21212121 11111111 ( )( )( )( )( )( )( )( ) ( , )( , )( ),( , )( , )( ),( , )( , )( ),( , )( , )( ), yyyyyyyy fx yf x yyxfx yf x yyxfx yf x yyxfx yf x yyx a a a a =+=+=+=+ 其中其中 21212121 1 1 1 111111111 (0)(0)(0)(0)(0)(0)(0)(0) ( )( )(0),( )( )(0),( )( )(0),( )( )(0),xxxxxxxxxxxx a a a a =+=+=+=+ 21212121 2 2 2 221212121 ( )( )( )( )( )( )( )( ) ( )( )( ),( )( )( ),( )( )( ),( )( )( ), bbbbbbbb xxbxxxbxxxbxxxbx a a a a =+=+=+=+ 然后利用然后利用2.4 2.4 2.4 2.4 非齐次方程的解法：非齐次方程的解法： (1)(2)(1)(2)(1)(2)(1)(2) ( , )( , )( , ),( , )( , )( , ),( , )( , )( , ),( , )( , )( , ),v x yvx yvx yv x yvx yvx yv x yvx yvx yv x yvx yvx y=+=+=+=+ 2 (1)2 (1)2 (1)2 (1)2 (1)2 (1)2 (1)2 (1) 1 1 1 122222222 ( , ),0,0,( , ),0,0,( , ),0,0,( , ),0,0, vvvvvvvv f x yxaybf x yxaybf x yxaybf x yxayb xyxyxyxy +=+=+=+= ( ( ( () ) ) )( ( ( () ) ) ) (1)(1)(1)(1)(1)(1)(1)(1) ,00;,0,0.,00;,0,0.,00;,0,0.,00;,0,0.vxvx bxavxvx bxavxvx bxavxvx bxa= ( ( ( () ) ) )( ( ( () ) ) ) (1)(1)(1)(1)(1)(1)(1)(1) 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,vyva yybvyva yybvyva yybvyva yyb= ( ( ( () ) ) )( ( ( () ) ) ) (2)(2)(2)(2)(2)(2)(2)(2) 12121212 ,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.,0( );,( ),0.vxxvx bxxavxxvx bxxavxxvx bxxavxxvx bxxa= 2 (2)2 (2)2 (2)2 (2)2 (2)2 (2)2 (2)2 (2) 22222222 0,0,0,0,0,0,0,0,0,0,0,0, vvvvvvvv xaybxaybxaybxayb xyxyxyxy +=+=+=+= ( ( ( () ) ) )( ( ( () ) ) ) (2)(2)(2)(2)(2)(2)(2)(2) 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,vyva yybvyva yybvyva yybvyva yyb= ( ( ( () ) ) ) (2)(2)(2)(2) 1 1 1 1 ,()sin,()sin,()sin,()sin nnnnnnnn yyyyyyyy aaaaaaaa nnnnnnnn n n n n n n n n vx yA eB exvx yA eB exvx yA eB exvx yA eB ex a a a a = = = = =+=+=+=+ 用分离变量法求得用分离变量法求得: : : : ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) )( ( ( ( ) ) ) ) 1 1 1 1 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 2 sinsinsinsin 1, 21, 21, 21, 2 2 2 2 2 sin.sin.sin.sin. a a a a nnnnnnnn a a a annnnnnnn bbbbbbbb aaaaaaaa nnnnnnnn n n n n ABdABdABdABd aaaaaaaa n n n n n n n n A eB edA eB edA eB edA eB ed aaaaaaaa +=+=+=+= = = = = +=+=+=+= ( ( ( () ) ) ) (1)(1)(1)(1)(1)(1)(1)(1) 1 1 1 1 ,( ) sin,( ) sin,( ) sin,( ) sin n n n n n n n n n n n n vx yvyxvx yvyxvx yvyxvx yvyx a a a a = = = = = = = = 再令再令 22222222 1 1 1 1 22222222 ( , ),0,0,( , ),0,0,( , ),0,0,( , ),0,0, vvvvvvvv fx yxaybfx yxaybfx yxaybfx yxayb xyxyxyxy +=+=+=+= = 方法一方法一：利用拉普拉斯变换：利用拉普拉斯变换 对对 y y y y 进行拉普拉斯变换。设进行拉普拉斯变换。设 ( ( ( () ) ) )( ( ( () ) ) ),u x yU x pu x yU x pu x yU x pu x yU x p 则方程变为：则方程变为： ( ( ( () ) ) ) 22222222 2 2 2 2 1 1 1 1 , , , , d d d d pU x pxxpU x pxxpU x pxxpU x pxx dxdxdxdxp p p p = 2 2 2 2 3 3 3 3 2 2 2 2 . . . . dUxxdUxxdUxxdUxx dxppdxppdxppdxpp =+=+=+=+ 而 变为而 变为 1 1 1 1 |cos|cos|cos|cos x x x x uyuyuyuy = = = = = = = = ( ( ( () ) ) ) 1 1 1 12 2 2 2 ,|,|,|,|, 1 1 1 1 x x x x p p p p U x pU x pU x pU x p p p p p = = = = = = = = + + + + 解常微分方程得 解常微分方程得 ( ( ( () ) ) ) 32323232 323323323323 , , , , 1111111111111111 . . . . 313313313313 U x pU x pU x pU x p p p p p xxxxxxxx pppppppppppppppppppp = = = = + + + + + 对对 p p p p 取拉普拉斯逆变换，得取拉普拉斯逆变换，得 ( ( ( () ) ) ) 3222322232223222 11111111 ,cos1,cos1,cos1,cos1 66666666 u x yx yxyyu x yx yxyyu x yx yxyyu x yx yxyy=+=+=+=+ 1 1 1 1 ! ! ! ! (),0,1,(),0,1,(),0,1,(),0,1, n n n n n n n n n n n n L tnL tnL tnL tn p p p p + + + + = 方法二方法二：先求通解：先求通解： 32323232 12121212 1 1 1 1 ( , )( )( )( , )( )( )( , )( )( )( , )( )( ) 6 6 6 6 u x yx yCyCxu x yx yCyCxu x yx yCyCxu x yx yCyCx=+=+=+=+ 2 2 2 2 12121212 (0)( )(0)( )(0)( )(0)( )xCCxxCCxxCCxxCCx=+=+=+=+ 2 2 2 2 21212121 ( )(0)( )(0)( )(0)( )(0)CxxCCxxCCxxCCxxC= 令令 y y y y=0 =0 =0 =0 并利用边界条件得并利用边界条件得 令令 x x x x=1 =1 =1 =1 并利用边界条件得并利用边界条件得 2 2 2 2 12121212 1 1 1 1 cos( )(1)cos( )(1)cos( )(1)cos( )(1) 6 6 6 6 yyCyCyyCyCyyCyCyyCyC=+=+=+=+ 2 2 2 2 11111111 1 1 1 1 ( )1(0)( )1(0)( )1(0)( )1(0) 6 6 6 6 yCyCyCyCyCyCyCyC=+=+=+=+ 2 2 2 2 11111111 1 1 1 1 ( )cos1(0)( )cos1(0)( )cos1(0)( )cos1(0) 6 6 6 6 CyyyCCyyyCCyyyCCyyyC= += += += + 3222322232223222 11111111 ( , )cos1( , )cos1( , )cos1( , )cos1 66666666 u x yx yxyyu x yx yxyyu x yx yxyyu x yx yxyy=+=+=+=+ (F sinx查Fourier变换表) 方法二方法二：利用：利用 + + + + + + + + + + + + + + += = = = + + + + + + + + t t t tt t t ta a a ax x x x t t t ta a a ax x x x at at at atx x x x at at at atx x x x d d d dd d d df f f f a a a a d d d d a a a a at at at atx x x xat at at atx x x xt t t tx x x xu u u u 0 0 0 0 ) ) ) )( ( ( ( ) ) ) )( ( ( ( ) ) ) ), , , ,( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) )( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) )( ( ( () ) ) )( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) ), , , ,( ( ( ( ()()()() 0()0()0()0() 11111111 ( , )sinsin( , )sinsin( , )sinsin( , )sinsin 22222222 x ttxtx ttxtx ttxtx ttxt x txtx txtx txtx txt u x tdd du x tdd du x tdd du x tdd d + =+=+=+=+ sinsinsinsintxtxtxtx= = = = (也可对 x 作拉普拉斯变换) lim( , )0lim( , )0lim( , )0lim( , )0 y y y y U w yU w yU w yU w y = = = = ( ( ( (查表查表) ) ) ) 22222222 ,Re( )0,Re( )0,Re( )0,Re( )0 a wa wa wa w a a a a FeaFeaFeaFea axaxaxax = + + + + ( ( ( () ) ) ) 22222222 ( , )( )( , )( )( , )( )( , )( ) y y y y u x yf xu x yf xu x yf xu x yf x yxyxyxyx = + + + + 22222222 1 1 1 1 ( )( )( )( ) ()()()() y y y y fdfdfdfd yxyxyxyx = = = = + 注注：也可用下一章的格林函数法。：也可用下一章的格林函数法。 ( ( ( (利用拉普拉斯变换反演公式和留数定理利用拉普拉斯变换反演公式和留数定理) ) ) ) + + + + + + + + + + + + + + += = = = + + + + + + + + t t t tt t t ta a a ax x x x t t t ta a a ax x x x at at at atx x x x at at at atx x x x d d d dd d d df f f f a a a a d d d d a a a a at at at atx x x xat at at atx x x xt t t tx x x xu u u u 0 0 0 0 ) ) ) )( ( ( ( ) ) ) )( ( ( ( ) ) ) ), , , ,( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) )( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) )( ( ( () ) ) )( ( ( ( 2 2 2 2 1 1 1 1 ) ) ) ), , , ,( ( ( ( 解：解：代入下式，代入下式， 9 9 9 9 ：求解下面初值问题：求解下面初值问题：P83. 9P83. 9P83. 9P83. 9 22222222 2222222222222222 000000002 2 2 2 ,0,0,0,0, (1)(1)(1)(1) 1 1 1 1 0,.0,.0,.0,. 1 1 1 1 tttttttt uuxtuuxtuuxtuuxt xtxtxtxt txxtxxtxxtxx u u u u uxuxuxux txtxtxtx = = = = = + = += += += + + 得得 ()()()() 222222222222 0()0()0()0() 111111111111 ( , )( , )( , )( , ) 212(1)212(1)212(1)212(1) x ttxtx ttxtx ttxtx ttxt x txtx txtx txtx txt u x tdd du x tdd du x tdd du x tdd d + =+=+=+=+ + ()()()() 2 2 2 2 0 0 0 0()()()() 1111111111111111 arctan()arctan()arctan()arctan() 222(1)222(1)222(1)222(1) xtxtxtxt x tx tx tx t t t t t x tx tx tx t xtxtxtxt d d d d + + + + + =+ =+ =+ =+ + + + + 22222222 0 0 0 0 11111111 arctan()arctan()arctan()arctan()arctan()arctan()arctan()arctan() 22222222 1 1 1 1 41() 1() 41() 1() 41() 1() 41() 1() t t t t xtxtxtxtxtxtxtxt d d d d xtxtxtxtxtxtxtxt =+=+=+=+ + + + + + + + + + 22222222 0 0 0 0 11111111 arctan()arctan()arctan()arctan()arctan()arctan()arctan()arctan() 22222222 1()()()()1()()()()1()()()()1()()()() 41() 1() 41() 1() 41() 1() 41() 1() t t t t xtxtxtxtxtxtxtxt xtxtxtxtxtxtxtxtxtxtxtxtxtxtxtxt d d d d xtxtxtxtxtxtxtxt =+=+=+=+ + + + + + + + + + + + + + + + + + + + + + + + 2 2 2 2 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 0 0 11111111 arctan()arctan()arctan()arctan()arctan()arctan()arctan()arctan() 22222222 1