數(shù)學(xué)物理方程_谷超豪_第三章答案

1、第三章調(diào)和方程1建立方程定解條件1 設(shè) u(x , x, , x) = f (r)(r =x2+ + x2 ) 是 n 維調(diào)和函數(shù)（即滿足方程1 2n1n2u+2u= 0 ），試證明x 2x 21nf (r) = c1 +c2(n 2)r n-2f (r) = c + cIn1(n = 2)12r其中 c1 , c2 為常數(shù)。證：u = f (r) ,u= f (r) r= f (r) xixixir 2u = f (r) xi2 + f (r) 1 - f (r) xi2xi2r 2r3rnnn2uxi2nxi2n -1i=1i=1= f(r) + f(r) - f(r) = f(r) +f

2、(r)x 2r 2rr 3r若 n = 2 ，則f (r) =A1故 f (r) = c + A I n rr11即 n = 2 ，則f (r) = c + cIn112r2 證明拉普拉斯算子在球面坐標(biāo) (r,q,j) 下，可以寫成Du =1(r 2u) +1(sinqu) +12ur 2rr 2 sinqqqr 2 sin 2 qj 2r= 0證：球坐標(biāo) (r,q,j) 與直角坐標(biāo) (x, y, z) 的關(guān)系：x = r sinq cosj ， y = r sinq sinj ， z = r cosq（1）Du =2u+2u+2ux2y 2z2為作變量的置換，首先令 r = r sinq ，

3、則變換（1）可分作兩步進(jìn)行x = r cosj ，y = r sinj（2）r = r sinq ，z = r cosq（3）由（2）i=1i即方程Du = 0化為 f (r) +n -1f (r) = 0rf (r)= -n -1f (r)ru = u c o sj + urxys i nju+(r c o sj)y所以f (r) = A r -(n-1)1若 n 2 ，積分得f (r) =A1r -n+2 + c- n + 21即 n 2 ，則f (r) = c1+c2r n-2由此解出u=ucos j -usin jxrjr（4）uuucos j=sin j +yrjr再微分一次，并利用

4、以上關(guān)系，得392u=(ucosj -usin j)x 2xrjr= cos j r ( ru cos j - ju sinrj ) - sinrj j ( ru cos j - ju sinrj )22u2 s i jnc o js2us i 2nj2u= c o sj-+r 2rrjr 2j 2+ 2 s i jnc o js u + s i 2nj ur 2jr r 2u = ( u s i jn + u c o js)y 2 y rj r= sin j r ( ru sin j + ju cosrj ) + + cosrj j ( ru sin j + ju cosrj )= sin

5、22u+2 sin j cos j2u+cos 2 j2u-r 2rrjr 2j 2- 2 sin j cos j u + cos 2 j ur 2jr r所以2u+2u=2u+12u+1u（5）x 2y 2r 2r 2j 2rr2u+2u+2u=2u+2u+12u+1ux 2y 2z 2r2z2r 2j2rr2u+2u再用（3）式，變換。這又可以直接利用（5）式，得r2z 22u+2u=2u+12u+1ur 2z2r2r 2q2rr再利用（4）式，得ru = ur sinq + qu cosrq所以2u+2u+2u=2u+12u+1u+x 2y 2z2r 2r 2q 2rr+12u+1(us

6、inq +ucosq)r 2 sin 2 qj2r sinqrqr=2u+12u+12u+2u+1ctgqur 2r 2q2r 2 sin 2qj 2rrr 2q即Du =1(r 2u) +1(sinqu)+12u= 0rrqq2 sinj 2r2r 2 sinqr2q3 證明拉普拉斯算子在柱坐標(biāo) (r,q, z) 下可以寫成Du =1(ru) +12u+2urrrr 2q2z 2證：柱坐標(biāo) (r,q, z) 與直角坐標(biāo) (x, y, z) 的關(guān)系x = r cosq ，y = r s i nq ，z = z利用上題結(jié)果知 2u + 2u = 2u + 1 2u + 1 ux 2 y 2 r

7、2 r 2 q 2 r r= 1 (r u ) + 1 2u r r r r 2 q 2所以Du =1 (ru) +1 2u+2urrrr 2 q 2z2404. 證明下列函數(shù)都是調(diào)和函數(shù)（1） ax + by + c(a, b, c 為常數(shù))證：令 u = ax + by + c , 顯然2u= 0,2u= 0.x 2y 2故 Du = 0 ，所以 u 為調(diào)和函數(shù)(2) x2 - y 2和2xy2 u= 2,2u= 2, 。所以 Du = 0 。u 為調(diào)和函數(shù)x 2y 2令v = 2xy則2v= 0,2 v= 0 。所以 Dv = 0 。v 為調(diào)和函數(shù)x 2y 2(3) x3 - 3xy 2

8、和3x2 y - y3證： 令u = x3 - 3xy 22u= 6x,2u= -6x, 所以Du = 0 , u 為調(diào)和函數(shù)。x 2y 2令v = 3x2 y - y32 v= 6 y,2 v= -6 y 。所以 Dv = 0 ， v 為調(diào)和函數(shù)。x 2y 2(4) shny sin nx, shny cos nx, chny sin nx和chny cos nx(n為常數(shù))證： 因(shny) y = n2 shny(chny) y = n2chny(sin nx) x = -n2 sin nnx(cos nx) x = -n2coxnx所以(shny sin nx) xx = -(shn

9、y sin nx) yy即D(shny sin nx) = 0故 shny sin nx為調(diào)和函數(shù)同理，其余三個函數(shù)也是調(diào)和的（5）s h x(c h x+ c o sy)-1和s i ny(c h x+ c o sy)-1證： 令u = shx(chx + cos y)-1u= chx(chx + cos y)-1- sh2 x(chx + cos y)-2x= (chx + cos y)-2 (1 + chx cos y) 2u = (chx + cos y)-2 shx cos y - 2(chx + cos y)-3 shx(1 + chx cos y)x2= (chx + cos y

10、)-3 (shxcox 2 y - 2shx - shxchx cos y)uy = shx sin y(chx + cos y)-2 2u = shx sin y(chx + cos y)-2 + 2(chx + cos y)-3 shx sin 2 yy 2= (chx + cos y)-3 (shxchx cos y + shx cos 2 y + shxchx cos y) 2u + 2u = (chx + cos y)-3 (2shx cos 2 y - 2shx + 2shx sin 2 y)x 2 y 2= (chx + cos y)-32shx(cos2 y + sin 2 y

11、) - 2shx = 0令v = sin y(chx + cos y)-1vx = -shx sin y(chx + cos y)-2 2v = -sin ychx(chx + cos y)-2 + 2(chx + cos y)-3 sh2 x sin yx 2= (chx + cos y)-3 (2sh2 x sin y - sin ych 2 x - sin ychx cos y)41yv = cos y(chx + cos y)-1 + sin 2 y(chx + cos y)-2= (chx + cos y)-2 (1 + chx cos y)2u = -sin ychx(chx +

12、cos y)-2 + 2(chx + cos y)-3 sin y(1 + chx cos y)y 2= (chx + cos y)-3 (2sin y + sin y cos ychx - 2sin ych 2 x)2v + 2v = (chx + cos y)-3 (2 sin ysh2 x - 2 sin ych 2 x + 2 sin y)x 2y 2= (chx + cos y)-3-2sin y(ch2 x - sh2 x) + 2sin y = 0所以 u, v 皆為調(diào)和函數(shù)。（5）。證明用極坐標(biāo)表示的下列函數(shù)都滿足調(diào)和方程（1） ln r和q證： 令u = ln r,由第1題知

13、,u為調(diào)和函數(shù) 。令v = q ,則顯然2 v= 0,v=0,2 v= 0,故r 2rq 2Du =2 v+1 v+1 2 v= 0r2rrr 2 q2（2）r n cos nq和r n sin nq (n為常數(shù)證： u = r n cos nqu= nr n-1 cos nq2u= n(n -1)r n-2 cos nqrr 2 2u = -n2 r n cos nqq 2所以 Du = n(n -1)r n-2 + nr n-2 - n2 r n-2 cos nq = 0令 v = r n sin nq則Dv = n(n -1)r n-2 + nr n-2 - n2 r n-2 sin n

14、q = 0(3) r ln r cosq - rq sinq + rq cosq 和 r ln r sinq + rq cosq 證：令 u = r ln r c o sq - rq s i nq.ur = (ln r +1) cosq -q sinq 2u = 1 cosq.r 2 ruu = -r ln r sinq - r sinq - rq cosq2u= -r ln r cosq - 2r cosq + rq sinq.q 2Du =1cosq +1(ln r +1) cosq -qsinq -1ln r cosq -2cosq +qsinq = 0rrrrrr令v = r ln r

15、 sinq + rq cosq.v= (ln r +1) sinq + cosqr 2v = 1 sinqr 2 rqv = r ln r cosq + r cosq - rq sinq 2v = -r(ln r + 2) sinq - rq cosqq 2Dv = 1r sinq + 1r (ln r +1) sinq + qr cosq - 1r (ln r + 2) sinq - qr cosq = 0.6.用分離變量法求解由下述調(diào)和方程的第一邊界問題所描述的矩形平板（0 x a,0 y b) 上的穩(wěn)定溫度分布：2u+2u= 0x 2y 2u(0, y) = u(a, y) = 0pxu

16、(x,0)= sin, u(x, b) = 0.a解：令 u(x, y) = X (x)Y ( y) 代入方程 ，得42X (x)= -Y = -lX (x)Y再由一對齊次邊界條件 u(0, y) = u(a, y) = 0 得X (0) = X (a) = 0由此得邊值問題X + lX = 0X (0) = X (a) = 0由第一章討論知，當(dāng) l = l= (np)2 時，以上問題有零解nanpX n (x) = sinx.(n = 1,2, )npa又Y - ()2 Y = 0nan求出通解，得npy-npy+ BY = A e aeannnnpy + Bn e-npnpy ) sin所

17、以u(x，y) = ( An eaax.n=1a由另一對邊值，得pxnpsin= ( An+ Bn ) sinxan=1anpnpb-bnp0 = ( An e a+ Bn ea) sinxan=1由此得，A + B = 1，A+ Bn= 0n = 2,3, 11nnpe-npA eb + Bb = 0n = 1,2,aa nn-pbpb解得A =-1 eaB =1e a12p12 shpshbbaaAn = Bn = 0n = 2,3,代入 u(x, y) 的表達(dá)式得pppu(x, y) =11(b-y) - e-(b-y) ) sinx(e aap2shbaa=1shp(b - y) si

18、npxpxashba7在膜型扁殼渠閘門的設(shè)計中，為了考察閘門在水壓力作用下的受力情況，要在矩形區(qū)域0 x a,0 y b 上解如下的非齊次調(diào)和方程的邊值問題：Du = py + q( p 0常數(shù)）u= 0, u= 0x=axx=0y=0 = uy=b = 0u試求解之（提示：令 v = u + (x2 - a2 )( fy + g) 以引入新的未知函數(shù) v ,并選擇適當(dāng)?shù)?f , g 值，使 v 滿足調(diào)和方程，再用分離變量法求解。）解：令 v = u + x2 - a2 )( fy + g), 2u = 2u + 2( fy + g), 2v = 2u ,x 2 x 2y 2 y 2Dv =

19、Du + 2( fy + g)又Du = py + g, 故取 f = -p, g= -q, 則 v 滿足調(diào)和方程Dv = 022即v = u -1(x2- a 2 )( py + q)2代入原定解問題，得 v 滿足43Dv = 0v= 0, vx=0x=axy=0 = -q(x 2-v2= 0a 2 ), vy=b = -1( pb + q)(x 2 - a 2 )2v所以B =(-1)n16a 2n(2n +1)3p 32得v(x, y) =16a3pn=0+ ( pb + pb + q(1 - ch2n +1pb) / sh2a(-1)n2n +1qshp (b - y)(2n +1)3

20、2aq)sh2n +1pycos2n +1px /2a2a2n +1pb .2ash 2n2a+1pb用分離變量法求 v(x, y) ，令 v(x, y) =得X (x)Y ( y) 代入方程及邊值= 0, vx=a = 0 ，x x=0最后得116a 2(-1)n1u(x, y) =(x 2- a 2 )( py + q) +2p3(2n +1)3sh2n+1pbn=0X + lX = 0X (0) = X (a) = 0及Y - lY = 0求非零解 X (x) ，得 l = ln = (2n +1p )2 , n = 0,1,2,2aX(x) = cos2n +1px,Y= A ch2n

21、 +1py + B sh2n +1py .n2ann2an2a2n +12n +12n +1所以v(x, y) = ( Anchpy + Bn shpy) cospx2an=02a2a再由另一對邊值得- q (x 2 - a 2 ) = An cos 2n +1px2n=02a12n +12n +12n +1-( pb + q)(x 2- a 2 ) =( Anchpb + Bn shpb) cospx2n=02a2a2aA =q a- x 2 ) cos2n +1pxdx =16a 2 q(-1)n .(a 2na 2a(2n+1)3 p 30A ch2n +1pb + B sh2n +1pb =pb + ql(a 2 - x 2 ) cos2n +1pxdxn2an2aa2a0=16a 2( pb + q)(-1)n(2n +1)3p 32a qsh 2n +1p (b - y) + ( pb + q)sh 2n +1pycos 2n +1px2a2a2a8. 舉例與說明在二維調(diào)和方程的狄利克萊外問題，如對解 u(x, y) 不加在無窮遠(yuǎn)處為有界的限制，那末定解問題的解以不是唯一的。解：考