第三章拉普拉斯方程的圆的狄利克雷问
的付氏解
第九章 拉普拉斯方程的圆的狄利克雷问题的傅氏解
111.试证拉普拉斯方程在极坐标下的形式解为 u,u,u,0u,u,0rrr,,xxyy2rr
,,xrsin,,,,,解:因为 所以ux,y,urcos,,rsin, ,y,rcos,,
,u,u,x,u,y,u,u,,,,,,,,cos,sin,,r,x,r,y,r,x,y, ,,u,u,x,u,y,u,u,,,,,,,rsin,rcos,,,,x,,y,,x,y,,,,
由此解出
,,,,u,usin,u,sin,,,,,,cos,,cos,u,,,,,,x,rr,,rr,,,, ,,,coscos,u,u,u,,,,,,,sinsin,,,,u,,,,y,rr,,,rr,,,,,
,,,,sin,,,,,cos,,,,,,x,rr,,,,记算子为 ,,,,cos,,,,,,sin,,,,,y,rr,,,,,
从而
,2,u,,u,,u,usin,,,,,,,,,,cos,,,,,,,2,x,x,rr,,x,x,,,,,,
,,,,,u,usinsin,,,,,,,,,coscos,,,,,,,rr,,rr,,,,, 222,,,,,,u,u,u,usincossinsincos2,,,,,,cos22,,,r,rr,r,,rr
222,,,,,,u,u,usincossinsincos,,222,,r,,,rrr,,
通理得
2,,,,,,u,,u,,,ucos,,,,,,,,,,sin,,2,,,,,,y,y,rr,,y,y,,,,,,
222,,,,,,u,u,u,usincossincoscos2,,,,,, sin22,,,r,,rr,r,rr
222,,,,,,u,u,usincossincoscos,,222r,r,,,,rr,,所以
22,u,u11 ,,u,u,u,0rrr,,222r,x,yr
2.求解下列狄利克雷问题
11,u,u,u,0rrr,,2,rr, ,,,,A,,,,,,,u1,,,,0,,,,(,,,,,,),,,
其中为已知常数。 A,,
解法1:直接用分离变量法解得
n,,,Ar ,,ur,,,,,2sinn,cosn,,,,n,n,1,,
,An0,,,,ur,,,,Asinn,,Bcosn,r解法2:设解为 ,nn2n1,
,12A当时 A,Acosn,d,,sinn,n,1n,,,n,,
,,12A, ,,AAd0,,,,,
,1 B,Asinn,d,,0,n,1,2,3,?n,,,,
n,,,Ar所以 ,,ur,,,,,2sinn,cosn,,,,n,n,1,,3. 求解下列狄利克雷问题
11,u,u,u,0,rrr,,2 rr,
,,,u,A,,,,1,,cos(,,,),
其中A为已知常数.
解发1: 直接应用分离变量法
,An0,,,,ur,,,,Asinn,,Bcosn,r解法2 设形式解为 ,nn2n1,
,1 A,Acos,d,,00,,,,
,12 A,Acos,d,,A1,,,,
,1 A,Acos,cosn,d,,0(n,0)n,,,,
,1 B,Acos,sinn,d,,0(n,1,2,3,?)n,,,,
,,ur,,,Arcos,所以
解法3, 公式法
222,,1L,rd,,,, ur,,Acos,,,220,2L,2rLcos(,,,),r
L,1因为
21,1r所以 ,,cos,,ur,A,d,2,2,,1,2cos,,rr,,,
i,在内应用复变
的流数定理设,则 z,ez,1
11,,,cos,z, ,,2z,,
dz11,,,i,i,,,,,,,ez,ed,,cos ,,ziz2,,
所以
,,,22cosd,11,zdz,,,,,,,,,,2i22i02,r,,rzerz,,rz,re12cos1,,,,,,
,,,1,,,i,2,,,,,,ResFzResFz2ii,,,z,0,,z,re ,i22,,,re11,,i,,,e,,,i,2rrer,,,1,,
r2,cos,`,,2r,1
2,1,r2r,,ur,,,A,cos,,Arcos, 221,r,
4.求解下列定解问题
11,,,,0uuurrr,,2,rr,, ,0,0,,,0(0,,)u,,,,rurrL,
,,,,,uL,,,f,(0,,,,),
,
,,f,其中为已知的连续函数. 解: 应用分离变量法
,,,,,,ur,,,Rr,,(1)变量分离 设代入泛定方程得
,,,,,,,,",,,0,
,,,,,,0 ,
,,,,0,0,
2,,,,,,rR"r,rR'r,,Rr,0 (2)解特征值问题
,,",,,0,
,,,,,,0 ,
,,,,0,0,
,n解得特征函数为 ,,,,,sin,(n,1,2,?)n,
22,,,,,,rR"r,rR'r,,Rr,0(3)解常微分方程 解得
,,nn,,,,,Rr,Cr,Dr(n,1,2,?) nnn
D,0为了解得有界解,必须使, n
n,,,n,(4)根据叠加原理得到解 ,,ur,,,asin,,r,n,n1,
a(5)由傅氏级数确定系数代入边界条件 n
n,,,n, 从而得到解的系数为 ,,f,,asin,,L,n,n1,
,,2n,,a,f,sin,d, nn,,0,,L,
5.考察由下列定解问题描述的矩形平板(0,x,a,0,y,b)上的温度分布
u,u,0,xxyy,u,,,,0,y,0,ua,y,0 ,
,,,,,,,ux,0,fx,ux,b,0,
,,fx其中为已知的连续函数. 解: 此定解问题的边界为矩形,圆的狄利克雷问题,应用极坐标求解.因此我们可以把极坐标
系的下边界:
,,,r0,r, ,,,0,,,,,
看做矩形,应用分离变量法:
(1)分离变量 设,,,,代入泛定方程得 u,XxYy
,X",X,0,X"Y" 即 ,,,,,Y",,Y,0XY,
,,,,代入边界条件:得X0,0,Xa,0
,X",X,0,
,,,Xx,0(?) ,
,,,Xa,0,
解特征值问题(?)
(a)若时,方程只有平凡解 ,,0X",,X,0
22(b)若时,令,则方程变为其通解为 ,,,kX",kX,0,,0
,,,,Xx,Acoskx,BsinkxX0,0,A,0
所以 A,0
,,Xa,0,0,Bsinka
n,于是得特征函数为 X,sinxna
2(3)解方程 得 y",ky,0n
ky,kynn,,Yy,Ce,De nnn
n,其中 ,,k,n,1,2,?na
(4)由叠加原理得
,n,kyky,nn,,,,ux,y,Ce,Desinx ,nnan1,
C,D(5)由傅氏级数确定系数代入边界条件 nn
,,,n,,,,,,,,,ux,0fxCDsinx,nn,a,1n,, ,
,
,,,