数理方程部分作业题目参考答案(31-36).pdf

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1、ên•§Ü©Š’K8땉YSK3.6µpage761.¦)½)¯Kµ8>>u=a2ub2u;00>:uj=u1x2:t=0L2Ù¥§bÚu1þ•~ê")u(x;t)=v(x;t)+u1(8µòšÙg>.^‡ujx=L=u1àgz)“½)•§¥§k8>>v=a2vb2vb2u;00>:vj=u1x2u:t=0L21du(1.1)´šàgàg>Š¯K(šàgà‘b2u)§¤±ò(1.1)ª©)•Xeü‡•

2、1§5?n§=µ8>>v1=a2v1b2v1;00>xx=0x=L:v1j=u1x2u:t=0L21†8>>v2=a2v2b2v2b2u;00>xx=0x=L:v2jt=0=0:k)•§(1.2)"•§(1.2)´àg>Š¯K§Œk©lCþ{§(1.2))•µ111v(x;t)=w(x)
T(t):òþª“(1.2)¥k§?6FÏ:2014-09-23?6<:cc_˜11121211w(x)Tt(t)+bT(t

3、)=awxx(x)T(t):T1(t)+b2T1(t)w1(x)txx)==:a2T1(t)w1(x)
1221)Tt(t)+b+aT(t)=0:11wxx(x)+w(x)=0:(v1(0;t)=w1(0)
T1(t)=0;d>Š^‡µxxw1(0)=0;w1(L)=0"v1(L;t)=w1(L)
T1(t)=0:x¤±Xe>Š¯K§(w1(x)+w1(x)=0;xxw1(0)=0;w1(L)=0:xppÙ)/ª•µw1(x)=acos(x)+bsin(x)"´•nnn()bn=0;12n+1p2n+1
2)wn(x)=

4、ancos2Lx:ancos(L)=0)=;n=0;1;


2L?k!2122n+11Tt(t)+b+aT(t)=0:2Lb2+(2n+1a)2t)ƒµT1(t)=de2L"nnu´•§(1.2)?ê)•µ11XXb2+(2n+1a)2t2n+1v1(x;t)=w1(x)
T1(t)=d0e2L
cosx:nnn2Ln=0n=0P1
dЊ^‡v1(x;0)=d0cos2n+1x=u1x2u§n2LL21n=0ZL02u122n+1dn=xu1cosxdxL0L22LZL

5、=4u1x2u
sin2n+1xL2u1x
sin2n+1xdx(2n+1)L212L0L22L0ZL8u12n+1=x
sinxdx(2n+1)L202L32u(1)n+11=:(2n+1)33P1n+1b2+(2n+1a)2t
v1(x;t)=32u(1)e2L
cos2n+1x"1(2n+1)332Ln=0e¡)•§(1.3)"•§(1.3)Œ±^k¼ê{¦)"(1.3))•µ222v(x;t)=w(x)
T(t):
ÓnŒµw2(x)=a0cos2n+1x;n=0;1;


nn2L

6、òv2(x;t);b2u¤k¼ê?ê/ªµ11X2n+12bu1=cn
cosx2Ln=0ZL222n+1)cn=(bu1)cosxdxL02L4b2u(1)n+11=:(2n+1)1X2n+12v(x;t)=mn(t)
cosx:2Ln=0òv2(x;t);b2u?ê/ª“•§(1.3)¥k§18hi2222n+1

7、Lm(t)=e4L21:n(2n+1)(4b2L2+(2n+1)22a2)l •§(1.3))•µ!22X1nb2+(2n+1)22a2216bLu1(1)4L22n+1v(x;t)=e1cosx:(2n+1)(4b2L2+(2n+1)22a2)2Ln=0?Œ•½•§)•µ12u(x;t)=v(x;t)+v(x;t)+u1X1n+12(1)b2+(2n+1a)t2n+1=32ue2L
cosx1(2n+1)332Ln=0!22X1nb2+(2n+1)22a216bLu1(1)2

8、n+1+e4L21cosx(2n+1)(4b2L2+(2n+1)22a2)2Ln=0+u1:2.¦)½)¯Kµ