v(r,?)??cnrnsinn?
满足方程及齐次边值和自然边界条件，再由
v|r?a?u0?u1
得 u0?u1??cnansinn?
n?1?因此 Cn?2??an??(u0?u1)sinn?d??02(u0?u1)n?an[1?(?1)n]
所以u(r,?)?u1?2(u0?u1)rn?n?(a)[1?(?1)n]sinn? n?13.2 傅立叶变换与初值问题
1． 求下述函数的傅立叶变换： (1) f(x)?e?ax,a?0; (2) f(x)?e(3)
?ax2,a?0;
x1,, (a?0,k为自然数) 22k22k(a?x)(a?x)?ax2?解：(1)F[e]?????e?ax2?e?ipxdx???ip2p2)?4?2??e?a(x2?ip?x)dx
=
???e?a(x?dx?e?p2?4a???p2?4a?aue?du （柯西定理） p24a2 =1ea????v?edv?2?ae?
?或者 F[e???x2]????e??x2(cospx?isinpx)dx?2?e0???x2cospxdx?2I(p)
?222dI1PP???xe??xsinpxdx?e??xsinpx?? e??xcospxdx?I(P)
002?dP2?2?0
积分得I(P)?Ce?p2?4?
又I(0)=e0???x2dx?1? 2?故 C=
1? 2?所以 F[e??x2]=2I(P)=
?e??p2?4?
0???0?0(2) F[e?ax]???0?e?axe?ipxdx??eax?ipxedx??e?axe?ipxdx
=
???e(a?ip)xdx??e?(a?ip)xdx?01e(a?ip)x+
??a?ip?1?(a?ip)x?112a e???220a?ipa?ipa?pa?ip或F[e?ax?]=
???e?axe?ipx?dx=?e???ax(cospx?isinpx)dx
? =2e0??axcospxdx??2aa?pe?ipx22
(3) F[
1(a?x)22k]=
??(a?2?x)2kdx?2?iResz?aie?ipz(a?z)22k
1dk?1e?ipz因Res=limk?1[]
22kkz?aiz?ai(a?z)?k?1?!dz(z?ai)k?1m1 =lim?C[(z?ai)?k](m)(e?ipz)(k?m?1)
(k?1)!z?aim?0k?1k?1m1 =lim?C(?1)mk(k?1).....(k?m?1)
(k?1)!z?aim?0k?1e?ipz(z?ai)?k?m(?ip)k?m?1e?ipz 1k?1mm?k?mk?m?1ap?(?1)k(k?1)?(k?m?1)(2ai)(?ip)e ?ck?1(k?1)!m?01k?1(k?m?1)!(?1)k?m?1k?m?1ap?pe ?k?m(k?1)!m?0m!(k?m?1)!i(2a)所以
??11k?1(k?m?1)!F?2?2?i.??2k?(k?1)!m?0m(k?m?1)!?(a?x)???(?1)k?m?1
i(2a)k?mpk?m?1eap2?k?1(k?m?1)!(?1)k?m?1k?m?1ap?pe ?k?m(k?1)!m?0m(k?m?1)!(2a)?xF?2?(a?x2)k???1d?1??F??2? 2kidp???(a?x)???2?k?2(k?m?1)!(?1)k?m?1?i? ?k?m?(k?1)!m?0m!(k?m?1)!(2a)[(k?m?1)pk?m?2eap?apk?m?1eap]?i?(2k?2)![(k?1)!]2(2a)?2k?2eap
??(2k?2)![(k?1)!]2(2a)?2k?2ape2?k?2(k?m?1)!?i? ?(k?1)!m?0m!(k?m?1)!(2a)?k?m(?1)k?m?1pk?m?2eap(ap?k?m?1)
2．证明当f(x)在(??,?)内绝对可积时，F[f]为连续函数。
?证：因F(t)?????f(x)e?ipxdx?g(p)对任何实数p有
|F(f)|?|g(p)|????|f(x)|dx
即关于p绝对一致收敛，因而可以在积分下取极限，故(g(p)关于p为连续函数。
3. 用傅立叶变换的方法求解下列定解问题
22???ut?auxx(t?0,???x??)?ut?auxx(t?0,???x??) (1)?;(2)? 2u?sinxu?x?1???t?0?t?0(3) 用延拓法求解半有界直线上热传导方程（3.22）,假设
?u(x,0)??(x)(0?x??) ??u(0,t)?0解：（1）sinx有界，故
u(x,t)?12a?t????(x??)2?2sin?e4atd???(x??)??14a2t?
12a?t????sin(x??)e1???2d?
???????2???2=cos?d??cosx?esin?d?? ?sinx?e2a?t??????????4?1sinxe?sinx=
?2a?t2a?t121?12a?te?a2t?e?a2tsinx
(2) 1+x无界， 但表达式
u(x,t)?12a?t???(x??)2?22(1??)e4atd??(x??)2?22(1??)e4atd?
仍收敛，且满足方程。因此
u(x,t)?12a?t????x????1??24at?
????12a?1?(x??)?e?t2???2d?
????2?2???2???1?x22a?t??1???e??????2d??2x??2??ed???????e2?d?? ????1?x2=
2a?t??1????????1?????????e|?ed??? ???????2????2????2?1?x2a?t?1??1???2???22??1?x?2at ??易验它也满初始条件.
4. 用傅立叶变换求解三维热传导方程的柯西问题
222???u?u?u?u2???a?2?2?2???x： ??t?y?z????u|t?0??(x,y,z)??解：令 F[u(x,y,z)]?对问题作傅立叶变换得
???i(xs1?ys2?zs3)~(s,s,s,t) u(x,y,z,t)edxdydz?u123???~du?222~2??as?s?su123?dt? ???i(xs1?ys2?zs3)~~?u|t?0?????(x,y,z)edxdydz??(s1,s2,s3)??????222?a2(s1?s2?s3)t~~解之得u??(s1,s2,s3)e
因F?1[e?a2(s1?s2?s3)t222]?1(2?)3??????e?a2(s1?s2?s3)t222?
ei(xs1?ys2?zs3)ds1ds2ds3
=
1(2?)3????e??a2s1t?ixs12?ds1?e???a2s2t?iys22?ds2???e?a2s3t?izs32ds3
?12a?tx2?2e4at12a?ty2?2e4at?12a?tz2?2e4at?1????2a?t????3x2?y2?z2?2e4at
再由卷积定理得
?1??u?x,y,z,t??????2a?t? 5. 证明函数
3????????,?,??e(x??)2?(y??)2?(z??)2?4a2ted?d?d?
v(x,y,t,?,?,?)?14?a(t??)2(x??)2?(y??)2?4a2(t??)
对于变量(x,y,t)满足方程
2?v?2v2?v?a(2?2) ?t?x?y对于变量(?,?,?)满足方程
2?v?2v2?v?a(2?2)?0 ?????? 证：验证即可。因
?v11(x??)2?(y??)2?[??]e223?t4?a(t??)4a(t??)(x??)2?(y??)2?4a2(t??)