偏微分方程數(shù)值解法答案2188

1.2.3.4.課本p有證明2課本p,p有說(shuō)明812課本p,p有說(shuō)明1520u是u的n維子空間，,...是u的一組基底，u中的任一元素u可nnnn12nRit2法，設(shè)nunnc，則J(u)n12a(u,u)(f,u)1na(,)ccc(f,)是表為ii2nnnijijji1ji,j1j1J(u)a(,)a(,)，令，從而得到0c,c...c的二次函數(shù)，2c,c...c滿足12nnc1nijjijnc，na(,)c(f,),j1,2...n，通過(guò)解線性方程組，求的c，代入uijijiniii1i1從而得到近似解u的過(guò)程稱為Rit2法n簡(jiǎn)而言之，Rit2法：為得到偏微分方程的有窮維解，構(gòu)造了一個(gè)近似解，nc，uniii1nJ(u)1利用2na(u,u)(f,u)1a(,)ccc(f,)n確定c，求得近似解u的in2nnnijijjji,j1j1過(guò)程nVu，滿足nuGalerkin法：為求得nc形式的近似解，在系數(shù)ic使u關(guān)于niii1a(u,V)(f,V)，對(duì)任意Vu或（取V,1jn）nnjna(,)c(f,),j1,2...n的情況下確定c，從而得到近似解nuc的過(guò)程稱ijijiniii1i1Galerkin法為Rit2-Galerkin法方程：a(,)c(f,)nijiji15.有限元法：將偏微分方程轉(zhuǎn)化為變分形式，選定單元的形狀，對(duì)求解域作剖分，進(jìn)而構(gòu)造基函數(shù)或單元形狀函數(shù)，形成有限元空間，將偏微分方程轉(zhuǎn)化成了有限元方程，利用有效的有限元方程的解法，給出偏微分方程近似解的過(guò)程稱為有限元法。6.行網(wǎng)格剖分，節(jié)點(diǎn)axx...x...xb得到相鄰節(jié)點(diǎn)x,xi1解：對(duì)求解區(qū)間進(jìn)01ini1I[x,x]，hxx，由節(jié)點(diǎn)上的一組值u0,u,u...u，按線之間的小區(qū)間ii1iiii1012lxxxx性插值公式i1u○xI,i1,2...n確定試探空間1u(x)nuu，令ihi1hiiniixxF(x)2○i1hii參考但愿[0,1]令N()1,N()01把Ii變到軸上的則：U(x)N()uN()uxIn0i11i，i○3將○1帶入該函數(shù)J(u)1b(puqu2fu)dx222a得到：J(u)1fudxb(pu2qu22fu)dx12(pu2qu2)dxnn2nnnnaIIii1i1i帶入○2可得(uu)J(u)11[p(xh)i2hq(xh)(N()uN()u)d2nii1ni1hii1i0i11i20i1inh1f(xh)(N()uN()u)dii1i0i11i0J(u)i1○4auauaub0nuj1,jj1jjjj1,jj1j令○5j其中a1[h1p(xh)]hq(xh)(1)]dj1j1,jjj1jjj0a1[h1p(xh)]hq(xh)(1)]dj1j1j1,jj1jij01[h1p(xh)]hq(xh)]d[h1p(xh)]hq(xh)(1)]d2j1j11aj,jjj1jjj1jj1jj1i00bh1f(xh)dh1f(xh)(1)djjj1jj10jj10從而得到u,u,...,u12n的線性方程組！7.矩形剖分假定區(qū)域C1可以分割成為有限個(gè)互不重疊的矩形的和，且每個(gè)小任意兩個(gè)矩形或者不相交或者有公共的邊和公共的頂點(diǎn)，成如此的矩形的邊和坐標(biāo)軸平行，分割為矩形剖2xxi)(1yy(1i),(x,y)Rx0,(others)yijij分基函數(shù)的取法其中R是以R(x,y)為頂點(diǎn)的矩形單元x,y0為的底和商的長(zhǎng)度。ijijij8.何為三角剖分，基函數(shù)怎樣??？G是多邊形域（否則可用多邊形域逼近它），將三角剖分：設(shè)G分割成有限個(gè)三角形之和，使不同三角形無(wú)重疊的內(nèi)部，且任一三角形的頂點(diǎn)不屬于其他三角形的內(nèi)部，這樣就把G分割成三角形網(wǎng)，稱為G的三角剖分?；瘮?shù)的取法：通過(guò)構(gòu)造Lagrange型插值公式可以得到基函數(shù)的取法。不妨以P(x,y)是1P(x,y)LLL，其中L是相應(yīng)于節(jié)點(diǎn)1的基函數(shù)在1一次多項(xiàng)式為例，得到1112233△上的限制（具體的過(guò)程，可參考課本：PP）57589.題，參考課后習(xí)題P的第一題，具體過(guò)程可參考積分插值的推導(dǎo)過(guò)程9210，11題不會(huì)。在此將14題推導(dǎo)過(guò)程介紹如下：12.對(duì)Possion方程f(x,y)，建立五點(diǎn)差分格式，并估計(jì)截?cái)嗾`差。取定沿x軸和y軸方向的步長(zhǎng)h和h，沿x,y方向分別用二階中心差商代替，則21i1,j22i,j1]f（五點(diǎn)差分格式）[hi,jh2i1,ji,j1i,jh2ijij12式中表示節(jié)點(diǎn)(i,j)上的網(wǎng)函數(shù)。i,j令(x,y)f(x,y)ff(x,y)nijijnijijij利用Taylor展式有6(x,y)2(x,y)(x,y)(x,y)(x,y)(x,y)h4(h6)2h214i1jih2ji1jijijij1x212x4360x611(x,y)2(x,y)(x,y)(x,y)h22(x,y)h24(x,y)246(h6)2ij1ih2jij1ijijijy212y4360y62截?cái)嗾`差為)(h4)(h2)14h24R()(x,y)(x,y)(h212x42y4ijijnij1ij313.對(duì)possion方程建立，極坐標(biāo)形式的差分格式poission方程的極坐標(biāo)形式為12)22]f(,)-----①1[(y其中02x2y2tan0x利用中心差商公式()11i1,ji,ji1,ji12i12i12i12[()]-----②h2(,i)ji12212[2]i,j1i,j1-----③i,j2(,h2)jii將②③兩式代入①式得()12i,j1]f(,)i1,ji,ji1,j21i12i12i12i12[i,j1i,jhh22iji即poission方程極坐標(biāo)形式的差分方程。i14.解：將uk1,uk1,uk,uk按照Taylor在uk處展開(kāi)整理得到其截?cái)嗾`差為j1jjj1jn1(在Richardson格式（4.1.10）中以n1)代入，便得DuFortFrankel格式：n12jjj(2n()nn1nj1)an1n1jj1jjj1h2t2t26t324t4121314n1(x,t)-----①j1nn234jjt2t26t324t4121314n1(x,t)-----②n234jjj2nn1n1jt6t33（省去了2的商階無(wú)窮?。?1①-②得j213從而得到了微分方程左邊的誤差26t3同理可得微分方程右邊的誤差：t212t412t4a(h22h2t212t4h44214)ah24a()2a44224ht212h2t44ht224從而得到ea()(h)()222h415.用Fourier方法討論向前差分格式的穩(wěn)定性。解：向前差分格式uk1ruk(12r)ukruk。以u(píng)kvkexp(ijh)代入得j1jj1jjvk1exp(ijh)rexp(i(j1)h)(12r)exp(ijh)rexp(i(j1)h)vk消去exp(ijh)則知增長(zhǎng)因子hC(x,)(12r)r(exp(ih)exp(ih))12r(1cosh)14rsin2由于21phphl()在[0，π]中分布稠密，(隨h0)為使C(x,)滿足vonNeu-Mann條件，21p1必須且只須網(wǎng)比r所以向前差分格式的穩(wěn)定性條件是r21216.用Fourier方法討論向后差分格式的穩(wěn)定性。un1解：對(duì)向后差分方程unjun1j12un1un1j1利用Fourier方法分析器穩(wěn)定性，整理ajjh2aaaa得：u(12)un1un1j1un1。令r，將unvnexp(ijh)代入得到：njhh2h2h2jj1j2vnexp(ijh)(12r)vn1exp(ijh)rvn1exp(i(j1)h)rvn1exp(i(j1)h)消去11exp(ijh)。則增長(zhǎng)因子為12rr(exp(ih)(exp(ih)))1。所以14rsin2h2向后差分方程是恒穩(wěn)定的。17.用Fourier方法討論六點(diǎn)對(duì)稱格式的穩(wěn)定性。un1解：六點(diǎn)對(duì)稱方程的格式為unjun2uunun1j12un1un1j1)。令1a2(2njj1jj1jh2h2unvnexp(ijh)代入得vn1exp(ijh)jvnexp(ijh)=a2[vnexp(i(j1)h)2vnexp(ijh)vnexp(i(j1)hvn1exp(i(j1)h)2h22vn1exp(ijh)vn1exp(i(j1)h]。消去exp(ijh)得增長(zhǎng)因子為5a2a2sin22h2(cosh1)11h2h2a2sin22h(cosh1)11。所以六點(diǎn)對(duì)稱格式是無(wú)條件恒定的。a21h2h2218.證明：利用Fourier方程將兩端同時(shí)做變換。ukvkexp(ijh)得jvk1exp(ijh)-vkexp(ijh)vkexp(i(j+1)h)-2vkexp(ijh)+vkexp(i(j-1)h)vkexp(i(j+1)h)-vkexp(i(j-1)h)-cvkexp(ijh)2h=a消1-+bh2去exp(ixjh)得增長(zhǎng)因子為2a+sinh-c，由von-Neumann條件可得a2hb1.即差分格式h22sinh22huk1-ukj=auk1-2uk+ukj1+bukj12hukj1+cuk(a>0)的充要條件是ja1jjjh2lh2219.討論三維熱傳導(dǎo)方程向前差分格式的穩(wěn)定性n1u-unajkm=(2un+2un+2un)解：三位熱傳導(dǎo)方程為（向前差分格式）.jkmh2xjkmyjkmzjkm2p2q2g,h=代到上式消去公因子取通項(xiàng)unjkmvnexp(i(x2hyhz)),=k,=llljm2h2hha)vn,r。從而增長(zhǎng)因子為h得vn1(14rsin4rsin4rsin2222hhhhsin2)為使|c|=1+O()必須且只須cc(h,h,hh)14r(sin21sin222211116r。當(dāng)r時(shí)三維熱傳導(dǎo)方程的向前差分格式穩(wěn)定。620.討論三維熱傳導(dǎo)方程向前后分格式的穩(wěn)定性n1u-unajkm=(2un1+2un1+2un1)解：三維熱傳導(dǎo)方程的向后差分格式為：jkmh2xjkmyjkmzjkm2p2q2g取通項(xiàng)Un=vnexp(i(x+y+z)),=,=,=,帶入上式，消去共jkmjkmlll1)1。恒成立因子得：14r(sin2hsin2hsin2h2226所以三維傳導(dǎo)方程向后差分格式是無(wú)條件穩(wěn)定的。21．三維傳導(dǎo)方程的PR格式為：(unun)jkmr2)13jkm1(1=(h2)Un（1）jkm222xxyz2l2un13jkmun23jkm1=h2y(unun)(2)jkm23jkml2un1un212(un1un)(3)jkmjkm3jkm=jkmh2zl2(1)(2)(3)合稱PR格式。22.將(1rr2)un=(1)un13jkm222xyjkm(1r=(1r2)un2)un23jkm13jkm22yz(1r2=(1r2)un2x2)un1jkm23zjkmyz)帶入上式得將Un=vnexpi(xjkmjkm(12rsin2h)(12rsin2h)(12rsin2h)G(h,h)=222對(duì)任何r>0(12rsin2h)(12rsin2h)(12rsin2h)222ununr1.所以13jkm1丨G丨(12)xjkm=(h2)Un絕對(duì)收斂。zjkm222lxy223．解：un12ul33ul4ulul2(x,t)(1)4u24t4nt2t26t3jjjn(x,t)(2)un12ul33ul4ulul24u24t4nt2t26t3jjj2n(x,t)(3)unj1uhxu2uh33uh4h24u2x26x324x4nj3jn(x,t)(4)4jnunj1uhxuh22x22uh33uh44u6x324x4nj(1)+(2)得un12unun1t2u2=+o(l2)jjjl27u2ununjh22un（3）+（4）得j1=x2+o(h2)j1o(lh2)。所以其截?cái)嗾`差為224.證明：用Fourier法證明：作變化un=vnexp(ijh)。得0時(shí)jvn1exp(jjh)vnexp(jjh)exp(jjh)vexp(j(j1)h)vn1n1=lh1消去exp(ijh)得：vn1=1l(1exp(ih))vnh11al(1exp(ih))1所以G(h,l)=丨丨1丨丨=alcosh)2sin2ha2l2(1alhhhh2un1unjlun1un1j所以當(dāng)a0時(shí),aj+j1=0絕對(duì)穩(wěn)定。h當(dāng)a<0時(shí)，vn1exp(ijh)vnexp(ijh)vn1exp(i(j1)h)vnexp(ijh)=alh1alh1alexp(ih)消去exp(ijh)得：vn1=vnh1alh1alexp(ih)丨1.丨=h1al1h(1alcosh)2sin2h22alhh2un1unjln1j1un1jhu所以當(dāng)a<0時(shí)j+a=0絕對(duì)穩(wěn)定8vexp(ijh)vnexp(ijh)a.(vn1exp(i(j1)h)vn1exp(i(j1)h)0n125解：作變換vexp(ijh)jFourieru得nn22h1aexp(ijh)v消去得vn1ihsinhvn,n11iaha221isin2hh2h2sin2h1-iasinha211h)2h21|(1a=22a22sin22（1asin2h221sin2h）a22sin2h21a22sin2hhh2h21h22n+1u-un+aun+1-un+10絕對(duì)穩(wěn)定。j1jjj12h26變系數(shù)方程的差分格式為un+1-unaununj10，當(dāng)a（01式）jjjhjjun+1-unaununj0，當(dāng)a（02式），其中aa(x)jjj+1hjjjj式）穩(wěn)定的充要條件為a0,|ar|1.rha0,|a|1,(2利用Fourier方法，類似24題得到（1式）穩(wěn)定的充要條件為jjjj27un11(unun)fnfn2格式：0,其中fn=f(x,n)uj1j1Laxjj1j1j2hjjuuuunjBox格式：（1)(1ar)n1(1ar)+a