齊次化原理講義

齊次化原理講義疊加原理和齊次化原理從通解求特解的方法對(duì)多數(shù)偏微分方程定解問題不適用。求解線性偏微分方程定解問題較多采用以下做法：先求方程的一族特殊解，再將這族特殊解作適當(dāng)線性疊加求得定解問題的特解。接下將介紹這種做法的理論根據(jù)——線性疊加原理，以及由疊加原理導(dǎo)出的處理非齊次方程初值問題的齊次化原理。1.線性疊加原理n個(gè)自變量的二階線性偏微分方程∑i,j=1naij?2u?xi?xj+∑j=1nbj?u?xj+cu=f\sum_{i,j=1}^na_{ij}\frac{\partial^2u}{\partialx_i\partialx_j}+\sum_{j=1}^nb_j\frac{\partialu}{\partialx_j}+cu=fi,j=1∑n?aij??xi??xj??2u?+j=1∑n?bj??xj??u?+cu=f引入偏微分算子L=∑i,j=1naij?2?xi?xj+∑j=1nbj??xj+c(1)L=\sum_{i,j=1}^na_{ij}\frac{\partial^2}{\partialx_i\partialx_j}+\sum_{j=1}^nb_j\frac{\partial}{\partialx_j}+c\tag{1}L=i,j=1∑n?aij??xi??xj??2?+j=1∑n?bj??xj???+c(1)則可簡(jiǎn)單表示為L(zhǎng)u(x)=f(x),x=(x1,x2,???,xn)Lu(\boldx)=f(\boldx),\quad\boldx=(x_1,x_2,···,x_n)Lu(x)=f(x),x=(x1?,x2?,???,xn?)一般地，稱從一個(gè)函數(shù)類（定義域）到另一個(gè)函數(shù)類（值域）的映射T為一個(gè)算子。如果定義域與值域都是數(shù)域Λ\LambdaΛ上的線性空間，對(duì)定義域中任意兩個(gè)函數(shù)u1,u2u_1,u_2u1?,u2?和Λ\LambdaΛ中任意兩個(gè)數(shù)λ1,λ2\lambda_1,\lambda_2λ1?,λ2?有T(λ1u1+λ2u2)=λ1Tu1+λ2Tu2T(\lambda_1u_1+\lambda_2u_2)=\lambda_1Tu_1+\lambda_2Tu_2T(λ1?u1?+λ2?u2?)=λ1?Tu1?+λ2?Tu2?，則稱T是線性算子。顯然，上述二階偏微分算子（1）式是函數(shù)空間C2C^2C2到CCC的線性算子，拉普拉斯算子Δ3=?2?x2+?2?y2+?2?z2\Delta_3=\frac{\partial^2}{\partialx^2}+\frac{\partial^2}{\partialy^2}+\frac{\partial^2}{\partialz^2}Δ3?=?x2?2?+?y2?2?+?z2?2?是其特殊情況。Fourier變換F[f(x)]=∫?∞+∞f(x)e?iλxdxF[f(x)]=\int_{-\infty}^{+\infty}f(x)e^{-i\lambdax}dxF[f(x)]=∫?∞+∞?f(x)e?iλxdx，Laplace變換L[f(t)]=∫0+∞f(t)e?ptdtL[f(t)]=\int^{+\infty}_0f(t)e^{-pt}dtL[f(t)]=∫0+∞?f(t)e?ptdt都是線性積分算子。若記L1=limt→0,L2=limt→0??t,L3=limx→x0(α+β??n)L_1=lim_{t\to0},L_2=lim_{t\to0}\frac{\partial}{\partialt},L_3=lim_{x\tox_0}(\alpha+\beta\frac{\partial}{\partialn})L1?=limt→0?,L2?=limt→0??t??,L3?=limx→x0??(α+β?n??)，則初始條件、邊界條件均可表示為線性算子形式Lju(x)=φ(x)L_ju(x)=\varphi(x)Lj?u(x)=φ(x)由線性算子的定義，我們有。定理1：(疊加原理)設(shè)L是關(guān)于x=(x1,x2,???,xn)\boldx=(x_1,x_2,···,x_n)x=(x1?,x2?,???,xn?)的任意階線性微分算子（?；蚱瑒t有。（1）有限疊加原理：若uj(x)u_j(\boldx)uj?(x)滿足Luj(x)=fj(x)(j=1,2,???,m)Lu_j(\boldx)=f_j(\boldx)(j=1,2,···,m)Luj?(x)=fj?(x)(j=1,2,???,m)，則當(dāng)u(x)=∑j=1mλjuj(x)u(\boldx)=\sum_{j=1}^m\lambda_ju_j(x)u(x)=∑j=1m?λj?uj?(x)時(shí)，有Lu(x)=∑j=1mλjfj(x),λj∈Λ,j=1,2,???,mLu(\boldx)=\sum_{j=1}^m\lambda_jf_j(x),\quad\lambda_j\in\Lambda,\quadj=1,2,···,mLu(x)=∑j=1m?λj?fj?(x),λj?∈Λ,j=1,2,???,m。（2）級(jí)數(shù)疊加原理：若uj(x)u_j(x)uj?(x)滿足Luj(x)=fj(x)(j=1,2,???),Lu_j(\boldx)=f_j(\boldx)(j=1,2,···),Luj?(x)=fj?(x)(j=1,2,???),則當(dāng)u(x)=∑j=1∞λjuj(x)u(\boldx)=\sum_{j=1}^\infty\lambda_ju_j(x)u(x)=∑j=1∞?λj?uj?(x)時(shí)，有Lu(x)=∑j=1∞λjfj(x),λj∈Λ，j=1,2,???Lu(x)=\sum_{j=1}^\infty\lambda_jf_j(\boldx),\space\lambda_j\in\Lambda，j=1,2,···Lu(x)=∑j=1∞?λj?fj?(x),λj?∈Λ，j=1,2,???。（3）積分疊加原理：若u(x;ξ)u(\boldx;\xi)u(x;ξ)滿足Lu(x;ξ)=f(x;ξ),ξ∈VLu(\boldx;\bold\xi)=f(\boldx;\bold\xi),\quad\xi\inVLu(x;ξ)=f(x;ξ),ξ∈V，則當(dāng)U(x)=∫Vλ(ξ)u(x;ξ)dξU(\boldx)=\int_V\lambda(\bold\xi)u(\boldx;\bold\xi)d\xiU(x)=∫V?λ(ξ)u(x;ξ)dξ時(shí)，有LU(x)=∫Vλ(ξ)f(x;ξ)dξ,λ(ξ)∈Λ,ξ∈VLU(\boldx)=\int_V\lambda(\xi)f(\boldx;\bold\xi)d\xi,\lambda(\xi)\in\Lambda,\xi\inVLU(x)=∫V?λ(ξ)f(x;ξ)dξ,λ(ξ)∈Λ,ξ∈V。有限疊加原理是線性算子定義的直接推廣，而級(jí)數(shù)疊加原理及積分疊加原理分別要求有關(guān)函數(shù)項(xiàng)級(jí)數(shù)、含參量積分收斂，算子L與求和號(hào)、積分號(hào)可交換次序。在經(jīng)典意義下這些條件要求很高，實(shí)際問題不一定能滿足，但是在推廣意義下，這種交換總是可進(jìn)行的。今后，我們將不受限制地使用這些疊加原理。在物理問題中，方程或定解條件中的非齊次項(xiàng)通常反映引起物理過(guò)程的源，線性疊加原理則說(shuō)明多個(gè)源共同作用的結(jié)果等于各個(gè)源單獨(dú)作用的總和，這在物理上稱為獨(dú)立作用原理，這是線性問題的基本特征。利用疊加原理我們總可以把一個(gè)較復(fù)雜的線性問題分解成若干簡(jiǎn)單線性問題來(lái)求解。例1：求解狄利克雷問題{?2u?x2+?2u?y2=1,x2+y2=1u∣x2+y2=1=x2\begin{cases}\frac{\partial^2u}{\partialx^2}+\frac{\partial^2u}{\partialy^2}=1,\quadx^2+y^2=1\\u|_{x^2+y^2=1}=x^2\end{cases}{?x2?2u?+?y2?2u?=1,x2+y2=1u∣x2+y2=1?=x2?解：易見u0=14(x2+y2)u_0=\frac{1}{4}(x^2+y^2)u0?=41?(x2+y2)是非齊次泛定方程的一個(gè)特解。令v=u?u0v=u-u_0v=u?u0?，由疊加原理知，v滿足邊值問題{?2v?x2+?2v?y2=0v∣x2+y2=1=x2?14\begin{cases}\frac{\partial^2v}{\partialx^2}+\frac{\partial^2v}{\partialy^2}=0\\v|_{x^2+y^2=1}=x^2-\frac{1}{4}\end{cases}{?x2?2v?+?y2?2v?=0v∣x2+y2=1?=x2?41??解析函數(shù)zn(z=x+iy)z^n(z=x+iy)zn(z=x+iy)的實(shí)、虛部均為調(diào)和函數(shù)，故1,x,y,x2?y2,xy1,x,y,x^2-y^2,xy1,x,y,x2?y2,xy為次數(shù)不超過(guò)2的調(diào)和多項(xiàng)式，故可設(shè)v=v0+a1x+a2y+a3(x2?y2)+a4xyv=v_0+a_1x+a_2y+a_3(x^2-y^2)+a_4xyv=v0?+a1?x+a2?y+a3?(x2?y2)+a4?xy由疊加原理知，對(duì)任意一組{aj,j=0,1,???,4}\{a_j,j=0,1,···,4\}{aj?,j=0,1,???,4}，v都是拉普拉斯方程的解。代入邊界條件v∣x2+y2=1=a0+a1x+a21?x2+a3(2x2?1)+a4x1?x2=x2?14v|_{x^2+y^2=1}=a_0+a_1x+a_2\sqrt{1-x^2}+a_3(2x^2-1)+a_4x\sqrt{1-x^2}=x^2-\frac{1}{4}v∣x2+y2=1?=a0?+a1?x+a2?1?x2?+a3?(2x2?1)+a4?x1?x2?=x2?41?比較系數(shù)，可得v=14+12(x2?y2)v=\frac{1}{4}+\frac{1}{2}(x^2-y^2)v=41?+21?(x2?y2)從而u=34x2?14y2+14u=\frac{3}{4}x^2-\frac{1}{4}y^2+\frac{1}{4}u=43?x2?41?y2+41?對(duì)于非線性方程，自然不存在線性疊加原理，因此求解要困難得多。對(duì)某些特殊的非線性方程，如果能發(fā)現(xiàn)某種形式的疊加原理，則對(duì)此非線性問題的研究有重大意義。2.齊次化原理（沖量原理）例1中通過(guò)求非齊次方程的一個(gè)特解將方程齊次化。齊次化原理將解決一般非齊次發(fā)展方程齊次化的問題。仍以理想弦的橫振動(dòng)為例，考慮在外力作用下弦的純受迫振動(dòng){?2u?t2=a2?2u?x2+f(t,x),t>0,?∞<x<+∞u∣t=0=0,?u?t∣t=0=0(2)\begin{cases}\frac{\partial^2u}{\partialt^2}=a^2\frac{\partial^2u}{\partialx^2}+f(t,x),\quadt>0,\quad-\infty<x<+\infty\\u|_{t=0}=0,\quad\frac{\partialu}{\partialt}|_{t=0}=0\end{cases}\tag{2}{?t2?2u?=a2?x2?2u?+f(t,x),t>0,?∞<x<+∞u∣t=0?=0,?t?u?∣t=0?=0?(2)這里，f(t,x)f(t,x)f(t,x)是作用在弦身單位質(zhì)量上的外力，u(t,x)u(t,x)u(t,x)是由此外力從初始時(shí)刻t=0t=0t=0持續(xù)到t時(shí)刻作用引起的位移。由獨(dú)立作用原理,u(t,x)u(t,x)u(t,x)可看成前后相繼的瞬間外力作用沖量f(τ,x)dτ(0≤τ≤t)f(\tau,x)d\tau(0\leq\tau\leqt)f(τ,x)dτ(0≤τ≤t)在t時(shí)刻引起的位移ω(t,x;τ)dr\omega(t,x;\tau)drω(t,x;τ)dr關(guān)于drdrdr的疊加，即u(t,x)=∫0tω(t,x;τ)dτu(t,x)=\int_0^t\omega(t,x;\tau)d\tauu(t,x)=∫0t?ω(t,x;τ)dτ。x顯然，當(dāng)t<τt<\taut<τ時(shí)ω（t,x;τ）≡0\omega（t,x;\tau）\equiv0ω（t,x;τ）≡0，當(dāng)t>τt>\taut>τ時(shí)，τ\tauτ時(shí)刻瞬間沖量的作用已轉(zhuǎn)化為從t=τ?0t=\tau-0t=τ?0到t=τ+0t=\tau+0t=τ+0動(dòng)量的增加，故w(t,x;τ)w(t,x;\tau)w(t,x;τ)應(yīng)滿足{?2ω?t2=a2?2ω?x2,t>τ,?∞<x<+∞w∣t=0=0,?w?t∣t=τ=f(τ,x)(3)\begin{cases}\frac{\partial^2\omega}{\partialt^2}=a^2\frac{\partial^2\omega}{\partialx^2},\quadt>\tau,\space-\infty<x<+\infty\\w|_{t=0}=0,\quad\frac{\partialw}{\partialt}|_{t=\tau}=f(\tau,x)\end{cases}\tag{3}{?t2?2ω?=a2?x2?2ω?,t>τ,?∞<x<+∞w∣t=0?=0,?t?w?∣t=τ?=f(τ,x)?(3)而（2）的解則為u(t,x)=∫0tw(t,x;τ)dτ(4)u(t,x)=\int_0^tw(t,x;\tau)d\tau\tag{4}u(t,x)=∫0t?w(t,x;τ)dτ(4)實(shí)際上，當(dāng)w(t,x;τ)w(t,x;\tau)w(t,x;τ)是問題（3）的解時(shí)，由（4）可驗(yàn)證u∣t=0=0?u?t=w(t,x;τ)∣τ=t+∫0t?ω(t,x;τ)?tdτ=∫0t?ω(t,x;r)?tdτ?u?t∣t=0=0?2u?t2=?ω(t,x;r)?t∣τ=t+∫0t?2ω(t,x;τ)?t2dr=f(t,x)+a2∫0t?2w(t,x;r)?x2dr=f(t,x)+a2?2u?x2u|_{t=0}=0\\\frac{\partialu}{\partialt}=w(t,x;\tau)|_{\tau=t}+\int_0^t\frac{\partial\omega(t,x;\tau)}{\partialt}d\tau=\int_0^t\frac{\partial\omega(t,x;r)}{\partialt}d\tau\\\frac{\partialu}{\partialt}|_{t=0}=0\\\frac{\partial^2u}{\partialt^2}=\frac{\partial\omega(t,x;r)}{\partialt}|_{\tau=t}+\int_0^t\frac{\partial^2\omega(t,x;\tau)}{\partialt^2}dr=f(t,x)+a^2\int_0^t\frac{\partial^2w(t,x;r)}{\partialx^2}dr=f(t,x)+a^2\frac{\partial^2u}{\partialx^2}u∣t=0?=0?t?u?=w(t,x;τ)∣τ=t?+∫0t??t?ω(t,x;τ)?dτ=∫0t??t?ω(t,x;r)?dτ?t?u?∣t=0?=0?t2?2u?=?t?ω(t,x;r)?∣τ=t?+∫0t??t2?2ω(t,x;τ)?dr=f(t,x)+a2∫0t??x2?2w(t,x;r)?dr=f(t,x)+a2?x2?2u?故式（4）給出了純受迫振動(dòng)（2）的解。在（3）中令t′=t?τ,ω(t,x;τ)t'=t-\tau,\omega(t,x;\tau)t′=t?τ,ω(t,x;τ)可由達(dá)朗貝爾公式得到，故純受迫振動(dòng)u(t,x)=12a∫0tdτ∫x?a(t?τ)x+a(t?τ)f(τ,ξ)dξu(t,x)=\frac{1}{2a}\int_0^td\tau\int_{x-a(t-\tau)}^{x+a(t-\tau)}f(\tau,\xi)d\xiu(t,x)=2a1?∫0t?dτ∫x?a(t?τ)x+a(t?τ)?f(τ,ξ)dξ若作變量代換r=a(t?τ)r=a(t-\tau)r=a(t?τ)，則u(t,x)=12a2∫0atdr∫x?rx+rf(t?ra,ξ)dξu(t,x)=\frac{1}{2a^2}\int_0^{at}dr\int_{x-r}^{x+r}f(t-\frac{r}{a},\xi)d\xiu(t,x)=2a21?∫0at?dr∫x?rx+r?f(t?ar?,ξ)dξ由此可見純外力引起的x點(diǎn)t時(shí)刻的位移受[x?at,x+at][x-at,x+at][x?at,x+at]內(nèi)各點(diǎn)所受外力的影響，而在[x?r,x+r](0≤r<at)[x-r,x+r](0\leqr<at)[x?r,x+r](0≤r<at)上的外力是t?rat-\frac{r}{a}t?ar?時(shí)刻的。由疊加原理可寫出一般弦振動(dòng)方程初值問題{?2u?t2=a2?2u?x2+f(t,x),t>0,?∞<x<+∞u∣t=0=φ(x),?u?t∣t=0=ψ(x)\begin{cases}\frac{\partial^2u}{\partialt^2}=a^2\frac{\partial^2u}{\partialx^2}+f(t,x),\quadt>0,-\infty<x<+\infty\\u|_{t=0}=\varphi(x),\quad\frac{\partialu}{\partialt}|_{t=0}=\psi(x)\end{cases}{?t2?2u?=a2?x2?2u?+f(t,x),t>0,?∞<x<+∞u∣t=0?=φ(x),?t?u?∣t=0?=ψ(x)?的解u(t,x)=12[φ(x?at)+φ(x+at)]+12a∫x?atx+atψ(ξ)dξ+12a∫0tdr∫x?a(t?τ)x+a(t?τ)f(τ,ξ)dξu(t,x)=\frac{1}{2}[\varphi(x-at)+\varphi(x+at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(\xi)d\xi+\frac{1}{2a}\int_0^tdr\int_{x-a(t-\tau)}^{x+a(t-\tau)}f(\tau,\xi)d\xiu(t,x)=21?[φ(x?at)+φ(x+at)]+2a1?∫x?atx+at?ψ(ξ)dξ+2a1?∫0t?dr∫x?a(t?τ)x+a(t?τ)?f(τ,ξ)dξ將此齊次化方法推廣到一般發(fā)展方程的初值問題，有如下原理。定理：（齊次化原理）設(shè)L是關(guān)于t與x=(x1,x2,???,xn)x=(x_1,x_2,···,x_n)x=(x1?,x2?,???,xn?)中各分量的線性偏微分算子，其中關(guān)于t的最高階導(dǎo)數(shù)不超過(guò)m?1m-1m?1階。若ω(t,x;τ)\omega(t,x;\tau)ω(t,x;τ)滿足齊次方程初值問題{?mw?tm=Lw,t>τ>0,x∈Rnw∣t=τ=?w?t∣t=τ=???=?m?2ω?tm?2∣t=r=0,?m?1w?tm?1∣t=τ=f(τ,x)(5)\begin{cases}\frac{\partial^mw}{\partialt^m}=Lw,\quadt>\tau>0,x\inR^n\\w|_{t=\tau}=\frac{\partialw