格林函数习题解答

Assign7,mathematicalMethodFirstname:LIANGLastname:ZHANGOct-25-201210.5.1FindtheGreenfunctionforthisoperatorandboundarycondition:Ly(x)≡d2ydx2,and{y(0)=0;y′(1)=0Whereweassumethattheendsoftheintervalareatx=0andx=1.Solution:First,∵Ly(x)≡d2ydx2∴thegeneralsolutionisy=kx+by(0)=0⇒y1=xy′(1)=0⇒y2=1Let’sGreenfunctiontobe,G(x;t)={B(t)y1;tx61A(t)y2;06xtAsGreenfunctioniscontinuous,soA(t)y2(t)=B(t)Y1(t)⇒A(t)=B(t)tOnetheotherhand,A(t)y′2(t)−B(t)y′1(t)=−1p(t)=−11=−1⇒0−B(t)=−1B(t)=1A(t)=t1⇒G(x;t)={x;06xtt;tx6110.05.1DGreen#1FindtheGreenfunctionG=G(x;t)fortheequationd2ydx2+dydx=f(x)giventhaty(0)=0anddydxx=0=0(isanon-zeroconstant.)Solution:d2ydx2+dydx=f(x)(x)d2ydx2+(x)dydx=(x)f(x)ddx=⇒(x)=ex⇒ddx(exdydx)=exf(x)Ly=ddx(exdydx)=0y(0)=0dydxx=0=0ThegeneralsolutionforLyisy=C1e−x+C2,Astheboundaryconditionisy(0)=0;y′(0)=0,then⇒G(x;t)=0;06xtC1e−x+C2;tx+∞AstheGreenfunctionG(x;t)shouldbecontinuousatpointt,then,C1e−t+C2=0⇒C2=−C1e−t2Ontheotherhand,thedifferentialofGreenfunctionG′(x;t)shouldsatisfyG′(x;t)|t=−1p(t)=−1et⇒−C1e−t=−e−t⇒C1=1;C2=−C1e−t=−e−t⇒G(x;t)=0;06xte−x−e−t;tx+∞⇒G(x;t)=e−x−e−t;06tx0;xt+∞⇒y(x)=−∫+∞0G(x;t)etf(t)dt=−∫x0e−x−e−tetf(t)dtComment:ItisapparentthatG(x;t)̸=G(t;x).Astheoperator′L′issecondorderdifferential,soifthetwoboundaryconditionsfocusononesidex=a,thentheGreenfunctionforaxtisdefinitelydetermined,and@G@t=0.Fortheintervaltxb,thetwoboundaryconditionsalsofocusatoneside,thatis,y2(t)=y1(t);y′2(t)=y′1(t)−1p(t).WecangettheGreenfunctionfortxband@G@t̸=0.Inshort,theGreenfunctionwon’tbesymmetricalifthetwoboundaryconditionsfocusononeside.10.05.1DGreen#2FindtheGreenfunctionG=G(x;t)fortheequationd2ydx2−y=f(x)giv-enthaty(0)=0andy(1)=0.Solution:3Ly=d2ydx2−y=0y(0)=0y(1)=0ThegeneralsolutionforLy=0isy=C1sinx+C2cosx,Accordingtotheboundarycondition,wehavey1=sinxy2=sinxtan1−cosxW[y1;y2]|t=y1(t)y′2(t)−y′1y2=sint×(costtan1+sint)−cost×(sinttan1−cost)=1p(t)=1⇒A=−1W[y1;y2]p(t)=−1G(x;t)=Ay2(t)y1(x);06xtAy1(t)y2(x);tx61⇒G(x;t)=−(sinttan1−cost)sinx;06xt−sint(sinxtant1−cosx);tx61⇒G(x;t)=−(sinxtant1−cosx)sint;06tx−sinx(sinttan1−cost);xt61y(x)=−∫10G(x;t)f(t)dt4