1.简答题 (1)行波法:u(x,)=f(x+t/b)+g(x-t/b),但g(x-t/b)不满足泛定方程 故:u(x,t)=f(x+t/b) (4分 比较初条得:u(x,t)=v(x+t/b) (4分) (2)(n+1)zn+x21=xzm-1, (n+2)2n+1+x2n+1=x2n (2分 (2分) a n+2 n+[az2-n(n+1)in=0 (4分 (3)原式=/P(x)B()dx 分 故:当l≠0时,原式=0,l=0时,原式=2。 分 (4)-kuz(L, t)=Ku(l, t) (4分) x(0,t)=q (4分) (5) (4分) u(0, t)=0, YSur(L, t)+mutt(L,t)+mg=0 (1+1分 a(x,0)=0,ut(x,t)=0 (1+1分 (6)u=Alnr+B 分 Alna+B=l0, Aln 2a+B= 2uo (1+1分) A=uo/In 2, B=uo-uo In a/In 2 (2分) (3分 YSu(0, t)=F, YSur(1, t)=-ku(l,t), (3分) (x,0)=,t(x,0)=h(1 (2+2分) (x,t)=F(x-1)/(YS)-F/k (5分 +)丌x Cn COS exp-(n+是)2x2t/12] (5+5分 n=0 (n+)丌x (5分) (5分) +
1. {K: (1) 1Å{µu(x, t) = f(x + t/b) + g(x − t/b)§ g(x − t/b) Ø÷v½§§ µu(x, t) = f(x + t/b) (4©) 'Ð^µu(x, t) = ψ(x + t/b) (4©) (2) (n + 1)zn + xz0 n = xzn−1, =⇒ (n + 2)zn+1 + xz0 n+1 = xzn (2©) −nzn + xz0 n = −xzn+1, =⇒ xz00 n + z 0 n = nz0 n − zn+1 − xz0 n+1 (2©) =⇒ x 2 z 00 n + 2xz0 n + [x 2 − n(n + 1)]zn = 0 (4©) (3) ª = Z 1 −1 Pl(x)P0(x)dx (4©) µ l 6= 0 §ª = 0§l = 0 §ª = 2" (4©) (4) −kux(l, t) = Ku(l, t) (4©) −kux(0, t) = q (4©) (5) utt = a 2uxx (4©) u(0, t) = 0, Y Sux(l, t) + mutt(l, t) + mg = 0 (1 + 1©) u(x, 0) = 0, ut(x, t) = 0 (1 + 1©) (6) u = A ln r + B (4©) A ln a + B = u0, A ln 2a + B = 2u0 (1 + 1©) A = u0/ ln 2, B = u0 − u0 ln a/ ln 2 (2©) 2. utt = a 2uxx (3©) Y Sux(0, t) = F, Y Sux(1, t) = −ku(l, t), (3©) ut(x, 0) =, u(x, 0) = h(1 − x) (2 + 2©) v(x, t) = F(x − l)/(Y S) − F/k (5©) 3. u = X∞ n=0 cn cos (n + 1 2 )πx l exp[−(n + 1 2 ) 2π 2 t/l2 ] (5 + 5©) X∞ n=0 cn cos (n + 1 2 )πx l = h(1 − x l ) ( 5©) cn = 2h (n + 1 2 ) 2π 2 ( 5©)
4.V2a=0 (2分 有限 (1分) (1分) d+2 lo 有限 (1分) uo cos 6, cos 0> h (2分) 6h (3分) l=1 h Wo cos 0 sin 0de lo sin gde 4(h+1 (5分) 5. ut=a V2u, u=uo (1+1+1分 u=t+t,U=u1,v满足齐次方程、齐次边条 0 (3分) w=R(r)(t) r2F"(r)+r()+P2B()=0→B2(m)=/≈ b (4分) T"(2)+a22T(2)=0=>T(t)=exp (4分) w(r, t eXD (2分) 应用初条 r|=(uo-a1) (2分 (u0-t1)/J (2分)
4. ∇2u = 0 (2©) u θ=0, π k (1©) u φ+2π = u φ (1©) u r=0 k (1©) u r=1 = ( u0 cos θ, cos θ > h −u0, cos θ h −u0, cos θ ^ wt = a 2∇2w, w t=0 = u0 − u1, w r=b = 0 (3©) w = R(r)T(t) r 2R00(r) + rR(r) + µ 2 r 2R(r) = 0 =⇒ Rn(r) = J0 x (0) n b r ! (4©) T 00(z) + a 2µ 2T(z) = 0 =⇒ T(t) = exp − x (0) n b a !2 t (4©) w(r, t) = X∞ n=1 cnJ0 x (0) n b r ! exp − x (0) n b a !2 t (2©) A^Ð^µ X∞ n=1 cnJ0 x (0) n b r ! = (u0 − u1) (2©) cn = (u0 − u1) Z b 0 J0 x (0) n b r ! rdr Z b 0 J 2 0 x (0) n b r ! rdr (2©)