数学物理方程(王明新等，清华大学出版社)_课后习题答案.pdf

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k/9#b6 x℄e ie?WY?4v?w?Zd/ dZXZG ?. i e? <3k? . v℄?

<>.:82AaYE_.:f"R\ §1.1CG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . §1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

f (x)dx = 0, f (x)v(x)dx = 0, = ?/;93BbZF`/;g#S℄ §1.1 y BvÆy <5Zb℄| —e yb,? ℄Æa Fu- biniE y< 66YWÆ yZ,;E": 1.1 Ω ⊂ Rn, f^ Ω eTXRj Ω′ ⊂ Ω,RW _^ Ω f ≡ 0. 1.2 Ω ⊂ Rn, f^ Ω eTXR v ∈ C∞0 (Ω),RW _^ Ω f ≡ 0. 1.3 ( Stokes /) Ω ⊂ Rm1aW|di!Z,TX C1 m?eg3 ~v,D [/A g ~n1 ∂Ω

2 Z b a a a t = t2TZd} − t = t1TZd} = w [a, b] × [t1, t2] 4ZA} w .4ZA} + Z b a Z t2 t1 sin αa = sin αb = uxp1 + u2 uxp1 + u2 ρ(x) [ut(x, t2) − ut(x, t1)] dx a [T (b) sin αb − T (a) sin αa] dt. [T (b)ux(b, t) − T (a)ux(a, t)]dt f0 dx dt +Z t2 t1 ρut|t=t2 dx −Z b xx=b ≈ ux|x=b, f0 dx dt +Z t2 f0 dx dt +Z t2 t1 Z b t1 ρ(x)utt dt dx = Z b = Z t2 t1 Z b = Z t2 t1 Z b ρut|t=t1 dx =Z t2 t1 Z b xx=a ≈ ux|x=a, 0 Y eY [ FubiniE y< / a, b, t1, t2ZR8eY ,g ρ(x) = ρ℄6,w T (x) = (l − x)ρg,wg f0(x, t) = 0.F (1.2)YY Zwd <: 1.2! wd,; Z!w_ g:{ aVCZwd < aU℄!,x ρ(x) = ρ0 (6), T (x) = T0 (6), f0(x, t) = −γut , γ > 0℄!w ,}"Z!w_ g ..F (1.2)YY 5 a2 = T0/ρ0, k2 = γ/ρ0,Ye2: 1.3`h!wY"BZ .Æ/wd aVCÆax Z < ( 1.2) ρ0utt = (T0ux)x − γut. utt = a2uxx − k2ut. ρ(x)utt = (T (x)ux)x + f0(x, t). (1.2) (T (x)ux)x dx dt. a utt = g[(l − x)ux]x. a a u(x,t) 1.2 b Ta a Tb x

a t1 ρutt = (Eux)x + f (x, t). utt. 3 (1.3) (ES(x)ux)x dx dt, f0(x, t)dx dt. t1 a a ∆P =Z b a utt dt dx. ρ(x)S(x)utt = (ES(x)ux)x + f0(x, t). ρ(x)S(x)Z t2 (ES(x)ux|x=b − ES(x)ux|x=a)dt I1 = Z t2 = Z t2 t1 Z b I2 = Z t2 t1 Z b ρ(x)S(x) [ut(x, t2) − ut(x, t1)] dx =Z b aCZwd5.g℄Z,5 ρ(x).Z ℄Z,5 S(x).Zxfw T (x) = ES(x)ux, E℄E`}. uZ{ .R7gZ wg℄ f0(x, t). [a, b] × [t1, t2]XZwd1j.Zd} A}k|y:w u .ZA} I1w u .ZA} I2, Zd}e I1 + I2 = ∆P,/ a, b, t1, t2ZR8,eY UZ℄`h!,x ρ(x) = ρ (6), S(x) = S0 (6).CzF (1.3)YYZÆ /wd <: 1.4[d6 ( 1.3), h,gE`}y#6 ρ E.a|wd < a 1.3Z5,6 ρ(x) = ρ (6),wg f0(x, t) = 0.YB[2Z ,eVC Z _Z: S0d6Z\ .F (1.3)YYd6Zwd <: 1.5aVCZwd <,y#k (1)Æ (2)yfWwZY(3) Æ T8}=,v1j2C A 1.3 S(x) = S01 − E1 − = ρ1 − x h2 = ρ1 − x h2 uxx x h2 uxx x h2 E1 − x h2 , utt.

4 f1 dx dy dt. f1 dx dy dt, Z t2 t1 ZD ρutt dt dx dy. t1 Z t2 t1 Z∂D T∇u · ~n dS dt =Z t2 t1 ZD ∇ · (T∇u)dx dy dt. Z t2 t1 ZD ZD ρut|t2 dx dy −ZD ρut|t1 dx dy =ZDZ t2 ρutt dx dy dt =Z t2 t1 ZD ∇ · (T∇u)dx dy dt +Z t2 t1 ZD a6/i D,Ti [t1, t2],5 DZ ∂D. ∂D6/i dS, dSfWZ w T dS ~n,w uZX℄ T dS~n · ∇u = T∇u · ~n dS.UGw4ZA} 5wg f1,w4ZA} d}Z}℄ [eY Z D, t1, t2ZR8Y ρutt = ∇ · (T∇u) + f1, utt = a2∆u + f. BA℄ u(x, y, 0) = ϕ(x, y), ut(x, y, 0) = ψ(x, y), (x, y) ∈ Ω. Ay#℄ 1.6.wdT,; (1)h_Æ (2)h_Z(3)h_Æ T8}= aVCZ wd <,2Cv1j"kWZA a5E`} E,g ρ,Z S.6/i7 [a, b],Ti [t1, t2]. a, b_E Zwy# ESux(a, t) ESux(b, t),/xfZw ESux(b, t) − ESux(a, t),A} ρSut|t1 dxZ b t1 t2TfZd}y#Z b Z a, b, t1, t2ZR8Y Ay#: ρSut|t2 dx.[e, [ESux(b, t) − ESux(a, t)]dt =Z t2 t1 Z b (ESux)x dx dt =Z b ρSut|t1 dx =Z b a Z t2 t1 ρSut|t2 dx −Z b a (x, y) ∈ ∂Ω, t ≥ 0, (x, y) ∈ ∂Ω, t ≥ 0, (x, y) ∈ ∂Ω, t ≥ 0. (ESux)x dx dt. a (3) αu + = 0, ∂u ∂~n (1) u(x, y, t) = g(x, y, t), (2) = g(x, y, t), ∂u ∂~n ρSutt = (ESux)x. Z t2 t1 Z b a a a a ρSutt dt dx. Z t2 t1

5 (1) u(0, t) = u(l, t) = 0; (2) ux(0, t) = ux(l, t) = 0; Q =Z b a Scρ (u|t=t2 − u|t=t1) dx =Z b a Z t2 t1 (3) (αu − βux)|x=0 = (αu + βux)|x=l = 0. t1 Z b S(kux|x=b − kux|x=a)dt +Z t2 Q∗ = Z t2 t1 Z b Skuxx dx dt +Z t2 t1 Z b = Z t2 1.7[W r,g ρZ`hd ;dZZg℄Z,d Z/_Æ [>E,/ >E .5dZ>℄ c,>FV℄ k, Z g℄ u06,>E℄ k1.aVCgy u ZÆy < ad . [a, b]iZ4`5 V .X4` V/Ti [t1, t2]>}Z 1j.Z }e~,[t1, t2]Ti V:Z>}[^ [t1, t2]Ti ∂VF VZ>}. [t1, t2]Ti Vg :Z>}℄ ∂VF VZ>} Q∗|y —dZ>FV/_S7>ExbZ>}: S = πr2℄d ,l = 2πr℄dÆ7.kZ Q = Q∗/ a, b, t1, t2ZR8,eY 1.8h ,g u(x, y, z, t), Nernst ,0yid g_g Z~g:{ .~v = −D∇u, D Zg c aVC ux Z Æy < aB Ze[t1, t2]Ti V:ZB Z } m1[^ [t1, t2]Ti ∂V F VZB Z } m2.~ [ m1 = m2/ t1, t2, VZR8Y 1.9`hd Zz"e℄_>Z t = 0Tg℄ rZ,rd Z_d 3Z℄ a|d Zg u(r, t) ZÆy <℄ (cu|t=t2 − cu|t=t1)dx dy dz =ZV Z t2 m1 = ZV m2 = −Z t2 t1 Z∂V ~v · ~n dS dt =Z t2 t1 Z∂V = Z t2 t1 ZV ∇ · (D∇u)dx dy dz dt. (cu)t = ∇ · (D∇u). lk1(u0 − u)dx dt. lk1(u0 − u)dx dt a 2k1 crρ (u0 − u). ut = uxx + k cρ t1 D ∂u ∂~n dS dt (cu)t dt dx dy dz, Scρut dt dx. t1 a a ut = a2urr + 1 r ur .

6 r2 O r1 S2 Ω S1 2d kW^ r1 ≤ r ≤ r2Z4` Ω ( 1.4). 1.4 Z}e~, [t1, t2]Ti Ω:Z>} Q[^ ∂Ω = S1S S2F ΩZ>} Q∗.5d Z>gy# c ρ (ze℄6).U t = 0Tg℄ rZ,x u = u(r, t)℄ r, t Z.^℄ 5>FV k,>g S1,. ~n = −(cos θ, sin θ) = −~n0. S2,. ~n = (cos θ, sin θ) = ~n0.ZGY [t1, t2] Ti ∂ΩF ΩZ>}℄ Y Q = Q∗/ t1, t2, r1, r2ZR8,Y 5 a2 = ,Ye2: 1.10`h!V,ÆR7Z`! R, Z` I,V"_ÆggZ S7 >E aVCVZg u Z (BXg|hge℄g). Q = ZΩ = cρZ 2π 0 Z r2 t1 Z r2 = 2πcρZ t2 ~q = −k(ux, uy) = −k(urrx, urry) = −kur(cos θ, sin θ) , −kur~n0, t1 Z∂Ω (−~q · ~n)dS dt =Z t2 t1 Z∂Ω Q∗ = Z t2 t1 Z 2π = Z t2 = 2πZ t2 t1 Z r2 r1 0 (krur|r=r2 − krur|r=r1)dθ dt ut = a2urr + 1 r ur . (cρu|t=t2 − cρu|t=t1)dx dy kur~n0 · ~n dS dt (u|t=t2 − u|t=t1)r dr dθ r1 rut dr dt. r1 k(rur)r dr dt. cρrut = k(rur)r. k cρ

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数学物理方程(王明新等，清华大学出版社)_课后习题答案.pdf

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