// //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du
|
|
- へいぞう いなおか
- 5 years ago
- Views:
Transcription
1 [ 1 ]< 33, 34 > 1 (the first law of thermodynamics) U du = δw + δq (1) (internal energy)u (work)w δw rev = PdV (2) P (heat)q 1 1. U ( U ) 2. 1 (perpetuum mobile) 3. du 21
2 // //( ) (Helmholtz, Hermann Ludwig Ferdinand von: ) [ ]< 35, 36 > δq =0 du = δw rev = PdV (3) ( ) du = C V (T )dt (4) C V ( C V =(3/2)Nk B ) γ = C P /C V C V dt = PdV = Nk BT dv (5) V C V ln T = ln V (6) Nk B T 0 V 0 TV 2/3 = const. (7) PV 5/3 = const. (8) T P 2/5 = const. (9) γ 1 γ (γ 1)/γ C V (3/2)Nk B 2 2 C V (5/2)Nk B 22
3 [ ]< 36, 37 > 1 du = δw rev + δq rev = δw irr + δq irr (10) δq irr =0 du du? δw (δw) δw irr >δw rev = PdV. (11) du δq δq irr <δq rev. (12) [ ]< 37 > (working material) (cycle) du = 0 (13)
4 2.2 ( ) [ ]< > (Carnot s cycle) ( 1824 ) (heat engine). P 1 IV I 2 4 II III 3 V Figure 2: I) II) III) IV) 1. (T h,v 1,P 1 ) (T h,v 2,P 2 ) 2. (T h,v 2,P 2 ) (T c,v 3,P 3 ) V 2 V 1 = P 1 P 1 (14) U I = 0 (15) Q I = W I = Nk B T h ln V 2 V 1 > 0 (16) V 3 V 2 = 24 ( Th T c ) 3/2 (17)
5 Q II = 0 (18) W II = U II = C V (T c T h ) < 0 (19) 3. (T c,v 3,P 3 ) (T c,v 4,P 4 ) 4. (T c,v 4,P 4 ) (T h,v 1,P 1 ) V 4 V 3 = P 3 P 4 (20) U III = 0 (21) Q III = W III = Nk B T c ln V 4 V 2 < 0 (22) V 1 V 4 = ( Tc T h ) 3/2 (23) Q IV = 0 (24) W IV = U IV = C V (T h T c ) > 0 (25) Q I Q III W = Q I + Q III (17) (23) V 3 /V 2 = V 4 /V 1 V 3 /V 4 = V 2 /V 1 (16) (22) Q I T h + Q III T c = 0 (26) (26) ( ) ( 1848 ) // //( Home ) (Carnot, Nicolas Leonard: ) 2 L.N.M (1824) ( ) 25
6 ( ) 2 B.E. [ ]< > (26) ( ) Q ( ) δqrev T = 0 (27) 2 δq rev (28) 1 T δq rev /T δq 1/T δq rev /T (entropy) ds = δq 1 rev T, S δq rev 1 S 0 = 0 T (29) [ ] W = Nk B (T h T c )ln V 2 V 1 = Q I + Q III (30) (effciency) η = W = Q I Q III. (31) δq 1 Q I (16) (22) η = T h T c T h =1 T c T h. (32) 26
7 2.3 [ ]< 41 > 2 ( ) 2 1. ( ) (26) (12) ( ) Q I T h (29) [ ]< 41 > δq T + Q III T c 0 (33) δq rev T = ds (34) A B B A A B (34) B δq B rev δq S(B) S(A) = A T A T. (35) δq =0 : ds > 0 (36) 27
8 // //( ) (Clausius, Rudolf Julius Emmanuel: ) [ ]< 41, 42 > (1) 1 ( ) ( ) 1 du = δq rev + δw rev = TdS PdV + µdn ( φdq) (37) (S, V, N, q, ) U T = U P = U µ = U (38) S V,N,q,, V S,N,q,, N S,V,q,, φ = U (39) q S,V,N,, T V U = T S = dq = C T V T V dt V, (40) V U = T S P, (41) V T V T U(S, V, N) S S(U, V, N) 28
9 ds = 1 T du + P T dv µ dn (42) T (ds =0) [ ]< 42, 43 > U =(3/2)Nk B T PV = Nk B T ds = 3 2 Nk du B U + Nk dv B V, (43) S(U, V ) = [ ( )3/2 ( ) ] U V S(U 0,V 0 )+Nk B ln U 0 V 0 (44) = 3 2 Nk B ln U + Nk B ln V + const. (45) U V S(T,V ) = [ ( )3/2 ( ) ] T V S(T 0,V 0 )+Nk B ln T 0 V 0 (46) = 3 2 Nk B ln T + Nk B ln V + const., (47) [ ( T S(T,P) = S(T 0,P 0 )+Nk B ln T 0 ) 5/2 ( ) ] P0 (48) P = 5 2 Nk B ln T Nk B ln P + const. (49) 3 δq = du + PdV. (50) δq = Nk B ( 3 2 dt + T V dv ) (51) 1/T ( δq 3 T = Nk dt B 2 T + dv V ds S (47) ) (52) 29
10 2.4 [ ]< 52, 53 > 1 W (11) (12) W (δw rev )! [ ]< > T h T c (<T h ) W Q h Q c W + Q h + Q c = 0 (53) ( W <0, Q h > 0, Q c < 0) η = W Q h (54) // //( ) (Kelvin, Willam Thomson, The Baron Kelvin of Larg: ) = Largs Kelvin Kelvin Largs 30
11 2.5 2 [ H ]< 43, 44 > (Boltzmann s eta theorem) f( v, t) ( v 1, v 2 ) ( v 3, v) f( v, t) = [f( v 1 )f( v 2 ) f( v 3 )f( v)] σ( v 1, v 2, v 3, v)d 3 v 3 d 3 v 1 (55) t σ( v 1, v 2, v 3, v) v 1 v 2 v 3 v 2 σ ( E ) H(t) f( v, r, t)lnf( v, r, t)d 3 vd 3 r. (56) dh(t) = d 3 v [ f( v, t)lnf( v, t)+ f( v, dt t) ] (57) f (55) dh(t) dt = 1 d 3 v d 3 v 3 d 3 v 1 [f( v 1 )f( v 2 ) f( v 3 )f( v)] [ln(f( v 3 )f( v)) ln(f( v 1 )f( v 2 ))] 4 σ( v 1, v 2, v 3, v). (58) (x y)(ln x ln y) 0 dh(t) 0 (59) dt H H f( v 1,t)f( v 2,t)=f( v 3,t)f( v, t) H [ ] i f i ( ) i j w i j ( ) df i dt = j f j w j i j 31 f i w i j. (60)
12 w j i = w i j. (61) d dt f i = 0 (62) i S k B f i ln f i (63) i w ji = w ij ( ) i j ds dt = k B = k B 2 = k B 2 ( f i ln f i + f ) i i i j i j [(f j w ji f i w ij )lnf i +(f i w ij f j w ji )lnf j ] (f i f j )w ij (ln f i ln f j ) (64) ds(t) dt 0 (65) f i = const. (66) [ ]< 45 > (Loschmidt 1876 ) (Zermero 1986 ) (Poincaré) H H 32
13 (Ostwald) (Mach) // //( ) (Ostwald, Friedrich Wilhelm: ) Ostwald Wolfgang (Mach, Ernst: ) [ ]< > (microstate) 6N (phase space) (q ν,p ν ) (q ν (t 0 ),p ν (t 0 )) (q ν (t f ),p ν (t f )) (q ν (t f ), p ν (t f )) 33
14 ( ) (q ν (t f ), p ν (t f )) Ω 6N A 3N ( V ) Ω(V )=AV N. (67) Ω(V/2) = A(V/2) N Ω(V/2)/Ω(V ) (1/2) N N S Ω? Ω tot =Ω 1 Ω 2, (68) S tot = S 1 + S 2 (69) S ln Ω (70) ( ) 34
15 2.6 [ ]< 50, 51 > U 1 + U 2 = U = const., V 1 + V 2 = V = const., N 1 + N 2 = N = const. (71) S = S 1 + S 2. (72) ds 1 = 1 T 1 du 1 + P 1 T 1 dv 1 µ 1 T 1 dn 1 (73) ds 2 = 1 T 2 du 2 + P 2 T 2 dv 2 µ 2 T 2 dn 2 (74) (71) du 2 = du 1 dv 2 = dv 1 dn 2 = dn 1 ds = ds 1 + ds 2 = ( 1 1 ) ( P1 du 1 + P ) ( 2 µ1 dv 1 µ ) 2 dn 1 (75) T1 T2 T 1 T 2 T 1 T 2 ds =0 T 1 = T 2, P 1 = P 2, µ 1 = µ 2 (76) (V 1,N 1 = const.) (75) dv 1 =0 dn 1 =0 T 1 = T 2. (P 1 P 2, µ 1 µ 2 ) (77) S = S 1 +S 2 ds 2 = ds 1 (75) du = du 1 + du 2 =(T 1 T 2 ) ds 1 (P 1 P 2 ) dv 1 +(µ 1 µ 2 ) dn 1 (78) (S 1,N 1 = const.) (78) ds 1 =0 dn 1 =0 P 1 = P 2. (T 1 T 2, µ 1 µ 2 ) (79) 35
16 [ ]< 51, 52 > (partial equilibrium) (local equilibrium) (global equilibrium) ( ) 36
17 2.7 [ ]< 59, 60 > U S V N du = TdS PdV + µdn. (80) K du = TdS PdV + µ i dn i (81) i=1 U(αS, αv, αn 1,,αN K )=αu(s, V, N 1,,N K ). (82) α =1+ɛ ɛ (Euler s quation) K U = TS PV + µ i N i. (83) i=1 du (Gibbs-Duhem relation) K SdT VdP + N i dµ i = 0 (84) i=1 (81) [ ]< > (84) dµ(p, T) = S(P, T) N V (P, T) dt + dp. (85) N [ ( T S(P, T) =S(P 0,T 0 )+Nk B ln T 0 ) 5/2 ( ) ] P0 (86) P 37
18 V (P, T) =Nk B T/P ( [ ( T dµ(p, T) = s(p 0,T 0 )k B + k B ln T 0 ) 5/2 ( ) ]) P0 dt + k BT dp. (87) P P dt dp dµ (P 0,T 0 ) (P 0,T) (P, T) (s 0 s(p 0,T 0 )) [ ( ) T 5/2 ( ) ] ( P0 5 µ(p, T) =µ(p 0,T 0 ) k B T ln + 0) P 2 s k B (T T 0 ). (88) T 0 µ(p 0,T 0 )=( 5 2 s 0 ) k B T 0 (89) ( [ ( µ(p0,t 0 ) T µ(p, T) =k B T ln k B T 0 T 0 ) 5/2 ( ) ]) P0 (90) P 38
3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3) P = U V S,N,, µ = U N. (4) S,V,, ( ) ds = 1 T du + P T dv µ dn +, (5) T 1 T = P U V,N,, T
3 3.1 [ ]< 85, 86 > ( ) ds > 0. (1) dt ds dt =0, S = S max. (2) ( δq 1 = TdS 1 =0) (δw 1 < 0) (du 1 < 0) (δq 2 > 0) (ds = ds 2 = TδQ 2 > 0) 39 3.2 [ ]< 86, 87 > ( ) T = U V,N,, du = TdS PdV + µdn +, (3)
More information1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
More informationflMŠÍ−w−î‚b
23 6 30 i 2 1980 2001 1979 K. 1971 ii 1992 iii 1 1 2 5 2.1 : : : : : : : : : : : : : : : : : : : : : : : : : : 5 2.2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8 2.3 : : : : : : : : :
More informationuntitled
1 Physical Chemistry I (Basic Chemical Thermodynamics) [I] [II] [III] [IV] Introduction Energy(The First Law of Thermodynamics) Work Heat Capacity C p and C v Adiabatic Change Exact(=Perfect) Differential
More informationSeptember 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1
September 25, 2017 1 1.1 1.2 p = nr = 273.15 + t : : K : 1.3 1.3.1 : e.g. 1.3.2 : 1 intensive variable e.g. extensive variable e.g. 1.3.3 Equation of State e.g. p = nr X = A 2 2.1 2.1.1 Quantity of Heat
More information70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain) (grain boundary) 5. 1
5 0 1 2 3 (Carnot) (Clausius) 2 5. 1 ( ) ( ) ( ) ( ) 5. 1. 1 (system) 1) 70 5. (isolated system) ( ) E N (closed system) N T (open system) (homogeneous) (heterogeneous) (phase) (phase boundary) (grain)
More information現代物理化学 2-1(9)16.ppt
--- S A, G U S S ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r S -- ds = d 'Q r / ΔS = S S = ds =,r,r d 'Q r r d Q r e = P e = P ΔS d 'Q / e (d'q / e ) --3,e Q W Q (> 0),e e ΔU = Q + W = (Q + Q ) + W = 0
More information6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1
6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) (e) Γ (6.2) : Γ B A R (reversible) 6-1 (e) = Clausius 0 = B A: Γ B A: Γ d Q A + d Q (e) B: R d Q + S(A) S(B) (6.3) (e) // 6.2 B A: Γ d Q S(B) S(A) = S (6.4) (e) Γ (6.5)
More informationP F ext 1: F ext P F ext (Count Rumford, ) H 2 O H 2 O 2 F ext F ext N 2 O 2 2
1 1 2 2 2 1 1 P F ext 1: F ext P F ext (Count Rumford, 1753 1814) 0 100 H 2 O H 2 O 2 F ext F ext N 2 O 2 2 P F S F = P S (1) ( 1 ) F ext x W ext W ext = F ext x (2) F ext P S W ext = P S x (3) S x V V
More informationall.dvi
I 1 Density Matrix 1.1 ( (Observable) Ô :ensemble ensemble average) Ô en =Tr ˆρ en Ô ˆρ en Tr  n, n =, 1,, Tr  = n n  n Tr  I w j j ( j =, 1,, ) ˆρ en j w j j ˆρ en = j w j j j Ô en = j w j j Ô j emsemble
More information/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat
/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiation and the Continuing Failure of the Bilinear Formalism,
More information現代物理化学 1-1(4)16.ppt
(pdf) pdf pdf http://www1.doshisha.ac.jp/~bukka/lecture/index.html http://www.doshisha.ac.jp/ Duet -1-1-1 2-a. 1-1-2 EU E = K E + P E + U ΔE K E = 0P E ΔE = ΔU U U = εn ΔU ΔU = Q + W, du = d 'Q + d 'W
More information6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2
1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a
More information30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
More informationUntitled
II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
More informationiBookBob:Users:bob:Documents:CurrentData:flMŠÍ…e…L…X…g:Statistics.dvi
4 4 9............................................... 3.3......................... 4.4................. 5.5............................ 7 9..................... 9.............................3................................4..........................5.............................6...........................
More informationi 18 2H 2 + O 2 2H 2 + ( ) 3K
i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................
More information( ) ( )
20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
More informationMicrosoft Word - 11問題表紙(選択).docx
A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx
More information20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................
More informationmaster.dvi
4 Maxwell- Boltzmann N 1 4.1 T R R 5 R (Heat Reservor) S E R 20 E 4.2 E E R E t = E + E R E R Ω R (E R ) S R (E R ) Ω R (E R ) = exp[s R (E R )/k] E, E E, E E t E E t E exps R (E t E) exp S R (E t E )
More information0201
2018 10 17 2019 9 19 SI J cal 1mL 1ºC 1999 cal nutrition facts label calories cal kcal 1 cal = 4.184 J heat capacity 1 K 1 J K 1 mol molar heat capacity J K mol (specific heat specific heat capacity) 1
More informationII ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
More information2013 25 9 i 1 1 1.1................................... 1 1.2........................... 2 1.3..................................... 3 1.4..................................... 4 2 6 2.1.................................
More information1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h
IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r
More information理想気体ideal gasの熱力学的基本関係式
the equipartition law of energy ( )kt k Boltzmann constant 5 Longman Dictionary of Physics (/)kt q Bq (/)kt equipartition law of energy mol (3/)kT (3/)RT (3/)R (5/)R 3R kt - equipartition of energy The
More information5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1
4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1
More informationδf = δn I [ ( FI (N I ) N I ) T,V δn I [ ( FI N I ( ) F N T,V ( ) FII (N N I ) + N I ) ( ) FII T,V N II T,V T,V ] ] = 0 = 0 (8.2) = µ (8.3) G
8 ( ) 8. 1 ( ) F F = F I (N I, T, V I ) + F II (N II, T, V II ) (8.1) F δf = δn I [ ( FI (N I ) N I 8. 1 111 ) T,V δn I [ ( FI N I ( ) F N T,V ( ) FII (N N I ) + N I ) ( ) FII T,V N II T,V T,V ] ] = 0
More informationOHP.dvi
t 0, X X t x t 0 t u u = x X (1) t t 0 u X x O 1 1 t 0 =0 X X +dx t x(x,t) x(x +dx,t). dx dx = x(x +dx,t) x(x,t) (2) dx, dx = F dx (3). F (deformation gradient tensor) t F t 0 dx dx X x O 2 2 F. (det F
More informationKENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
More informationI-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co
16 I ( ) (1) I-1 I-2 I-3 (2) I-1 ( ) (100 ) 2l x x = 0 y t y(x, t) y(±l, t) = 0 m T g y(x, t) l y(x, t) c = 2 y(x, t) c 2 2 y(x, t) = g (A) t 2 x 2 T/m (1) y 0 (x) y 0 (x) = g c 2 (l2 x 2 ) (B) (2) (1)
More information.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T
NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
More information変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
More informationTEL URL B HP A4 pdf pdf ( )
EL 484-2295 482-1799 URL http://taruibunkalacoocanjp E-mail bkit45@rinkuzaqnejp 48 1988 5 1989 B4 211 HP 4 pf pf () 4 24 2 26 28 3 219/4/28 2:44 7 28 5 11 1956 21 1 1961 8 6 1 () 4 () 1 1 374 372 373 ()
More informationi
mailto: tomita@physhkyoto-uacjp 2000 3 2000 8 2001 7 2002 9 2003 9 2000 2002 9 i 1 1 11 { : : : : : : : : : : : : : : : : : : : : : : : : 1 12 : : : : : : : : : : : : : : : : : : : : : : : : : : : : 3
More informationDecember 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
More information21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........
More informationNo δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
More information( ) ,
II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00
More information構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
More informationII
II 28 5 31 3 I 5 1 7 1.1.......................... 7 1.1.1 ( )................ 7 1.1.2........................ 12 1.1.3................... 13 1.1.4 ( )................. 14 1.1.5................... 15
More information. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n
003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........
More informationThe Physics of Atmospheres CAPTER :
The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(
More information5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E
5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 2, S 1 N 1 = S 2 N 2 2 (chemical potential) µ S N
More information(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x
Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k
More informationC A B A = B (conservation of heat) (thermal equilibrium) Advanced m A [g], c A [J/(g K)] T A [K] A m B [g], c B [J/(g K)] T B [K] B T E [K] T
27 (2015 ) 5 5.1 1 1 0 C 100 C 100 C 0.01 0.006 3 (phase diagram) (triple point) 1: Topic 0.64 88 C 1.5 57 1.5 115 C A B A = B (conservation of heat) (thermal equilibrium) Advanced m A [g], c A [J/(g K)]
More informationt = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
More information“‡Łª”qŠn›tflMŠÍ−w
21 7 2 i Flory-Huggins van t Hoff ( ) ii iii 1 1 1.1.............................. 1 1.2........................ 4 1.3........................ 6 1.4......................... 8 1.5 ( ).......................
More information44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
More informationi ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
More information9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P
9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
More informationhousoku.dvi
: 1 :, 2002 07 14 1 3 11 3 12 : 3 13 : 5 14 6 141 6 142 8 143 8 144 8 145 9 2 10 21 10 : 2 22 11 23 11 24 11 A : 12 B 14 C 15 ( ) ( ) ( ),, (,, ) 2,, : 1 3 1 11 (t),, ρv 0 (1) t (t) ( A ) ( ) ρ (t) t +ρ
More informationall.dvi
72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G
More information2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =
1 1 1.1 I R 1.1.1 c : I R 2 (i) c C (ii) t I c (t) (0, 0) c (t) c(i) c c(t) 1.1.2 (1) (2) (3) (1) r > 0 c : R R 2 : t (r cos t, r sin t) (2) C f : I R c : I R 2 : t (t, f(t)) (3) y = x c : R R 2 : t (t,
More informationi Γ
018 4 10 i 1 1.1.............................. 1.......................... 3 1.3............................ 6 1.4............................ 7 8.1 Γ.................................... 8.......................
More information20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................
More information1 1 x y = y(x) y, y,..., y (n) : n y F (x, y, y,..., y (n) ) = 0 n F (x, y, y ) = 0 1 y(x) y y = G(x, y) y, y y + p(x)y = q(x) 1 p(x) q(
1 1 y = y() y, y,..., y (n) : n y F (, y, y,..., y (n) ) = 0 n F (, y, y ) = 0 1 y() 1.1 1 y y = G(, y) 1.1.1 1 y, y y + p()y = q() 1 p() q() (q() = 0) y + p()y = 0 y y + py = 0 y y = p (log y) = p log
More informationgr09.dvi
.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
More informationH 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [
3 3. 3.. H H = H + V (t), V (t) = gµ B α B e e iωt i t Ψ(t) = [H + V (t)]ψ(t) Φ(t) Ψ(t) = e iht Φ(t) H e iht Φ(t) + ie iht t Φ(t) = [H + V (t)]e iht Φ(t) Φ(t) i t Φ(t) = V H(t)Φ(t), V H (t) = e iht V (t)e
More information18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α
18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t
More information2007 5 iii 1 1 1.1.................... 1 2 5 2.1 (shear stress) (shear strain)...... 5 2.1.1...................... 6 2.1.2.................... 6 2.2....................... 7 2.2.1........................
More informationMaxwell
I 2018 12 13 0 4 1 6 1.1............................ 6 1.2 Maxwell......................... 8 1.3.......................... 9 1.4..................... 11 1.5..................... 12 2 13 2.1...................
More informationchap9.dvi
9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =
More information24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x
24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),
More informationD v D F v/d F v D F η v D (3.2) (a) F=0 (b) v=const. D F v Newtonian fluid σ ė σ = ηė (2.2) ė kl σ ij = D ijkl ė kl D ijkl (2.14) ė ij (3.3) µ η visco
post glacial rebound 3.1 Viscosity and Newtonian fluid f i = kx i σ ij e kl ideal fluid (1.9) irreversible process e ij u k strain rate tensor (3.1) v i u i / t e ij v F 23 D v D F v/d F v D F η v D (3.2)
More information201711grade1ouyou.pdf
2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2
More information,,,17,,, ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,,
14 5 1 ,,,17,,,194 1 4 ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,, 1 4 1.1........................................ 4 5.1........................................ 5.........................................
More information医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.
医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987
More information25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3
More informationN/m f x x L dl U 1 du = T ds pdv + fdl (2.1)
23 2 2.1 10 5 6 N/m 2 2.1.1 f x x L dl U 1 du = T ds pdv + fdl (2.1) 24 2 dv = 0 dl ( ) U f = T L p,t ( ) S L p,t (2.2) 2 ( ) ( ) S f = L T p,t p,l (2.3) ( ) U f = L p,t + T ( ) f T p,l (2.4) 1 f e ( U/
More information) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
More informationS I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d
S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
More informationB 1 B.1.......................... 1 B.1.1................. 1 B.1.2................. 2 B.2........................... 5 B.2.1.......................... 5 B.2.2.................. 6 B.2.3..................
More informationuntitled
1 25/5/3-6/3 1 1 1.1.................................. 1 1.2.................................. 4 2 5 2.1.............................. 5 2.2.............................. 6 3 Black Scholes 7 3.1 BS............................
More informationall.dvi
38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
More information読めば必ずわかる 分散分析の基礎 第2版
2 2003 12 5 ( ) ( ) 2 I 3 1 3 2 2? 6 3 11 4? 12 II 14 5 15 6 16 7 17 8 19 9 21 10 22 11 F 25 12 : 1 26 3 I 1 17 11 x 1, x 2,, x n x( ) x = 1 n n i=1 x i 12 (SD ) x 1, x 2,, x n s 2 s 2 = 1 n n (x i x)
More informationC (q, p) (1)(2) C (Q, P ) ( Qi (q, p) P i (q, p) dq j + Q ) i(q, p) dp j P i dq i (5) q j p j C i,j1 (q,p) C D C (Q,P) D C Phase Space (1)(2) C p i dq
7 2003 6 26 ( ) 5 5.1 F K 0 (q 1,,q N,p 1,,p N ) (Q 1,,Q N,P 1,,P N ) Q i Q i (q, p). (1) P i P i (q, p), (2) (p i dq i P i dq i )df. (3) [ ] Q αq + βp, P γq + δp α, β, γ, δ [ ] PdQ pdq (γq + δp)(αdq +
More informationS I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
More informationi
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
More information~nabe/lecture/index.html 2
2001 12 13 1 http://www.sml.k.u-tokyo.ac.jp/ ~nabe/lecture/index.html nabe@sml.k.u-tokyo.ac.jp 2 1. 10/ 4 2. 10/11 3. 10/18 1 4. 10/25 2 5. 11/ 1 6. 11/ 8 7. 11/15 8. 11/22 9. 11/29 10. 12/ 6 1 11. 12/13
More informationK E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................
More information2 1 x 2 x 2 = RT 3πηaN A t (1.2) R/N A N A N A = N A m n(z) = n exp ( ) m gz k B T (1.3) z n z = m = m ρgv k B = erg K 1 R =
1 1 1.1 1827 *1 195 *2 x 2 t x 2 = 2Dt D RT D = RT N A 1 6πaη (1.1) D N A a η 198 *3 ( a =.212µ) *1 Robert Brown (1773-1858. *2 Albert Einstein (1879-1955 *3 Jean Baptiste Perrin (187-1942 2 1 x 2 x 2
More informationuntitled
0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.
More informationn (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz
1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x
More informationNote.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
More informationTOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
More informationx,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2 = ( 2, b 2, c 2 ) v
12 -- 1 4 2009 9 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8 4-9 4-10 c 2011 1/(13) 4--1 2009 9 3 x,, z v = (, b, c) v v 2 + b 2 + c 2 x,, z 1 i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) v 1 = ( 1, b 1, c 1 ), v 2
More informationAHPを用いた大相撲の新しい番付編成
5304050 2008/2/15 1 2008/2/15 2 42 2008/2/15 3 2008/2/15 4 195 2008/2/15 5 2008/2/15 6 i j ij >1 ij ij1/>1 i j i 1 ji 1/ j ij 2008/2/15 7 1 =2.01/=0.5 =1.51/=0.67 2008/2/15 8 1 2008/2/15 9 () u ) i i i
More information7-12.dvi
26 12 1 23. xyz ϕ f(x, y, z) Φ F (x, y, z) = F (x, y, z) G(x, y, z) rot(grad ϕ) rot(grad f) H(x, y, z) div(rot Φ) div(rot F ) (x, y, z) rot(grad f) = rot f x f y f z = (f z ) y (f y ) z (f x ) z (f z )
More informationtomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.
tomocci 18 7 5...,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. M F (M), X(F (M)).. T M p e i = e µ i µ. a a = a i
More informationDynkin Serre Weyl
Dynkin Naoya Enomoto 2003.3. paper Dynkin Introduction Dynkin Lie Lie paper 1 0 Introduction 3 I ( ) Lie Dynkin 4 1 ( ) Lie 4 1.1 Lie ( )................................ 4 1.2 Killing form...........................................
More information(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t
6 6.1 6.1 (1 Z ( X = e Z, Y = Im Z ( Z = X + iy, i = 1 (2 Z E[ e Z ] < E[ Im Z ] < Z E[Z] = E[e Z] + ie[im Z] 6.2 Z E[Z] E[ Z ] : E[ Z ] < e Z Z, Im Z Z E[Z] α = E[Z], Z = Z Z 1 {Z } E[Z] = α = α [ α ]
More informationBayesfiI‡É“ÅfiK‡È−w‘K‡Ì‡½‡ß‡ÌChow-Liu…A…‰…S…−…Y…•
1 / 21 Kruscal V : w i,j R: w i,j = w j,i i j Kruscal (w i,j 0 ) 1 E {{i, j} i, j V, i i} 2 E {} 3 while(e = ϕ) for w i,j {i, j} E 1 E E\{i, j} 2 G = (V, E {i, j}) = E E {i, j} G {i,j} E w i,j 2 / 21 w
More informationn ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
More informationC : q i (t) C : q i (t) q i (t) q i(t) q i(t) q i (t)+δq i (t) (2) δq i (t) δq i (t) C, C δq i (t 0 )0, δq i (t 1 ) 0 (3) δs S[C ] S[C] t1 t 0 t1 t 0
1 2003 4 24 ( ) 1 1.1 q i (i 1,,N) N [ ] t t 0 q i (t 0 )q 0 i t 1 q i (t 1 )q 1 i t 0 t t 1 t t 0 q 0 i t 1 q 1 i S[q(t)] t1 t 0 L(q(t), q(t),t)dt (1) S[q(t)] L(q(t), q(t),t) q 1.,q N q 1,, q N t C :
More informationIntroduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math))
Introduction to Numerical Analysis of Differential Equations Naoya Enomoto (Kyoto.univ.Dept.Science(math)) 2001 1 e-mail:s00x0427@ip.media.kyoto-u.ac.jp 1 1 Van der Pol 1 1 2 2 Bergers 2 KdV 2 1 5 1.1........................................
More informationb3e2003.dvi
15 II 5 5.1 (1) p, q p = (x + 2y, xy, 1), q = (x 2 + 3y 2, xyz, ) (i) p rotq (ii) p gradq D (2) a, b rot(a b) div [11, p.75] (3) (i) f f grad f = 1 2 grad( f 2) (ii) f f gradf 1 2 grad ( f 2) rotf 5.2
More information