CAE ( 6 ) 2 k ε LES (Large Eddy Simulation) DNS (Direct Numerical Simulation) Pr Pr 1 1 (Pr = 0.7) Pr 1 Pr 1

Size: px
Start display at page:

Download "CAE ( 6 ) 2 k ε LES (Large Eddy Simulation) DNS (Direct Numerical Simulation) Pr Pr 1 1 (Pr = 0.7) Pr 1 Pr 1"

Transcription

1 CAE ( 6 ) 2 k ε LES (Large Eddy Simulation) DNS (Diret Numerial Simulation) Pr Pr 1 1 (Pr =.7) Pr 1 Pr 1

2 T u Pr < 1 Pr ~ 1 Pr > 1 Tw 1: Pr Tmi y x Tmo TFD HFD x 2: (Hydrodynamially Fully Developed = HFD) Thermally Fully Developed = TFD 3 3 (q w = λ( T/ y) y= = onst.) (T w = onst.) 3 2

3 u (y) Tmi Tw Tw T (x, y) 3: (1) (2) x y u, v u x + v y = ( p 2 ) x + ν u x + 2 u 2 y 2 ( y = 1 p 2 ) ρ y + ν v x + 2 v 2 y 2 u u x + v u y = 1 ρ u v x + v v (1) (2) ρ, ν [kg/m 3 ] [m 2 /s] (3) [4] u T x + v T y = α ( 2 T x + 2 T 2 y 2 ) α [m 2 /s] u(x, y), v(x, y) T (x, y) T (1), (2) (3) 3

4 y x u 2H 4: y = 2H y = ±H u x =, v = u x v (4) (4) (2) = 1 p ρ x + ν 2 u y 2 = 1 (5) p ρ y u(x, y) u(y), p(x, y) p(x) (4) ν d2 u dy = 1 dp 2 ρ dx (6) (6) y 2 u = 1 ( ) dp y 2 + C 1 y + C 2 (7) 2µ dx y = ±H u = (7) = 1 ( ) dp H 2 ± C 1 H + C 2 (8) 2µ dx C 1, C 2 C 1 = C 2 = 1 2µ ( ) dp H 2 dx (9) 4

5 (7) u(y) u = 1 ( ) dp (y 2 H 2 ) = 1 ( dp ) (H 2 y 2 ) (1) 2µ dx 2µ dx y = (dp/dx) > µ y = u max u max = u y= = 1 2µ ( dp ) H 2 (11) dx u max (1) ( ) u = u max 1 y2 H 2 (12) u m u m 1 H udy 2H H = u [ ] H max y y3 = 2 2H 3H 2 3 u max (13) H u max u m T m T (y) y T m = H H H u(y)t (y)dy H u(y)dy = H H u(y)t (y)dy 2u m H (14) u m (14) ρ p (14) u(y) T av = H H T (y)dy 2.4 Θ Θ(x, y) = T w(x) T (x, y) T w (x) T m (x) (15) 5

6 T w, T m Θ(x, y) Θ(y) (15) y Θ(x, y) x = (15) x [( Θ x = 1 dtw (T w T m ) 2 dx T ) ( dtw (T w T m ) (T w T ) x dx dt )] m dx = (16) (17) (T w T m ) T x = (T w T m ) dt w dx (T w T ) dt w dx + (T w T ) dt m dx = (T T m ) dt w dx + (T w T ) dt m dx x T/ x T x = T T ( ) m dtw + T ( ) w T dtm (19) T w T m dx T w T m dx (19) h q w h(t w T ) T T w T m h (18) q w h(t w T m ) (2) q w q w = λ T = λ T λ (21) y y y= H y=h T (x, y) 6

7 dq w dx = h (22) (2) dt w dx dt m dx = (22) (23) dt w dx = dt m dx (19) T/ x x T x = dt w dx = dt m dx (24) = F untion(x) (25) T/ x T w T m x (25) 5(a) T m W x x + dx x ṁ p T m (x + dx) = ṁ p T m (x) + 2q w W dx (26) ṁ, p [kg/s] [J/(kg K)] (26) ṁ = ρu m W 2H dt m dx = q w ρ p u m H (27) 5(a) q w = onst. dt m /dx = onst. dt w dx = (19) T x = T w T dt m T w T m dx (28) = F untion(x, y) (29) T/ x T/ x x y (29) 5(b) 7

8 TFD Tw Tm = onst. Tw Tm Tmi (a) x Tw Tw =onst. Tm Tmi TFD (Thermally Fully Developed) (b) x 5: x 6 6 T w T T w (25) (3) 2 T x 2 T 2 y 2 (3) v = (3) u T x = T α 2 y 2 (31) (25) T/ x = dt m /dx (31) 2.6 T/ x 8

9 Tw Tw y qw x qw (a) (qw = onst.) Tw Tw y qw x qw (b) (Tw = onst.) 6: 2 T/ x 2 = (31) (31) (12), (13) 2 T = 3u ( ) ( ) m dtm 1 y2 (32) y 2 2α dx H 2 (32) A x A 3u ( ) m dtm 2α dx (33) y ( ) T = A y y3 + C y 3H 2 1 (34) ( ) y 2 T = A 2 y4 + C 12H 2 1 y + C 2 (35) C 1, C 2 y = T y = y = H T = T w (36) 1 y = 2 y = H T T w (36) (35) 9

10 C 1 = T w = A ( ) H 2 2 H4 12H 2 + C 2 = A 5H C 2 T (x, y) ( ) y 2 T (x, y) = A 2 y4 12H 5H2 + T 2 w (x) 12 [ ( = T w (x) AH2 )4 ( ) ] y y H H q w q w = λ T = 2λAH y y=h 3 (37) (38) (39) T m (x) (14) T m (x) = 1 H u T dy 2u m H H ( = 1 H 3u m 2u m H H 2 1 y2 H 2 ) { T w (x) AH2 12 [ ( )4 ( ) ]} y y dy (4) H H (4) η y/h η dy = Hdη (4) T m (x) = = 3T w(x) 4 ( 1 η 2 ) [ T w (x) AH ( 1 η 2 ) dη + AH2 16 ( η 4 6η )] dη 1 1 (η 2 1) ( η 4 6η ) dη = T w (x) 34AH2 15 (2), (39), (41) h h = q w T w T m = 2λAH/3 34AH 2 /15 = 35 λ 17 H (41) (42) Nu 2H Nu h (2H) λ = 35λ 17H 2H λ = 7 17 = (43) 1

11 1 2 Nu 1 (HFD, TFD) x / (D Re Pr) 7: (W ) D h D h = 4 = lim 2W H 4 W 2(W + 2H) = 4H (44) Nu = 8.24 Nu Nu = x D Gz (Re Pr D)/x Gz 7 Gz 1 >

12 Inlet y x Flow 2H 8: 3. Nu = u, v, p, T u = u, v = v, p = p, θ = (T T in) (45) U in U in ρuin 2 T in U in, T in x, y t x = x 2H, y = y 2H, t = t (2H/U in ) (46) u u u + u + v t x y = p x + 1 ( 2 u Re x + 2 u ) 2 y 2 v v v + u + v t x y = p y + 1 ( 2 v Re x + 2 v ) 2 y 2 θ θ θ + u + v t x y = 1 RePr ( 2 ) θ x + 2 θ 2 y 2 (47) (48) Re U in 2H/ν, Pr = ν/α (48) Re Pr Pe 12

13 3.2 q w q w = λ T ( ) Tin θ = λ y y= 2H y y = (49) λ [W/(m K)] (49) q w ( ) 2H qw = q w (5) λt in qw qw = θ y = θ i, j=1 θ i, j= (51) y y = (51) θ(i, ) θ(i, ) = θ(i, 1) + q w y (52) [ ] θ(i, N y + 1) q w h q w = h(t w T m ) (53) q w = Nu(θ w θ m ) (54) Nu = h 2H/λ x x Nu Nu m [ ] (54) (54) Nu = q w θ w θ m x θ w θ m (55) θ w = θ i, j= + θ i, j=1 2 (56) 13

14 (14) u θ (i, j) u θ y θ m [ ] θ m θ m = 1 1 u θdy u dy = 1 u θdy (57) 3.3 Nu = % y x SMAC (original version by Dr. T. Ushijima) Simplified Marker and Cell method Solving Heat Transfer in flow between parallel walls. impliit double preision(a-h,o-z) n parameter(n=2, nx=n*1, ny=n) p: u, v: phi: divup: up, vp: psi: the: (e.g. ) dimension p(:nx+1,:ny+1) +,u(:nx,:ny+1),v(:nx+1,:ny) +,phi(:nx+1,:ny+1),divup(1:nx,1:ny) +,up(:nx,:ny+1),vp(:nx+1,:ny) +,psi(:nx,:ny+1) +,the(:nx+1,:ny+1),the(:nx+1,:ny+1) loop loop=2 re re=5.d pr ( ) pr=.7d q_w=1.d 14

15 dx(=dy) dy=1.d/dble(n) dx=dy dx=dy*2.d dt dt=.2d dt=min(dt,.25*dx) ( ) dt=min(dt,.2*re*dx*dx) write(6,*) dt =,dt ddx=1.d/dx ddy=1.d/dy ddx2=ddx*ddx ddy2=ddy*ddy ddt=1.d/dt initial ondition iont=1 iont= if (iont.eq.1) then open(unit=9,file= fort.21 +,form= unformatted, status= unknown ) read(9) u,v,p lose(9) else do 131 j=,ny do 132 i=,nx u(i,j)=1.d v(i,j)=.d p(i,j)=.d the(i,j)=.d 132 ontinue 131 ontinue end if do 133 i=,nx p(i,ny+1)=.d u(i,ny+1)=.d the(i,ny+1)=.d 133 ontinue do 134 j=,ny p(nx+1,j)=.d v(nx+1,j)=.d the(nx+1,j)=.d 134 ontinue un=.d uw=1.d us=.d ue=.d vn=.d vw=.d vs=.d 15

16 ve=.d do 1 it=1,loop boundary ondition do 135 j=,ny right wall (east) or outlet v(nx+1,j)=v(nx,j) u(nx,j)=u(nx-1,j) p(nx+1,j)=. the(nx+1,j)=2.*the(nx,j)-the(nx-1,j) left wall (west) or inlet v(,j)=vw u(,j)=uw p(,j)=p(1,j) the(,j)=.d 135 ontinue u(,ny/2+1)=uw do 136 i=,nx lower wall (south) u(i,)=2.*us-u(i,1) v(i,)=vs p(i,)=p(i,1) the(i,)=the(i,1)+q_w*dy!heat flux onstant upper wall (north) u(i,ny+1)=2.*un-u(i,ny) v(i,ny)=vn p(i,ny+1)=p(i,ny) 136 ontinue the(i,ny+1)=the(i,ny)+q_w*dy!heat flux onstant do 81 j=1,ny do 82 i=1,nx nvt=(ddx*((the(i,j)+the(i-1,j))*u(i-1,j) + -(the(i,j)+the(i+1,j))*u(i,j)) + +ddy*((the(i,j)+the(i,j-1))*v(i,j-1) + -(the(i,j)+the(i,j+1))*v(i,j)))/2.d dift=(ddx2*(the(i+1,j)-2.*the(i,j)+the(i-1,j)) + +ddy2*(the(i,j+1)-2.*the(i,j)+the(i,j-1)))/(pr*re) the(i,j)=the(i,j)+dt*(nvt+dift) 82 ontinue 81 ontinue do 83 j=1,ny do 84 i=1,nx the(i,j)=the(i,j) 84 ontinue 83 ontinue 16

17 preditor step for u_ij (up-u)/dt=-dp/dx-duu/dx-duv/dy+(nabla)ˆ2 u write(6,*) up do 125 j=1,ny do 126 i=1,nx-1 vij=(v(i,j)+v(i,j-1)+v(i+1,j)+v(i+1,j+1))*.25 nvu=.5*ddx*(u(i,j)*(u(i+1,j)-u(i-1,j)) + -abs(u(i,j))*(u(i-1,j)-2.*u(i,j)+u(i+1,j))) + +.5*ddy*(vij*(u(i,j+1)-u(i,j-1)) + -abs(vij)*(u(i,j-1)-2.*u(i,j)+u(i,j+1))) nvu: nvu=ddx*((u(i+1,j)+u(i,j))**2 + -(u(i-1,j)+u(i,j))**2)/4.d + +ddy*((u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j)) + -(u(i,j)+u(i,j-1))*(v(i,j-1)+v(i+1,j-1)))/4.d fij=-ddx*(p(i+1,j)-p(i,j))-nvu + +ddx2*(u(i+1,j)-2.d*u(i,j)+u(i-1,j))/re + +ddy2*(u(i,j+1)-2.d*u(i,j)+u(i,j-1))/re up(i,j)=u(i,j)+dt*fij 126 ontinue write(6,63) (up(i,j),i=1,nx-1) 125 ontinue do 124 j=1,ny up(,j)=uw up(nx,j)=up(nx-1,j) 124 ontinue up(,ny/2+1)=uw*1.1 do 224 i=1,nx-1 up(i,)=2.*us-up(i,1) up(i,ny+1)=2.*un-up(i,ny) 224 ontinue for v_ij (vp-v)/dt=-dp/dy-duv/dx-dvv/dy+(nabla)ˆ2 v write(6,*) vp do 122 j=1,ny-1 do 123 i=1,nx uij=.25*(u(i,j)+u(i+1,j)+u(i,j+1)+u(i+1,j+1)) nvv=.5*ddx*(uij*(v(i+1,j)-v(i-1,j)) + -abs(uij)*(v(i-1,j)-2.*v(i,j)+v(i+1,j))) + +.5*ddy*(v(i,j)*(v(i,j+1)-v(i,j-1)) + -abs(v(i,j))*(v(i,j-1)-2.*v(i,j)+v(i,j+1))) nvv: nvv=ddx*((u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j)) + -(u(i-1,j+1)+u(i-1,j))*(v(i-1,j)+v(i,j)))/4.d + +ddy*((v(i,j+1)+v(i,j))**2 + -(v(i,j)+v(i,j-1))**2)/4.d gij=-ddy*(p(i,j+1)-p(i,j))-nvv + +ddx2*(v(i+1,j)-2.d*v(i,j)+v(i-1,j))/re + +ddy2*(v(i,j+1)-2.d*v(i,j)+v(i,j-1))/re vp(i,j)=v(i,j)+dt*gij 123 ontinue write(6,63) (vp(i,j),i=1,nx) 122 ontinue 17

18 do 121 i=1,nx vp(i,)=vs vp(i,ny)=vn 121 ontinue do 221 j=1,ny-1 vp(,j)=vw vp(nx+1,j)=vp(nx,j) 221 ontinue evaluate ontinuity write(6,*) evaluate ontinuity i= div=. do 112 j=1,ny do 111 i=1,nx divup(i,j)=ddx*(up(i,j)-up(i-1,j)) + +ddy*(vp(i,j)-vp(i,j-1)) div=div+divup(i,j)**2 i=i ontinue write(6,63) (divup(i,j),i=1,nx) 112 ontinue write(6,*) sqrt(div/dble(i)) solve the poisson equation (nabla)2 p=(nabla)up/dt by SOR write(6,*) solve the poisson equation for pressure initialisation do 17 i=,nx+1 do 18 j=,ny+1 phi(i,j)=.d 18 ontinue 17 ontinue eps=1.d-6 maxitr maxitr=nx*ny C C maxitr=nx*ny/1 alpha= alpha=1.7 do 1 iter=1,maxitr error=.d do 11 j=1,ny do 12 i=1,nx rhs=ddt*divup(i,j) resid=ddx2*(phi(i-1,j)-2.d*phi(i,j)+phi(i+1,j)) + +ddy2*(phi(i,j-1)-2.d*phi(i,j)+phi(i,j+1)) + -rhs den=2.d*(ddx2+ddy2) dphi=alpha*resid/den error=max(abs(dphi),error) phi(i,j)=phi(i,j)+dphi 12 ontinue 11 ontinue do 13 j=1,ny 18

19 phi(,j)=phi(1,j) phi(nx+1,j)=. 13 ontinue do 14 i=1,nx phi(i,)=phi(i,1) phi(i,ny+1)=phi(i,ny) 14 ontinue if (error.lt.eps) goto ontinue 998 ontinue write(6,*) iter =, iter, it, error pause if (iter.ge.maxitr) write(6,*) maximum iteration exeeded! orretor step do 15 j=1,ny do 151 i=1,nx-1 u(i,j)=up(i,j)-dt*ddx*(phi(i+1,j)-phi(i,j)) 151 ontinue 15 ontinue do 152 j=1,ny-1 do 153 i=1,nx v(i,j)=vp(i,j)-dt*ddy*(phi(i,j+1)-phi(i,j)) 153 ontinue 152 ontinue do 16 j=1,ny do 161 i=1,nx p(i,j)=p(i,j)+phi(i,j) 161 ontinue 16 ontinue hek the ontinuity for n+1 th step write(6,*) hek the ontinuty for n+1 th step i= div=. do 155 j=1,ny do 156 i=1,nx divup(i,j)=ddx*(u(i,j)-u(i-1,j))+ddy*(v(i,j)-v(i,j-1)) i=i+1 div=div+divup(i,j)**2 156 ontinue write(6,63)(divup(i,j),i=1,nx) 63 format(2(1x,e1.2)) 155 ontinue write(6,*) sqrt(div/dble(i)) ********************************************* i=nx ********************************************* in=nx tm=.d um=.d do 17 j=1,ny tm=tm+.5d*(u(in,j)+u(in-1,j))*the(in,j)*dy 19

20 um=um+.5d*(u(in,j)+u(in-1,j))*dy 17 ontinue tm=tm/um tw=.5d*(the(in,)+the(in,1)) s_nu=q_w/(tw-tm) write(6,*) s_nu 1 ontinue output do 491 i=,nx psi(i,)=.d do 492 j=1,ny+1 psi(i,j)=psi(i,j-1)+.5*dy*(u(i,j-1)+u(i,j)) 492 ontinue 491 ontinue do 51 j=1,ny do 52 i=1,nx x=dx*(dble(i)-.5) y=dy*(dble(j)-.5) u=.5*(u(i,j)+u(i-1,j)) v=.5*(v(i,j)+v(i,j-1)) p=p(i,j) divu=divup(i,j) psi=.5*(psi(i,j)+psi(i-1,j)) u,v P write(1,699) x,y,u,v,p,divu,psi,the(i,j) 699 format (8(1X,E12.5)) 52 ontinue write(1,*) 51 ontinue do 53 j=1,ny-1 do 54 i=1,nx-1 x=dx*dble(i) y=dy*dble(j) u=.5*(u(i,j)+u(i,j+1)) v=.5*(v(i,j)+v(i+1,j)) omega=dx*(v(i+1,j)-v(i,j))-dy*(u(i,j+1)-u(i,j)) psi=.5*(psi(i,j)+psi(i,j+1)) C x y u,v write(11,699) x,y,u,v,omega,psi 54 ontinue write(11,699) 53 ontinue write(21) u,v,p end 2

21 y/(2h)=.5 u / Uin 1..5 y/(2h)=.1 y/(2h)= x / (2H) : (T - Tin)/Tin.6.4 y/(2h)=.5 y/(2h)= x / (2H) : 5(a) 21

22 y / (2H) x / (2H) 1 11: u y / (2H) x / (2H) : θ 1., (1995), 2. JSME, (25), 3. White, F. M. Heat and Mass Transfer, (1988), Addison-Wesley. 4., , (24.1), pp ( 22

Untitled

Untitled 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 information

KENZOU Karman) x

KENZOU Karman) x KENZO 8 8 31 8 1 3 4 5 6 Karman) 7 3 8 x 8 1 1.1.............................. 3 1............................................. 5 1.3................................... 5 1.4 /.........................

More information

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) du (L) = f (9.3) dx (9.) P

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) 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 information

D 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

D 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 information

20 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 information

untitled

untitled 20010916 22;1017;23;20020108;15;20; 1 N = {1, 2, } Z + = {0, 1, 2, } Z = {0, ±1, ±2, } Q = { p p Z, q N} R = { lim a q n n a n Q, n N; sup a n < } R + = {x R x 0} n = {a + b 1 a, b R} u, v 1 R 2 2 R 3

More information

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi) 0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

More information

n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x

n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

21 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 information

25 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 information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

note1.dvi

note1.dvi (1) 1996 11 7 1 (1) 1. 1 dx dy d x τ xx x x, stress x + dx x τ xx x+dx dyd x x τ xx x dyd y τ xx x τ xx x+dx d dx y x dy 1. dx dy d x τ xy x τ x ρdxdyd x dx dy d ρdxdyd u x t = τ xx x+dx dyd τ xx x dyd

More information

Evoltion of onentration by Eler method (Dirihlet) Evoltion of onentration by Eler method (Nemann).2 t n =.4n.2 t n =.4n : t n

Evoltion of onentration by Eler method (Dirihlet) Evoltion of onentration by Eler method (Nemann).2 t n =.4n.2 t n =.4n : t n 5 t = = (, y, z) t (, y, z, t) t = κ (68) κ [, ] (, ) = ( ) A ( /2)2 ep, A =., t =.. (69) 4πκt 4κt = /2 (, t) = for ( =, ) (Dirihlet ondition) (7) = for ( =, ) (Nemann ondition) (7) (68) (, t) = ( ) (

More information

II 2 II

II 2 II II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................

More information

微分積分 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, 初版 1 刷発行時のものです.

微分積分 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, 初版 1 刷発行時のものです. 微分積分 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. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)

More information

PowerPoint プレゼンテーション

PowerPoint プレゼンテーション Nagoya Institute of Tehnology 中部 CAE 懇話会 流体伝熱基礎講座 第 1 回午後 名古屋工業大学大学院 創成シミュレーション工学専攻 後藤俊幸 Nagoya Institute of Tehnology 数値計算の基礎 x=a の近傍での Taylor 展開 Nagoya Institute of Tehnology Nagoya Institute of Tehnology

More information

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10% 1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n

More information

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63> 基礎からの冷凍空調 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/067311 このサンプルページの内容は, 初版 1 刷発行当時のものです. http://www.morikita.co.jp/support. 03-3817-5670FAX 03-3815-8199 i () () Q&A

More information

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

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 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 information

Z: Q: R: C:

Z: Q: R: C: 0 Z: Q: R: C: 3 4 4 4................................ 4 4.................................. 7 5 3 5...................... 3 5......................... 40 5.3 snz) z)........................... 4 6 46 x

More information

橡3章波浪取扱.PDF

橡3章波浪取扱.PDF 2 2-2 * -* *EM EM -1- (2.1) Dally x κ ( F cos θ ) + ( F sin θ ) = ( F F s ) 2.1 y d 2-7 F F Stable wave energy flux d F (2.2) F s = E s gs = 1 8 ρgh 2 s gh = 1 8 ρg(γ h) 2 gh 2.2 η (2.3) ds xx = ρgd d

More information

Title 混合体モデルに基づく圧縮性流体と移動する固体の熱連成計算手法 Author(s) 鳥生, 大祐 ; 牛島, 省 Citation 土木学会論文集 A2( 応用力学 ) = Journal of Japan Civil Engineers, Ser. A2 (2017), 73 Issue

Title 混合体モデルに基づく圧縮性流体と移動する固体の熱連成計算手法 Author(s) 鳥生, 大祐 ; 牛島, 省 Citation 土木学会論文集 A2( 応用力学 ) = Journal of Japan Civil Engineers, Ser. A2 (2017), 73 Issue Title 混合体モデルに基づく圧縮性流体と移動する固体の熱連成計算手法 Author(s) 鳥生, 大祐 ; 牛島, 省 Citation 土木学会論文集 A2( 応用力学 ) = Journal of Japan Civil Engineers, Ser. A2 (2017), 73 Issue Date 2017 URL http://hdl.handle.net/2433/229150 Right

More information

4 2016 3 8 2.,. 2. Arakawa Jacobin., 2 Adams-Bashforth. Re = 80, 90, 100.. h l, h/l, Kármán, h/l 0.28,, h/l.., (2010), 46.2., t = 100 t = 2000 46.2 < Re 46.5. 1 1 4 2 6 2.1............................

More information

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

More information

8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J

8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J 26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK

More information

September 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1

September 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 information

X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2

More information

v er.1/ c /(21)

v er.1/ c /(21) 12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,

More information

1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx

1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx 1 1 1 1 1. U(x, t) U(x, t) + c t x c, κ. (1). κ U(x, t) x. (1) 1, f(x).. U(x, t) U(x, t) + c κ U(x, t), t x x : U(, t) U(1, t) ( x 1), () : U(x, ) f(x). (3) U(x, t). [ U(x, t) Re u k (t) exp(πkx). (4)

More information

Microsoft Word - 演習5_蒸発装置

Microsoft Word - 演習5_蒸発装置 1-4) q T sat T w T S E D.N.B.(Departure from Nucleate Boiling)h(=q/ T sat ) F q B q D 3) 1 5-7) 4) T W [K] T B [K]T BPR [K] B.P.E.Boiling Pressure Rising T B T W T BPR (3.1) Dühring T BPR T W T W T B T

More information

D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j

D = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j 6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..

More information

A

A A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................

More information

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A .. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.

More information

1 (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

1 (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 information

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z

More information

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k 63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5

More information

simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =

simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a = II 6 ishimori@phys.titech.ac.jp 6.. 5.4.. f Rx = f Lx = fx fx + lim = lim x x + x x f c = f x + x < c < x x x + lim x x fx fx x x = lim x x f c = f x x < c < x cosmx cosxdx = {cosm x + cosm + x} dx = [

More information

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

I A A441 : April 15, 2013 Version : 1.1 I   Kawahira, Tomoki TA (Shigehiro, Yoshida ) I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17

More information

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5. A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx i B5 7.8. p89 4. ψ x, tψx, t = ψ R x, t iψ I x, t ψ R x, t + iψ I x, t = ψ R x, t + ψ I x, t p 5.8 π π π F e ix + F e ix + F 3 e 3ix F e ix + F e ix + F 3 e 3ix dx πψ x πψx p39 7. AX = X A [ a b c d x

More information

(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b)

(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 74 Re, bondar laer (Prandtl) Re z ω z = x (5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 76 l V x ) 1/ 1 ( 1 1 1 δ δ = x Re x p V x t V l l (1-1) 1/ 1 δ δ δ δ = x Re p V x t V

More information

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,. 9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,

More information

i

i i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,

More information

1. A0 A B A0 A : A1,...,A5 B : B1,...,B

1. A0 A B A0 A : A1,...,A5 B : B1,...,B 1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A B f : A B 4 (i) f (ii) f (iii) C 2 g, h: C A f g = f h g = h (iv) C 2 g, h: B C g f = h f g = h 4 (1) (i) (iii) (2) (iii) (i) (3) (ii) (iv) (4)

More information

The Physics of Atmospheres CAPTER :

The 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 information

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r 2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2

More information

Microsoft Word - 11問題表紙(選択).docx

Microsoft 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 information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

i 18 2H 2 + O 2 2H 2 + ( ) 3K

i 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

gr09.dvi

gr09.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 information

1 u t = au (finite difference) u t = au Von Neumann

1 u t = au (finite difference) u t = au Von Neumann 1 u t = au 3 1.1 (finite difference)............................. 3 1.2 u t = au.................................. 3 1.3 Von Neumann............... 5 1.4 Von Neumann............... 6 1.5............................

More information

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0, .1.1 Y K L Y = K 1 3 L 3 L K K (K + ) 1 1 3 L 3 K 3 L 3 K 0 (K + K) 1 3 L 3 K 1 3 L 3 lim K 0 K = L (K + K) 1 3 K 1 3 3 lim K 0 K = 1 3 K 3 L 3 z = f(x, y) x y z x-y-z.1 z = e x +xy y 3 x-y ( ) z 0 f(x,

More information

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)

More information

2009 2 26 1 3 1.1.................................................. 3 1.2..................................................... 3 1.3...................................................... 3 1.4.....................................................

More information

1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +

1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 + 1.3 1.4. (pp.14-27) 1 1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 + i2xy x = 1 y (1 + iy) 2 = 1

More information

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II ( ) (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 information

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

More information

(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou

(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou (Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fourier) (Fourier Bessel).. V ρ(x, y, z) V = 4πGρ G :.

More information

3. :, c, ν. 4. Burgers : u t + c u x = ν 2 u x 2, (3), ν. 5. : u t + u u x = ν 2 u x 2, (4), c. 2 u t 2 = c2 2 u x 2, (5) (1) (4), (1 Navier Stokes,.,

3. :, c, ν. 4. Burgers : u t + c u x = ν 2 u x 2, (3), ν. 5. : u t + u u x = ν 2 u x 2, (4), c. 2 u t 2 = c2 2 u x 2, (5) (1) (4), (1 Navier Stokes,., B:,, 2017 12 1, 8, 15, 22 1,.,,,,.,.,,,., 1,. 1. :, ν. 2. : u t = ν 2 u x 2, (1), c. u t + c u x = 0, (2), ( ). 1 3. :, c, ν. 4. Burgers : u t + c u x = ν 2 u x 2, (3), ν. 5. : u t + u u x = ν 2 u x 2,

More information

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

More information

b3e2003.dvi

b3e2003.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

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s ... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z

More information

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + ( IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1

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

(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 information

数値計算:有限要素法

数値計算:有限要素法 ( ) 1 / 61 1 2 3 4 ( ) 2 / 61 ( ) 3 / 61 P(0) P(x) u(x) P(L) f P(0) P(x) P(L) ( ) 4 / 61 L P(x) E(x) A(x) x P(x) P(x) u(x) P(x) u(x) (0 x L) ( ) 5 / 61 u(x) 0 L x ( ) 6 / 61 P(0) P(L) f d dx ( EA du dx

More information

成長機構

成長機構 j im πmkt jin jim π mkt j q out j q im π mkt jin j j q out out π mkt π mkt dn dt πmkt dn v( ) Rmax bf dt πmkt R v ( J J ), J J, J J + + T T, J J m + Q+ / kt Q / kt + ( Q Q+ )/ ktm l / ktm J / J, l Q Q

More information

1 4 1 ( ) ( ) ( ) ( ) () 1 4 2

1 4 1 ( ) ( ) ( ) ( ) () 1 4 2 7 1995, 2017 7 21 1 2 2 3 3 4 4 6 (1).................................... 6 (2)..................................... 6 (3) t................. 9 5 11 (1)......................................... 11 (2)

More information

,,,,., = (),, (1) (4) :,,,, (1),. (2),, =. (3),,. (4),,,,.. (1) (3), (4).,,., () : = , ( ) : = F 1 + F 2 + F 3 + ( ) : = i Fj j=1 2

,,,,., = (),, (1) (4) :,,,, (1),. (2),, =. (3),,. (4),,,,.. (1) (3), (4).,,., () : = , ( ) : = F 1 + F 2 + F 3 + ( ) : = i Fj j=1 2 6 2 6.1 2 2, 2 5.2 R 2, 2 (R 2, B, µ)., R 2,,., 1, 2, 3,., 1, 2, 3,,. () : = 1 + 2 + 3 + (6.1.1).,,, 1 ,,,,., = (),, (1) (4) :,,,, (1),. (2),, =. (3),,. (4),,,,.. (1) (3), (4).,,., () : = 1 + 2 + 3 +,

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 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

all.dvi

all.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 information

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,. 24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)

More information

t = 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

t = 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

数値計算:常微分方程式

数値計算:常微分方程式 ( ) 1 / 82 1 2 3 4 5 6 ( ) 2 / 82 ( ) 3 / 82 C θ l y m O x mg λ ( ) 4 / 82 θ t C J = ml 2 C mgl sin θ θ C J θ = mgl sin θ = θ ( ) 5 / 82 ω = θ J ω = mgl sin θ ω J = ml 2 θ = ω, ω = g l sin θ = θ ω ( )

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

pdf

pdf http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

More information

( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1

( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1 (3.5 3.8) 03032s 2006.7.0 n (n = 0,,...) n n = δ nn n n = I n=0 ψ = n C n n () C n = n ψ α = e 2 α 2 n=0 α, β α n n (2) β α = e 2 α 2 2 β 2 n=0 =0 = e 2 α 2 β n α 2 β 2 n=0 = e 2 α 2 2 β 2 +β α β n α!

More information

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c)   yoshioka/education-09.html pdf 1 2009 1 ( ) ( 40 )+( 60 ) 1 1. 2. Schrödinger 3. (a) (b) (c) http://goofy.phys.nara-wu.ac.jp/ yoshioka/education-09.html pdf 1 1. ( photon) ν λ = c ν (c = 3.0 108 /m : ) ɛ = hν (1) p = hν/c = h/λ (2) h

More information

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2) 3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)

More information

all.dvi

all.dvi 29 4 Green-Lagrange,,.,,,,,,.,,,,,,,,,, E, σ, ε σ = Eε,,.. 4.1? l, l 1 (l 1 l) ε ε = l 1 l l (4.1) F l l 1 F 30 4 Green-Lagrange Δz Δδ γ = Δδ (4.2) Δz π/2 φ γ = π 2 φ (4.3) γ tan γ γ,sin γ γ ( π ) γ tan

More information

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2 3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1

More information

構造と連続体の力学基礎

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 information

= π2 6, ( ) = π 4, ( ). 1 ( ( 5) ) ( 9 1 ( ( ) ) (

= π2 6, ( ) = π 4, ( ). 1 ( ( 5) ) ( 9 1 ( ( ) ) ( + + 3 + 4 +... π 6, ( ) 3 + 5 7 +... π 4, ( ). ( 3 + ( 5) + 7 + ) ( 9 ( ( + 3) 5 + ) ( 7 + 9 + + 3 ) +... log( + ), ) +... π. ) ( 3 + 5 e x dx π.......................................................................

More information

量子力学 問題

量子力学 問題 3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,

More information

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k 7 b f n f} d = b f n f d,. 5,. [ ] ɛ >, n ɛ + + n < ɛ. m. n m log + < n m. n lim sin kπ sin kπ } k π sin = n n n. k= 4 f, y = r + s, y = rs f rs = f + r + sf y + rsf yy + f y. f = f =, f = sin. 5 f f =.

More information

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g( 06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,

More information

1 1.1 / Fik Γ= D n x / Newton Γ= µ vx y / Fouie Q = κ T x 1. fx, tdx t x x + dx f t = D f x 1 fx, t = 1 exp x 4πDt 4Dt lim fx, t =δx 3 t + dxfx, t = 1

1 1.1 / Fik Γ= D n x / Newton Γ= µ vx y / Fouie Q = κ T x 1. fx, tdx t x x + dx f t = D f x 1 fx, t = 1 exp x 4πDt 4Dt lim fx, t =δx 3 t + dxfx, t = 1 1 1.1......... 1............. 1.3... 1.4......... 1.5.............. 1.6................ Bownian Motion.1.......... Einstein.............. 3.3 Einstein........ 3.4..... 3.5 Langevin Eq.... 3.6................

More information

( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................

More information

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1) 1. 1.1...,. 1.1.1 V, V x, y, x y x + y x + y V,, V x α, αx αx V,, (i) (viii) : x, y, z V, α, β C, (i) x + y = y + x. (ii) (x + y) + z = x + (y + z). 1 (iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y

More information

1.3 (heat transfer with phase change) (phase change) (evaporation) (boiling) (condensation) (melting) (solidification) 1.4 (thermal radiation) 4 2. 1

1.3 (heat transfer with phase change) (phase change) (evaporation) (boiling) (condensation) (melting) (solidification) 1.4 (thermal radiation) 4 2. 1 CAE ( 6 ) 1 1. (heat transfer) 4 1.1 (heat conduction) 1.2 (convective heat transfer) (convection) (natural convection) (free convection) (forced convection) 1 1.3 (heat transfer with phase change) (phase

More information

u = u(t, x 1,..., x d ) : R R d C λ i = 1 := x 2 1 x 2 d d Euclid Laplace Schrödinger N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3

u = u(t, x 1,..., x d ) : R R d C λ i = 1 := x 2 1 x 2 d d Euclid Laplace Schrödinger N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3 2 2 1 5 5 Schrödinger i u t + u = λ u 2 u. u = u(t, x 1,..., x d ) : R R d C λ i = 1 := 2 + + 2 x 2 1 x 2 d d Euclid Laplace Schrödinger 3 1 1.1 N := {1, 2, 3,... } Z := {..., 3, 2, 1,, 1, 2, 3,... } Q

More information

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10 1 2007.4.13. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 0. 1. 1. 2. 3. 2. ɛ-δ 1. ɛ-n

More information

KENZOU

KENZOU KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................

More information

第5章 偏微分方程式の境界値問題

第5章 偏微分方程式の境界値問題 October 5, 2018 1 / 113 4 ( ) 2 / 113 Poisson 5.1 Poisson ( A.7.1) Poisson Poisson 1 (A.6 ) Γ p p N u D Γ D b 5.1.1: = Γ D Γ N 3 / 113 Poisson 5.1.1 d {2, 3} Lipschitz (A.5 ) Γ D Γ N = \ Γ D Γ p Γ N Γ

More information

3. :, c, ν. 4. Burgers : t + c x = ν 2 u x 2, (3), ν. 5. : t + u x = ν 2 u x 2, (4), c. 2 u t 2 = c2 2 u x 2, (5) (1) (4), (1 Navier Stokes,., ν. t +

3. :, c, ν. 4. Burgers : t + c x = ν 2 u x 2, (3), ν. 5. : t + u x = ν 2 u x 2, (4), c. 2 u t 2 = c2 2 u x 2, (5) (1) (4), (1 Navier Stokes,., ν. t + B: 2016 12 2, 9, 16, 2017 1 6 1,.,,,,.,.,,,., 1,. 1. :, ν. 2. : t = ν 2 u x 2, (1), c. t + c x = 0, (2). e-mail: iwayama@kobe-u.ac.jp,. 1 3. :, c, ν. 4. Burgers : t + c x = ν 2 u x 2, (3), ν. 5. : t +

More information

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P 1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A

More information

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =

More information