Integrtion by prts Integrting the both sides of yields (uv) u v + uv, (1) or b (uv) dx b u vdx + b uv dx, (2) or b [uv] b u vdx + Eqution (4) is the 1-D formul for integrtion by prts. Eqution (4) cn be extended to 2D s Exmple (Green s identity) b b uv dx, (3) b u vdx [uv] b uv dx. (4) uvd Proof: The one-dimensionl version of Eq.(6) is uvndl ( u v + u v) d which cn be esily proven by using integrtion by prts s This cn be extended to 2-D s b b u vd u vd. (5) u v dl. (6) n (u v + uv )dx [uv ] b, (7) uv dx [uv ] b b u vd u v ndl u v dx. (8) u vd u v n dl u vd. (9) Note tht most of the usge of the divergence theorem is to convert boundry integrl tht contins the norml to the boundry into volume (re) integrl by replcing the norml (n) by nbl ( ) to be plced in front of the expression. 1
Exmples... n... dl... d. (1) 1. Het conduction The blnce of energy is stted s T ρc p t d ( n) hdl, (11) where ρ is the mss density, C p is the specific het, h is the het flux cross the boundry of the control surfce. Using the Guss theorem, the right hnd side of Eq.(11) becomes so it follows or Using Fourier s lw, ( n) hdl hd, (12) T ρc p t d hd, (13) ρc p T t where k is the therml conductivity, Eq.(14) becomes + h. (14) h k T, (15) ρc p T t (k T ). (16) 2. Equilibrium eqution in sttic elsticity The blnce of force for continu is stted s tdl + bd, (17) where t is the surfce trction force nd b is the body force. The surfce trction force t is the contribution of the stress tensor, σ, in the direction of n s t σ n, (18) so Eq.(17) becomes σ ndl + bd, (19) 2
or or σd + bd, (2) σ + b, (21) which is known s the eqution of equilibrium. For smll elstic deformtion, σ C u, (22) where C is the elstic modulus nd u is the displcement, Eq.(21) becomes which is known s the Nvier s eqution 1. 3. Green s second identity (C u) + b, (23) (u v v u)d (u v n v u n )dl. From the first Green s identity, V V ( u v + u 2 v)dv ( v u + v 2 u)dv u v d, (24) n v u d. (25) n ubtrcting Eq.(25) from Eq.(24) gives the second Green s identity. This identity is used to derive solutions to Poisson s eqution ( u ρ). Line integrls Exmples: y (1,2) y2x (,) x 1 Ech quntity is tensor. 3
(1) where W (1,2) (,) F dr, F (xy, x 2 ), (26) dr (dx, dy). (27) W xydx + x 2 dy x(2x)dx + x 2 2dx 2 1 1 (2x 2 + 2x 2 )dx 4x 2 dx 4 3. (28) y (1,2) y2x 2 (,) x (2) W xydx + x 2 dy x(2x 2 )dx + x 2 (4xdx) 3 1 1 (2x 3 + 4x 3 )dx 6x 3 dx 3 2. (29) It cn be seen tht the vlues of line integrls generlly depend on the integrl pths. 4
5π/2 y r2θ x (3) F (2xy, x 2 1). W 2xydx + (x 2 1)dy 5/2π {2(r cos θ)(r sin θ)(dr cos θ r sin θdθ) +((r cos θ) 2 1)(dr sin θ + r cos θdθ)}... 5/2π, (3) where x r cos θ, (31) y r sin θ, (32) dx dr cos θ r sin θdθ, (33) dy dr sin θ + r cos θdθ, (34) ws used. Alterntively, if one defines G(x, y) x 2 y y, F dr dg 4 [G] (x,y)(,5/2π) (x,y)(,) 5 2 π. While the line integrls of Exmples 1 nd 2 depend on the integrl pth, the line integrl of Exmple 3 is independent of the pth. This difference will be clrified in the next section. 5
tokes theorem m d dr Proof (not shown in clss) trt with F dr m ( F)d. I F dr, (36) n where the integrl pth is shown below. (x, y+ y) C (x+ x, y+ y) B D (x, y) A (x+ x, y) tegrting long A: tegrting long B: so F dr F x (x, y)dx, (37) A x+ x x F x (x, y) x. F x (x, y)dx F dr F y (x + x, y)dy, (38) 6
so tegrting long C: so tegrting long D: so B C y+ y y F y (x + x, y)dy F y (x + x, y)dy ( F y (x, y) + F ) y x x y. F dr F x (x, y + y)dx, (39) D x x+ x F x (x, y + y)dx F x (x, y + y) x ( F x (x, y) + F ) x y y x. F dr F y (x, y)dy, (4) y y+ y F y (x, y)dy F y (x, y) y. Hence, F dr F dr A+B+C+D ( Fy x F ) x x y y m ( F) x y. (41) Repeting this for finite region, one obtins F dr m ( F) dxdy. (42) The term rottion for F comes from the left hnd side, i.e., the mount of quntity tht flows long A B C D. Exmple The velocity field of vortex tht rottes with the ngulr velocity of ω is given s v rωe θ rω ωy ωx 7 sin θ cos θ, (43)
so Another proof (shown in clss) From the figure bove, so v 2ωk. (44) dr m ndr, m ( F)dA m (n F)dr F (m n)dr 5 F dr dr dr If F (P, Q, ), m (,, 1) nd the tokes theorem is rewritten s Irrottionl Field P dx + Qdy F dr. (45) ( Q x P ) da. (46) y Definition: If the vector field, v, stisfies v, v is clled n irrottionl field. Theorem: The following four sttements re ll equivlent. (1) (2) v ϕ. v. (3) v dr. (4) v dr, is independent of the integrl pth. 1. 1 2. ( ϕ) 2. 2 3 v dr m ( v)da 8
3. 3 4 (shown in clss) 4. 4 1 Define ϕ s then ϕ(r) r v dr, ϕ(r + dr) ϕ(r) ϕ dr 6 (48) On the other hnd, ϕ(r + dr) ϕ(r) r+dr v dr r v dr v dr, (49) so v ϕ. clr potentil As shown in clss, n irrottionl field, v, hs sclr potentil ϕ, such tht v ϕ, (5) i.e. if v is rottion-free, v, then, v cn be derived from sclr function, ϕ, which simplifies the nlysis (three components of v vs. single component of ϕ). Exmple 1: v r (position vector) (51) As r hs sclr potentil. Verify tht ϕ 1 2 r, ( x 2 + y 2 + z 2). (52) 6 ϕ(x + dx, y + dy, z + dz) ϕ(x, y, z) ϕ(x, y, z) + ϕ x ϕ ϕ dx + dx + ϕ y ϕ dy + dz ϕ(x, y, z) z ϕ dy + x y z dz ϕ dr. (48) 9
Exmple 2: (Grvittionl field) F G mmr r 3. (53) Verify tht ( ) r F GmM r ( ) 3 1 GmM( r + 1 r 3 r r) ( ) 3 r 3 GmM r r 6 GmM 3r2 r r 6 GmM 3 r 5 r r r. (54) ϕ GmM. (55) r Exmple 3: The vector field, v (2xy, x 2 1, ) is n irrottionl field (verify tht v ) so there must exist sclr potentil, ϕ, such tht v ϕ, i.e. By integrting Eq.(56) with respect to x, one obtins 2xy ϕ x, (56) x 2 1 ϕ y, (57) ϕ z. (58) ϕ x 2 y + f(y), (59) where f(y) is function of y lone. Eqution (59) is substituted into Eq.(57) s x 2 1 x 2 + f (y), (6) which is solved for f(y) s f(y) y + c where c is constnt so the sclr potentil for v is ϕ(x, y, z) x 2 y y + C. (61) Exmple 4: Compute C (3dx + 6y 2 z 3/2 dy + (3y 3 z + z)dz) where C is curve x sin t, y cos 3t, z 2t s t goes from to 2π. 1
(olution) The vector field, v ( 3, 6y 2 z 3/2, 3y 3 z + z ) is n irrottionl field (verify!). olving 3 ϕ x, 6y2 z 3/2 ϕ y, 3y3 z + z ϕ z, (62) one gets Therefore, C ( 3dx + 6y 2 z 3/2 dy + (3y 3 z + z)dz ) ϕ 3x + 2y 3 z 3/2 + z2 2. (63) C d ( 3x + 2y 3 z 3/2 + z2 2 [3x + 2y 3 z 3/2 + z2 ) 2 ](,1,4π) (,1,) 16π 3/2 + 8π 2. (64) Note tht dϕ ϕ ϕ ϕ dx + dy + x y z dz dx dy dz ϕ x ϕ y ϕ z ϕ dr v dr. (65) o tht line integrl, v dr, (see n exmple in the lst lecture) cn be evluted s v dr dϕ [ϕ] B A ϕ B ϕ A. (66) Note lso tht sclr potentils re not unique. If n irrottionl field, v, is derived from different sclr potentils, ϕ 1 nd ϕ 2, i.e. It follows v ϕ 1, (67) v ϕ 2. (68) ϕ 1 ϕ 2 (ϕ 1 ϕ 2 ), (69) 11
so ϕ 1 ϕ 2 C, (7) where C is constnt, i.e. the difference of two sclr potentils which yield the sme irrottionl field is t most constnt. Vector potentil When vector field, v, is divergence-free, i.e. v, (71) in simply connected region, there exists vector, ψ, whose rottion is v, i.e. v ψ. (72) The vector, ψ, is clled vector potentil. The necessry condition for the existence of vector potentil is proven esily s ( ψ). 7 (74) The converse (the sufficient condition) is more difficult to prove nd is not presented here. Just like sclr potentils, vector potentils re not unique, i.e. if vector v is derived from two different vector potentils, it follows which implies tht the vector, ψ 1 ψ 2, is irrottionl. Thereby, v ψ 1 ψ 2, (75) (ψ 1 ψ 2 ), (76) ψ 1 ψ 2 ϕ, (77) i.e. the difference between vector potentils tht yield the sme vector filed is t most ϕ. Unlike sclr potentils, vector potentils re more difficult to derive nd less useful s three components of ψ re needed to yield three components of v. 7 ( ψ) ( ) ψ (74) 12
Helmholtz theorem If vector field, v, is neither rottion free nor divergence free, it cn be still expressed by both vector potentil nd sclr potentil, i.e. Proof: Consider the Poisson eqution, v ϕ + ψ. (78) Provided this eqution is solved for ϕ, Eq.(79) cn be written s ϕ v. (79) (v ϕ), (8) which implies tht v ϕ is divergence free. Hence, it hs vector potentil, i.e or v ϕ ψ, (81) v ϕ + ψ. (82) 13