Vector Calculus / Numerical Analysis

Transcription

1 % Mathematics 2 % Vasil Zlatanov Vector alculus / Numerical Analysis 1 Triple products Scalar triple product yclic rule a -> b -> c -> a.. a (b c) b (c a) c (a b) Vector triple product a (b c) Placement of the brackets is important a b c Doesn t mean anything! 2 Scalar Fields ψ ψ(x, y, z, t)... It s a function that takes values at every point in space and time. A good example of this would be the temperature of a room. Also electric potential is also a sclar field V E φ(x, y, z) 2.1 Visulations z φ(x, y) We can 3d plot it with the help of a computer, alternatively we can plot a countour line. We do this by setting phi to a constant. 1

2 φ c 1 12 y3 y 1 4 x emember at saddle points the contour lines intersect. If we set k 7/2 we get: 1 4 y y3 y Here is some MATALB code to work do these. v-5:.2:5; [x,y]meshgrid(v); z y.^3/12-y-x.^2/4+3.5; figure; surf(x,y,z); figure; contour(x,y,z); Figure 1: surf plot Example 2 φ(x, y, z) x + y + z 2

3 Figure 2: contour plot 3

4 We can t easily graph this so we have to go for: φ c x + y + z If we use phi to represent an electric potential, taking phi c we obtain socalled equipotentials. If we have a function with 2 variables, this will give us equipotential lines, for 3 it would be a surface. Example 3 are given by φ(x, y, z) x 2 + y 2 + z2 4 φ(x, y, z) c x 2 + y 2 + z2 4 Gives us equipotential ellipsoids. equipotential surface. ecall: the surface of a conductor is an 2.2 Scalar fields varying with time f(x, t) A cos(2π(f t 1 x)) A cos(wt kx) λ w 2πF k 2π λ Taking t at different values we get a clear picture of how the wave progragates. We can even graph the function in 3d with time as one of the axis to appreciate this. We can time a snapshot of a a time varying 3d surface. For examples: Does: φ(x, y, t) e (x2 +y 2) cos(wt) wt we get a hill wtpi/4 we have a small hill wtpi/2 we have nothing (as cos(pi/2) ) wt 3pi/4 small hill down wt pi we have a hill max down 4

5 3. Vectors φ(x, y, z, t) Is a snapshot of level surfaceas at t... So for electric field B ib x (x, y, z, t) + +jb y (x, y, z, t) + kb z (x, y, z, t) And magnetic E ie x (x, y, z, t) + +je y (x, y, z, t) + ke z (x, y, z, t) H ih x (x, y, z, t) + +jh y (x, y, z, t) + kh z (x, y, z, t) 3.1 Visualasations The relationship given a 2d field E(x, y) yi xj F F x i + F y jy y(x) satisfy dy dx F y F x If we integrate the field lines we will get circles. So to get the field lines we simly have to integrate Example 2 Given a 3d coordinate system, we place a charged line on the z-axis. At any point in the space P(x,y) the electrical field is going to be radial and calculated based on the distance from the line. r x 2 + y 2 E q 2πε r r 5

6 Figure 3: plot 6

7 So we write E in terms of cartesian components. let k E E x i + E y j q r cos θi + sin θje k (cos θi + sin θj) 2 π ε r Thus givens ( x cos θ x/r sin θ y/re k r 2 i + y ) r 2 j k ( x x 2 + y 2 i + ) y x 2 + y 2 j This models gravitational field lines. The further we go from the origin the smaller the magnitude. 4. Vector operators: grad, div, and curl Definition of gradient operator onsider a field such as Varies as φ φ(x, y, z) iδ δx + jδ δy + kδ δz So nabla makes phi from a scalr to a vector. 4.2 Differenetial derivative If we want the rate of change of phi in a particular direction. We do this by taking the derivative. δφ φ m δ m This is called the direction derivative of phi in the direction m hat. 7

8 4.3 The gradient of a scalar field This is for example the electric voltage (potential) in a 2-D plate.. We can do this by drawing contour lines where the voltage is constant. On these equipotential lines we have something specific: nabla phi is perpendicular to the tangent or: φ m Where φ ( i δ δx + j δ δy k δ ) φ n δφ δz δn n is perpendicular to the contour lines Example, lets take a scalar field: Which means we get circle! Now obtain: φ φ φ(x, y) e (x2 +y 2 ) φ (x 2 + y 2 ) log cx 2 + y 2 k φ 2xe (x2 +y 2 ) î 2xe (x2 +y 2 ) ĵ Dir n r 3.3 Definition of divergence of a vector field div B Vector algebra allows us to do scalar and vector products. Likewise there is two ways to operate grad on a vector field divb B ( î δ δx + ĵ δ δy + k δ ) (îb 1 + ĵb 2 + δz kb 3 ) Since and Follows: i i j j k k i j i k k j divb B δb 1 δx + δb 2 δy + δb 3 δz 8

9 divb is a scalar field (because it is dot product). divb is a meassure of compression or expansions trough the faces of a cube Note: B B δ B B 1 δx B δ 2 δy B δ 3 δz Example: B ix 2 + jy 2 + kz 2 divb B δ δx x2 + δ δy y2 + δ δz z2 2x + 2y + 2z Figure 4: plot Example: Taking the derivative again F (xi + yj + zk) F Another example: F yi xj + k 9

10 * heck if right Diveregence and Electric fields There are three possible ways to interpret a vector field: source - going out : div F > sink - going in div F < none - net flow in and out is zero, div F This relates to the elctric flux density D, which was related to the electric field E: D ε E This is determined by the electrostatic charge which is distributed by: ρ(x, y, z) There is a proof with taylor series and a cube that is confusing and hard to understand. It boils down to: ( δdx δx + δdy δy + δdz ) ρ(x, yz) δz 1

11 Which is the same as: D ρ(x, y, z) E ρ ε Another Maxwell equation we in an area enclosing a magnet: B Similar reasoning can also be done when working with fluid motion. Definition of curl of a vector It is the measure of curvature or rotation. For instance a shpere turning around itself. How rapidly the vectors rotate is the magnitude of curl. Example: ir ix + jy + kz r ( δ i ( δ +j δx (z) δ δz (x) i j k δ x δ y δ z x y z δy (z) δ ) + k ) δz (y) ( δ δx (y) δ δy (x) ) epeated use of grad operator div(gradφ) ( φ) ) ( i δ δx + j δ δy + δ δz ( δφ δx i + δφ δy j + δφ δz k δ2 φ δx 2 + δ2 φ δy 2 + δ2 phi δz 2 ) And written shortly ( φ) 2 φ Laplaces equation can then be written as: 2 φ 11

12 Whre nabla2 is called Laplacian Applying curl on the gradient of a sclar field: ( φ) i j k δ x δ y δ z δφ δx ( δ 2 φ i δyδz δ2 φ δzδy δφ δy δφ δz ) + j(... ) + k(... ) So if we have an electric fields which is a gardient of a potential E φ Then we immediately can tell that it is curl free as: E Vector identities 1. Gradient of the product of two scalars (φψ) ψ φ + φ ψ 2. The divergence of the product of a scalar with a vector div(ψb) ψdivb + ( ψ) B 3. The curl of the product of a scalar and a vector curl(ψb) ψcurlb + ( ψ) B or (ψb) ψ B + ( ψ) B 4. The curl of the gardient of any scalar curl( ψ) ψ 5. The divergence of curl and a vector div(curlb) ( B) 12

13 Irotational and solenioidal vecotr fieldso Any vector B(x,y,z) that can be written as the gradient of a scalar phi(x,y,z) is call irotational or solenoidal and has curl B Matheamatically: curlb φ B ± φ ** missed out do at home, wtf? So vector fields A where div A are called solenoidal or divergence free A curlb B is called the vector potential Line (path) integration b There are 2 types of line integrals: f(x)dx a i1 N f(x i )δx i Scalar field ds dt dx dt 2 ψ(x, y, z) ds 1 ds length + dy 2 ds x dt 2 + y 2 dt Arc length is given by : ds b a x 2 + y 2 dt Example: Lets show that: x 2 y ds 1/3 Where is the circular arc in the first quadrant of the unit circle oriented anti-clockwise. 13

14 x cos t x sin t y sin t y cos t ds x 2 + y 2 dt sin t 2 t + cos 2 tdt dt x 2 yds π/2 cos 2 t sin tdt [ ] π/2 cos3 t 3 Example 2: y -3x where x -1 to 1 ds is an element on the line y -3x. We parmetrize this with: x t, y 3t 1 t 1 ds ( 3) 2 dt 1dt 1 1 Example 3: Show 1(t)( 3t) 3 dt In a right angle triangle with legs of 1 On leg 1: 1 x 2 y ds 2/12 y ds dt dt t 4 dt On other leg: x ds dt dt On base y 1 xy tx 1 tds ( 1) dt 2dt (1 t) 2 t 2 dt 14

15 Vector field F (x, y, z)dr Example is F as a force on a particle. In that case the work is δw F δr Example: Given F ix 2 + j(x z) + kxyz and we travel the paths 1 and 2, Where 1 is y x2 and 2 is y x in the plane z 2 from (,,2) -> (1,1,2) Evaluate the F dr F dr (F 1, F 2, F 3 ) (δ x, δ y, δ z ) 1 1 F 1 dx + F 2 dx + F 3 dz 1 x 2 ydx + (x z)dy + xyzdz 1 As z 2 and constant then dz x 2 ydx + (x 2)dy 1 Using: Follows: 1 y x 2 dy dx 2x x 2 (x 2 )dx + (x 2)2xdx 1 x 4 + 2x 2 4xdx Now along the sraith line _2: y x dy dx dz 1 F dr x 2 y dx+(x 2) dx x 2 (x)+(x 2) dx x 3 +x 2 dx 5 4 So this shows that if you have the same start points the integral can differ if the route is different. 15

16 Independece of Path taken this is to show when the line integral of F is the same now mater the path taken: F. dr is an exact differential: dr dφ Starting from A and B of the urve : F dr dφ B A φ(a) φ(b) So this is independent of the path taken. dφ δφ δφ δφ dx dy + δx δy δz φ dr dφ [φ] B A Therefore F φ Which means it is a curl free vector: curl(f ) φ F ± φ Where phi is the scalar potential. F is also known as a conservative vector field. So is indendent of the path only if F dr curlf Also A A F dr dφ φ(a) φ(a) 16

17 Example: Is this line integral independent? (2xy 2 dx + 2x 2 y dy) Solving: F F (2xy 2, 2x 2 y) i j k δ x δ y δ z 2xy 2 2x 2 y Thus the line integral is independent of the path (as curl is ) and we should be able to calculate phi from: F φ (F x, F y, F z ) (φ x, φ y, φ z ) F x δφ δx F y δφ δy 2xy2 2x2 y [δ x (2x 2 y) δ y (2xy 2 )]k4(xy 4xy)k So with partial integration: φ φ 2xy 2 dx x 2 y 2 + A(y) 2x 2 y dy x 2 y 2 + B(x) Therefore A B const φ x 2 y 2 + Example 2: With the same method: F (yz, xz, xy) φ xyz + Using this you can find the work: (3,3,2) w dφ [ φ] (1,1,1) (1,1,1) [xyz](3,3,2) (1,1,1) (1,1,1) 17

18 Double integral I ψ(x, y) dx dy b [ yh(x) a yg(x) ψ(x, y) dy ] dx Area covered : Area() yh(x) yg(x) I ψ(x, y) dy P (x) b a b [ yh(x) P (x) dx 1 dy ] b a yg(x) a h(x) g(x) dx So in other words if you have a scalar field in terms of x and y you can integrate it with respect to both and you will get the area covered by it within the limits. Example: x 2 dy dx where is the region in the first quadrant between: y x 2 y x We do: 1 1 [ x ] x 2 dy dx yx 2 1 [x 2 y] x yx 2 dx x 2 (x x 2 ) dx Example 2: A cicrcle with radius a in the first quadrant. a [ ] a 2 x 2 A 1 dy dx a a2 x 2 dx 18

19 To solve this one integral: x a cos θ dx a sin θ dθ So a a2 x 2 dx a 2 π/2 πa 2 /4 sin 2 θ dθ If you want to change the oredr of integration yohave to change the limits. Jacobian We can also change the coordinates to polar coordinates For polar coordinates: x r cos θ We take Then So in essence: y r sin θ u r v J r,θ (x, y) [ cos θ r sin θ sin θ r cos θ r cos 2 θ + r sin 2 θ r J r,θ r δ x δ y r dr dθ ] Example, calculate the volume of a shpere: V 2 a2 x 2 y 2 dx dy Let: 2 V 2 a2 r 2 r dr dθ a [ 2π 4π a ] a2 r 2 r dθ dr r a 2 r 2 dr w a 2 r 2 19

20 dw 2r dr 4πr3 3 Example: alculate the Area of regino between the lines x + y 2 x + y 1 x 3y 2 2x 3y 5 Pairs of lines -> change of variables let 1 dx dy a x + y u V 2x 3y V Solving > x 3 5 u vy 2 5 u 1 5 v Now the Jacobian: J u,v (x, y) [ [ δx δu δy δu δx δv δy δv ] 1 5 ] > u,v plane Ex: I 1 J du dv 1 [ : region between the curves I (x 2 + y 2 ) 3 dx dy x 2 y 2 1 xy 1 ] [ 5 ] 1 du 1 dv

21 x 2 y 2 4 xy 3 hange of variables, let: ( x 2 y 2 uxy v ) (u v ) Jacobian: J u,v (x, y) J u,v (x, y) 1 J x,y (u, v) [ δx δu δy δu δx δv δy δv ] [ ] 2x 2y 2x 2 + 2y 2 y x J(u, v)(x, y) 1 2(x 2 + y 2 ) I (x 2 + y 2 ) 3 1 (x 2 + y 2 ) 2 du dv 2 1 2(x 2 du dv + y2 (x 2 + y 2 ) 2 x 4 + 2x 2 y 2 + y 4 x 4 2x 2 y 2 + y 4 + 4x 2 y 2 u 2 + 4v 2 I v1 3 v1 [ 4 ] u 2 + 4v 2 du u1 [ ] u v2 u dv 1 2 (21 + v2 ) dv 1 dv DIY Greens Theorem in a plane (P dx + Q dy) (Q x P y ) dx dy This theorem tells us how the integral of the bounary of a close curve relates to the area enclosed in the curve. Where: Q x δq δx P y δp δy 21

22 δp dx dy δy b [ yh(x) b a b a a b a yg(x) ] δp δy dy dx [P (x, y)] yh(x) yg(x) dx P (x, H(x)) P (x, g(x)) dx P (x, g(x)) dx P (x, y) dx 1 a b P (x, h(x)) dx P (x, y) dx 2 Examples from problem sheets: Find (x 2 + y 2 + z 2 ) ds Where is the helix: From (1,, to (1,, 2pi)o This is a cylinder Setting it up: ds x 2 + y 2 + z 2 dt r i cos t + j sin t + kt x cos ty sin t z t P dx sin 2 t + cos 2 t dt 2 dt x 2 + y 2 + z 2 cos 2 t + sin 2 t + t t 2 Question 2: I 2pi (1 + t 2 ) 2 dt 2(2π + 8π3 3 xy ds a on the triangle from (,) to (1,) to (,1) 1 : y 3 : x 2 : y 1 x 22

23 So _2 is the only non-zero ones. Where y 1-x from (1,) to (,1) Let t y t... 1 x 1 y 1 t 2 xy ds Where Question 4: or ds x 2 + y 2 dt ( 1) dt 1 (1 t)t 2 dt t t 2 dt 6 y 2 cos x dx + 2y sin x dy f dr Path independent? F So the integral is independent of path. i j k δ x δ y δ z y 2 cos x 2y sin x Doing a -> for the line from (,) to (pi/2, 1) Doing b -> from (,) to (pi/2, ) to (pi/2, 1) 1 : y dy 2 : x π/2 dx So _1 and: 2 2y(sin π 1 2 ) dy 2 1 y dy 1 Doing if for the diagonal _3 should also give 1. 23

24 Begin Nunzio -> Picking up again with Green s theorem: (P dx + Q dy) (Q x P y ) dx dy We attempt: (x y) dx x 2 dy Along a square with sides 2 counter-clockwise. Using the theorem: 2 Q x P Y dx dy dy 2 ( 2x + y) dx dy 2x + 1 dx 2[ x 2 + x] 2 4 We can also directly evaluate (rather than using the theorem: 2 So they are the same. x dx dy (x 2) dx + Now lets use green theorem on: y 3 dx + (x 3 + 3xy 2 ) dy 3/2 Along y x 2 from (,) to (1,1) and back to zero via y x I Q x P y dx dy 3x 2 + 3y 2 3y 2 dx dy 3x 2 [ yx yx 2 ] dy dx x 2 [x x 2 ] dx dy

25 Divergence and Stoke s Theroems We want to compute the component on the vector field in the direction normal to the surface with the notation: n ds Where n-hat is the normal to the surface at any point. We can use these to find the flux trough a surface. (Or if we have fluids it s the volume flowing out) Divergence (Gauss ) Theorem We defined the unit tangent vector with ds being the arc: T i dx ds + j dy ds The normal and tangent are perpendicular so: n T We can defined a 2D vector: u iq jp and so: u n (iq jp ) (i dy ds j dx ds P dx ds + Qdy ds u n ds P dx + Q dy divu u δq δx δp δy Q x P y Q x P y dx dy u dx dy And using Greens theoroem: P dx + Q dy So: u dx dy u n ds 25

26 So the line integral expresses the normal component of u around the bounary. The 3D version of this is: V u dv S u n ds Stokes theorem Defining v ip + jq V dr V T ds (ip + jq)(i dx ds + j dy ) ds P dx + Q dy ds Also: curlv K(Q x P y ) K curlv Q x P y Using green theorem (k curlv) dx dy v dr The 3D version is: v dr curlv n ds ( v) n ds s So Strokes theorem equations the surface integral of the curl of a vector field to the line integral of the vector field along the boundary of the surface. We can use the right hand rule to determine the direction of the bounary. So now if we have a complex surface integral we can convert it to a line integral if we can write the field as F Or alternatively we can express a line integral as any the integral of any surface. So if we have a complex surface with a given buonary, we can convert it into a simpler surface with the same boundary. This is akin to how the flux of F trough a surface is indenpend of the surface you pick if they are bound by the same curve. S 26