ENGI 5432 Chapter 1 Complete Notes
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ENGI 5432 1. Vector Fields and the Gradient Operator Page 1-01
1. Vector Fields and the Gradient Operator
In this chapter, a review of vectors from previous courses is followed by the introduction of lines of force. The gradient operator is extended to divergence, curl and Laplacian in both Cartesian and general orthonormal curvilinear coordinate systems. Conversion of components of vectors between Cartesian and other coordinate systems is also covered.
Contents of this Chapter:
1.1 Review of Vectors 1.2 Lines of Force 1.3 The Gradient Vector 1.4 Divergence and Curl 1.5 Conversions between Coordinate Systems 1.6 Basis Vectors in Other Coordinate Systems 1.7 Gradient Operator in Other Coordinate Systems ENGI 5432 1.1 Review of Vectors Page 1-02
1.1 Review of Vectors
This first section is a brief review of concepts that were introduced during terms 1 and 2.
The displacement vector of a particle can be represented in Cartesian components by rvt x t, y t , z t where t is a parameter (time or angle or distance along a curve, etc.)
The distance of the particle from the origin at any value of t is given by the scalar function 2 2 2 r( t) =rv( t) =( x( t)) +( y( t)) + ( z( t))
Note the various alternative conventions for a vector and its magnitude: r rv rand r rv rv
Scalar product (dot product): v v ug v = uv cosq
The component of vector uv in the direction of vector v is
v uv =ug vˆ = u cosq v v vv v v v v v v ug v=0� u = 0 or v ^ 0 or u v ("u at right angles to v ")
The scalar product is commutative: v v v v v v vg u= u g v " u, v
Vector product (cross product):
The vector product of two vectors uv, v is in a direction nˆ at right angles to both uv and v .
uv v= (uvsinq ) nˆ The area of the parallelogram is
A = uv v and h= u sinq = uv vˆ ENGI 5432 1.1 Review of Vectors Page 1-03
The orientation of the plane containing vectors uv and v is defined by the direction of the vector product nv = uv v . v vv v v v v v v v v u v= 0� u = 0or v 0 or uP v ("u parallel to v ")
The vector product is anti-commutative: v v创 u v� 藓 uv v v v 0
The product rule of differentiation is valid for scalar and vector products: v v dv v du v v d v (ug v) = g v + u g and dt dt dt d duv d v d uv d v (uv创 v) = v + u v 创 = - v + u v dt dt dt dt dt
Example 1.1.1
For the vectors uv = 4, 3, 2 and v =2, - 1, - 2 v v v v v v find ug v,uv , u v , q (= the angle between u and v ) and the equation of the plane parallel to uv and v that passes through the origin. u =uv =16 + 9 + 4 = 29 v =v =4 + 1 + 4 = 3 v v ug v =4( 2) + 3( - 1) + 2( - 2) = 1 v v v ug v 1 uv =ug vˆ = = v 3
ˆi ˆj kˆ uv v =4 3 2 = - 4,12, - 10 = - 2 2, - 6, 5 142 43 2- 1 - 2 nv uv v 1 cosq=g = 晦 0.0619 q 话 86.5 uv 3 29 ENGI 5432 1.1 Review of Vectors Page 1-04
Example 1.1.1 (continued)
A normal to the plane is nv =2, - 6, 5 . v v v A point on the plane is a= 0, 0, 0� ag n 0 . v v v v The equation of the plane is rg n= a g n � 2x + 6 y = 5 z 0
The arc length s is the distance travelled along the curve. It is related to the displacement vector r by ds drv = dt dt
The distance along a curve between two points whose parameter values are t0 and t1 is t t t 2 2 2 1 1ds 1 骣 dx 骣 dy 骣 dz L= ds = dt =琪 + 琪 + 琪 dt 蝌 dt桫 dt 桫 dt 桫 dt t0 t 0 t 0
A unit vector û has a magnitude of 1: | û | = 1 Any non-zero vector r can be decomposed into its magnitude r and its direction: rvr rˆ , where r rv 0
The unit tangent at any point on the curve is drv d rv d rv T dt dt ds ENGI 5432 1.1 Review of Vectors Page 1-05
The unit principal normal at any point on the curve is dT d T d T N ds dt dt where = radius of curvature (SI units: metre) = curvature (SI units: m1) 1 and Of all circles that touch a curve on the “inside” at a particular point (and which therefore all share the same unit tangent vector [or its negative] there), the one whose radius is is the best fit to that curve at that point. In general, curvature varies along most curves.
At points of inflexion, = 0
The unit binormal is B T N The three vectors form an orthonormal set (they are mutually perpendicular and each one is a unit vector; that is, magnitude = 1)
B constant the curve lies entirely in one plane (the plane defined by Tˆand N ˆ ).
Note that Nˆand B ˆ are undefined at points of inflexion. All three unit vectors are undefined where the curve suddenly reverses direction, (such as at a cusp). ENGI 5432 1.1 Review of Vectors Page 1-06
Velocity and Acceleration
If the parameter t is the time, then the velocity of a particle [a vector function] is drv v t dt and its speed [a scalar function] is drv ds v( t) =v( t ) = = dt dt and v v T
The acceleration [vector] is dv d2 rv d 2 x d 2 y d 2 z avt , , a T a N dt dt2 dt 2 dt 2 dt 2 T N where v v dv v ˆ vg a aT = =ag T = dt v and v av 2 2 2 v ˆ aN=k v = a - a T =ag N = v ENGI 5432 1.1 Review of Vectors Page 1-07
Example 1.1.2
The equation of a curve in 3 is given parametrically by rv =etsin t i - j + e t cos t k v v Findv , a ,v , aT , a N , , T , N , B . Show that the curve lies entirely in one plane and find the equation of that plane.
t t Let c = cos t and s = sin t then rv =e s, - 1, e c
drv � v = +et( s - c), 0, e t ( c s) dt
2 2 � v+ et( s + c) +02 -( c = s) + e t s 2 + 2 sc + c 2 - c 2 + 2 sc s 2 \v = et 2
v 2 Tˆ = =s + c, 0, c - s v 2 d v av = =et( s + c + c - s), 0,( c - s - s - c) = 2 e t c , 0, - s dt dv d a= = et2 = e t 2 T dt dt ( ) OR v v vg a v ˆ t 2 aT = =ag T =2 e c , 0, - s g s + c , 0, c - s v 2 =et2(( sc + c2) + 0 +( - cs + s 2 )) = e t 2 ENGI 5432 1.1 Review of Vectors Page 1-08
Example 1.1.2 (continued) a a2 a 2 et4 c 2 s2 2 e t 4 2 e t 2 N= - T =( +( -) ) - = - = OR 轾 ˆiˆ j kˆ v av 1 犏 a= =det et( s + c) 0 e t ( c - s) N v v 犏 犏 2et c 0- 2 e t s 臌 1 2e2t =0, 2e2t( c2 - cs + s 2 + sc) , 0 = = e t 2 v et 2
t aN e 2 2-t 1 t k=2 =2 =e� r = e 2 v (et 2) 2 k OR evaluate from a v av k =N = v2 v 3 dTˆ 2 =c - s, 0, - s - c dt 2 dTˆ 2 2 2 � -c2 +2 cs + s 2 + s 2 + 2 sc = c 2 = 1 dt 2 2 dTˆ d T ˆ 2 �Nˆ � - -c - s, 0, s c dt dt 2 OR vˆ ˆ ˆ1 v ˆ a=aT T + a N N� N - ( a a T T) aN
1骣 t t 2 1 =t 琪2e c , 0, - s -( e 2) s + c , 0, c - s = 2 c - s - c , 0, - 2 s - c + s e 2桫 2 2 2 � Nˆ -c - s, 0, - s c 2 ENGI 5432 1.1 Review of Vectors Page 1-09
Example 1.1.2 (continued)
2 轾 ˆi ˆj kˆ 骣 2 犏 Bˆ= T ˆ N ˆ =琪 dets + c 0 c - s 琪2 犏 桫 犏c- s0 - s + c 臌 ( ) 1 1 =0,(c2 - 2 cs + s 2 + s 2 + 2 sc + c 2 ) , 0 = 0, 2, 0 2 2 \Bˆ = ˆ j
dBˆ v � � 0 t 0 ds where is the torsion (the measure of the rate at which the curve is twisting out of the plane defined by Tˆand N ˆ ).
= 0 the curve lies in one plane, with plane normal nv = Bˆ = ˆ j.
A point on the curve can be found by setting the parameter t = 0: rv(0) = 0, - 1,1 = " av " The equation of the plane containing the curve is v v v v ng r= n g a � 0x + 1 y = 0 z + 0( 0) - 1( + 1) 0( 1) � y - 1
Frenet-Serret formulæ: d T 1. N ds d N 2. B T ds d B 3. N ds The proofs are in Problem Set 1. ENGI 5432 1.2 Lines of Force Page 1-10
1.2 Lines of Force
A vector function of n variables in n is a vector field.
F(x, y, z) = f1(x, y, z), f2(x, y, z), f3(x, y, z) (and the fi form scalar fields). The domain must be defined. If not explicitly mentioned, the domain is assumed to be n all of that part of for which all of the scalar fields fi are defined.
A vector field defines a vector F at each point in the domain. If these vectors are tangents to a family of curves, then those curves are streamlines or flow lines or lines of force.
Let F(x, y, z) be a vector field to a family of lines of force r(x, y, z). Then
drvv d rv v Tˆ = P F� k F( s) d s d s where k is some scalar.
� xⅱ( s), y( s) , z( s) k f1( s) , f 2( s) , f 3 ( s)
dx dy dz � =k f, = k f , k f ds1 ds 2 ds 3
dx dy dz � (k ds =) = f1 f 2 f 3
(provided f1, f2, f3 are all non-zero). 轾 dx f1 恨0 恨 0 x constant , etc. 臌犏 ds ENGI 5432 1.2 Lines of Force Page 1-11
Example 1.2.1
z 2 Find the lines of force associated with the vector field F =e, 0, - x and find the line of force that passes through the point (4, 2, 0).
drvv d rv v Tˆ = P F� k F ( s) d s d s dx dy dz =k ez , = 0 , = - k x2 � y A and ds ds ds
dx dz k ds= =� = x2 dx ez dz ez - x2 x3 �=x2 dx � ez dz + = B e z 蝌 3 1 3 � z- ln ( B3 x ) The lines of force are therefore v 1 3 r ( x, y , z) = x , A , ln ( B - 3 x )
We want the line of force through (4, 2, 0) 64 64 0 �4,A , ln( B=3) � 4, 2, 0 - A = 2 and B = 3 e 1 64 67 � B +1 = 3 3 The required line of force is 3 v 骣67 - x r ( x, y , z) = x , 2, ln 琪 桫 3 (forx < 3 67 only) ENGI 5432 1.2 Lines of Force Page 1-12
Example 1.2.2
Find the lines of force associated with the vector field F x2 , 2 y , 1 and find the line of force that passes through the point (1, 6, 2). drv v dx dy dz =kF � , , - k x2 , 2 y , 1 d s ds ds ds dx dy dz � k ds= = x 0 and y 0 x2 2 y - 1 ( ) -1 - 1 - 1 - 1 �dz= dx � dy = dz dx dy x22 y x 2 2 y 1 � z+ = cand z - + 1 ln y c x 12 2 1 � x=and ln y - - 2 z 2 c2 z- c1 v 1 -2z � r ,c3 e , z z- c1
But the required line of force passes through (–1, 6, 2):
1 -4 4 �,c3 e , 2- � 1, 6, 2 + c 1 = 2 = 1 3 , c 3 6 e 2 - c1 1 � rv , 6e4- 2z , z( z 3) z - 3 ENGI 5432 1.2 Lines of Force Page 1-13
Example 1.2.3
-y x Find the streamlines associated with the velocity field v = , , 0 x2+ y 2 x 2 + y 2 and find the streamline through the point (1, 0, 0).
drv dx dy dz- y x = kv � , , k , , 0 d s ds ds ds x2+ y 2 x 2 + y 2 dz x2+ y 2 x 2 + y 2 � =0 and k ds = dy dx ds x- y
\z = B .
Provided( x , y) �( 0,0) , = y dy � x dx = 蝌 y dy x dx y2 x 2 �= + � C = x2 y 2 A 2 2 2
The streamlines are therefore a family of circles, parallel to the x-y plane, centre the z-axis, lying on a concentric set of circular cylinders:
x2+ y 2 = A 2, z = B orrv = x ,� A 2 x 2 , B
( x, y , z) =( 1,0,0) � 12 = 0 2 = A 2 , 0 B The required streamline is therefore the unit circle
x2+ y 2 =1, z = 0 ENGI 5432 1.3 The Gradient Vector Page 1-14
1.3 The Gradient Vector
If a curve in 2 is represented by y = f (x) , then
dy y f x x f x lim lim dxQ P x x 0 x
If a surface in 3 is represented by z = f (x, y) , then in a slice y = constant,
z f( x+ D x, y , z + D z) - f( x , y , z) = lim 禗xDx 0 x Similarly, z f( x, y+ D y , z + D z) - f( x , y , z) = lim 禗yDy 0 y
In the plane of the independent variables:
f (P) = f f (Q) = f + df dr = dx, dy , 0
Chain rule: df f dx f dy dt x dt y dt f f df dx dy x y ENGI 5432 1.3 The Gradient Vector Page 1-15
抖f f v v � df= ,g dx , dy f gdr 抖x y where f (pronounced as “del f”) is the gradient vector.
At any point (x, y) in the domain, the value of the function f (x, y) changes at different rates when one moves in different directions on the xy-plane.
f is a vector in the plane of the independent variables (the xy-plane). The magnitude of f at a point (x, y) is the maximum instantaneous rate of increase of f at that point. The direction of f at that point is the direction in which one would have to start moving on the xy-plane in order to experience that maximum rate of increase, (which is also at right angles to the contour f (x, y) = constant at that point). v v Points where f = 0 are critical points of f, (maximum, minimum or saddle point).
The directional derivative of f in the direction of the unit vector û is v Duˆ f f guˆ Both vectors are in the plane of the independent variables. The directional derivative is the component of f in the direction of û. v Duˆ f f uˆ cos
v v 蜒 ˆ � max( Duˆ f) f when u P f
v ˆ and min Duˆ f= 0 when u ^ f (when û is tangential to the contours of f ).
The results above can be extended to functions of more than two variables. n+1 For the hypersurface z = f (x1, x2, ... , xn) in , the chain rule becomes v v dfv dr v 抖 f f f d r dx1 dx 2 dxn =蜒fg , where f = , ,⋯ , and = , , ⋯ , dt dt抖 x1 x 2 xn dt dt dt dt ENGI 5432 1.3 The Gradient Vector Page 1-16
Example 1.3.1
The electrostatic potential at a point P(x, y, z) in 3 due to a point charge Q at the origin is 1 Q , wherer x2 y 2 z 2 . 4 r Find the rate of change of at the point (1, 2, 2) in the direction k 2i . Find the maximum value of the directional derivative over all directions at any point. Find the level surfaces.
1Q抖f Q d骣 1 r f =邹 = 鬃 琪 4per抖 x 4 pe dr桫 r x 抖r 1/2 But =( x2 + y 2 + z 2 ) 抖x x 1 -1/2 =( x2 + y 2 + z 2) ( x 2 + y 2 + z 2 ) 2 x 1 x =r-1 (2 x + 0 + 0) = 2 r �f -Q x Qx � � x4pe r2 r 4 pe r 3 This is the x component of the gradient vector. 抖f f Obtain , by symmetry. 抖y z v -Q - Q � f = x, y , z rv 4per3 4 pe r 3
At (1, 2, 2) r =12 + 2 2 + 2 2 = 9 = 3 . v -Q � f 1, 2, 2 P 4pe 27
1 uv = kˆ -2 iˆ � uˆ - 2, 0,1 5 -Q1 - Q � Df - 1, 2, 2g = 2, 0,1 -( 1( + 2) + 2( 0) 2( 1)) uˆ P 4pe 27 5 4 27pe 5 � Df 0 uˆ P ENGI 5432 1.3 The Gradient Vector Page 1-17
Example 1.3.1 (continued)
v + Q Q max D f= f =r = ( uˆ ) 4per3 4 pe r 2 This occurs when -rv (Q > 0) uˆ P +rv (Q < 0)
Level surfaces (contours): Q Q f =c� � c = r A 4per 4 pe c But r = A is a sphere, centre O.
Therefore the level surfaces are a family of concentric spheres.
[Q > 0 case] ENGI 5432 1.3 The Gradient Vector Page 1-18
Change of coordinates:
z Suppose z = f (x, y) (where (x, y) are Cartesian coordinates) and is wanted, (where r (r, ) are plane polar coordinates). Then
z z x z y r x r y r But x = r cos , y = r sin z z z cos sin r x y z z z rsin r cos can be found in a similar way. x y
In matrix form, the chain rule can be expressed concisely as z x y z r r r x . z x y z y Note that x y x, y r r abs det is the Jacobian r, x y For the transformation from Cartesian to plane polar coordinates in 2, the Jacobian is ( x, y) cosq sin q = =rcos2q + r sin 2 q = r (r,q ) -rsinq r cos q Integrals over areas can therefore be transformed using the Jacobian: ( x, y) f( x, y) dx dy= f( x , y) dr dq = g( r , q) r dr d q 蝌蝌 (r,q ) 蝌 AA A where f (x, y) = g(r, ) at all points in the area A of integration. We shall return to this topic later. ENGI 5432 1.3 The Gradient Vector Page 1-19
Planes
Let n = a normal vector to a plane = A, B, C and ro = the position vector of a point on the plane = xo, yo, zo where A, B, C, xo, yo, zo are all constants. Let r = the position vector of a general point in 3 = x, y, z Then the point (x, y, z) is on the plane if and only if v v v v ng r n g ro which is a vector equation of the plane. Evaluating the scalar [dot] products generates the Cartesian equation of the plane: Ax By Cz D where D = Axo + Byo + Czo .
Lines
Let v = a vector parallel to the line = v1, v2, v3 and ro = the position vector of a point on the line = xo, yo, zo then the vector parametric form of the equation of the line is v v v r ro t v and the Cartesian equivalent, (in the case where none of v1, v2, v3 is zero) is x x y y z z o o o v1 v 2 v 3
Surfaces The general Cartesian equation of a surface in 3 is of the form f (x, y, z) = c At every point on the surface where f exists as a non-zero vector, f is orthogonal (perpendicular) to the level surface of the function f that passes through that point. Therefore, at every point on the surface f (x, y, z) = c, the gradient vector f is normal to the tangent plane.
The tangent plane at the point P(xo, yo, zo) to the surface f (x, y, z) = c has the equation
v nv rv nv rv, where nv f g go P ENGI 5432 1.3 The Gradient Vector Page 1-20
Example 1.3.2
Find the Cartesian equations of the tangent plane and normal line to the surface z = x2 + y at the point (1, 1, 2).
The implicit form of the equation of the surface is f = x2 + y – z = 0 v f=2 x ,1, - 1
v � f - -2,1, = 1 nv P v v ng ro = -2( - 1) + 1( 1) - 1( 2) = 1
Therefore the equation of the tangent plane is –2x + y – z = 1 or
2 x – y + z + 1 = 0
The equation of the normal line follows immediately:
In vector parametric form: rv = -1,1, 2 +t - 2,1, - 1 or, in Cartesian parametric form, x= -1 - 2 t , y = 1 + t , z = 2 - t
x+1 y - 1 z - 2 or, in Cartesian symmetric form, = = -2 1 - 1 or x+1 y - 1 z - 2 = = 2- 1 1 ENGI 5432 1.3 The Gradient Vector Page 1-21
Example 1.3.3
Find the angle between the surfaces x2 + y2 + z2 = 4 and z2 + x2 = 2 at the point (1, 2 , 1).
First note that these surfaces are a sphere, radius 2, centre the origin and a circular cylinder of radius 2 aligned along the y axis. Their intersection will therefore be a pair of parallel circles, equally spaced on either side of the x-z plane.
The point P (1, 2 , 1) is clearly on both surfaces [satisfies the equations of both surfaces].
2 2 2 v v f= x + y + z =4� f = 2 x , 2 y , 2 z = 2 x , y , z 2n1
v At P, n1 = 1, 2,1
2 2 v v g= x + z =2� g = 2 x , 0, 2 z = 2 x , 0, z 2n2
v At P, n2 = 1, 0,1
The angle between the surfaces equals the angle between the two normal vectors. v v n1g n 2 =1( 1) + 2( 0) + 1( 1) = 2 n1 =1 + 2 + 1 = 2 n2 =1 + 0 + 1 = 2 nv nv 2 1 p �cosq=1g 2 = � q ( 45 ) n1 n 2 2 2 2 4
v v Note: In the event that n1g n 2 < 0 , then the two normal vectors meet at an obtuse angle v v (they are pointing in approximately opposite directions). In that case use n1g n 2 . ENGI 5432 1.3 The Gradient Vector Page 1-22
Gradient Operator
For three independent variables (x, y, z), the gradient operator is the “vector” v ˆi ˆj kˆ , , x y z x y z It operates on anything immediately to its right; either a scalar function or scalar field, or, via a dot product or cross product, on a vector function or vector field. ENGI 5432 1.4 Divergence and Curl Page 1-23
1.4 Divergence and Curl
For an elementary area A in a vector field F , nˆ is an outward unit normal to the surface.
A is sufficiently small that F = f1, f2, f3 is approximately constant over A.
The element of flux from the vector field through A is approximately v Df 籇Fg nˆ A Now add up the elements of flux passing through the six faces of an elementary cuboid of sides x, y, z , volume V = x y z and with one corner at (x, y, z).
The front face is at (x + x) and the back face is at (x).
ˆ Back face: nˆ B = - i
DA = Dy 譊z
v v ˆ ˆ Fg nB = Fg(- i) = - f1 ( x, y , z)
DfB = - f1 ( x, y , z) D y譊 z
ˆ Front face: nˆ F = + i
DA = Dy 譊z
v v ˆ � Fg nˆ F Fg i =f1 ( x + D x, y , z) [front face is at x + x]
轉 fF = +f1 ( x + D x, y , z)譊 y 譊 z ENGI 5432 1.4 Divergence and Curl Page 1-24
Divergence (continued)
Df+ D f f( x+ D x, y , z) - f( x ,y , z) � F B 1 1 DV Dx
骣Df + D f f and lim 琪 F B = 1 DV 0 桫 DV x
Similarly, for the left and right faces,
骣Df + D f f lim 琪 L R = 2 DV 0 桫 DV y and for the top and bottom faces,
骣Df + D f f lim 琪 TOP BOT = 3 DV 0 桫 DV z
Summing over all six faces of the cuboid, the net rate of flux per unit volume out of a point (x, y, z) is
骣Dfd f 抖f f f lim 琪 = =1 + 2 + 3 DV 0 桫DV dV抖 x y z
v v v =蜒gf1, f 2 , f 3 = gF which is the divergence of F. v No net flux (no sources) � divF 0 ENGI 5432 1.4 Divergence and Curl Page 1-25
Example 1.4.1
Q Find the divergence of the vector F, given that F = , , 4r r x2 y 2 z 2 .
From example 1.3.1, v-Q v v + Q 蜒f=rv� F - = f rv 4per3 4 pe r 3
v Q抖 x y z � divF , ,g , , 4pe 抖x y z r3 r 3 r 3
抖r x r y r z Also from example 1.3.1, =, = , = 抖x r y r z r
3骣 2 x 1r- x琪 3 r 2 2 �骣x桫 r r( r- 3 x ) r23 x 2 � 琪 = = x桫 r3 r 6 r 6 r 5
The other two derivatives follow by symmetry.
2 2 v Q骣 r2-3 x 2 + r 2 - 3 y 2 + r 2 - 3 z 2 Q(3 r- 3 r ) � divF =琪 5 = 5 0 4pe桫 r 4 pe r (unless r = 0, where div F is undefined).
Therefore the ideal electrostatic field is source free except at the point charge itself. The flux passing in through any volume not enclosing the charge is balanced by the flux passing out. ENGI 5432 1.4 Divergence and Curl Page 1-26
Streamlines for Fluid Flow
Let v(r) be the velocity at any point (x, y, z) in an incompressible fluid. Because the fluid is incompressible, the flow in to any region must be matched by the flow out from that region (except when the region includes a source or a sink). This generates the continuity equation
div v = 0
Let us take the case of fluid flow parallel to the x-y plane everywhere, so that we can ignore the third dimension and consider the flow in two dimensions only. Then
v= v( x, y) = u( x , y)ˆ i + v( x , y) ˆj
The continuity equation then becomes
vv v 抖u v divv=g v = + = 0 抖x y
The fluid flows along streamlines, but never across any streamline. Consider the flow through a small region bounded partly by a streamline: streamline
轉 v x= u D y � v dx = u dy 0
This ODE is exact if and only if
抖 抖u v (v) =( - u) � = 0 抖y x 抖 x y which is automatically true (equation of continuity).
The solution to the ODE is then the stream function y ( x, y) , where
抖y y =vand = - u 抖x y ENGI 5432 1.4 Divergence and Curl Page 1-27
Example 1.4.2 (Example 1.2.3 repeated)
-y x Find the streamlines associated with the velocity field v = , x2+ y 2 x 2 + y 2 and find the streamline through the point (1, 0).
-y x u=, v = x2+ y 2 x 2 + y 2
Verify that the equation of continuity is satisfied: v v 抖u v� 骣 y 骣 x gv = + =琪 + 琪 抖x y� x桫 x2 � y 2 y 桫 x 2 y 2 -y( -2 x) x( - 2 y) = + =0 " x , y 0,0 2 2 ( ) ( ) ( x2+ y 2) ( x 2 + y 2 )
Therefore the stream function y ( x, y) exists, such that 抖yx y + y =v =and = - u = �x x2 � y 2 y x 2 y 2
1 2 2 2 22C 2 �y ( x, y) +2 ln ( x = y) � C = x = y e A
Therefore the streamlines in the x-y plane are x2+ y 2 = A 2
( x, y) =( 1,0) � 12 = 0 2 A 2 The required streamline is therefore the unit circle
x2+ y 2 =1
The flow is a vortex around the origin. ENGI 5432 1.4 Divergence and Curl Page 1-28
Divergence (a scalar quantity): vv v div F g F Curl (a vector quantity):
ˆiˆ j kˆ vv v F F F F F F curlF F det 3 2 , 1 3 , 2 1 x y z y z z x x y F1 F 2 F 3 curl F = 0 everywhere F is an irrotational vector field.
Example 1.4.3
Find curl F for F = cos y, sin x, 0 . Also find the lines of force for the vector field F.
ˆi ˆj kˆ v v 抖 汛 F = =0 - 0, 0 - 0, - cosx + sin y 抖x y z cosy- sin x 0
v � curlF- ( siny cos x) kˆ
Lines of force: drv v dx dy dz =kF � , , - k cos y , sin x , 0 ds ds ds ds
dz dx dy � =0 and k ds = dscos y- sin x
� z- Band蝌 = sin x dx cos y dy The lines of force are therefore siny= A + cos x , z = B In order to be real, -2＃A 2
It then follows that curl F = A k and is constant along any one line of force. ENGI 5432 1.4 Divergence and Curl Page 1-29
Example 1.4.3 (continued)
Direction field plot for the vector field F = cos y, sin x, 0 :
curl F = (sin y cos x) k
Lines of force: sin y = A + cos x (2 A 2)
and curl F = A k .
Along the highlighted lines (at 45° angles), A = 0 (and therefore curl F = 0). Where those lines cross, F = 0 also, (in addition to A = 0 and curl F = 0). Half way along the lines between those intersections, | F | is at a maximum ( 2 ). At the highlighted dots, | curl F | achieves its maximum value of 2 and F = 0 and where the lines of force near a dot are in an anticlockwise direction, curl F = +2 k; where the lines of force near a dot are in a clockwise direction, curl F = –2 k. ENGI 5432 1.4 Divergence and Curl Page 1-30
All differentiable gradient-vector fields are irrotational:
v v v curl grad 0 Proof:
ˆi ˆj kˆ v v v 抖 v curlF=汛 f = = f - f , f - f , f - f = 0 抖x y z zy yz xz zx yx xy 抖f f f 抖x y z
Also: vv v v div curlFg F 0 Proof:
v f LetF =f , f , f and f = 1 etc., then 1 2 3 1x x v v v 抖 蜒gF =, , g f3y - f 2 z , f 1 z - f 3 x , f 2 x - f 1 y 抖x y z
= ( f3yx - f2zx ) + ( f1zy - f3xy ) + ( f2xz - f1yz ) = 0
The Laplacian of a twice-differentiable scalar field is: 2 2 2 2 div grad , ,g , , x y z x y z x2 y 2 z 2 Laplace’s equation is 2 2 2 2 0 x2 y 2 z 2 ENGI 5432 1.4 Divergence and Curl Page 1-31
Some Vector Identities f curl grad f 0 F div curl F 0 Laplacian of V 2V V div grad V
2F F F grad div F curl curl F f g f g f g g F g F g F div (g F) = (grad g)F + g div F g F g F g F curl (g F) = (grad g)F + g curl F F G F G F G div (F G) = (curl F)G F(curl G)
F G G F F G GF FG , G1 G1 G1 ˆ where F G F1 F2 F3 i x y z G2 G2 G2 ˆ F1 F2 F3 j x y z G3 G3 G3 ˆ F1 F2 F3 k x y z so that F is the operator F1 F2 F3 x y z F G G F F G G F F G