ENGI Gradient, Divergence, Curl Page 5.01

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1 ENGI Gradient, Divergence, Curl Page The Gradient Operator A brief review is provided here for the gradient operator in both Cartesian and orthogonal non-cartesian coordinate systems. Sections in this Chapter: 5. Gradient, Divergence, Curl and Laplacian (Cartesian) 5. Differentiation in Orthogonal Curvilinear Coordinate Systems 5. Summary Table for the Gradient Operator 5.4 Derivatives of Basis Vectors 5. Gradient, Divergence, Curl and Laplacian (Cartesian) Let z be a function of two independent variables (x, y), so that z = f (x, y). The function z = f (x, y) defines a surface in. At any point (x, y) in the x-y plane, the direction in which one must travel in order to experience the greatest possible rate of increase in z at that point is the direction of the gradient vector, f ˆ f f i ˆj x y The magnitude of the gradient vector is that greatest possible rate of increase in z at that point. The gradient vector is not constant everywhere, unless the surface is a plane. (The symbol is usually pronounced del ). The concept of the gradient vector can be extended to functions of any number of variables. If u = f (x, y, z, t), then f f f f f x y z t. If v is a function of position r and time t, while position is in turn a function of time, then by the chain rule of differentiation, d v v v dy v dz v dr v v dt x dt y dt z dt t dt t d v v v v dt t which is of use in the study of fluid dynamics. T

2 ENGI Gradient, Divergence, Curl Page 5. The gradient operator can also be applied to vectors via the scalar (dot) and vector (cross) products: The divergence of a vector field F(x, y, z) is div F T F F F F F F F x y z x y z T A region free of sources and sinks will have zero divergence: the total flux into any region is balanced by the total flux out from that region. The curl of a vector field F(x, y, z) is ˆi curl F F ˆj F kˆ F F F x y z F F y z x F F F z x y In an irrotational field, curl Whenever or Proof: F. F for some twice differentiable potential function, curl F curl grad T T F F F F x y z ˆ i x x y z z y curl ˆ j y y z x x z ˆ k z z x y y x

3 ENGI Gradient, Divergence, Curl Page 5. Among many identities involving the gradient operator is div curl F F for all twice-differentiable vector functions F Proof: div curl F F F F F F F x y z y z x z x y F F F F F F x y x z y z y x z x z y The divergence of the gradient of a scalar function is the Laplacian: div grad f f f f f f x y z for all twice-differentiable scalar functions f. In orthogonal non-cartesian coordinate systems, the expressions for the gradient operator are not as simple.

4 ENGI Curvilinear Gradient Page Differentiation in Orthogonal Curvilinear Coordinate Systems For any orthogonal curvilinear coordinate system (u, u, u ) in the unit tangent vectors along the curvilinear axes are ˆ r e T, h u where the scale factors h i r u i., ˆi i i i The displacement vector r can then be written as r ueˆ ˆ ˆ ue ue, where the unit vectors ê i form an orthonormal basis for. i j eˆi e ˆ j i j i j The differential displacement vector dr is (by the Chain Rule) r r r dr du du du h du eˆ h du eˆ h du e ˆ u u u and the differential arc length ds is given by ds dr dr h du h du h du The element of volume dv is x, y, z dv h h h du du du du du du u, u, u Jacobian x y u u x y u u x y u u z u z u z u Example 5..: Find the scale factor h θ for the spherical polar coordinate system x, y, z r sin cos, r sin sin, r cos : T r x y z r cos cos r cos sin r sin r h r r r cos cos cos sin sin r cos cos sin sin cos sin du du du r r r T

5 ENGI Gradient Summary Page Summary Table for the Gradient Operator eˆ eˆ ˆ e Gradient operator h u h u h u eˆ V eˆ ˆ V e V Gradient V h u h u h u Divergence h h F h h F h h F F h h h u u u Curl F h h h h eˆ h F u h eˆ h F u h eˆ h F u Laplacian V = h h h h h u h V u u h h V h u u h h h V u Scale factors: Cartesian: h x = h y = h z =. Cylindrical polar: h = h z =, h =. Spherical polar: h r =, h = r, h = r sin. Example 5..: The Laplacian of V in spherical polars is sin V V r sin V V r sin r r sin or V V V cot V V V r r r r r sin

6 ENGI Gradient Summary Page 5.6 Example 5.. A potential function V r is spherically symmetric, (that is, its value depends only on the distance r from the origin), due solely to a point source at the origin. There are no other sources or sinks anywhere in V r.. Deduce the functional form of V r is spherically symmetric V r,, f r In any regions not containing any sources of the vector field, the divergence of the vector field F V (and therefore the Laplacian of the associated potential function V) must be zero. Therefore, for all r, div F V V But sin V r sin V r sin r r sin d V dv sin r r sin dr dr d dv dv dv r r B B r dr dr dr dr V Br A, where A, B are arbitrary constants of integration. Therefore the potential function must be of the form B V r,, A r This is the standard form of the potential function associated with a force that obeys the inverse square law F. r

7 ENGI Basis Vectors Page Derivatives of Basis Vectors d Cartesian: ˆ d ˆ d i j kˆ dt dt dt r x ˆi y ˆj z kˆ v x ˆi y ˆj z kˆ Cylindrical Polar Coordinates: x cos, y sin, z z d d ˆ ˆ r ˆ z kˆ dt dt d ˆ d ˆ v ˆ ˆ z kˆ dt dt d kˆ [radial and transverse components of v ] dt Spherical Polar Coordinates. The declination angle θ is the angle between the positive z axis and the radius vector r. < θ < π. The azimuth angle is the angle on the x-y plane, measured anticlockwise from the positive x axis, of the shadow of the radius vector. < < π. z = r cos θ. The shadow of the radius vector on the x-y plane has length r sin θ. It then follows that x = r sin θ cos and y = r sin θ sin. d d ˆ ˆ d r sin ˆ dt dt dt r r r ˆ d ˆ d d rˆ cos ˆ dt dt dt v r rˆ r ˆ r sin ˆ d ˆ d sin rˆ cos ˆ dt dt

8 ENGI Basis Vectors Page 5.8 Example 5.4. Find the velocity and acceleration in cylindrical polar coordinates for a particle travelling along the helix x = cos t, y = sin t, z = t. Cylindrical polar coordinates: x cos, y sin, z z y x y, tan x 9cos t 9sin t 9 sin t tan tan t t cost r ˆ z kˆ z t z dr v ˆ ˆ zkˆ ˆ ˆ kˆ 6 ˆ kˆ dt [The velocity has no radial component the helix remains the same distance from the z axis at all times.] d v a 6 ˆ kˆ 6 ˆ ˆ dt [The acceleration vector points directly at the z axis at all times.] Other examples are in the problem sets. END OF CHAPTER 5

9 ENGI 94 Lecture Notes 6 Calculus of Variations Page Calculus of Variations The method of calculus of variations involves finding the path between two points that provides the minimum (or maximum) value of integrals of the form b,, a F x y y Sections in this Chapter: 6. Introduction 6. Theory 6. Examples Sections for reference; not examinable: 6.4 Integrals with more than One Dependent Variable 6.5 Integrals with Higher Derivatives 6.6 Integrals with Several Independent Variables 6.7 Integrals subject to a Constraint

10 ENGI Introduction Page Introduction Example 6.. To find the shortest path, (the geodesic), between two points, we need to find an expression for the arc length along a path between the two points. Consider a pair of nearby points. The element of arc length Δs is approximately the hypotenuse of the triangle. s x y s x y x x x In the limit as the two points approach each other and x, we obtain ds dy ds dy The arc length s between any two points x = a and x = b along any path C in line integral dy s ds f x where C is the path y f x C C C is the The geodesic will be the path C for which the line integral for s attains its minimum value. Of course, in a flat space such as, that geodesic is just the straight line between the two points.

11 ENGI Theory Page Theory We wish to find the curve y(x) which passes through the points and which minimizes the integral x I F x, y x, yx x Consider the one parameter family of curves yx ux x x y and x, y,, where is a real parameter, η(x) is an arbitrary function except for the requirement x x and u(x) represents the (as yet unknown) solution. Every member of this family of curves passes through the points For any member of the family, x I F x, u x x, ux x x x y and,, we know that y(x) = u(x) minimizes I. di Therefore the minimum for I occurs when α =, so that. d Carrying out a Leibnitz differentiation of the integral I, di x F x, u x x, ux x d x x F F u x x ux x x y y At the minimum α =, so that y(x) = u(x) and y'(x) = u'(x). Therefore x F F x x x u u Also note, by the product rule of differentiation, that d x F x F x d F u u u Therefore the integral can be written as x F d F d F x x x x u u u x x F d F d F x x x u u x u x x F d F F x x x u u u x x y.

12 ENGI Theory Page 6.4 ( x) η( x) But η = = Therefore the minimizing curve u(x) satisfies x F d F η ( x) = x u u But η ( x) is an arbitrary function of x, which leads to F d F = u u Thus, if y = f ( x) is a path that minimizes the integral F ( x, y, y ), then y = f ( x) a and F (x, y, y') must satisfy the Euler equation for extremals b d F F = y y Euler s equation requires the assumption that F(x, y, y') has continuous second derivatives in all three of its variables and that all members of the family y x = u x + αη x have continuous second derivatives. ( ) ( ) ( ) Expansion of Euler s Equation: d F F ( xyx, ( ), y ( x) ) ( xyx, ( ), y ( x) ) = y y F F F F + y ( x) + y ( x) x y y y y y = or y y yy xy ( y ) y F + yf + F F = gz ( ) = F x z with respect to z is: f ( z) = g ( z) F( g( z), z) gz ( ) f ( z) F( f ( z), z) + F( x, z) f ( z) z d x f t dt a = f x. Note: Leibnitz differentiation of I ( z) (, ) di dz A special case of this is ( ) ( )