MA20010: Vector Calculus and Partial Differential Equations

Transcription

1 MA20010: Vector alculus and Partial Differential Equations Robert cheichl Department of Mathematical ciences University of Bath October 2006 January 2007 This course deals with basic concepts and results in vector integration and vector calculus, Fourier series, and the solution of partial differential equations by separation of variables. It is fundamental to most areas of applied Maths and many areas of pure Maths, a prerequisite for a large number (11!!) of courses in emester 2 and in Year 3/4. There will be two class tests: Tuesday, 31st October, 5.15pm (Line, surface, volume integrals) Tuesday, 28th November, 5.15pm (Vector calculus and integral theorems) Questions will be similar to the questions on the problem sheets! Please take this course seriously and do not fall behind with the problem sheets. The syllabus for this course is very dense and only through constant practice will you be able to grasp the multitude of methods and concepts. Please revise MA10005 ect. 6-8 Multivariate alculus MA10006 ect. 1 Vector Algebra The first AIM quiz is designed to help you focus on the relevant material in those courses.

2 ontents 1 Vector Integration Line Integrals Parametric Representation Arclength Line Integrals of calar Fields Line Integrals of Vector Fields Application to Particle Motion urface Integrals The urface Elements d and d urface Integrals of calar Fields Flux urface Integrals of Vector Fields Volume Integrals hange of Variables Reparametrisation Vector alculus Directional Derivatives and Gradients Application: Level urfaces and Grad Divergence and url; the Operator econd order derivatives the Laplace Operator Application: Potential Theory [Bourne, pp ] Differentiation in urvilinear oordinates Orthogonal urvilinear oordinates Grad, div, curl, 2 in Orthogonal urvilinear oordinates Integral Theorems The Divergence Theorem of Gauss Green s Theorem in the Plane tokes Theorem Fourier eries Periodicity of Fourier eries Fourier onvergence Gibbs Phenomenon Half-range eries Application: Eigenproblems for 2nd-order ODEs Partial Differential Equations lassification of Partial Differential Equations eparation of Variables Fourier s Method Elliptic PDEs Laplace s Equation

3 5.2.2 ircular Geometry Hyperbolic PDEs Wave Equation Parabolic PDEs Diffusion Equation Other olution Methods for PDEs (not examinable!) The Laplace Transform Method D Alembert s olution of the Wave Equation

4 hapter 1 Vector Integration In this course we will be dealing with functions in more than one unknown whose function value might be vector valued as well. Definition 1.1. A function F : R n R n is called a vector field on R n. A function f : R n R is called a scalar field on R n. (Usually n will be 2 or 3.) Example 1.2. Typical examples of vector fields are gravitational field G(x) electrical field E(x) magnetic field B(x) velocity field V (x) Example 1.3. Typical examples of scalar fields are speed V (x) kinetic energy 1 2 m V (x) 2 electrical potential ϕ E (x) We will now learn how to integrate such fields in R n. 1.1 Line Integrals [Bourne, pp ] & [Anton, pp ] Example 1.4. (Motivating Problem). y B What is the perimeter of the unit circle? replacements 1 A x Perimeter = 4 length s of arc AB Let us divide the interval [ 1, 0] into N elements of length x i = 1/N, i = 0,..., N 1: 1

5 Pfrag replacements y s i B y i s i y i x i A x eter = 4 length s of arc AB 1 x i Obviously, with s i as constructed in the above figure s = N 1 i=0 s i. Moreover 2

6 s = dx (1.1) 1 x 2 DIY Use the substitution x = cos θ to evaluate (1.1): Hence, the perimeter is Parametric Representation Arclength Let be a curve in R n with parametric representation r(t), i.e. = {r(t) : t 0 t t e } (1.2) where r : [t 0, t e ] R n is continuously differentiable. replacements 0 r(t) As t increases r(t) traces out the curve. Note. In Ex. 1.4 t was chosen to be x. In general, t can be any parameter (e.g. angle θ). 3

7 Example 1.5. Give a parametric representation of the ellipse x2 a 2 + y2 b 2 = 1. As in Example 1.4 we can now calculate the arclength s(t) of from r(t 0 ) to r(t) for every t [t 0, t e ]. Let t := t t 0 N for some N N and let t i := t 0 + i t, for i = 0,..., N 1. Therefore Pfrag replacements s i r(t i ) r(t i + t) (1.3) and (1.4) Letting N (which also means that t 0), the arclength is t ( s(t) = dr t ) ( dt dt = dx ) 2 ( dt + dy ) 2 ( dt + dz ) 2 dt dt in R 3. (1.5) t 0 t 0 Also, in the limit as t 0 equation (1.3) becomes ds = dr dt dt. (1.6) The length of the entire curve is given by L = s(t e ) = te t 0 dr dt dt. (1.7) 4

8 1.1.2 Line Integrals of calar Fields Now assume we want to calculate the integral of a scalar field f : R n R along the curve. Geometrically this means, to calculate the following area A: Pfrag replacements z = f(x, y) z f(r(t i )) A y (in R 2 ) x r(t 0 ) As in (1.4) with t = t e t 0 N we get s i r(t e ) (i.e. we approximate the area by the sum of the strips.) Hence for N A = te t 0 f(r(t)) dr dt dt (an ordinary integral of a scalar valued function of t). Definition 1.6. The line integral of a scalar field f : R n R along the curve in (1.2) is defined as te f ds := f(r(t)) dr dt dt. (1.8) Remark 1.7. t 0 (a) The value of f ds does not depend on the choice of parametrisation, even if the orientation of is reversed [Bourne, p. 148], [Anton, pp ]. (b) The length of the curve in (1.6) is a special line integral, i.e. L = 1 ds. Example 1.8. Integrate f(x, y) = 2xy around the first quadrant of a circle with radius a as shown: y Pfrag replacements a x 5

9 tep 1. Parametrise circle: tep 2. alculate the line element ds : DIY tep 3. Apply formula (1.8) for the line integral: Line Integrals of Vector Fields Definition 1.9. The work integral of a vector field F : R n R n along the curve in (1.2) is defined as te F dr := F (r(t)) dr dt. (1.9) dt (dot product!) t 0 Theorem If ˆT is the unit tangent vector to in (1.2) that points in the direction in which t is increasing, then F dr = (F ˆT ) ds, (1.10) i.e. the work integral of F = the line integral of the component of F parallel to. Proof. 6

10 Remark (a) It follows directly from Theorem 1.10 that a reversal of the orientation of does change the sign of the work integral (in contrast to the line integral of a scalar field, cf. Remark 1.7(a)) [Bourne, p. 153], [Anton, p. 1109]. (b) If F is a force field then F dr is the work done by moving a particle from r(t 0) to r(t e ) along the curve ; hence the term work integral. Example (in tutorial). x2 is the ellipse a + y2 = 1 in the upper half plane as shown: 2 b2 Pfrag replacements y Evaluate the integral F dr where F (x, y) = (2y, x2 ) T and b a x tep 1. Parametrise the curve (Example 1.5): r(t) = (a cos t, b sin t) T, t [0, π]. tep 2. alculate the vector line element dr = dr dt dt : tep 3. Apply formula (1.9) for the work integral: 7

11 Definition (in tutorial). Let F be a vector field such that F = F 1 i + F 2 j + F 3 k. The (vector valued) integrals F ds and F dr are defined as ( ) T F ds := F 1 ds, F 2 ds, F 3 ds F dr := te t 0 F (r(t)) dr dt dt. Definition A vector field F : R n R n is called conservative, if F dr = F dr (1.11) 1 2 for any two curves 1 and 2 connecting two points x 0 and x e in R n Pfrag replacements 2 x e 1 x Application to Particle Motion uppose F : R n R n is a force field and suppose that a particle of mass m moves along a curve through this field. uppose further that r(t) is the position of the particle at time t [t 0, t e ] (i.e. r(t) is a parametric representation of where the parameter t is time). Recall: particle velocity kinetic energy V (t) = dr dt K(t) = 1 m V (t) 2 2 Newton s econd Law states Therefore Force applied = mass * acceleration = F (r(t)) = m d2 r dt 2. F dr = te t 0 = 1 2 m te F (r(t)) dr dt dt = t 0 te ( d dr dt dt dr ) dt = 1 te dt 2 m t 0 t 0 ( ) m d2 r dr dt 2 dt dt d dt V (t) 2 dt and so by the Fundamental Theorem of alculus F dr = 1 2 m V (t e) m V (t 0) 2, (1.12) or in other words 8

12 Work done = hange in kinetic energy from t 0 to t e ( conservation of energy principle ). 1.2 urface Integrals [Bourne, pp ] & [Anton, pp ] In the previous section we discussed integrals along one-dimensional curves in R n. Now we want to look at integrals on two-dimensional surfaces in R n. You have already done the case n = 2 in MA10005, i.e. (flat) surfaces in R 2. Now we will consider surfaces in R 3, i.e. n = 3. Pfrag Recall replacements (for curves): parametric representation r(t) t 0 t e Now, let be a (two-dimensional) surface in R 3 with parametric representation r(u, v), i.e. = {r(u, v) : (u, v) D} (1.13) where r : D R 3 is continuously differentiable and D R 2. Example (tep 1). A cylindrical shell of height b and radius a frag replacements z a r(u, v) v u v b y b 2π D u x ric representation can be parametrised by 9

13 1.2.1 The urface Elements d and d Recall: the equation of a plane in R 3 is x = x 0 + λl + µm, λ, µ R (where l, m are two arbitrary non-parallel vectors on the plane). Pfrag replacements The unit normal to this plane is ˆn = l m l m. x 0 ˆn l m 0 Now, let = {r(u, v) : (u, v) D} as in (1.13) and let (u 0, v 0 ) D. Fixing v 0 we get Pfrag replacements r v r u u0 r(u 0, v 0 ) v0 imilarly, 0 Definition (a) The tangent plane of in (1.13) at the point r(u 0, v 0 ) is { T := x R 3 : x = r(u 0, v 0 ) + λ r u (u 0, v 0 ) + µ r } v (u 0, v 0 ), λ, µ R. (1.14) (b) The unit normal to at r(u 0, v 0 ) is ( r ˆn := u r ) / r v u r v (u 0, v 0 ) (1.15) Note (on orientation). Reversing the rôles of u and v, changes the sign of ˆn. This is a matter of convention. For closed surfaces (e.g. sphere, cylinder, torus) it is common to define ˆn to be the normal that points outward. (If it is not clear we will always specify the orientation of the normal/surface.) 10

14 We will now define a surface element d for the surface = {r(u, v) : (u, v) D} in (1.13): Recall the line element ds := dr dt dt which we defined in (1.6) for a Pfrag replacements (one-dimensional) curve. ds g replacements Let P 0 be a point on the surface, such that OP 0 = r(u, v), let P 1 be a neighbouring point with OP 1 = r(u + u, v). imilarly, let P 2 and P 3 be the points with OP 2 = r(u, v + v) and OP 3 = r(u + u, v + v). v + v v r(u, v) P 0 P 2 P 3 P 1 v + v v D 0 u u u + u u + u For u and v sufficiently small d = r u r v du dv. (1.16) d is called the (scalar) surface element. Example (tep 2). alculate d and find the outward unit normal of the cylindrical shell of height b and radius a. 11

15 DIY (Recall tep 1 (i.e. Example 1.15): r(u, v) := (a cos u, a sin u, v) T.) The vector surface element d is defined as d := d ˆn, (1.17) or using (1.15) and (1.16) d = ( r u r ) du dv. (1.18) v urface Integrals of calar Fields Definition The surface integral of a scalar field f : R n R on in (1.13) is defined as f d = f(r(u, v)) r u r v du dv. (1.19) D Example (tep 3). Find the surface area of a cylindrical shell of height b, radius a. [Recall: tep 1 (i.e. Example 1.15: r(u, v) := (a cos u, a sin u, v) T, 0 u 2π, 0 v b, tep 2 (i.e. Example 1.17: d = a du dv. ] ( surface integral 2D integral repeated integral. ) Note. Try always to parametrise a surface in such a way that either D is rectangular or D = {(u, v) : u [u 0, u e ] and v [v 0 (u), v e (u)]}. This makes the transition from the 2D-integral to the repeated integral easier (compare MA10005). 12

16 Pfrag replacements v v e (u) D v 0 (u) u u 0 u e Remark (a) Another way to define the surface integral would be to use a subdivision of into (little) surface elements i with centre x i (as for line integrals), i.e. f d = ee [Anton, p. 1130]. lim N N 1 i=0 f(x i ) Pfrag i. replacements (1.20) x i i (b) The area of a surface is a special surface integral, i.e. A = Flux urface Integrals of Vector Fields d (cf. Remark 1.7(b)). Definition The flux (or surface integral) of a vector field F : R n R n across a surface is defined as ( r F d = F (r(u, v)) u r ) du dv. (1.21) v Remark (a) It follows from (1.17) that F d = D (F ˆn) d, i.e. the flux of F across = surface integral of the normal component of F on. (b) (physical interpretation) If F is for example the velocity field of some fluid then the flux across is the net volume of fluid that passes through the surface per unit of time. For more details: DIY Example Find Please study [Anton, pp ] until the next lecture. x d, where is the sphere with radius a, i.e. := {x R 3 : x 2 + y 2 + z 2 = a 2 }. 13

17 tep 1. Parametrise : spherical polar coordinates tep 2. alculate r θ r ϕ : tep 3. Apply formula (1.21) for the surface integral: Remark (a) o far we have only considered smooth surfaces, i.e. surfaces with continuously differentiable parametric representation r. However, the notion of surface integrals can be extended to surface integrals on piecewise smooth surfaces: If := n, where i is smooth for each i = 1,..., n, then we define := n 14

18 For example: Let be the surface of the unit cube and let 1,..., 6 be the six faces (obviously Pfrag replacements each one of the i, i = 1,..., 6 is smooth), and so = (b) Analagously, for piecewise smooth curves := m, we define := m. 1.3 Volume Integrals [Bourne, pp ] & [Anton, pp ] Let f : Ω R be a scalar field defined on a bounded region Ω R 3. As in R 2 (recall MA10005) we can subdivide Ω into little volume elements V i with centre x i and define the volume integral N 1 Pfrag replacements f dv = lim f(x i ) V i N z Ω Ω i=0 (where V i 0 as N ). In rectangular artesian coordinates one might choose V i = x y z and hence obtain in the limit the volume element dv = dx dy dz. Thus Ω f dv = Ω x x i y V i x z f(x, y, z) dx dy dz. (1.22) If f is continuous, the right hand side can be evaluated as a repeated integral again (in any order you want). However the limits of these repeated integrals might be very akward. Example Find the volume of the ball of radius a in the first octant, i.e. Ω := {x R 3 : x 2 + y 2 + z 2 a 2, x, y, z 0}. y y acements z x a 15

19 Things may become even nastier if we want to integrate over the entire ball hange of Variables Reparametrisation Let Ω := {r(u, v, w) : (u, v, w) D} (1.23) where D R 3 and r : D Ω is continuously differentiable. This will be particularly uesful, if D is box-shaped: Pfrag replacements D r Ω (u, v, w) r(u, v, w) = (x, y, z) T Example The parametrisation r(ρ, θ, φ) := (ρ sin θ cos φ, ρ sin θ sin φ, ρ cos θ) T maps the (box-shaped) region D = {(ρ, θ, φ) R 3 : 0 ρ a, 0 θ π, 0 φ < 2π } to the ball with radius a. (ρ, θ, φ) are called spherical polar coordinates (ρ... radial distance, θ... polar angle, φ... azimuthal angle). ee handout! DIY eplacements Please study the handouts on spherical and cylindrical polar coordinates until the next lecture. Motivation. Let r(u, v, w + w). OP 0 = r(u, v, w), OP 1 = r(u + u, v, w), OP 2 = r(u, v + v, w), OP 3 = w+ w v+ v r P 3 P 2 V w u u+ u v D P 0 P1 Ω 16

20 For u, v, w sufficiently small V will be approximately a parallelepiped. Recall (MA10006): Pfrag replacements ˆn c b a A V and in the limit as u, v, w 0 ( dv = r u r ) r v w du dv dw (1.24) the so-called volume element. Definition The volume integral of a scalar field f : Ω R over the region Ω R 3 in (1.23) is defined as ( f dv = f(r(u, v, w)) r u r ) r v w du dv dw. (1.25) Ω D 17

21 Note. Definition 1.27 is no contradiction to the change of variables formula which you have learnt in MA10005, since r 1 r 2 r 3 ( r u r ) r u u u v w = det r 1 r 2 r 3 (x, y, z) v v v =: (1.26) (u, v, w) r 1 r 2 r 3 w w w i.e. the Jacobian determinant. [ Proof. (recall MA10006) Remark (a) u, v, w can be regarded as curvilinear coordinates on Ω. (ee the figure at the beginning of ection ) (b) The integral (1.22) ( in artesian coordinates is just a special case with r(x, y, z) = (x, y, z) T, and so r r ) r x y z = 1. (c) The volume of Ω in (1.23) is a special volume integral, i.e. V Ω = Ω dv. Example Redo Example 1.25 using spherical polar coordinates. tep 1. Parametrise (see handout): tep 2. alculate the volume element dv = ( r r ) u v r w du dv dw : 18

22 tep 3. Evaluate Formula (1.25) for the volume integral: 19