NST1A: Mathematics II (Course A) End of Course Summary, Lent 2011

Transcription

1 General notes Proofs NT1A: Mathematics II (Course A) End of Course ummar, Lent 011 tuart Dalziel (011) This course is not about proofs, but rather about using different techniques. The proofs presented in lectures were intended primaril to help ou understand wh and where techniques work. Eaminations There are two papers, each divided into two sections. ection A on each paper is compulsor and each carries 0 marks worth of short-answer questions. ection B on each paper contains ten 0-mark questions, of which ou ma onl attempt five on each paper. Two of the ten questions on each paper will carr a star (*) to indicate that the require a knowledge of B course material, thus there are eight questions in ection B of each paper that are accessible to those undertaking the A course. While the division of the siteen A-course questions across the two papers will var from ear-to-ear, it is reasonable to epect around si questions covering Lent Term material. These questions ma not be distributed evenl between the two papers. tuart Dalziel 1 Lent 011

2 1. Ordinar differential equations 1. First order equations 1..1 eparable equations g F, h g h, h h g constant of integration from initial condition (or boundar condition ) 1..3 Linear equations p f. Homogeneous equation complementar function 1 p p ln p c 1..4 Integrating factors p Ae. Inhomogeneous equation particular integral If I ep p d I p I I f then so I I f, ep p d f d c. ep pd tuart Dalziel Lent 011

3 The term involving the constant is the complementar function. The term not involving the constant is the particular integral. Linear can add complementar function to particular integral and still have solution of inhomogeneous equation 1..5 olution b substitution The idea is to convert the equation into something that is either linear or separable. Homogeneous equations H d du Let = u uu H H u du Bernoulli s differential equation H u u If n du H u u p q use z = 1n. Man other substitutions possible f() suggests u = ; ln suggests u = ln Look at terms that arise in the equation (f(,) sa) and tr u = f(,) Other strategies (from Chapter ) Can rewrite equation for () as equation for (): e.g. a ( ) b 1 0 a ( ) b ( ) Can also use the idea of eact differentials and integrating factors tuart Dalziel 3 Lent 011

4 trateg first order ordinar differential equations 1. Is equation separable?. Is equation linear? 3. Is equation homogeneous? Tr u = / 4. Is equation Bernoulli: p 5. Is it easier to solve / than /? n q? Tr z = 1n 6. Treat as differential and find integrating factor? (Chapter ) 7. Look for other substitution? (Eaminers will generall suggest what ou should tr.) 1.3 econd order equations d p q f Principal of linear superposition: d q 0 If 0 satisfies p and 1 satisfies p q f d d then = A satisfies p q f 1.3. Linear equations with constant coefficients d p q f. Homogeneous equation d p q 0 olutions of the form e ( ma be comple) tuart Dalziel 4 Lent 011

5 Auiliar equation: + p + q = 0, p p 4q 1,. Complementar functions (solutions to homogeneous equation) 1 Ae Be A, B (comple if comple) determined b initial/boundar conditions. 1 If comple, then p 1 1 e Aˆ cos q ˆ 4 p Bsin q 4 p. 1 If repeated root, then A Be Finding a particular integral Trial solution As with the first order inhomogeneous equation, we can pose trial solutions for the particular integral of the second order inhomogeneous equation. If f() polnomial, tr If f() = e eponential, tr If f() = cos, tr = a + b + c + = a e = a cos + b sin If f() is proportional to a complementar function, tr multipling eponential or trig function b trateg second order ordinar differential equations Constant coefficients + p + q = f() 1. Write down and solve auiliar equation for.. If 1, real and distinct, then C.F. = A ep( 1 ) + B ep( ) 3. If repeated root, then C.F. = (A + B) ep() 4. If comple roots 1, = a ib, then C.F. = e a (A cos(b) + B sin(b)) 5. Look for forms of PI matching right-hand side. If right-hand side has terms proportional to C.F.s, then tr multipling b, etc. 6. Impose boundar and/or initial conditions tuart Dalziel 5 Lent 011

6 Variable coefficients + p() = 0 You will not be asked to solve a general second-order linear equation if the coefficients are not constant unless it can be written as + p() = 0. In this case if we write = then we can treat this as the first order linear equation: d p 0 Aep p which can then be integrated again to obtain : 1.4 stems of equations Aep pd d B Covert a sstem of two first order equations into a single second order equation for each of the two variables. The two equations will be identical, and the solutions will share the same pair of complementar functions. trateg sstems of equations For () and z() 1. Eliminate z between equations (ou will still have dz/). Differentiate the equation containing / to obtain d / (the result will normall contain dz/) 3. Use the result of step 1 to eliminate dz/ from the result of step. The result will be a second order equation for. 4. The second order equation for z will be identical to that of. 5. olve the second order equations, imposing initial/boundar conditions. 1.5 Higher order equations The material covered in this section is not eaminable. tuart Dalziel 6 Lent 011

7 . Functions of several variables. Differentiation..1 Partial derivatives Treat other variables as constants. mmetr of mied partial derivatives (order of differentiation does not matter): f f f f f f. f = f = f = f = f = f. f f Gradient vector f(,) grad f f,, f f f f(,,z) grad f f,, z Gradient operator (3D),, z... Differentials f() df df f(,) f(p,q,r,s) f f df f f f f df dp dq dr ds p q r s qrs,, prs,, pqs,, pqr,,..3 The chain rule ˆ f uv,, uv, fuv (, ). tuart Dalziel 7 Lent 011

8 o and du dv, u v v u du dv, u v v u f f df f f du dv du dv u v u v v u v u f f f f du u v u v v u f f du dv u v v u v u dv Hence f f f u v u v. uv f f f v v v u u u Look for all the different was ou can differentiate...4 Eact differentials w = P(,) + Q(,).. For w to be eact (i.e. for some function f(,) we have then require f f wdf ), P Q (necessar condition since f f )...5 Integrating factors To make a differential epression eact, select () or () so that w = P + Q tuart Dalziel 8 Lent 011

9 is eact, requiring P Q This leads to an ode for. If selected () then ode must not contain : 1 d 1 P Q Q If selected (), then ode must not contain : trateg partial derivatives 1. Order of differentiation does not matter. The differential of f(,,z) is 1 d 1 P Q P f f f df dz z. z, z,, 3. The chain rule includes differentiation through all routes to independent variable of interest 4. w = P(,) + Q(,) is eact with w = df if and onl if P Q 5. It might be possible to make a differential epression eact b multipling b an integrating factor so that w = P + Q where is chosen so that P Q. Need to be able to choose either = () or = () to make progress. 6. Remember to emplo the definition of the differential (see ) and the condition for an epression to be eact (see 4) when working out relationships..3 tationar points.3. tationar points with more than one independent variable Require all first-order partial derivatives to vanish. If f(,), then (,) = ( 0, 0 ) stationar if and onl if tuart Dalziel 9 Lent 011

10 f,, f f and, f, Tpes: local minimum, local maimum, saddle point (and stationar point of inflection) Etrema must be separated b saddle points A saddle point net to an etremum must have one of its contours circling the etremum.3.3 Classification of stationar points ometimes it will be obvious from the form of the equations, so look first! Local minimum: f > 0, f > 0 and f f f > 0. Local maimum: f < 0, f < 0 and f f f > 0. addle point: f f f < 0 Contours circle local minima and local maima, and cross at saddle points. Contours at saddle points have slope 1 with the root of f f f 0 trateg 1. Identif and plot an eas contours, e.g. f(,) = 0.. Identif an smmetries in the problem. The derivative of an even function must be odd and must vanish on the ais about which it is even. 3. Aismmetric problems can onl have etrema at the origin. Locations where f/r = 0 for r > 0 will be circular ridges. There will be no saddle points. 4. If the function is odd in one direction, sa about = 0, then look for saddle points along = 0. The locations of other stationar points will be arranged smmetricall about the line of smmetr, with etrema taking the opposite character on each side of the line. 5. If the function is even in one direction, sa about = 0, then look for etrema along = 0, or saddle points that have contours perpendicular to = 0. tuart Dalziel 10 Lent 011

11 6. If possible, determine contours on which f/ = 0, and contours on which f/ = Remember that a. Contours loop around etrema. b. The contours cross at a saddle points. c. There will be a pair of contours around an neighbouring etrema that touch/meet at a saddle point. d. Contours from saddle points either open to infinit, join neighbouring saddle points, or loop around etrema..4 Partial differential equations Three generic eamples of partial differential equations given. Course is limited to solution b substitution. Eam questions could use equations of a form not covered eplicitl in lectures..4. The Poisson and Laplace equations Poisson equation: f f s Laplace equation: g g 0 Linearit: h = f + Ag is a solution of Poisson equation for arbitrar A. Need boundar conditions. Course limited to solution b substitution.4.3 The diffusion equation f t f, upports similarit solutions of the form f(,t) = f() with = /(t) 1/. Transform pde second order ode in. olve as sequence of first order odes. tuart Dalziel 11 Lent 011

12 .4.4 The wave equation General solution for infinite domain 1 f f c t f(,t) = F( ct) + G( + ct), where F() and G() are given b initial conditions. trateg partial differential equations You will onl be asked to solve b substitution. 1. Read the question carefull. You will be given guidance on how to proceed.. Use chain rule to differentiate and if there is a change of variables. 3. Don t be afraid to differentiate an equation if attempting to combine multiple equations. tuart Dalziel 1 Lent 011

13 3. Multiple integration 3. Double integrals 3..1 Definition V A h, da Area element da, often epressed as da = V b d h, a c Compute inner integral first, treating the other variables as constant. Order of integration arbitrar. If rectangular domain then trivial to interchange: b d d b V h, h, ac ca If integrand separable f(,) = g() h(), and limits are independent of variables: bd b d I f, g h a c ac g h g h b d b d a c a c tuart Dalziel 13 Lent 011

14 Watch out for other notation conventions V b d h, a c bd a c h, b d a c b d a c b a d h, h, h, c Non-rectangular region V 1 1 h, 0 0 Compute first the inner integral 1 1 I H, H, h,, then V I. Can reverse order if can epress boundaries on as a function of : V 1 1 h, 0 0. tuart Dalziel 14 Lent 011

15 3.3 Integration in D polar coordinates Epress area element in polar coordinates: da = = r dr d: 3.4 Triple integrals,, ˆ, A A. A V h da h h r r dr d V,,,, I f z dv f z dz Choose coordinate sstem that best conforms to geometr of volume V integrating over Element of volume V Cartesian: Clindrical polar: pherical polar: dv = dz dv = r dr d dz dv = r dr sin d d trateg 1. Decide on coordinate sstem (Cartesian, circular/clindrical polar, spherical polar); tpicall choose the one that best matches domain shape.. Determine limits, e.g. 0 () 1 () and 0 1, or 0 r a and 0 /. 3. Determine element of area/volume in coordinate sstem: da = = r dr d or dv = dz = r dr d dz = r sin dr d d 4. Transform integrand into selected coordinate sstem. 5. Choose order of integration; if integrand f(,,z) = X() Y() Z(z) and limits are constants then separate nested integrals into products 6. Integrate outwards from innermost integral. Each integration removes one variable from the problem. 3.5 The Gaussian integral I a a e a. tuart Dalziel 15 Lent 011

16 bounded as a. Instead of computing I e, compute I e e e e e r r rdr d d e rdr tuart Dalziel 16 Lent 011

17 4. calar and vector fields Concentrate on independent variables representing space 4. The gradient of a scalar field Ideas work in D or in 3D. v=grad i j k z The rate of change of in some direction q (unit vector) is d q sq ds The gradient is normal to contours/surfaces on which () = const. Unit normal: 4.3 Line integrals n Integrals along a path through space,, given b = (s) for s a s s b, s0 calar field I s s b ds. sa Parameter has phsical meaning and so it affects outcome. Will often choose s as distance along the path, or t as time. s tb ds dt dt b I ds s ds t sa ta Choice of parameterising of path does not matter: alwas integrating with respect to vector. Vector field Interested in component of field in the direction of the path tuart Dalziel 17 Lent 011

18 s b b J FFs ds t ds F dt dt s a The choice of parameterisation for does not matter as alwas integrating with respect to the vector. The integral J will generall depend on the path as well as the end points a and b. If the path starts and ends at the same point then the integral ma still be non-zero: 4.4 Conservative vector fields FF d ds s. t t a If F() = (). t1 F dt t0 dt t1 dz dt t 0 dt dt zdt d t1 1 t dt d t0 dt and the integral depends onl on the start and end points (not the route taken) provided the function () is single-valued. Hence F d 0 for all closed paths. To be able to write F() in terms of a scalar potential (()), we need F = 0. This (in D) is equivalent to the condition for F to be an eact differential. trateg line integrals Determine whether ou are integrating a scalar field or vector field tuart Dalziel 18 Lent 011

19 Integration of a scalar field s 1. Parameterise the path to integrate along, = (s).. The solution depends on the parameterisation of the path as well as the route taken b the path. Integration of a vector field For a line integral of the form F d F d : 1. Parameterise the path to integrate along, = (s). For a specified path, the solution is independent of the parameterisation. 3. Establish if the vector field F is conservative b testing F = If field is conservative, then ou ma use an path between the two end points. 3. If field is conservative, then the integral around an closed path is zero. 4. If field is not conservative, then calculate defines the path parametricall as (s). 4.5 urface integrals ds FF d ds s where Integrating on a D surface in 3D space. The D surface has an area and orientation. An element within this area is d = nd where d is an element of area and n is the unit normal to the surface. If A is a D region on a plane then If is a closed surface then Flu across surface d n d n d An. A A A d 0. Flu of a vector field F across a surface is tuart Dalziel 19 Lent 011

20 Fd Fn d. trateg integration on a surface 1. Alwas use the outward facing normal n.. Distinguish between integrating a scalar over a surface f of which is a vector) and the flu across a surface result). Integration of scalar field f d 3. The vector area d of a closed surface is alwas zero. d Fd (giving a scalar (the result 4. For an aismmetric surface, the vector area d will be equal to the projection of the area along the smmetr ais (e.g. a hemisphere of radius a will have a vector area a p, where p is the unit vector in the direction of the smmetr ais for the hemisphere). 5. For f d, look to see if f() is constant on the surface (f() = c). If it is, then the result will be that constant multiplied b the vector area, f d c d. Flu of vector field Fd Fn d 6. Determine normal vector n and calculate Fn on surface. 7. If Fn is constant (c) on surface, then Fd c d c, where is the surface area. 4.6 The gradient operator acting on vectors Vector operator i j k z z tuart Dalziel 0 Lent 011

21 Divergence Divergence of vector field U = ui + vj + wk: u div UU u v w,, v i j k z i j k z w u v w z We must be ver careful when computing the divergence in non-cartesian coordinates since the orientation of unit vectors are then themselves a function of space. Laplacian divgrad z Curl curluuu i j k z u v w w v u w v u i z j k z A vector field F that satisfies F = 0 is often referred to as an irrotational field. curl grad A conservative field F = = i / + j / + k /z is irrotational since div curl F = 0. (F) = 0. tuart Dalziel 1 Lent 011

22 trateg differential operators 1. Watch out for combinations that alwas give a zero result:. Equivalent notations For scalar field () For vector field U() = (u,v,w) 0 for all 0 F for all grad z u v w div UU z w u curluuu z v 4.7 The divergence theorem (statement) You do not need to know this for eamination purposes Flu of a vector field F across a surface V bounding a volume V : 4.8 tokes theorem (statement) V Fd FdV You do not need to know this for eamination purposes Vector field F at an open surface bounded b a closed curve C: d V. C F d F v z w u tuart Dalziel Lent 011