2.20 Fall 2018 Math Review

Transcription

1 2.20 Fall 2018 Math Review September 10, 2018 These notes are to help you through the math used in this class. This is just a refresher, so if you never learned one of these topics you should look more into it. Outline: 1. Einstein s Indicial Notation 2. Scalars, vectors and tensors 3. Vector Calculus 4. Taylor Series Expansion 5. Divergence (Gauss) Theorem 6. Leibniz Integral Rule 7. Differential Equations 8. Coordinate Systems 1

2 1 Einstein s Indicial Notation Simplifies the notation in vector calculus. Types of variables are determined by the number of indices required to describe them. Terminology: Free index: appears exactly once in each term of an expression Dummy index: appears exactly twice in each term of an expression Range convention Whenever a subscript appears only once in a term, the subscript takes all of the values. The range of indices should be given explicitly in the problem (i.e. i = 1, 2, 3 for 3D case, i = 1, 2 for 2D case). Example: i {1, 2, 3} x i = {x 1, x 2, x 3 } Example: i {1, 2, 3} x i = {x 1, x 2, x 3 } The name of the free index (e.g. i in the above example) MUST remain the same throughout the calculation (e.g. v i v j ) To facilitate writing of the coordinates in indicial notation, we will be interchanging {x 1, x 2, x 3 } and {x, y, z} Summation convention Repeated indices are summed over and no index can repeat more than twice. Example: i {1, 2, 3} a ii 3 a ii = a 11 + a 22 + a 33 i=1 Example: {i, j} {1, 2, 3} a ij b i c j = b 1 (a 11 c 1 + a 12 c 2 + a 13 c 3 )+ + b 2 (a 21 c 1 + a 22 c 2 + a 23 c 3 )+ + b 3 (a 31 c 1 + a 32 c 2 + a 33 c 3 ) Example (matrix multiplication): {i, j, k} {1, 2, 3} The first term of A is, b 11 b 12 b 13 c 11 c 12 c 13 A = B C = b 21 b 22 b 23 c 21 c 22 c 23 b 31 b 32 b 33 c 31 c 32 c 33 2

3 a 11 = b 11 c 11 + b 12 c 21 + b 13 c 31 Or more generally, 3 a ij = b ik c kj = b ik c kj k=1 Comma convention A comma indicates partial differentiation with respect to the index coordinate. Examples: Kronecker delta, δ ij a i,j a i x j a i,i a i x i = a 1 x 1 + a 2 x 2 + a 3 x 3 The Kronecker delta is basically the identity matrix: Example: i {1, 2, 3} δ ij = { 0, i j 1, i = j δ ii = δ 11 + δ 22 + δ 33 = = 3 it has the effect of substituting the free index of the term it multiplies. For example We can check by just doing the algebra, δ ij a j = a i Levi-Civita Permutation Symbol ε ijk δ 11 a δ 12 a δ 13 a 3 = a 1 1, (i, j, k) = (1, 2, 3), (2, 3, 1), or (3, 1, 2) ε ijk = 1, (i, j, k) = (1, 3, 2), (2, 1, 3), or (3, 2, 1) 0, otherwise (i.e. repeated indices) Used to write cross products in index notation a b = ε ijk a j b k 3

4 2 Scalars, Vectors and Tensors All of these quantities can be thought of as tensors of different order Scalar (e.g. α): No index (i.e. quantity expressed in 1 term) tensor of zeroth order. Only a magnitude. Examples: Temperature, pressure, density, energy Vector (e.g. a i ): One free index (3 terms in 3D) tensor of first order. Has magnitude and direction. Example: i {1, 2, 3} a i = {a 1, a 2, a 3 } Examples: Position, velocity, acceleration, force, momentum. Tensor (e.g b ij ): Two free indices (9 terms in 3D) tensor of second order. magnitude, direction and orientation. Example: {i, j} {1, 2, 3} b 11 b 12 b 13 b ij = b 21 b 22 b 23 b 31 b 32 b 33 Examples: Stress tensor, deformation tensor, momentum flux Has 3 Vector calculus Del/nabla ( ) operator: differential vector operator in 3D space (or 2D): xî + y ĵ + z ˆk = x, y, z (Often written as to better express its vector nature.) In indicial notation, Del operator is written as 4 x i

5 Important uses of Del/nabla ( ) operator: 1. Gradient: the derivative of a scalar function f(x, y, z) in the direction of the greatest change is called the gradient of f grad f f x, f y, f = f z The gradient is similar to a derivative, but for a function of more than one variable. Similar to the derivative, the gradient is representative of the slope of the function at a point, but since it is for a function of multiple variables, it is the slope of the direction of greatest rate of increase of the field at that point. Notice that the gradient of a scalar is a vector field, and the gradient of a vector is a tensor field. This is easier to see in the index notation grad a = x j a i = a i x j b ij operating with a gradient increases the dimension of the field by one! Finding the unit normal vector ˆn to a curve or a surface: For an implicitly given curve F (x, y) = C (or a surface F (x, y, z) = C), the unit normal at any point (x, y, z) is given by ˆn(x, y, z) = F (x,y,z) F (x,y,z) where, 2. Divergence: F = ( F x ) 2 ( ) F 2 ( ) F y z For a vector field V in the x, y, z coordinate space with components V = u, v, w : div V = V = u x + v y + w z = V i x i It can be thought of as the expansion of the field at a certain point, or the rate at which V leaves a volume. 5

6 It has the effect of decreasing the order of the field by 1 (it has to operate on vector fields at least) It can also be defined as V = lim V 0 A v ˆn da V 3. Curl: for a vector field V in the x, y, z coordinate space with components V = u, v, w : curl V Curl can also be defined as î ĵ ˆk x y z u v w ( V ) ˆn = lim A 0 = V = ε ijk V k x xj V d s A Describes the (infinitesimal) rotation or spin (strength and direction) of a vector field at a particular point. The direction of the curl defines the axis of rotation, while its magnitude defines the strength of the rotation. 4. Laplacian (second derivative): Scalar operator, so it can be applied to a tensor field of any order (scalar, vector and tensor fields) 2 = = 2 x y z 2 = 6 2 x i x i

7 (Why do we write x i x i and not x 2 i? Summation convention!) After applying the Laplacian, the tensor order remains the same Laplacian of a scalar field f(x, y, z): 2 f = 2 f = 2 f x i x i x f y f z 2 Laplacian of a vector field V (x, y, z) with components u, v, w : 2 V = 2 V x V y V z 2 = 2 u x u y u z 2, 2 v x v y v z 2, 2 w x w y w z 2 In indicial notation, Laplacian of a vector field u i is written as u i x j x j Notice the difference in the length of the expression! 5. Other second derivative operators: div(grad f) = ( f) = f div(curl v) = ( v) 0 curl(grad f) = ( f) 0 curl(curl v) = ( v) grad(div v) = ( v) Laplacian (identically) (identically) 4 Taylor Series Expansion Taylor series are a useful tool to calculate the value of a function in the neighborhood of a certain point. The general formula for the Taylor series of a one-dimensional function f(x) around a point x 0 is: f(x) = (x x 0 ) n d n f n! dx n = f(x 0 ) + (x x 0 ) df dx + (x x 0) 2 d 2 f x=x0 2 dx 2 + x=x0 x=x0 n=0 7

8 Often we want to linearize a function. This means that we are assuming that (x x 0 ) is very small, so we can ignore any terms of second order, (x x 0 ) 2, or larger: Two-dimensional Taylor expansion: f(x) f(x 0 ) + (x x 0 ) df dx x=x0 [ f(x, y) = f(x 0, y 0 ) + (x x 0 ) f x + (y y 0) f ] y (x 0,y 0 ) + 1 [ ] (x x 0 ) 2 2 f 2 x 2 + 2(x x 0)(y y 0 ) 2 f x y + (y y 0) 2 2 f y 2 + O( x 3, x 2 y, x y 2, y 3 ) 5 Divergence (Gauss) Theorem The divergence (Gauss) theorem is used many times in this class. The purpose of this theorem is to convert volume integrals to surface integrals (or vice versa). Theorem: If V is a vector field in {x, y, z}: S F ˆn ds = V F dv where ˆn is the unit normal vector to the surface S with volume V. notation, this reads F i F i n i ds = dv x i S V In indicial 8

9 If τ ij is a tensor field in {x 1, x 2, x 3 }: τ ij n j da = A V τ ij x j dv 6 Leibniz Integral Rule Gives an expression for how to differentiate an integral whose limits are not a constant 1D: for a function I(t) = b(t) a(t) F (x, t) dx we have Note that da dt Proof: and db dt ˆ di b(t) dt = F a(t) t db da dx + F (b, t) F (a, t) dt dt are speeds by which limits at x = a and x = b move along x-axis. At time t, I(t) is the area under F (x, t) for x [a(t), b(t)]. At time t+ t, I(t+ t) is the area under F (x, t + t) for x [a(t + t), b(t + t)]. To calculate I(t+ t), we first consider the integral with a and b fixed, then account for the modification due to the change of a and b I(t + t) = = ˆ b(t+ t) a(t+ t) ˆ b(t) a(t) F (x, t + t) dx F (x, t + t) dx + ˆ b(t+ t) b(t) F (x, t + t) dx Apply Taylor expansion and keep only the term up to O( t) F (x, t + t) = F (x, t) + t F t + O ( ( t) 2) a(t + t) = a(t) + t da dt + O ( ( t) 2) b(t + t) = b(t) + t db dt + O ( ( t) 2) ˆ a(t+ t) a(t) F (x, t + t) dx 9

10 I(t+ t) = ˆ b(t) a(t) By letting t to go to zero ( F (x, t) + t F ) dx+f (b, t) [b(t + t) b(t)] F (a, t) [a(t + t) a(t)] t di dt = lim I(t + t) I(t) = t 0 t ˆ b(t) a(t) F t db da dx + F (b, t) F (a, t). dt dt Generalizations to 3D x = (x, y, z), interval volume, boundary points (closed) boundary surface ˆ I(t) = F ( x, t) dv di ˆ dt = F t dv + F ( x, t) U n da V (t) where S(t) is the boundary surface of volume V (t), da its differential element, and U n ( x, t) = U ˆn the normal velocity of the moving boundary S(t). So rate of change of the integral is equal to the integral of how much F (x, t) changed inside the volume, plus what came in/out through the boundary. 7 Integral by Parts This very useful integration trick allows us to simplify an integrand by using derivatives and anti-derivatives. Consider two functions, u(x) and v(x), ˆb a V (t) ˆ u(x) v(x)dx = [u(x)v(x)] b a a b S(t) u(x)v(x)dx Integration by parts is of great help in solving integrals whose integrands are functions which can be simplified after a derivative. Examples would be products between polynomials and exponentials (or sines and cosines). Example: solve π/2 0 x sin xdx Apply integration by parts where u(x) = x and v(x) = sin x, so v(x) = cos x, ˆ π/2 8 Differential equations 0 x sin xdx = 0 [x cos x] π/2 0 + ˆ π/2 0 cos xdx = 1 In this class, you will be expected to know how to solve linear differential equations with constant coefficients. Remember, the order of the differential equation is given by the value of its highest derivative, which will also dictate the number of boundary conditions you will need to solve the problem. 10

11 1. Example: simple ODE d n y dx n = a This n-th order ODE (ordinary differential equation) is solved by integrating n times and applying n boundary conditions (B.C. s) 2. Example: 2 nd order inhomogeneous ODE with constant coefficients If y is a function of t, and a, b are constants, and we are given the equation ÿ + a ẏ + b y = F (t) The solution will be a summation of a homogeneous and a particular solution: Homogeneous solution Set F (t) = 0 and solve for y h y = y h + y p ÿ h + a ẏ h + b y h = 0 The solution takes the form of an exponential times a constant y h (t) = Ae m t Substituting the above into the equation for y h yields the characteristic equation, m 2 + a m + b = 0 And we can solve for m, which results in (up to) two solutions, m 1 and m 2. Then, the solution to the ODE will be a linear superposition of both, y h (t) = c 1 e m 1t + c 2 e m 2t. where c 1 and c 2 are constants that are determined by the B.C. s. Note how we need two B.C. s since we have two constants. Particular solution ÿ p + a ẏ p + b y p = F (t) To find the expression for y p we assume its shape to be that of F (t) up to a multiplicative constant. For example, say the forcing term is F (t) = e 5 t, then assume y p = C e 5t and substitute, (25 + 5a + b)ce 5t = e 5t 11

12 C = a + b The final solution is (for the example with F (t) = e 5 t ) y(t) = c 1 e m 1t + c 2 e m 2t + e 5 t a + b 3. Partial differential equations (PDEs): In this class you will be expected to know how to solve very simple PDEs, such as one where you can separate variables and simplify PDE to a collection of ODEs. Example: Solve the following PDE for φ(x, y) φ x = 2x, φ y = 2y Solution: ˆ φ x dx = x2 + f(y), ˆ φ y dy = y2 + g(x) φ(x, y) = x 2 + f(y) = y 2 + g(x) φ(x, y) = x 2 + y 2 + C If the PDE is too complicated, we will show you how to solve it, or give you the solution. 9 Coordinate systems Cartesian coordinates: standard (x, y, z) coordinates. 12

13 Cylindrical coordinates (r, θ, z) (polar coordinates (r, θ) if only 2D): r [0,, θ [0, 2π] Transformation from cylindrical to Cartesian coordinates x = r cos θ y = r sin θ z = z Transformation from Cartesian to cylindrical coordinates r = x 2 + y 2 θ = arctan y x We will sometimes want to convert integrals from Cartesian coordinates to cylindrical: f(x, y, z) dv = f(r cos θ, r sin θ, z) r dr dθ dz V z,θ,r 13

14 Spherical coordinates (ρ, θ, φ): ρ [0,, θ [0, 2π], φ [0, π] Transformation from spherical to Cartesian coordinates x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ Transformation from Cartesian to spherical coordinates ρ = x 2 + y 2 + z 2 θ = arctan y x z φ = arccos x 2 + y 2 + x 2 Similarly we will want to convert integrals from Cartesian coordinates to spherical: V f(x, y, z) dv = φ,θ,ρ f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ 2 sin φ dρ dθ dφ 14