PH2130. Week 1. Week 2. Week 3. Questions for contemplation. Why differential equations? Why usually linear diff eq n s? Why usually 2 nd order?

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1 PH130 week 3, page 1 PH130 Questions for contemplation Week 1 Why differential equations? Week Why usually linear diff eq n s? Week 3 Why usually nd order?

2 PH130 week 3, page Aims of Wk 3 Lect 1 Recognise diffusion eq n and wave eq n. Know the type of phenomena they describe Know the meaning and use of the symbol Understand the physical meaning of the laplacian operator

3 PH130 week 3, page One dimension: x and t independent variables Ψ 1 Ψ = 0 Diffusion eq n x D t describes diffusion, heat flow etc. D is the diffusion coefficient. Ψ 1 Ψ = 0 Wave eq n x v t describes vibrating string. v is the speed of propagation. Note different orders of time Connection with relativity.

4 PH130 week 3, page 4.3. Two dimensions: x, y and t independent variables + Ψ Ψ 1 Ψ = 0 Diff n eq n x y D t describes diffusion, heat flow etc. in two dimensions D is the diffusion coefficient. + Ψ Ψ 1 Ψ = 0 Wave eq n x y v t describes vibrating sheet -- a drum for example. v is the speed of propagation.

5 PH130 week 3, page Three dimensions: x, y, z and t independent variables + + Ψ Ψ Ψ 1 Ψ = 0 Diff n eq n x y z D t describes diffusion, heat flow etc. in three dimensions D is the diffusion coefficient. + + Ψ Ψ Ψ 1 Ψ = 0 Wave x y z v t eq n describes vibrations in 3d -- sound waves for example. v is the speed of propagation.

6 PH130 week 3, page The laplacian Have seen before. + + x y z Recall vector calculus in PH110 and the formula div grad = The laplacian operator, denoted by, is given (in cartesian coordinates) by = + + x y z. Some books denote by ; we don t

7 Ubiquity of the laplacian The laplacian appears in many differential equations: Diffusion equation 1 Ψ Ψ. D t PH130 week 3, page 7 Wave equation Ψ 1 v Ψ. t Even the Schrödinger equation = Ψ Ψ+ VΨ= i= m t recall from PH1530 Why is so common?

8 PH130 week 3, page Physical meaning of The laplacian gives the smoothness of a function. It measures the difference between the value of Ψ at a point and its mean value at surrounding points. A little to the left of x a Ψ Ψ x a = Ψ x a Ψ + x x +... while a little to the right a Ψ Ψ x a = Ψ x + a Ψ + x x +...

9 PH130 week 3, page 9 On taking the average Ψ = Ψ x a + Ψ x+ a 05 a Ψ = Ψ x + x or 05 a Ψ Ψ Ψ x = x The argument can be extended to d and 3d. Thus we conclude: The deviation from the value of Ψ at a point and its mean value in the surrounding region is proportional to Ψ. In the Schrödinger equation bending Ψ costs kinetic energy.

10 PH130 week 3, page Laplace s equation In the steady state i.e. / t, / t etc. = 0. Then both the wave equation and the diffusion equation reduce to (another equation to spot) Ψ = 0. Laplace s equation Will see this in Electromagnetism PH40. Physical interpretation of implies: In a region where Laplace s eq n holds, there can be no maxima or minima in Ψ.

11 .3.7 The d alembertian kjhkjhkjhk PH130 week 3, page 11

12 PH130 week 3, page 1 Aims of Wk 3 Lect Understand separation of variables method for solving PDEs Use separation of variables to convert PDEs into ODEs Boundary conds and Initial conds in solving real problems Solve simple ( indep. vars) PDEs, given BCs and ICs

13 PH130 week 3, page 13 3 Separation of Variables Look for solutions of PDEs which are a product of the independent variables. Converts PDEs into a number of ODEs. - So in 1d case : x, t indep. vars., look for solutions like Ψ xt, = X xtt

14 PH130 week 3, page d wave equation Ψ 1 Ψ = x v t Writing Ψ xt, = X xtt Then Ψ xt, d X x 05 = Tt x dx and 0 5 Ψ xt, d Tt = X x t dt has total derivatives. Put in wave equation

15 d X x dx Tt = 1 v Divide by X0505, x T t gives d Tt dt PH130 week 3, page 15 1 d X 1 1 = X dx v T d T dt LHS depends on x only RHS depends on t only But x and t are independent! So both sides must be constant Put const = k. Called separation constant.

16 PH130 week 3, page 16 Have ODEs: d X dx d T dt + k X = + vkt= 0 0 ( ) K KK * - Have turned 1 PDE into ODEs - Assuming k is positive, these are both SHO equations.

17 PH130 week 3, page Boundary conditions & Initial conditions Need some physical information to solve real problems. E.g. Piano string, length L, where Ψ xt, 0 5 is displacement of string. Fixed at both ends: Ψ 0, t = Ψ L, t = 0 for all t. Restriction on Ψ by the boundary, so called boundary condition Initial shape: Ψ x,0 = f x, Restriction on Ψ by the initial state called initial condition.

18 PH130 week 3, page 18 The Boundary Condition helps solve the X equation. BC is X = X L = 0. Gen. Sol n of d X + k X = 0 dx is X x = Asin kx + Bcos kx. Recall from PH1110 BC X 05= 0 0 B= 0 BC X05= L 0 restricts allowed values of k since sin kl must = 0; i.e kl = nπ for integer n. (See why k must be +ve now)

19 PICTURE of Piano string PH130 week 3, page 19

20 PH130 week 3, page 0 Recall particle in a box in PH530. There we saw you needed an integer n o of ½ waves to fill L. Same thing. n = 1 n = n = 3 We label the allowed values of k: k L n n = π Then X solutions are: X x = A sin k x n n n underermined as yet See that Boundary conditions Quantisation

21 PH130 week 3, page 1 The Initial Condition helps solve the T equation 05 d Tt 0 + kvt = dt another SHO equation since we know k is positive. Solution is T t = P cos k vt + Q sin k vt n n n n n Solution for Ψ(x, t) for given n is then xt X xt t = sin k x P cos k vt + Q sin k vt Ψ n, = n n ; @ n n n n n (have subsumed the A n into the P n, Q n )

22 PH130 week 3, page Linearity allows us to write the general solution as a linear superposition 0 5 ; @ Ψ x, t = sinknx Pncos knvt + Qnsin knvt n again have subsumed coeffs into the P n and Q n. Satisfying the initial condition will determine the P n and Q n. so Ψ x,0 = f x Pnsin knx = f05. x n This is a Fourier sine series. Remember from PH110

23 PH130 week 3, page 3 The Fourier components P n are found 05 from f x using the inversion formula: IL Pn = f05sin x knxdx L 0

24 PH130 week 3, page 4 So: Solution to vibrating string obeying Ψ 1 Ψ = x v t and subject to: BC: Ψ 0, t = Ψ L, t = 0 for all t (fixed at both ends) and IC: Ψ x,0 = f x (shape at t = 0) Is Ψ xt, = Pnsin kx n cos kvt n where and P n n k L n n = π IL = f05sin x knxd x. L 0

25 PH130 week 3, page 5 Summary of S.V method: 1 Express Ψ as a product ODEs plus separation constant Solve ODEs 3 Boundary conditions determine allowed spatial solutions, values of separation constant 4 Make linear superposition of X n T n solutions. 5 Initial conditions allow determination of superposition coefficients.