PH2130 Questions for contemplation Week 1 Why differential - PDF document

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

PH2130 week 3, page 2 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 ∇ 2 symbol • Understand the physical meaning of the laplacian operator

PH2130 week 3, page 3 2.3.1 One dimension: x and t independent variables ∂ Ψ ∂Ψ 2 1 − = Diffusion eq n 0 ∂ ∂ 2 x D t describes diffusion, heat flow etc. D is the diffusion coefficient. ∂ Ψ ∂ Ψ 2 2 1 − = Wave eq n 0 ∂ ∂ 2 2 2 x v t describes vibrating string. v is the speed of propagation . Note different orders of time Connection with relativity.

PH2130 week 3, page 4 2.3.2 Two dimensions: x, y and t independent variables ∂ Ψ + ∂ Ψ ∂Ψ 2 2 1 − = Diff n eq n 0 ∂ ∂ ∂ 2 2 x y D t describes diffusion, heat flow etc. in two dimensions D is the diffusion coefficient. ∂ Ψ + ∂ Ψ ∂ Ψ 2 2 2 1 − = Wave eq n 0 ∂ ∂ ∂ 2 2 2 2 x y v t describes vibrating sheet -- a drum for example. v is the speed of propagation .

PH2130 week 3, page 5 2.3.3 Three dimensions: x, y, z and t independent variables ∂ Ψ + ∂ Ψ + ∂ Ψ ∂Ψ 2 2 2 1 − = Diff n eq n 0 ∂ ∂ ∂ ∂ 2 2 2 x y z D t describes diffusion, heat flow etc. in three dimensions D is the diffusion coefficient. ∂ Ψ + ∂ Ψ + ∂ Ψ ∂ Ψ 2 2 2 2 1 − = 0 Wave ∂ ∂ ∂ ∂ 2 2 2 2 2 x y z v t eq n describes vibrations in 3d -- sound waves for example. v is the speed of propagation .

PH2130 week 3, page 6 2.3.4 The laplacian Have seen ∂ + ∂ + ∂ 2 2 2 ∂ ∂ ∂ 2 2 2 x y z before.  Recall vector calculus in PH1120 and the formula div grad = ∇ 2  The laplacian operator, denoted by ∇ 2 , is given (in cartesian coordinates) by ∇ = ∂ + ∂ + ∂ 2 2 2 2 z . ∂ ∂ ∂ 2 2 2 x y  Some books denote ∇ 2 by ∆ ; we don’t 

PH2130 week 3, page 7 Ubiquity of the laplacian The laplacian appears in many differential equations: Diffusion equation ∂Ψ 1 ∇ Ψ − 2 t . ∂ D Wave equation ∂ Ψ 2 1 ∇ Ψ − 2 . ∂ 2 2 v t Even the Schrödinger equation ∂Ψ 2 � − ∇ Ψ + Ψ = 2 V i � ∂ 2 m t  recall from PH1530  Why is ∇ 2 so common?

PH2130 week 3, page 8 2.3.5 Physical meaning of ∇ 2 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 ∂Ψ ∂ Ψ 2 2 a Ψ − = Ψ − + + � � � � ... x a x a ∂ ∂ 2 x 2 x while a little to the right ∂Ψ ∂ Ψ 2 2 a Ψ − = Ψ + + + � � � � x a x a ... ∂ ∂ 2 x 2 x

PH2130 week 3, page 9 On taking the average 1 Ψ = Ψ − + Ψ + � � � � x a x a 2 ∂ Ψ 2 2 a = Ψ + � � x ∂ 2 2 x or ∂ Ψ 2 2 a Ψ − Ψ = � � x ∂ 2 2 x The argument can be extended to 2d 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 ∇ 2 Ψ . In the Schrödinger equation bending Ψ costs kinetic energy.

PH2130 week 3, page 10 2.3.6 Laplace’s equation In the steady state i.e. ∂ / ∂ t , ∂ 2 / ∂ t 2 etc. = 0. Then both the wave equation and the diffusion equation reduce to (another equation to spot) ∇ Ψ = 2 . Laplace’s equation 0  Will see this in Electromagnetism PH2420.  Physical interpretation of ∇ 2 implies: In a region where Laplace’s eq n holds, there can be no maxima or minima in Ψ .

PH2130 week 3, page 11 2.3.7 The d’alembertian kjhkjhkjhk

PH2130 week 3, page 12 Aims of Wk 3 Lect 2 • 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 (2 indep. vars) PDEs, given BCs and ICs

PH2130 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 x t = � � � � � � , X x T t

PH2130 week 3, page 14 3.1 1-d wave equation ∂ Ψ ∂ Ψ 2 2 1 = ∂ ∂ 2 2 2 x v t Writing Ψ x t = � � � � � � , X x T t Then ∂ Ψ x t � � � � � � 2 2 , d X x = T t ∂ 2 2 x d x and ∂ Ψ x t � � � � 2 2 , d T t = � � X x ∂ 2 2 t dt has total derivatives. Put in wave equation ⇒

PH2130 week 3, page 15 � � � � � � 2 2 d X x 1 d T t = T t 2 2 2 d x v d t � � � � , gives Divide by X x T t 2 2 1 d X 1 1 d T = 2 2 2 X d x v T d t 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 2 . Called separation constant.

PH2130 week 3, page 16 Have 2 ODEs: � 2 d X + = 2 k X 0 � � 2 d x � 2 d T � + = 2 2 v k T 0 � � 2 d t - Have turned 1 PDE into 2 ODEs - Assuming k 2 is positive, these are both SHO equations.

PH2130 week 3, page 17 3.1.1 Boundary conditions & Initial conditions Need some physical information to solve real problems. E.g. Piano string, length L , where Ψ x t � � is displacement of string. , • Fixed at both ends: Ψ = Ψ = for all t . � � � � 0 , t L t , 0 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.

PH2130 week 3, page 18 The Boundary Condition helps solve the X equation. = = . � � � � BC is X 0 X L 0 Gen. Sol n of 2 d X + = 2 k X 0 2 d x is = + � � � � � � X x A sin kx B cos kx .  Recall from PH1110  � � = ⇒ = BC X 0 0 B 0 � � = 0 restricts allowed values BC X L = π of k since sin kL must = 0; i.e kL n for integer n . (See why k 2 must be +ve now)

PH2130 week 3, page 19 PICTURE of Piano string

PH2130 week 3, page 20 Recall particle in a box in PH2530. There we saw you needed an integer n o of ½ waves to fill L . – Same thing. n = 1 n = 2 n = 3 We label the allowed values of k : n = π k Ln Then X solutions are: = � � � � X x A sin k x n n n ↑ underermined as yet See that Boundary conditions ⇒ Quantisation

PH2130 week 3, page 21 The Initial Condition helps solve the T equation � � + 2 d T t = 2 2 k v T 0 2 d t another SHO equation – since we know k 2 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 Ψ n = � � � � � � � x t , X x T t n n = + � � � � � sin k x P cos k vt Q sin k vt n n n n n (have subsumed the A n into the P n , Q n )

PH2130 week 3, page 22 Linearity allows us to write the general solution as a linear superposition ∑ Ψ x t = + � � � � � � � � , sin k x P cos k vt Q sin k vt n n n n n n again have subsumed coeffs into the P n and Q n . Satisfying the initial condition will determine the P n and Q n . Ψ x = � � � � ,0 f x so ∑ = � � . P sin k x f x n n n This is a Fourier sine series.  Remember from PH1120 

PH2130 week 3, page 23 The Fourier components P n are found � � using the inversion formula : from f x = � 2 L 0 � � sin d P f x k x x n n L

PH2130 week 3, page 24 So: Solution to vibrating string obeying ∂ Ψ ∂ Ψ 2 2 1 = ∂ ∂ 2 2 2 x v t and subject to: BC: Ψ = Ψ = for all t � � � � 0 , t L t , 0 (fixed at both ends) and IC: Ψ x = � � � � ,0 f x (shape at t = 0) Is = ∑ Ψ x t � � � � � � , P sin k x cos k vt n n n n where n = π k Ln and = � 2 L 0 � � sin P f x k x x d . n n L

PH2130 week 3, page 25 Summary of S.V method: 1 Express Ψ as a product ⇒ ODEs plus separation constant 2 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.