Cossey L. Introduction to Mathematical Physics

Подождите немного. Документ загружается.

First Edition, 2009

ISBN 978 93 80168 33 3

Published by:

Global Media

1819, Bhagirath Palace,

Chandni Chowk, Delhi-110 006

Email: globalmedia@dkpd.com

Table of Contents

1. Some mathematical problems and their solution

2. N body probl

em and matte

r description

3. Relativity

4. Electrom

agn

etism

5. Quantum m

chanics

6. N body probl

em in quantu

7. Statistical physics

8. N body probl

ems an

d statistical equilibrium

9. Contin

uous approximation

10. Energy in con

inuous media

11. Appendix

Boundary, spectral and evolution

problems

A quick classification of various mathematical problems encountered in the modelization

of physical phenomena is proposed in the present chapter. More precisely, the problems

considered in this chapter are those that can be reduced to the finding of the solution of a

partial differential equation (PDE). Indeed, for many physicists, to provide a model of a

phenomenon means to provide a PDE describing this phenomenon. They can be

boundary problems, spectral problems, evolution problems. General ideas about the

methods of exact and approximate solving of those PDE is also proposed\footnote{The

reader is supposed to have a good knowledge of the solving of ordinary differential

equations. A good reference on this subject is ).}. This chapter contains numerous

references to the "physical" part of this book which justify the interest given to those

mathematical problems.

In classical books about PDE, equations are usually classified into three categories:

hyperbolic, parabolic and elliptic equation. This classification is connected to the proof of

existence and unicity of the solutions rather than to the actual way of obtaining the

solution. We present here another classification connected to the way one obtains the

solutions: we distinguish mainly boundary problems and evolution problems.

Let us introduce boundary problems:

Problem: (Boundary problem) Find u such that:

1. u satisfies the boundary conditions.

This class of problems can be solved by integral methods section chapmethint) and by

variational methods section chapmetvar). Let us introduce a second class of problems:

evolution problems. Initial conditions are usually implied by time variables. One knows

at time t

the function u(x,y,t )

and try to get the values of u(x,y,t) for t greater than t

This our second class of problem:

Problem: (Evolution problem) Find such that:

1. u satisfies the boundary conditions.

2. u satisfies initial conditions.

Of course in some problems time can act "like" a space variable. This is the case when

shooting problems where u should satisfy a condition at t = t

and another condition at t =

The third class of problem (spectral problem) often arises from the solving section

chapmethspec) of linear evolution problems (evolution problems where the operator L is

linear):

Problem: (Spectral problem) Find u and λ such that:

1. u satisfies the boundary conditions.

The difference between boundary problems and evolution problem rely on the different

role played by the different independent variables on which depend the unknown

function. Space variables usually implies boundary conditions. For instance the elevation

u(x,y) of a membrane which is defined for each position x,y in a domain Ω delimited by a

boundary should be zero on the boundary. If the equation satisfied by u at equilibrium

and at position (x,y) in Ω is:

Lu = f,

then L should be a differential operator acting on space variables which is at least of the

second order.

Let us now develop some ideas about boundary conditions. In the case of ordinary

differential equations (ODE) the unicity of solution is connected to the initial conditions

via the Cauchy theorem. For instance to determine fully a solution of an equation :

one needs to know both u(0) and . Boundary conditions are more subtle, since the

space can have a dimension greater than one. Let us consider a 1-D boundary problem.

The equation is then an ODE that can be written:

Lu = f

Boundary conditions are imposed for two (at least) different values of the space variables

x. Thus the operator should be at least of second order. Let us take . Equation

eqgenerwrit is then called Laplace equation\index{Laplace equation}. The elevation of a

string obey to such an equation, where f is the distribution of weight on the string. A

clamped string Fig. figcordef) corresponds to the case where the boundary conditions

are:

Those conditions are called Dirichlet conditions. A sliding string Fig. figcordeg)

corresponds to the case where the boundary conditions are:

Those conditions are called Neumann conditions. Let us recall here the definition of

adjoint operator\index{adjoint operator}:

Definition: The adjoint operator in a Hilbert space of an operator L is the operator L

such that for all u and h in , one has .

File:Cordef

The elevation of a clamped string verify a Laplace equation with Dirichlet boundary

conditions.

One can show that , the space of functions zero in 0 and L is a Hilbert space for the

scalar product

. (One speaks of Sobolev space, see

section secvafor). Moreover, one can show that the adjoint of L is :

As L = L

, L is called self-adjoint .

Cordeg

The elevation of a sliding string verify a Laplace equation with Neumann boundary

conditions.

One can also show that the space of function with derivative zero at x = 0 and at x = L

is a Hilbert space for the scalar product: . Using

equation eqadjoimq, one shows that L is also self-adjoint.

The form \index{form (definite)}

is negative definite in the case of the clamped string and negative (no definite) in the case

of the sliding string. Indeed, in this last case, function u = constant makes < Lu,u > zero.

Let us conclude those remarks about boundary conditions by the compatibility between

the solution u of a boundary problem

Lu = f

and the right hand member f ).

Theorem:

If there exists a solution v

for the homogeneous adjoint problem :

L v = 0

then a necessary condition that the problem Lu = f has a solution is that f is orthogonal to

Proof:

Let us assume that u

is solution of:

Lu = f

Then

thus

If there exists a function v

such that L v = 0

, then previous equation implies:

< v ,f > = 0

Example:

In the case of the sliding string, function v = 1

over [0,L] is solution of the homogeneous

adjoint problem. The previous compatibility equation is then: f should be of average zero:

For the clamped string the homogeneous adjoint problem has no non zero solution.

Linear boundary problems, integral

methods

Problem: Problem P(f,φ,ψ)) : Find such that:

Lu = f in Ω

with .

Let us find the solution using the Green method. Several cases exist:

Nucleus zero, homogeneous problem

Definition: The Green solution\index{Green solution} of problem P(f,0,0) is the function

solution of:

The Green solution of the adjoint problem P(f,0,0) is the function

solution of

where horizontal bar represents complex conjugation.

Theorem: If

and exist then

Proof:

By definition of the adjoint operator L

of an operator L

Using equations eqdefgydy and eqdefgydyc of definition defgreen for

and , one

achieves the proof of the result.

In particular, if L = L

then .

Theorem:

There exists a unique function such that

is solution of Problem P(f,0,0) and

The proof\footnote{In particular, proof of the existence of .} of this result is not given

here but note that if exists then:

Thus, by the definition of the adjoint operator of an operator L:

Using equality Lu = f, we obtain:

and from theorem theogreen we have:

This last equation allows to find the solution of boundary problem, for any function f,

once Green function is known.

Kernel zero, non homogeneous problem

Solution of problem P(f,φ,ψ) is derived from previous Green functions:

Using Green's theorem, one has:

Using boundary conditions and theorem theogreen, we get:

This last equation allows to find the solution of problem P(f,φ,ψ), for any triplet (f,φ,ψ),

once Green function

is known.

Non zero kernel, homogeneous problem

Let's recall the result of section secchoixesp :

Theorem:

If L u = 0

has non zero solutions, and if f isn't in the orthogonal of Ker (L )

, the

problem P(f,0,0) has no solution.

Proof: