Cossey L. Introduction to Mathematical Physics

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Let u such that Lu = f. Let v

be a solution of L v = 0

({\it i.e} a function of Ker (L )

Then:

Thus

However, once f is projected onto the orthogonal of Ker (L )

, calculations similar to the

previous ones can be made: Let us assume that the kernel of L is spanned by a function u

and that the kernel of L

is spanned by v

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

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

where horizontal bar represents complex conjugaison.

Theorem: If and exist, then

Proof:

By definition of the adjoint L

of an operator L

Using definition relations defgreen2 of and , one obtains the result.

In particular, if L = L

then .

Theorem:

There exists a unique function

such that

is solution of problem P(f,0,0) in

and

Proof\footnote{In particular, the existence of .} of this theorem is not given here.

However, let us justify solution definition formula. Assume that

exists. Let u

be the

projection of a function u onto

This can also be written:

From theorem theogreen2, we have:

Resolution

Once problem's Green function is found, problem's solution is obtained by simple

integration. Using of symmetries allows to simplify seeking of Green's functions.

Images method

\index{images method}

Let U be a domain having a symmetry plan:

\partial UbetheborderofU</math>. Symmetry

plane shares U into two subdomains: U

and U

(voir la figure figsymet).

Symet

Domain U is the union of U

and U

symetrical with respect to plane x = 0.} LABEL

figsymet

Let us seek the solution of the folowing problem:

Problem: Find G

(x) such that:

(x) = δ(x) in U

and

knowing solution of problem

Problem: Find such that:

and

Method of solving problem p em probconnu is called robori by using solution of probl

images method ). Let us set . Function verifies

and

Functions and G (x) the same equation. Green function

verify x) is thus simply

the restriction of function

(

to U

. Problem probori is thus solved.

Invariance by translation

riant b unction's dWhen problem P(f,0,0) is inva y translation, Green f efinition relation can

be simplified. Green function becomes a function that depends only on

difference x - y and its definition relation is:

where * is the convolution product appendix{appendconvoldist})\index{convolution}.

Function is in this case called elementary solution and noted e. Case where P(f,0,0) is

translation invariant typically corresponds to infinite boundaries. Here are some examples

of well known elemantary solutions:

Example:

Laplace equation in R

. Considered operator is:

L = ∆

Elementary solution is:

Example:

Helmholtz equation in R

. Considered operator is:

L = ∆ + k

Elementary solution is:

Boundary problems, variational methods

==Variational formulation==LABEL secvafor

Let us consider a boundary problem:

Problem:

Find

such that:

Let us suppose that is a unique solution u of this problem. For a functional space V

sufficiently large the previous problem, may be equivalent to the following:

Problem:

Find such that:

This is the variational form of the boundary problem. To obtain the equality

eqvari, we have just multiplied scalarly by a "test function" < v |

equation eqLufva. A "weak form" of this problem can be found using Green formula type

formulas: the solution space is taken larger and as a counterpart, the test functions space

is taken smaller. let us illustrate those ideas on a simple example:

Example:

Find u in U = C

such that:

The variational formulation is: Find u in U = C

such that:

Using the Green equality appendix secappendgreeneq)\index{Green formula} and the

boundary conditions we can reformulate the problem in: find , such that:

One can show that this problem as a unique solution in

where

is the Sobolev space of order 1 on Ω and is the adherence of H (Ω)

in the space

(space of infinitely differentiable functions on Ω, with compact support in Ω).

It may be that, as for the previous example, the solution function space and the test

function space are the same. This is the case for most of linear operators L encountered in

physics. In this case, one can associate to L a bilinear form a. The (weak) variational

problem is thus:

Problem: LABEL provari2 Find

such that

Example:

In this previous example, the bilinear form is:

There exist theorems (Lax-Milgram theorem for instance) that proove the existence and

unicity of the solution of the previous problem

provari2 , under certain

conditions on the bilinear form a(u,v).

==Finite elements approximation==LABEL secvarinum

Let us consider the problem provari2 :

Problem:

Find

such that:

The method of approximation consists in choosing a finite dimension subspace V

of V

and the problem to solve becomes:

Problem: LABEL provari3 Find

such that

A base of of V

is chosen to satisfy boundary conditions. The problem is reduced to

finding the components of u

If a is a bilinear form

and to find the 's is equivalent to solve a linear system (often close to a

diagonal system)that can be solved by classical algorithms

which can be direct methods (Gauss,Cholesky) or iterative (Jacobi, Gauss-Seidel,

relaxation). Note that if the vectors of the basis of V

are eigenvectors of L, then the

solving of the system is immediate (diagonal system). This is the basis of the spectral

methods for solving evolution problems. If a is not linear, we have to solve a nonlinear

system. Let us give an example of a basis .

Example:

When V = L ([0,1])

, an example of V

that can be chosen

Finite difference approximation

Finite difference method is one of the most basic method to tackle PDE problems. It is

not strictly speaking a variational approximation. It is rather a sort of variational method

where weight functions w

are Dirac functions δ

. Indeed, when considering the boundary

problem,

instead of looking for an approximate solution u

which can be decomposed on a basis of

weight functions w

∑

< w

| u

> w

the action of L on u is directly expressed in terms of Dirac functions, as well as the right

hand term of equation eqfini:

\begin{rem} If L contains derivatives, the following formulas are used : right formula,

order 1:

right formula order 2:

Left formulas can be written in a similar way. Centred formulas, second order are:

Centred formulas, fourth order are:

\end{rem} One can show that the equation eqfini2 is equivalent to the system of

equations:

LABELeqfini3

∑

(Lu)

= f

One can see immediately that equation eqfini2 implies equation

eqfini3 in choosing "test" functions v

of support

[x − 1 / 2,x + 1 / 2]

i i

and such that v(x ) = 1

i i

Minimization problems

A minimization problem can be written as follows:

Problem: LABEL promini Let V a functional space, a J(u) a functional. Find ,

such that:

The solving of minimization problems depends first on the nature of the functional J and

on the space V.

As usually, the functional J(u) is often approximated by a function of several variables

where the u

's are the coordinate of u in some base E

that approximates

V. The methods to solve minimization problems can be classified into two categories:

One can distinguish problems without constraints Fig. figcontraintesans) and problems

with constraints Fig. figcontrainteavec). Minimization problems without constraints can

be tackled theoretically by the study of the zeros of the differential function dF(u) if

exists. Numerically it can be less expensive to use dedicated methods. There are methods

that don't use derivatives of F (downhill simplex method, direction-set method) and

methods that use derivative of F (conjugate gradient method, quasi-Newton methods).

Details are given in

Problems with constraints reduce the functional space U to a set V of functions that

satisfy some additional conditions. Note that those sets V are not vectorial spaces: a linear

combination of vectors of V are not always in V. Let us give some example of constraints:

Let U a functional space. Consider the space

where φ (v)

are n functionals. This is a first example of constraints. It can be solved

theoretically by using Lagrange multipliers ),\index{constraint}. A second example of

constraints is given by

where φ (v)

are n functionals. The linear programming example exmplinepro) problem is

an example of minimization problem with such constraints (in fact a mixing of equalities

and inequalities constraints).

Example:

Let us consider of a first class of functional J that are important for PDE problems.

Consider again the bilinear form introduced at section secvafor. If this bilinear form

a(u,v) is symmetrical, {\it i.e.}

the problem can be written as a minimization problem by introducing the functional:

One can show that (under certain conditions) solving the variational problem:\\ Find

such that:

is equivalent to solve the minimization problem:\\ Find , such that:

Physical principles have sometimes a natural variational formulation (as natural as a PDE

formulation). We will come back to the variational formulations at the section on least

action principle section secprinmoindreact ) and at the section

secpuisvirtu on the principle of virtual powers.

Example: LABEL exmplinepro Another example of functional is given by the linear

programming problem. Let us consider a function u that can be located by its coordinates

in a basis e

, :

In linear programming, the functional to minimize can be written

and is subject to N primary constraints:

and M = m + m + m

1 2 3

additional constraints: