Introduction to Mathematical Physics/Some mathematical problems and their solution/Linear boundary problems, integral methods

the following problem:

probpfppgreen

Problem: Problem ) : Find such that:

with .

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

Nucleus zero, homogeneous problemEdit

defgreen

Definition: The Green solution\index{Green solution} of problem is the function solution of:

eqdefgydy

The Green solution of the adjoint problem is the function solution of

eqdefgydyc

where horizontal bar represents complex conjugation.

theogreen

Theorem: If and exist then

Proof:

By definition of the adjoint operator of an operator

Using equations eqdefgydy and eqdefgydyc of definition defgreen for and , one achieves the proof of the result.

In particular, if then .

Theorem:

There exists a unique function such that

is solution of Problem and

The proof[1] of this result is not given here but note that if exists then:

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

Using equality , we obtain:

and from theorem theogreen we have:

This last equation allows to find the solution of boundary problem, for any function , once Green function is known.

Kernel zero, non homogeneous problemEdit

Solution of problem 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 , for any triplet , once Green function is known.

Non zero kernel, homogeneous problemEdit

Let's recall the result of section secchoixesp :

Theorem:

If has non zero solutions, and if isn't in the orthogonal of , the problem has no solution.

Proof:

Let such that . Let be a solution of ({\it i.e} a function of ). Then:

Thus

However, once is projected onto the orthogonal of , calculations similar to the previous ones can be made: Let us assume that the kernel of is spanned by a function and that the kernel of is spanned by .

defgreen2

Definition: The Green solution of problem is the function solution of

The Green solution of adjoint problem is the function solution of

where horizontal bar represents complex conjugaison.

theogreen2

Theorem: If and exist, then

Proof:

By definition of the adjoint of an operator

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

In particular, if then .

Theorem:

There exists a unique function such that

is solution of problem in and

Proof[2] of this theorem is not given here. However, let us justify solution definition formula. Assume that exists. Let be the projection of a function onto .

This can also be written:

or

From theorem theogreen2, we have:

  zlxuc olzkyc bjnc oihxc vz8cg

ResolutionEdit

secresolinv

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 methodEdit

secimage

\index{images method}

Let   be a domain having a symmetry plan:  . Let   be the border of  . Symmetry plane shares   into two subdomains:   and   (voir la figure figsymet).

figsymet

 
Domain   is the union of   and   symetrical with respect to plane  .}

Let us seek the solution of the folowing problem:

probori

Problem: Find   such that:

 

and

 

knowing solution of problem

probconnu

Problem: Find   such that:

 

and

 

Method of solving problem probori by using solution of problem probconnu is called images method ([ma:equad:Dautray1]). Let us set  . Function   verifies

 

and

 

Functions   and   verify the same equation. Green function   is thus simply the restriction of function   to  . Problem probori is thus solved.

Invariance by translationEdit

When problem   is invariant by translation, Green function's definition relation can be simplified. Green function   becomes a function   that depends only on difference   and its definition relation is:

 

where   is the convolution product (see appendix{appendconvoldist})\index{convolution}. Function   is in this case called elementary solution and noted  . Case where   is translation invariant typically corresponds to infinite boundaries ([ma:distr:Schwartz65]).

Here are some examples of well known elementary solutions:

Example:

Laplace f equation in  . Considered operator is:

 

Elementary solution is:

 


Example:

Helmholtz equation in  . Considered operator is:

 

Elementary solution is:

 

  1. In particular, proof of the existence of  .
  2. In particular, the existence of  .