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

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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 problem[edit]

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 problem[edit]

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 problem[edit]

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

Resolution[edit]

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 method[edit]

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 translation[edit]

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 .