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

## Contents

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

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: