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Introduction to Mathematical Physics/Some mathematical problems and their solution/Linear boundary problems, integral methods
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the following problem:
probpfppgreen
Problem ) :
Find such that:
with .
Let us find the solution using the Green method. Several cases exist:
Nucleus zero, homogeneous problem
defgreen
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
If and exist then
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 .
There exists a unique function such that
is solution of Problem and
The proofIn 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 :
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
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
Let's recall the result of section secchoixesp :
If has non zero solutions, and if isn't in the orthogonal of
, the problem has no solution.
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
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
If and exist, then
By definition of the adjoint of an operator
Using definition relations defgreen2
of and , one obtains the result.
In particular, if then .
There exists a unique function such that
is solution of problem in and
ProofIn particular, the existence of . 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
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
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
center | frame |Domain is the union of and symetrical with respect to plane .}
Let us seek the solution of the folowing problem:
probori
Find such that:
and
knowing solution of problem
probconnu
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
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:
Laplace f equation in . Considered operator is:
Elementary solution is:
Helmholtz equation in .
Considered operator is:
Elementary solution is: