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

the following problem:

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

Nucleus zero, homogeneous problem
In particular, if $$L=L^*$$ then $${\mathcal G}^*_x(y)= {\mathcal G}_x(y)$$.

The proof of this result is not given here but note that if $${\mathcal G}_y$$ exists then:

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

Using equality $$Lu=f$$, we obtain:

and from theorem theogreen we have:

This last equation allows to find the solution of boundary problem, for any function $$f$$, once Green function $${\mathcal G}_x(y)$$ is known.

Kernel zero, non homogeneous problem
Solution of problem $$P(f,\phi,\psi)$$ 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 $$P(f,\phi,\psi)$$, for any triplet $$(f,\phi,\psi)$$, once Green function $${\mathcal G}_x(y)$$ is known.

Non zero kernel, homogeneous problem
Let's recall the result of section secchoixesp :

However, once $$f$$ is projected onto the orthogonal of $$\mbox{ Ker }(L^*)$$, calculations similar to the previous ones can be made: Let us assume that the kernel of $$L$$ is spanned by a function $$u_0$$ and that the kernel of $$L^*$$ is spanned by $$v_0$$.

In particular, if $$L=L^*$$ then $${\mathcal G}^*_x(y)= {\mathcal G}_x(y)$$.

Proof of this theorem is not given here. However, let us justify solution definition formula. Assume that $${\mathcal G}_y$$ exists. Let $$u_c$$ be the projection of a function $$u$$ onto $$\mbox{Ker}(L)^\perp$$.

This can also be written:

or

From theorem theogreen2, we have:

zlxuc olzkyc bjnc oihxc vz8cg

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: $$\forall x \in U, -x\in U$$. Let $$\partial U$$ be the border of $$U$$. Symmetry plane shares $$U$$ into two subdomains: $$U_1$$ and $$U_2$$ (voir la figure figsymet).

Let us seek the solution of the folowing problem:

knowing solution of problem

Method of solving problem probori by using solution of problem probconnu is called images method. Let us set $${\mathcal G}_y(x)={G}^U_y(x)-{G}^U_y(-x)$$. Function $${\mathcal G}_y(x)$$ verifies

and

Functions $${\mathcal G}_y(x)$$ and $$G_y(x)$$ verify the same equation. Green function $$G_y(x)$$ is thus simply the restriction of function $${\mathcal G}_y(x)$$ to $$U_1$$. Problem probori is thus solved.

Invariance by translation
When problem $$P(f,0,0)$$ is invariant by translation, Green function's definition relation can be simplified. Green function $${\mathcal G}_y(x)$$ becomes a function $${\mathcal G}(x-y)$$ that depends only on difference $$x-y$$ and its definition relation is:

where $$*$$ is the convolution product (see appendix{appendconvoldist})\index{convolution}. Function $${\mathcal G}$$ 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 elementary solutions: