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

the following problem:

probpfppgreen

Problem: Problem $P(f,\phi ,\psi )$ ) : Find $u\in U$ such that:

$Lu=f{\mbox{ in }}\Omega$ $u=\phi {\mbox{ in }}\partial \Omega _{1}$ ${\frac {\partial u}{\partial n_{L}}}=\psi {\mbox{ in }}\partial \Omega _{2}$ with $\partial \Omega _{1}\cup \partial \Omega _{2}=\partial \Omega$ .

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

### Nucleus zero, homogeneous problem

defgreen

Definition: The Green solution\index{Green solution} of problem $P(f,0,0)$ is the function ${\mathcal {G}}_{y}(x)$ solution of:

eqdefgydy

$L{\mathcal {G}}_{y}=\delta _{y}$ The Green solution of the adjoint problem $P(f,0,0)$ is the function ${\mathcal {G}}_{y}^{*}(x)$ solution of

eqdefgydyc

${\overline {L^{*}{\mathcal {G}}^{*}}}_{y}=\delta _{y}$ where horizontal bar represents complex conjugation.

theogreen

Theorem: If ${\mathcal {G}}_{y}$ and ${\mathcal {G}}_{y}^{*}$ exist then ${\overline {{\mathcal {G}}^{*}}}_{y}(x)={\mathcal {G}}_{x}(y)$ Proof:

By definition of the adjoint operator $L^{*}$ of an operator $L$ $\int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(r){\mathcal {G}}_{x}(r)dr=\int {\overline {{\mathcal {G}}^{*}}}_{y}(r)L{\mathcal {G}}_{x}(r)dr$ Using equations eqdefgydy and eqdefgydyc of definition defgreen for ${\mathcal {G}}_{y}$ and ${\mathcal {G}}_{y}^{*}$ , one achieves the proof of the result.

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

Theorem:

There exists a unique function ${\mathcal {G}}_{y}$ such that

$u(y)=\int {\mathcal {G}}_{x}(y)f(x)dx$ is solution of Problem $P(f,0,0)$ and

$L{\mathcal {G}}_{y}=\delta _{y}$ The proof of this result is not given here but note that if ${\mathcal {G}}_{y}$ exists then:

$u(y)=\int \delta _{y}(x)u(x)=\int {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}(x)u(x)dx.$ Thus, by the definition of the adjoint operator of an operator $L$ :

$u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx.$ Using equality $Lu=f$ , we obtain:

$u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)f(x),$ and from theorem theogreen we have:

$u(y)=\int {\mathcal {G}}_{x}(y)f(x)dx$ 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:

$u(y)=\int \delta _{y}(x)u(x)dx=\int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(x)u(x)dx.$ Using Green's theorem, one has:

$u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx+\int _{\Gamma }({\bar {\mathcal {G}}}_{y}^{*}(x){\frac {\partial u}{\partial n}}(x)-u(x){\frac {\partial {\bar {\mathcal {G}}}_{y}^{*}}{\partial n}}(x))dx,$ Using boundary conditions and theorem theogreen, we get:

$u(y)=\int {\mathcal {G}}_{x}(y)f(x)dx+\int _{\Gamma }({\mathcal {G}}_{x}(y)\phi (x)-\psi (x){\frac {\partial {\mathcal {G}}_{x}(y)}{\partial n}})dx,$ 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 :

Theorem:

If $L^{*}u_{0}=0$ has non zero solutions, and if $f$ isn't in the orthogonal of ${\mbox{ Ker }}(L^{*})$ , the problem $P(f,0,0)$ has no solution.

Proof:

Let $u$ such that $Lu=f$ . Let $v_{0}$ be a solution of $L^{*}v_{0}=0$ ({\it i.e} a function of ${\mbox{ Ker }}(L^{*})$ ). Then:

${\begin{matrix}{\mathrel {<}}f|v_{0}{\mathrel {>}}&=&{\mathrel {<}}Lu,v_{0}{\mathrel {>}}\\&=&{\mathrel {<}}u,L^{*}v_{0}{\mathrel {>}}.\end{matrix}}$ Thus ${\mathrel {<}}f|v_{0}{\mathrel {>}}=0$ 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 $v_{0}$ .

defgreen2

Definition: The Green solution of problem $P(f,0,0)$ is the function ${\mathcal {G}}_{y}(x)$ solution of

$L{\mathcal {G}}_{y}=\delta _{y}-u_{0}(y)u_{0}$ The Green solution of adjoint problem $P(f,0,0)$ is the function ${\mathcal {G}}_{y}^{*}(x)$ solution of

${\overline {L^{*}{\mathcal {G}}^{*}}}_{y}=\delta _{y}-v_{0}(y)v_{0}$ where horizontal bar represents complex conjugaison.

theogreen2

Theorem: If ${\mathcal {G}}_{y}$ and ${\mathcal {G}}_{y}^{*}$ exist, then ${\overline {{\mathcal {G}}^{*}}}_{y}(x)={\mathcal {G}}_{x}(y)$ Proof:

By definition of the adjoint $L^{*}$ of an operator $L$ $\int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(r){\mathcal {G}}_{x}(r)dr=\int {\overline {{\mathcal {G}}^{*}}}_{y}(r)L{\mathcal {G}}_{x}(r)dr$ Using definition relations defgreen2 of ${\mathcal {G}}_{y}$ and ${\mathcal {G}}_{y}^{*}$ , one obtains the result.

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

Theorem:

There exists a unique function ${\mathcal {G}}_{y}$ such that

$u(y)=\int {\mathcal {G}}_{x}(y)f(x)dx$ is solution of problem $P(f,0,0)$ in ${\mbox{Ker}}(L)^{\perp }$ and

$L{\mathcal {G}}_{y}=\delta _{y}-u_{0}(y)u_{0}$ 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 ${\mbox{Ker}}(L)^{\perp }$ .

$u_{c}(y)=u(y)-u_{0}(y)\int {\bar {u}}_{0}(x)u(x)dx.$ This can also be written:

$u_{c}(y)=\int \delta _{y}(x)u(x)-u_{0}(y)\int {\bar {u}}_{0}(x)u(x)dx,$ or

$u_{c}(y)=\int {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}(x)u(x)dx=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx=\int {\bar {\mathcal {G}}}_{y}^{*}(x)f(x).$ From theorem theogreen2, we have:

$u_{c}(y)=\int {\mathcal {G}}_{x}(y)f(x)dx.$ 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 $U$  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).

figsymet

Domain $U$  is the union of $U_{1}$  and $U_{2}$  symetrical with respect to plane $x=0$ .}

Let us seek the solution of the folowing problem:

probori

Problem: Find $G_{y}(x)$  such that:

$LG_{y}(x)=\delta (x){\mbox{ in }}U_{1}$

and

$G_{y}(x)=0{\mbox{ on }}\partial U_{1}$

knowing solution of problem

probconnu

Problem: Find $G_{y}^{U}(x)$  such that:

$LG_{y}^{U}(x)=\delta (x){\mbox{ in }}U$

and

$G_{y}^{U}(x)=0{\mbox{ on }}\partial U$

Method of solving problem probori by using solution of problem probconnu is called images method ([ma:equad:Dautray1]). Let us set ${\mathcal {G}}_{y}(x)={G}_{y}^{U}(x)-{G}_{y}^{U}(-x)$ . Function ${\mathcal {G}}_{y}(x)$  verifies

$L{\mathcal {G}}_{y}(x)=\delta (x){\mbox{ in }}U_{1}$

and

${\mathcal {G}}_{y}(x)=0{\mbox{ on }}\partial U_{1}$

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:

$u(x)=\int {\mathcal {G}}(x-y)f(y)dx={\mathcal {G}}*f$

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 ([ma:distr:Schwartz65]).

Here are some examples of well known elementary solutions:

Example:

Laplace f equation in $R^{3}$ . Considered operator is:

$L=\Delta$

Elementary solution is:

$e(r)={\frac {1}{4\pi r}}$

Example:

Helmholtz equation in $R^{3}$ . Considered operator is:

$L=\Delta +k^{2}$

Elementary solution is:

$e(r)={\frac {e^{jkr}}{4\pi r}}$
1. In particular, proof of the existence of ${\mathcal {G}}_{y}$ .
2. In particular, the existence of ${\mathcal {G}}_{y}$ .