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

the following problem:

probpfppgreen

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

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

with ${\displaystyle \partial \Omega _{1}\cup \partial \Omega _{2}=\partial \Omega }$.

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

eqdefgydy

${\displaystyle L{\mathcal {G}}_{y}=\delta _{y}}$

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

eqdefgydyc

${\displaystyle {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}=\delta _{y}}$

where horizontal bar represents complex conjugation.

theogreen

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

Proof:

By definition of the adjoint operator ${\displaystyle L^{*}}$ of an operator ${\displaystyle L}$

${\displaystyle \int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(r){\mathcal {G}}_{x}(r)dr=\int {\overline {{\mathcal {G}}^{*}}}_{y}(r)L{\mathcal {G}}_{x}(r)dr}$

Using equations eqdefgydy and eqdefgydyc of definition defgreen for ${\displaystyle {\mathcal {G}}_{y}}$ and ${\displaystyle {\mathcal {G}}_{y}^{*}}$, one achieves the proof of the result.

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

Theorem:

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

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

is solution of Problem ${\displaystyle P(f,0,0)}$ and

${\displaystyle L{\mathcal {G}}_{y}=\delta _{y}}$

The proof[1] of this result is not given here but note that if ${\displaystyle {\mathcal {G}}_{y}}$ exists then:

${\displaystyle u(y)=\int \delta _{y}(x)u(x)=\int {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}(x)u(x)dx.}$

Thus, by the definition of the adjoint operator of an operator ${\displaystyle L}$:

${\displaystyle u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx.}$

Using equality ${\displaystyle Lu=f}$, we obtain:

${\displaystyle u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)f(x),}$

and from theorem theogreen we have:

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

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

### Kernel zero, non homogeneous problem

Solution of problem ${\displaystyle P(f,\phi ,\psi )}$ is derived from previous Green functions:

${\displaystyle u(y)=\int \delta _{y}(x)u(x)dx=\int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(x)u(x)dx.}$

Using Green's theorem, one has:

${\displaystyle u(y)=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx+\int _{\Gamma }({\bar {\mathcal {G}}}_{y}^{*}(x){\frac {\partial u}{\partial n}}(x)-u(x){\frac {\partial {\bar {\mathcal {G}}}_{y}^{*}}{\partial n}}(x))dx,}$

Using boundary conditions and theorem theogreen, we get:

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

This last equation allows to find the solution of problem ${\displaystyle P(f,\phi ,\psi )}$, for any triplet ${\displaystyle (f,\phi ,\psi )}$, once Green function ${\displaystyle {\mathcal {G}}_{x}(y)}$ is known.

### Non zero kernel, homogeneous problem

Let's recall the result of section secchoixesp :

Theorem:

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

Proof:

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

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

Thus ${\displaystyle {\mathrel {<}}f|v_{0}{\mathrel {>}}=0}$

However, once ${\displaystyle f}$ is projected onto the orthogonal of ${\displaystyle {\mbox{ Ker }}(L^{*})}$, calculations similar to the previous ones can be made: Let us assume that the kernel of ${\displaystyle L}$ is spanned by a function ${\displaystyle u_{0}}$ and that the kernel of ${\displaystyle L^{*}}$ is spanned by ${\displaystyle v_{0}}$.

defgreen2

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

${\displaystyle L{\mathcal {G}}_{y}=\delta _{y}-u_{0}(y)u_{0}}$

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

${\displaystyle {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}=\delta _{y}-v_{0}(y)v_{0}}$

where horizontal bar represents complex conjugaison.

theogreen2

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

Proof:

By definition of the adjoint ${\displaystyle L^{*}}$ of an operator ${\displaystyle L}$

${\displaystyle \int {\overline {L^{*}{{\mathcal {G}}^{*}}}}_{y}(r){\mathcal {G}}_{x}(r)dr=\int {\overline {{\mathcal {G}}^{*}}}_{y}(r)L{\mathcal {G}}_{x}(r)dr}$

Using definition relations defgreen2 of ${\displaystyle {\mathcal {G}}_{y}}$ and ${\displaystyle {\mathcal {G}}_{y}^{*}}$, one obtains the result.

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

Theorem:

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

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

is solution of problem ${\displaystyle P(f,0,0)}$ in ${\displaystyle {\mbox{Ker}}(L)^{\perp }}$ and

${\displaystyle L{\mathcal {G}}_{y}=\delta _{y}-u_{0}(y)u_{0}}$

Proof[2] of this theorem is not given here. However, let us justify solution definition formula. Assume that ${\displaystyle {\mathcal {G}}_{y}}$ exists. Let ${\displaystyle u_{c}}$ be the projection of a function ${\displaystyle u}$ onto ${\displaystyle {\mbox{Ker}}(L)^{\perp }}$.

${\displaystyle u_{c}(y)=u(y)-u_{0}(y)\int {\bar {u}}_{0}(x)u(x)dx.}$

This can also be written:

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

or

${\displaystyle u_{c}(y)=\int {\overline {L^{*}{\mathcal {G}}^{*}}}_{y}(x)u(x)dx=\int {\bar {\mathcal {G}}}_{y}^{*}(x)Lu(x)dx=\int {\bar {\mathcal {G}}}_{y}^{*}(x)f(x).}$

From theorem theogreen2, we have:

${\displaystyle u_{c}(y)=\int {\mathcal {G}}_{x}(y)f(x)dx.}$
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## 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 ${\displaystyle U}$ be a domain having a symmetry plan: ${\displaystyle \forall x\in U,-x\in U}$. Let ${\displaystyle \partial U}$ be the border of ${\displaystyle U}$. Symmetry plane shares ${\displaystyle U}$ into two subdomains: ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ (voir la figure figsymet).

figsymet

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

Let us seek the solution of the folowing problem:

probori

Problem: Find ${\displaystyle G_{y}(x)}$ such that:

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

and

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

knowing solution of problem

probconnu

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

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

and

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

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

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

and

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

Functions ${\displaystyle {\mathcal {G}}_{y}(x)}$ and ${\displaystyle G_{y}(x)}$ verify the same equation. Green function ${\displaystyle G_{y}(x)}$ is thus simply the restriction of function ${\displaystyle {\mathcal {G}}_{y}(x)}$ to ${\displaystyle U_{1}}$. Problem probori is thus solved.

### Invariance by translation

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

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

where ${\displaystyle *}$ is the convolution product (see appendix{appendconvoldist})\index{convolution}. Function ${\displaystyle {\mathcal {G}}}$ is in this case called elementary solution and noted ${\displaystyle e}$. Case where ${\displaystyle P(f,0,0)}$ 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 ${\displaystyle R^{3}}$. Considered operator is:

${\displaystyle L=\Delta }$

Elementary solution is:

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

Example:

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

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

Elementary solution is:

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