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:

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

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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 $U$ 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:

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