# 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[1] 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[2] 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 $G^U_y(x)$ such that:

$LG^U_y(x)=\delta(x) \mbox{ in } U$

and

$G^U_y(x)=0 \mbox{ on } \partial U$

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