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

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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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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.