Why boundary condition on Green's function?

Question

Say I have a linear operator $L$ and I define its Green's function by $LG(x,y)=\delta(x-y)$. Then the problem of finding $f(x)$ such that $Lf(x)=g(x)$ for given $g(x)$ admits as solution $f(x)=\int G(x,y)g(y)dy$.

Now, I read everywhere that I must impose on $G$ the same boundary conditions that are imposed on $f$. So, if we want $f(0)=0$, say, then we must also impose $G(0,y)=0$.

I don't understand this. I mean, we have $f(0)=\int G(0,y)g(y)dy$. This integral can be zero without forcing $G(0,y)=0$, which seems as overkill.

What is the deal?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The solution is given by \begin{align} \mrm{f}\pars{x} & = \ell_{p}\pars{x} + \int\mrm{G}\pars{x,y}\,\dd y\,,\qquad \overbrace{\qquad\mrm{f}\pars{a} = \varphi_{a}\qquad} ^{\mbox{Boundary Condition}} \end{align}

where $\ds{L\,\ell_{p}\pars{x} = 0}$ and $\ds{\ell_{p}\pars{x}}$ satisfies the given boundary condition at $\ds{x = a}$. Namely, $\ds{\ell_{p}\pars{a} = \varphi_{a}}$. In order that $\ds{\mrm{f}\pars{x}}$ satisfies that boundary condition you'll impose $\ds{\mrm{G}\pars{a,y} = 0}$ because

$$ \mrm{f}\pars{a} = \ell_{p}\pars{a} + \int\mrm{G}\pars{a,y}\,\dd y = \varphi_a + \int\mrm{G}\pars{a,y}\,\dd y $$

Namely, the boundary condition $\ds{\mrm{f}\pars{a} = \varphi_{a}}$ is satisfied by imposing $\ds{\mrm{G}\pars{a,y} = 0}$.

Answer 2

At least when the condition is $f(a)=0$ I think I understand. The point is that the prescription $f(a)=\int G(a,y)g(y)dy=0$ must be true for any function $g(x)$. This is only possible if $G(a,y)=0$.

For more general conditions, the situation is still not clear to me.

Answer 3

A possible point of view for many linear problems of the form $Lf = g$ on $\Omega$ and $f=\phi$ on $ \partial \Omega$. Two observations: (1) Set $\phi=0$, then it is easier to construct a space with zero boundary conditions where the inverse $L^{-1}$ is welldefined. (2) If you can find a harmonic function $h_\phi$ for the harmonic problem ($g=0$), then linearity allows to built a solution $f:= h_\phi+L^{-1}g$.

For example if $L=\frac d{dx}$ then (1) one can use the Bounded Inverse Theorem to show the existence of $L^{-1}(=\int_0^xdy)$, by constructing an appropriate norm $\|\cdot\|_{L}=\|\cdot\|_{C^1}$ on an appropriate space \begin{align} L:& C^1[0,1]\cap\{f(0)=0\} \to C[0,1], \end{align} and (2) pick $h_\phi=\phi$.

Alternatively you could define the Green's function distributionally to include boundary conditions as $G(x,y)=\mathbf 1(x\ge y)+\phi\delta_0(x-y)$.