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What is the green function of the triangular kernel $K$: $$ K(x,y)=1-|x-y| $$ where $x,y\in R$ such that $|x-y|<1$?

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  • $\begingroup$ At the very least, you'd have to say something about dimensionality, domain ... $\endgroup$ Commented Jul 17, 2020 at 14:46
  • $\begingroup$ Yes sorry I added the details. Thanks. $\endgroup$ Commented Jul 17, 2020 at 15:09
  • $\begingroup$ Do you mean the domain to be $R \times R$, as you first say, or $\{(x, y) \in R \times R \mathrel: \lvert x - y\rvert < 1\}$, as you later say? $\endgroup$ Commented Jul 17, 2020 at 15:47
  • $\begingroup$ I am interested in the case where $|x-y|<1$, I corrected, Thanks. $\endgroup$ Commented Jul 17, 2020 at 16:03

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This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = $$

$$ \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) = $$

$$ \int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = $$

$$ \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ as well as $$ \left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} = $$

$$ \left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} = $$

$$ -\int_{-1/2}^{1/2} dy f(y) =0 $$ Of course, the same is true for acting with the inverse.

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  • $\begingroup$ Thank you! Would the differential operator $D$ then be a solution in the sense: $\int dx^{''}K(x,x^{''})D(x^{''},x')=\delta(x-x')$? $\endgroup$ Commented Jul 20, 2020 at 15:37
  • $\begingroup$ Yes - even though this is not manifest in the usual short-hand notation "$-(1/2) d^2 /dx^2$". Just write it as "$D(x^{''} ,x) =\delta (x^{''} - x) (-1/2) d^2 /dx^2 $", then you have the standard two-variable form. The appearance of the $\delta $-function makes the operator local, and one usually just immediately integrates it out, making the notation somewhat less transparent. $\endgroup$ Commented Jul 20, 2020 at 15:58
  • $\begingroup$ Thank you now is much clearer! Why do the functions f have to be odd? Could you please point me to a reference book/paper/notes about inverting kernels? $\endgroup$ Commented Jul 20, 2020 at 18:12
  • $\begingroup$ Good that you asked your question - I had in fact neglected to write another boundary condition. I've added it now. The boundary conditions (including the oddness) are necessary for $-(1/2) d^2 /dx^2 $ to work as an inverse - otherwise, the things I said in my answer don't all work out, you can try it. Any decent mathematical methods book should tell you about Green's functions, say, Riley, Hobson and Bence. $\endgroup$ Commented Jul 20, 2020 at 19:45
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    $\begingroup$ If you try to include even functions, the operator certainly can't be a simple second derivative anymore. First of all, you'd need to exclude the constant function, which has zero eigenvalue and thus would make the operator non-invertible. And even then, $K$ would need to have an extra term quadratic in $(x-y)$. So the operator would have to be more complicated - I don't know how much more off-hand. $\endgroup$ Commented Jul 23, 2020 at 14:01

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