Let $k :[0,1]^2 \rightarrow \bf{R} $ be a kernel function definded by $ k(x,y)= (1- max(x,y))^2 .$ Now, let $ L $ be a linear operator defined on $ L^2 [0,1] $ by $$ Lf(x):=\int_0^1 k(x,y)f(y)dy$$. Can we find eigenfunctions and the associated eigenvalues? (I'm looking for eigenfunctions which forms an orthonormal basis. cf. Hilbert-Schmidt Theorem)
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1$\begingroup$ I've edited the title to try to make it more descriptive, but you could probably improve it further. Please don't post questions with titles that don't give any hint as to what the question is about. $\endgroup$– user21349Commented Aug 27, 2019 at 23:19
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2$\begingroup$ Your operator was a perfectly well defined (compact, self-adjoint) operator on $L^2(0,1)$ before you added these boundary conditions. These aren't helping at all. Quite on the contrary: If you do include them, then you now have a non-closed operator, so probably you'd want to take its closure, which gets you back to the original one. $\endgroup$– Christian RemlingCommented Aug 29, 2019 at 13:43
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$\begingroup$ I suspect you got confused about the boundary conditions for the inverse: this is an unbounded (as it has to be, since $0\in\sigma (L)$) differential operator whose domain is therefore a subspace of $L^2$, and the bc's are part of the description of this domain. They follow from the form of $L$, as given in the original version of the question. $\endgroup$– Christian RemlingCommented Aug 29, 2019 at 13:45
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$\begingroup$ I probably contributed to the confusion by not thinking it all the way through in the initial version of my answer. $\endgroup$– Michael EngelhardtCommented Aug 29, 2019 at 13:49
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The inverse operator is $L^{-1} = 1/2 \cdot (x-1)^{-1} d^2/dx^2 - 1/2 \cdot (x-1)^{-2} d/dx$. Its eigenfunctions are derivatives of Airy functions, $Ai^{\prime } ((2\lambda )^{1/3} (x-1))$, $Bi^{\prime } ((2\lambda )^{1/3} (x-1))$, where $\lambda $ are the corresponding eigenvalues. Without further specification of the space of functions $f$, i.e., boundary conditions, no further restriction is placed on $\lambda $.
EDIT: Spurred by Christian Remling's comments, the boundary conditions and the resulting quantization of eigenvalues can be further specified: Since $k(1,y)=0$, solutions must vanish at $x=1$, implying that we have to combine the Airy function derivatives as $\sqrt{3} Ai^{\prime } ((2\lambda )^{1/3} (x-1)) + Bi^{\prime } ((2\lambda )^{1/3} (x-1))$. The allowed eigenvalues $\lambda $ are then determined by the condition that $-(2\lambda )^{1/3} $ must correspond to one of the extrema of the function $\sqrt{3} Ai^{\prime } + Bi^{\prime } $.
These eigenfunctions of $L^{-1} $ are also the eigenfunctions of $L$, with eigenvalues $1/\lambda $.
answered Aug 28, 2019 at 2:37
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1$\begingroup$ Your sentence starting with "Without further specification ..." is quite confusing (to me, maybe I'm misinterpreting it). It almost sounds as if you're saying that any $\lambda\in\mathbb R$ could be an eigenvalue, but of course the original operator $L$ is a well defined (compact, self-adjoint) operator that has some equally well defined spectrum. $\endgroup$ Commented Aug 28, 2019 at 3:30
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$\begingroup$ Rather, the appropriate boundary conditions are delivered to us from the form of the kernel: the functions $u=1$ and $v=(1-x)^2$ are in the domain of $L^{-1}$ near the left and right endpoints, respectively. (So we have Neumann boundary conditions $y'(0)=0$ at $x=0$, and at $x=1$ it's more complicated since this endpoint is singular.) $\endgroup$ Commented Aug 28, 2019 at 3:41
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$\begingroup$ Ah, ok, I didn't try to read the boundary conditions off the kernel, I was expecting them to be stated. But you are right, combined with the statement that we're on $L^2 $, not any boundary behavior is allowed, I didn't follow through on that. OK, it remains to impose the boundary conditions on the given solutions. Thank you! $\endgroup$ Commented Aug 28, 2019 at 3:47
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$\begingroup$ just out of curiosity, how did you find this form of the inverse operator? $\endgroup$– user69109Commented Aug 29, 2019 at 6:21
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2$\begingroup$ $max(x,y)$ exhibits the "active crease" typical of Green's functions at $x=y$. One can write it in terms of step functions, and taking derivatives w.r.t. $x$ generates $\delta $-functions. So I wrote down the first two derivatives of $k$ and cobbled them together with appropriate prefactors such as to leave an isolated $\delta $-function. $\endgroup$ Commented Aug 29, 2019 at 14:04