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It's my first post here, but I worked very hard to find solution and I failed.

Hereinafter, I skip physical background and directly proceed to my mathematical problem.

No matter how, you know the functional $\mathcal{F} ( f ) = \frac{\displaystyle \int\limits_{-1}^1 \int\limits_{-1}^1 \left( \left(\frac{\pi}{2}\right)^2 f(x) f(y) - g(x) g (y) \right) \frac{\cos \left(\frac{\pi}{2} |x - y|\right)}{|x - y|} \mathrm{d} x \mathrm{d} y}{\displaystyle \int\limits_{-1}^1 \int\limits_{-1}^1 \left( \left(\frac{\pi}{2}\right)^2 f(x) f(y) - g(x) g (y) \right) \frac{\sin \left(\frac{\pi}{2} |x - y|\right)}{|x - y|} \mathrm{d} x \mathrm{d} y} = \displaystyle\frac{\mathcal{W}(f)}{\mathcal{P}(f)},\qquad(1)$

where the function $g(x)$ has a form of
$g(x) = \displaystyle\frac{\partial f(x)}{\partial x}.\qquad(2)$
and similarly for $g(y)$.

Moreover, you know the boundary condition
$f(-1) = f(1) = 0 \qquad(3)$
(Dirichlet B.C.).

Expression (1) has a form of energetic functional which can be discretized to the generalized eigenvalue problem (GEP):
$F(f) = \displaystyle\frac{\langle f, \mathbf{X} f \rangle}{\langle f, \mathbf{R} f \rangle} = \frac{W(f)}{P(f)}, \qquad(4)$
where $\mathbf{R}$ and $\mathbf{X}$ are square $(N \times N)$ symmetric matrices. Some of you probably know the Method of moments in electrodynamics, where the resultant impedance matrix is $\mathbf{Z} = \mathbf{R} + \jmath \mathbf{X}$.

Problem is,that the GEP (4) says that there is a function $f_0(x)$ in natural resonance for which the numerator and denumerator in (1) are:

$W(f_0) = 0$ and $P(f_O) = \max $, respectively.

So, my question is:
Can you find analytical function $f_0$ that causes $\mathcal{W}(f_0) = 0$ and $\mathcal{P}(f_0) = \max$. Note, that the second condition quarantees the uniqueness of solution.

And additionally:
Do you know any transformation of (1) which transforms the double integration (convolution) to the single integration?

Many thanks.

Edit1: Is it possible to prove that (1) converges only for the function $f(x) = \sin(\pi x/2)$?

Edit2 /Based on latest comments/: Well, here's the transition from the 3D to the 1D functional (1). We started from the following functional:

$\mathcal{F} ( \mathbf{J} ) = \frac{\displaystyle \iiint\limits_{\Omega} \iiint\limits_{\Omega'} \left(k^2 \mathbf{J} (\mathbf{r}) \cdot \mathbf{J} (\mathbf{r}') - \nabla \cdot \mathbf{J} (\mathbf{r}) \nabla ' \cdot \mathbf{J} (\mathbf{r}') \right) \frac{\cos \left(k R \right)}{R} \, \mathrm{d} \mathbf{r} \, \mathrm{d} \mathbf{r}'}{\displaystyle \iiint\limits_{\Omega} \iiint\limits_{\Omega'} \left(k^2 \mathbf{J} (\mathbf{r}) \cdot \mathbf{J} (\mathbf{r}') - \nabla \cdot \mathbf{J} (\mathbf{r}) \nabla ' \cdot \mathbf{J} (\mathbf{r}') \right) \frac{\sin \left(k R \right)}{R} \, \mathrm{d} \mathbf{r} \, \mathrm{d} \mathbf{r}'} = \displaystyle\frac{\mathcal{W}(f)}{\mathcal{P}(f)},\qquad(5)$

where $R = ||\mathbf{r}, \mathbf{r}'||$ is the Euclidean distance, $\Omega$ is region containing all sources (i.e., only here is $\mathbf{J}\neq 0$), $\mathbf{J}$ is arbitrary vector function with Dirichlet B.C. on boundary of $\Omega$ (noted as $\partial\Omega$) and tangential to the surface $\partial\Omega$ ($\mathbf{J}$ is actually current density on a perfect electric conductor).

Functional (5) is derived from Maxwell theory, Electric Field Integral Equation (EFIE), Poynting and some other theorems. We tested it numerically with great success.

If we stress the region $\Omega$ to straight thin-wire (1D object) and define as yet-unknown function

$\mathbf{J} = I(x) \delta(y) \delta(z) \, \mathbf{x}_0 \sim f(x),\qquad(6)$

where $\delta$ is delta function and $\mathbf{x}_0$ is unit vector pointing in $x$-direction, then according to the definition of divergence

$\displaystyle \nabla \cdot \mathbf{J} = \frac{\partial _k J_k} {\partial_k x_k} = \frac{\partial I(x)}{\partial x}\,\delta (y)\,\delta (z),\qquad (7)$

from (6) we have

$\displaystyle \nabla \cdot \mathbf{J} = \frac{\partial I(x)}{\partial x} \delta(y) \delta(z) \sim \frac{\partial f(x)}{\partial x},\qquad(8)$.

After assumptions (5)-(8) we can conclude, that 1D variation of (5) is (1). Note, that $k$ is "wave number", which is equal to $k = \pi / 2$ for our case of thin-wire object.

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    Erm... Doesn't it bother you that the kernel in the numerator has a bad singularity on the diagonal $x=y$ (so bad, that formally the integral $W(f)$ doesn't converge at all)2013-02-14
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    Good question at all. I'm aware of $\cos(kR)/R$ singularity. But, however, this singularity is removable. For example of following function, which is $f(x) = \cos\left(\frac{\pi x}{2}\right)$, the whole functional could be transformed to the simplex coordinates where the singularities are automatically removed. The presence of the singularity makes finding of proper solution extremely difficult.2013-02-14
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    I choose to humbly disagree with you here. If you try to remove it in the usual way ($\lim_{\delta\to 0}\int_{|x-y|>\delta}$), then you'll need $\left(\frac\pi 2\right)^2\int f^2=\int g^2$ to avoid divergence. The only function for which it is possible is the one you mentioned (otherwise Poincare's inequality kills the whole business and you get $-\infty$ with no right of appeal). So, what do you really mean?2013-02-14
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    Thank you! So (please, feel free to correct me - I'm rather physicist than pure mathematician), the $\sin\left(n\pi x / 2\right)$ base is the only one that have finite solution of $\mathcal{F(f)}$? While I cannot contradict this conclusion, it sounds strange to me, because we are sure that there is a function in nature, that provides $\mathcal{F(f)} = 0$. Of course, this function can be written as a Fourier series. However, this series is infinite (but convergent).2013-02-14
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    Please, could you give me some hints regarding Poincare's inequality? What about triangle function $f_2 (x) = 1 - |x|$? Still hoping that the triangle function is also solvable. On the other hand, it will be nice to have exact proof that only $\sin$ is finite (and thus only thinkable) function.2013-02-14
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    Sure. You know that $\cos\frac\pi 2(2n-1) x$, $n=1,2,3,\dots$ is a complete orthonormal system on $[-1,1]$ (usually it is stated as "$\sqrt 2\sin\pi nx$ is ... on $[0,1]$". Decomposing $f=\sum_n a_n\cos\frac\pi 2(2n-1) x$, differentiating, and using the orthogonality of the corresponding $\sin$ system, we get $\int_{-1}^1f^2=\sum_n |a_n|^2$ and $\int_{-1}^1|g|^2=\left(\frac\pi 2\right)^2\sum_n(2n-1)^2|a_n|^2$, so the only chance to get the equality that prevents divergence is to have $a_n=0$ for $n\ge 2$, i.e., to have $f$ a pure $\cos$ wave at the lowest frequency.2013-02-14
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    Thank you for that. I was thinking about this functional during the weekend. And still I doubt about rareness of the $\sin$ solution. While I haven't any exact explanation, it is clear from the observation that there are also other possible solutions. Maybe I made a mistake in 3D to 1D functional's transformation. Are you absolutely sure, that your prove is complete and without any exception? Note, that the double (convolution) integration may help with handling of singularities.2013-02-18
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    I'm absolutely sure that what I wrote is correct for the case of reasonably good $f$ ($W^0_{1,2}([-1,1)]$ is the exact class to talk about). Why don't you post the original 3D version of the problem and your derivation of the 1D one? Then we can see if something went astray.2013-02-18
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    Dear _fedja_ (and others), please find completed **Edit2** in my question. Many thanks.2013-02-18
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    And what is the reason for the value $k=\frac\pi 2$? As soon as $k$ is larger than that number the story gets interesting but with exactly that number you, indeed, cannot get anything good for zero wire thickness though, of course, have something for any finite thickness. Note also that the divergence is merely logarithmic, so it may happen that your real wire is just too thick to see that blow-up in the experiment. Something is strange, indeed. I'll try to think of it but a bit later...2013-02-19
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    Well, parameter $k$ has strictly physical meaning - wave number is defined as $k = 2 \pi f / c_0 = 2 \pi / \lambda$ where $\lambda$ is wavelength, $c_0$ is velocity of light and $f$ is frequency (so, for self-resonant half-wave thin-wire dipole in vacuum the $k$ is equal to $k = \pi / 2$, this case is the most interesting because it's widely misinterpreted in the literature). However, of course, the $k$ is optional in general.2013-02-19
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    By the way, the solution for any 2D region of form $\Omega \in $ in which (5) is defined for function $f(x,y) = f(x)$ ($f$ doesn't vary with $y$) and $k = \pi / 2$ should be exactly the same as in 1D case. This approach may excludes some problems with singularity in case of 1D problem $, a\rightarrow 0$. Please, in case of any doubts or questions about physical background just ask me :)2013-02-19

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