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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.

share|improve this question
    
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) –  fedja Feb 14 '13 at 5:26
    
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. –  Miloslav Capek Feb 14 '13 at 12:54
2  
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? –  fedja Feb 14 '13 at 14:46
    
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). –  Miloslav Capek Feb 14 '13 at 16:14
    
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. –  Miloslav Capek Feb 14 '13 at 16:17
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