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.