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2

This does not answer the question asked (see the other answer for a good counterexample) and I don't know if it's relevant to the paper. However, if $u$ is sufficiently far from $\varphi_1$, the first eigenfucntion, you do get a positive answer. Below I assume $\Omega$ is connected (you'd have to work out what happens more carefully if it's not). First, we ...


2

The first is not true, and probably also the others. Take $L^2(0, \pi)$ and $u_1=\sin x$, $u_2=\sin (2x)$, so that $E_2=\{u=a\sin x+b \sin (2x)\}$ and $u \geq 0$ iff $a \geq 0$ and $2|b| \leq a$. If $v=\alpha sin x+\beta \sin (2x)$, then $\|u-v\|_2^2=\frac{\pi}{2} \left((a-\alpha)^2+(b-\beta)^2\right)$ and, if $v_\epsilon=\sin x-\frac{1+\epsilon}{2}\sin (2x)$...


0

A simple observation which proves that the Lagrangian does not represent the reality of physical phynomemes, we can divide the Lagrangian into two parts, the part which contains the free field (without the presence of charges) and the part which contains the interaction part between the EM field and the material: A simple observation that proves that the ...


-1

In the Lagrangian he got, there is a term that reflects the interaction of the electromagnetic field with matter which has the form of equation (2) in this article ( to within a multiplicative factor : derivation of the energy per unit of volume)) but with a different factor which contains the tensor and its Hermitian conjugate. we do not know how he ...


-1

It seems to me that your Lagrangian does not reflect the reality of physical phenomena and especially the absorption phenomenon of an EM wave, in the case of the equality of your result, there is no absorption (see, Landau, Lifchitz, tome VIII and also tome V), and all the results derive from the relations of Kramers – Kronig :https://en.wikipedia.org/wiki/...


3

Expanding on Carlo Beenakker's comment, one can't just expect to substitute in a complex dielectric function to properly describe absorption. Rather, the relevant question is how to structure a Lagrangean such as to generate the desired damping term in the equation of motion. For example, to generate linear damping in a harmonic oscillator, one can use $$ L=...


2

Consider $$ f(x, y) = \max \biggl\{ \frac{3 x}{2^n} - \frac{1}{4^n} + \frac{(-1)^n y}{n + 1} : n = 0, 1, 2, \ldots \biggr\} . $$ Then $f$ is clearly convex, and $$ \partial_y f(x, 0) = \frac{(-1)^n}{n + 1} \qquad \text{for } x \in [2^{-n-1}, 2^{-n}] , $$ so that $\partial_y f(x, 0)$ is a function of unbounded variation near $x = 0$. Clearly, $f$ is not $C^1$,...


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