In the context of a realistic model for soap films (Almgren's minimal sets), prove the existence of one of least area with a given boundary. ...

let us moving to telescopic sum using exponent ,Assume we have this sequence: $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$ with $n\geq1$ , This sequence can be written as power of sequences :${x_n} ^ {{{y_n}^{C_n}}^{......}} $ such that all them value are in $(0,1)$, We want to know if the titled sequence should converge to $1$ ? ...

Let ${\sigma}_x(n) =\sum_{d\div n} d^x$ is the sum divisor function. After a few computations in wolfram alpha about the divisor function for some values of $n$ to look the behavior of $\sigma_x(n)\bmod n$ for $\,n=6,\,$ we got this result : $\sigma_x(6)=0 \bmod 6$ for $x$ odd and $2 \bmod 6$ if $x$ is even. One can ask this question: Is $n=6$ the only integer satisfies $\sigma_x(n) \equiv 0\bmod n$ for every odd integer $x > 0$ and $2 \bmod n$ if $x$ is even integer? One can answer this question using divisibility and Congruence properties for sum power divisor function, you can see the proof in this paper. The answer of this question yield to reformulate the following problem regarding periodicity on iterative of sum power divsior function. ...

Suppose we have a second order elliptic differential operator $$ L(v) = -\text{div}(A(x) \nabla v) $$ $A(x)$ is a bounded and strictly positive definite matrix with Hölder continuous entries. And suppose $\Omega$ is a $C^1$ domain (can be considered more regular if required). ...

Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain. Let $N$ be a compact convex set and assume that $\sigma: N\to \mathbb{R}$ is a bounded convex function such that $u\in C^1(N^\circ)$ and strictly convex on $N^\circ$. Consider the functional \begin{align*} E[u]=\int_{\Omega}\sigma(\nabla u(x))dx, \end{align*} defined for all Lipschitz functions $u\in C^{0,1}(\Omega)$ such that $\nabla u(x)\in N$ for a.e. $x\in \Omega$ and called this closed convex subset of Lipschitz functions $\mathscr{A}_N(\Omega)$. Furthermore, chose a $\phi \in \mathscr{A}_N(\Omega)$ and let \begin{align*} \mathscr{A}_N(\Omega,\phi):=\{u\in \mathscr{A}_N(\Omega): u\vert_{\partial \Omega}=\phi\}. \end{align*} ...

The question is to prove that the solution to the tuned KPZ $$ \partial_{t}h(x,t)=\partial_{x}^{2}h(x,t)+\delta(\partial_{x}h(x,t))^{2}+\delta^{1/2}\xi, $$ ...

The corresponding minimization problem with the area constraint replaced by a perimeter constraint (whatever notion of perimeter one would consider) is in general not well-posed. Indeed, for every $L>0$, one can construct a sequence of smooth connected and closed sets $K_n\subset\overline\Omega$ of perimeter $L$ approaching a subset of $\partial \Omega$ so that $\lambda_1(\Omega\setminus K_n)\downarrow \lambda_1(\Omega)$ as $n\to\infty$ (notice that by regularity there is no doubt on the notion of perimeter of $K_n$). Therefore, as in Problem 2, we restrict the class of admissible obstacles to convex sets. For a fixed $L\in (0,P(\overline{\Omega}))$, consider the {minimization} problem \begin{equation}\label{prob3} \min \{ \lambda_1(\Omega\setminus K) : \; \text{$K\subset \overline{\Omega}$, $K$ closed and convex, $P(K)=L$}\}. \end{equation} The existence of a minimizer is a consequence of the compactness of the class of convex sets and of the continuity of the perimeter w.r.t Hausdorff convergence of convex sets. Notice that, for particular domains $\Omega$ and small values $L$, it is still possible to have trivial solutions. For example, if the boundary $\partial\Omega$ contains a segment and if $L$ is smaller than twice the length of such a segment, then every segment contained in $\partial \Omega$ of perimeter $L$ is a minimizer. On the other hand, if $L$ is large enough, every minimizer has positive Lebesgue measure, since minimizing sequences will not be able to degenerate to a segment. In any case, one expects that every minimizer touches the boundary $\partial \Omega$. ...