Found problems: 79
Let $ U\subset\mathbb R^n$ be an open set, and let $ L: U\times\mathbb R^n\to\mathbb R$ be a continuous, in its second variable first order positive homogeneous, positive over $ U\times (\mathbb R^n\setminus\{0\})$ and of $ C^2$-class Langrange function, such that for all $ p\in U$ the Gauss-curvature of the hyper surface
\[ \{ v\in\mathbb R^n \mid L(p,v) \equal{} 1 \}\]
is nowhere zero. Determine the extremals of $ L$ if it satisfies the following system
\[ \sum_{k \equal{} 1}^n y^k\partial_k\partial_{n \plus{} i}L \equal{} \sum_{k \equal{} 1}^n y^k\partial_i\partial_{n \plus{} k} L \qquad (i\in\{1,\dots,n\})\]
of partial differetial equations, where $ y^k(u,v) : \equal{} v^k$ for $ (u,v)\in U\times\mathbb R^k$, $ v \equal{} (v^1,\dots,v^k)$.
A subalgebra $\mathfrak{h}$ of a Lie algebra $\mathfrak g$ is said to have the $\gamma$ property with respect to a scalar product ${\langle \cdot,\cdot \rangle}$ given on ${\mathfrak g}$ if ${X \in \mathfrak{h}}$ implies ${\langle [X,Y],X\rangle =0}$ for all ${Y \in \mathfrak g}$. Prove that the maximum dimension of ${\gamma}$-property subalgebras of a given ${2}$ step nilpotent Lie algebra with respect to a scalar product is independent of the selection of the scalar product.
[i]Proposed by Péter Nagy Tibor[/i]
Let $ p$ be a prime and let \[ l_k(x,y)\equal{}a_kx\plus{}b_ky \;(k\equal{}1,2,...,p^2)\ .\] be homogeneous linear polynomials with integral coefficients. Suppose that for every pair $ (\xi,\eta)$ of integers, not both divisible by $ p$, the values $ l_k(\xi,\eta), \;1\leq k\leq p^2 $, represent every residue class $ \textrm{mod} \;p$ exactly $ p$ times. Prove that the set of pairs $ \{(a_k,b_k): 1\leq k \leq p^2 \}$ is identical $ \textrm{mod} \;p$ with the set $ \{(m,n): 0\leq m,n \leq p\minus{}1 \}.$
Let $ C$ be a simple arc with monotone curvature such that $ C$ is congruent to its evolute. Show that under appropriate differentiability conditions, $ C$ is a part of a cycloid or a logarithmic spiral with polar equation $ r\equal{}ae^{\vartheta}$.
[i]J. Szenthe[/i]