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Третья Российско-Китайская научная конференция по комплексному анализу, алгебре, алгебраической геометрии и математической физике
12 мая 2016 г. 11:20–12:10, г. Москва, МИАН, ул. Губкина, д. 8
 


New versions of the Cauchy-Kovalevskaya theorem and the Weierstrass preparation theorem

A. K. Tsikh

Institute of Mathematics and Computer Science, Siberian Federal University, Krasnoyarsk
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A. K. Tsikh
Фотогалерея



Аннотация: In the traditional formulation of the Cauchy-Kovalevsky theorem one assumes solvability with respect to maximal pure derivative, say $\partial^m u / \partial x^m_n$, where $m$ is the order of the differential equation. For linear differential equations with analytic coefficients this means that the point $(0,\dots,0,m)\in \mathbb{Z}^n\subset \mathbb{R}^n$ is a vertex of the Newton polytope of the characteristic polynomial $P(x_0,\xi)$ for the equation
$$ P(x,D)u=f,\ \hbox{where }x=(x_1,\ldots, x_n),\ D=\left(\frac{\partial}{\partial x_1},\ldots, \frac{\partial}{\partial x_n}\right). $$

Hörmander gave a version of the Cauchy-Kovalevsky theorem where solvability with respect to some arbitrary derivative $D^{\beta}u$, with $|\beta|=\beta_1+\ldots+\beta_n$ equal to $m$, was assumed. However, in this case one had to require the coefficients of the other highest order derivatives to be small in the point $x_0$. Moreover, the initial values should be given not just on one coordinate plane, but on a union, or cross, of several such planes. Such initial values correspond to a Cauchy-Goursat problem.
It was pointed out in a paper by Palamodov (1968) that the Cauchy-Kovalevsky theorem is intimately related with the Weierstrass preparation and division theorems.
In the talk I will tell how to relax the Hörmander condition for a Cauchy-Goursat problem for an equation in two variables with constant coefficients. The solvability conditions are obtained in terms of an amoeba of a characteristic equation. In parallel to this we formulate the corresponding version of the Weierstrass preparation theorem.

Язык доклада: английский
 
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