"Differential Galois theory"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
Liouville extension==
imported>Pythagoras0 잔글 (찾아 바꾸기 – “</h5>” 문자열을 “==” 문자열로) |
imported>Pythagoras0 |
||
| 191번째 줄: | 191번째 줄: | ||
* [http://gigapedia.info/1/Galois%27+dream http://gigapedia.info/1/Galois'+dream] | * [http://gigapedia.info/1/Galois%27+dream http://gigapedia.info/1/Galois'+dream] | ||
* http://gigapedia.info/1/differntial+algebra | * http://gigapedia.info/1/differntial+algebra | ||
| + | [[분류:개인노트]] | ||
2012년 10월 28일 (일) 15:50 판
introduction
- differential galois theory
- Liouville
- 2008년 12월 9일 MCF 'differential Galois theory'
historical origin
- integration in finite terms
- quadrature of second order differential equation (Fuchsian differential equation)
solution by quadrature
- 일계 선형미분방정식
\(\frac{dy}{dx}+a(x)y=b(x)\)
\(y(x)e^{\int a(x)\,dx}=\int b(x)e^{\int a(x)\,dx} \,dx+C\) - \(y''-2xy'=0\)
\(y=\int e^{x^2}\, dx\) - note that the integral of an exponential naturally shows up in expression solutions
differential field
- a pair \((F,\partial)\) such that
- \(\partial(a+b)=\partial a+\partial b\)
- \(\partial(ab)=(\partial a)b+a(\partial b)\)
- \(C_F=\ker \partial\)
solvable by quadratures
- basic functions : basic elementary functions
- allowed operatrions : compositions, arithmetic operations, differentiation, integration
- examples
- an elliptic integral is representable by quadrature
elementary extension
- it is allowed to take exponentials and logarithms to make a field extension
- elementary element
- difference between Liouville extension
- exponential+ integral <=> differentiation + exponential of integral
- in elementary extension, we are not allowed to get an integrated element
Liouville extension==
- an element is said to be representable by a generalized quadrature
- we can capture these properties using the concept of Liouville extension
- to get a Liouville extension, we can adjoin
- integrals
- exponentials of integrals
- algebraic extension (generalized Liouville extension)
- from these we can include the following operations
- exponential
- logarithm
- For\(K_{i}=K_{i-1}(e_i)\) , one of the following condition holds
- \(e_i'\in K_{i-1}\), i.e. \(e_i=\int e_i'\in K_i\)
- \(e_{i}'/e_{i}\in K_{i-1}\) i.e. \((\log e_i)' \in K_{i-1}\)
- \(e_{i}\) is algebraic over \(K_{i-1}\)
- remark on exponentiation
- Let \(a,a'\in F\). Is \(b=e^a\in K\) where K is a Liouville extension?
- \(b'=a' e^a=a'b\) implies \(a'=\frac{b'}{b}\in F\).
- the exponential of the integral of a' i.e. \(e^{\int a'}=e^a+c\) must be in the Liouville extension. So \(b=e^a\in K\).
- remark on logarithm
- \(b=\log a\) is the integral of \(a'/a\in F\). So \(b\in K\)
- a few result
- K/F is a Liouville extension iff the differential Galois group K over F is solvable.
- K/F is a generalized Liouville extension iff the differential Galois group K over F has the solvable component of the identity
Picard-Vessiot extension
- framework for linear differential equation
- field extension is made by including solutions of DE to the base field (e.g. rational function field)
- consider monic differential equations over a differential field F
\(\mathcal{L}[Y]=Y^{(n)}+a_{n-1}Y^{(n-1)}+\cdots+a_{1}Y^{(1)}+a_0\), \(a_i\in F\)
- \((E,\partial_E)\supseteq (F,\partial_F)\) is a Picard-Vessiot extension for \(\mathcal{L}\) if
- E/F is generated by n linear independent solution to \(\mathcal{L}\), i.e. adjoining basis of \(V=\mathcal{L}^{-1}(0)\) to F
- \(C_E=C_F\), \(\partial_E\mid_F=\partial_F\)
- this corresponds to the concept of the splitting fields(or Galois extensions)
- examples
- algebraic extension
- adjoining an integral
- adjoining the exponential of an integral
- we can define a Galois group for a linear differential equation
\(\operatorname{Gal}(E/F)=\{\sigma\in\operatorname{Aut}E|\partial(\sigma(n))=\sigma(\partial a), \sigma\mid _F=\operatorname{id} \}\)
- the action of an element of the Galois group is determined by its action on a basis of V
theorem
If a Picard-Vessiot extension is a Liouville extension, then the Galois group of this extension is solvable.
Fuchsian differential equation
- differential equation with regular singularities
- indicial equation
\(x(x-1)+px+q=0\)
theorem
A Fuchsian linear differential equation is solvable by quadratures if and only if the monodromy group of this equation is solvable.
solution by quadrature
- Differential Galois Theory and Non-Integrability of Hamiltonian Systems
- Integrability and non-integrability in Hamiltonian mechanics
- [1]http://www.caminos.upm.es/matematicas/morales%20ruiz/libroFSB.pdf
- http://andromeda.rutgers.edu/~liguo/DARTIII/Presentations/Khovanskii.pdf
- http://www.math.purdue.edu/~agabriel/topological_galois.pdf
- integrals
- exponentials of integrals
- algebraic extension (generalized Liouville extension)
- from these we can include the following operations
- exponential
- logarithm
- from these we can include the following operations
- \(e_i'\in K_{i-1}\), i.e. \(e_i=\int e_i'\in K_i\)
- \(e_{i}'/e_{i}\in K_{i-1}\) i.e. \((\log e_i)' \in K_{i-1}\)
- \(e_{i}\) is algebraic over \(K_{i-1}\)
- Let \(a,a'\in F\). Is \(b=e^a\in K\) where K is a Liouville extension?
- \(b'=a' e^a=a'b\) implies \(a'=\frac{b'}{b}\in F\).
- the exponential of the integral of a' i.e. \(e^{\int a'}=e^a+c\) must be in the Liouville extension. So \(b=e^a\in K\).
- \(b=\log a\) is the integral of \(a'/a\in F\). So \(b\in K\)
- K/F is a Liouville extension iff the differential Galois group K over F is solvable.
- K/F is a generalized Liouville extension iff the differential Galois group K over F has the solvable component of the identity
\(\mathcal{L}[Y]=Y^{(n)}+a_{n-1}Y^{(n-1)}+\cdots+a_{1}Y^{(1)}+a_0\), \(a_i\in F\)
- E/F is generated by n linear independent solution to \(\mathcal{L}\), i.e. adjoining basis of \(V=\mathcal{L}^{-1}(0)\) to F
- \(C_E=C_F\), \(\partial_E\mid_F=\partial_F\)
- algebraic extension
- adjoining an integral
- adjoining the exponential of an integral
\(\operatorname{Gal}(E/F)=\{\sigma\in\operatorname{Aut}E|\partial(\sigma(n))=\sigma(\partial a), \sigma\mid _F=\operatorname{id} \}\)
- the action of an element of the Galois group is determined by its action on a basis of V
\(x(x-1)+px+q=0\)