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Today, there are several other types of web development and blogging software available to design and host your website blogs online and that too in minutes, if not hours. What I advise you do next is save the backup data file to a remote place like a CD-ROM, external disk drive if you have one or a provider such as Dropbox. The effect is to promote older posts by moving them back onto the front page and into the rss feed. If you're using Wordpress and want to make your blog a "dofollow" blog, meaning that links from your blog pass on the benefits of Google pagerank, you can install one of the many dofollow plugins available. If you are happy with your new look then click "Activate 'New Theme'" in the top right corner. <br><br>Always remember that an effective linkwheel strategy strives to answer all the demands of popular search engines while reacting to the latest marketing number trends. Some of the Wordpress development services offered by us are:. It sorts the results of a search according to category, tags and comments. So if you want to create blogs or have a website for your business or for personal reasons, you can take advantage of free Word - Press installation to get started. As soon as you start developing your Word - Press MLM website you'll see how straightforward and simple it is to create an online presence for you and the products and services you offer. <br><br>It is very easy to install Word - Press blog or website. Word - Press has ensured the users of this open source blogging platform do not have to troubleshoot on their own, or seek outside help. Those who cannot conceive with donor eggs due to some problems can also opt for surrogacy option using the services of surrogate mother. You or your web designer can customize it as per your specific needs.  If you have any issues with regards to exactly where and how to use [http://www.enriqueonline.org/guestbook/go.php?url=https://wordpress.org/plugins/ready-backup/ wordpress backup plugin], you can get hold of us at our own web site. For any web design and development assignment, this is definitely one of the key concerns, specifically for online retail outlets as well as e-commerce websites. <br><br>The disadvantage is it requires a considerable amount of time to set every thing up. * Robust CRM to control and connect with your subscribers. Thus it is difficult to outrank any one of these because of their different usages. If you are looking for Hire Wordpress Developer then just get in touch with him. Look for experience: When you are searching for a Word - Press developer you should always look at their experience level. <br><br>As a open source platform Wordpress offers distinctive ready to use themes for free along with custom theme support and easy customization. Visit our website to learn more about how you can benefit. In simple words, this step can be interpreted as the planning phase of entire PSD to wordpress conversion process. This is because of the customization that works as a keystone for a SEO friendly blogging portal website. Customers within a few seconds after visiting a site form their opinion about the site.
 
==Introduction==
The factorization of a linear partial differential operator (LPDO) is an important issue in the theory of integrability, due to the Laplace-Darboux transformations,<ref>Weiss (1986)</ref> which allow to construct integrable LPDEs.  [[Laplace]] solved factorization problem for a '''bivariate hyperbolic operator of the second order''' (see [[Hyperbolic partial differential equation]]), constructing two Laplace invariants. Each [[Laplace invariant]] is an explicit polynomial condition of factorization; coefficients of this polynomial are explicit functions of the coefficients of the initial LPDO. The polynomial conditions of factorization are called '''invariants''' because they have the same form for equivalent (i.e. self-adjoint) operators.
 
'''Beals-Kartashova-factorization''' (also called BK-factorization) is a constructive procedure to factorize '''a bivariate operator of the arbitrary order and arbitrary form'''. Correspondingly, the factorization conditions in this case also have polynomial form, are invariants and '''coincide with Laplace invariants''' for bivariate hyperbolic operator of the second order. The factorization procedure is purely algebraic, the number of possible factirzations depends on the number of simple roots of the [[Characteristic polynomial]] (also called symbol) of the initial LPDO and reduced LPDOs appearing at each factorization step. Below the factorization procedure is described for a bivariate operator of the  arbitrary form, of the order 2 and 3. Explicit factorization formulas for an operator of the order <math> n </math> can be found in<ref>R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. [http://www.springerlink.com/content/yx664142514k0217/ Theor. Math. Phys. '''145'''(2), pp. 1510-1523 (2005)] </ref> General invariants are defined in<ref>
E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. [http://www.springerlink.com/content/lp81238030114354/ Theor. Math. Phys. '''147'''(3), pp. 839-846 (2006)] </ref> and invariant formulation of the Beals-Kartashova factorization is given in<ref>  E. Kartashova, O. Rudenko. Invariant Form of BK-factorization and its Applications. Proc. GIFT-2006, pp.225-241, Eds.: J. Calmet, R. W. Tucker,  Karlsruhe University Press  (2006); [http://arxiv.org/abs/math-ph/0607040/ arXiv]</ref>
 
==Beals-Kartashova Factorization==
===Operator of order 2===
 
Consider an operator
:<math>
\mathcal{A}_2 = a_{20}\partial_x^2 + a_{11}\partial_x\partial_y + a_{02}\partial_y^2+a_{10}\partial_x+a_{01}\partial_y+a_{00}.
</math>
 
with smooth coefficients and look for a factorization
:<math>
\mathcal{A}_2=(p_1\partial_x+p_2\partial_y+p_3)(p_4\partial_x+p_5\partial_y+p_6).
</math>
 
Let us write down the equations on <math> p_i</math> explicitly, keeping in
mind the rule of '''left''' composition, i.e. that  
:<math>  \partial_x (\alpha
\partial_y) = \partial_x (\alpha) \partial_y +
\alpha \partial_{xy}.</math>
 
Then in all cases
 
:<math> a_{20} =  p_1p_4, </math>
 
:<math>  a_{11} =  p_2p_4+p_1p_5, </math>
 
:<math>  a_{02} =  p_2p_5, </math>
 
:<math>  a_{10} =  \mathcal{L}(p_4) + p_3p_4+p_1p_6, </math>
 
:<math>  a_{01} =  \mathcal{L}(p_5) + p_3p_5+p_2p_6, </math>
 
:<math>  a_{00} =  \mathcal{L}(p_6) + p_3p_6, </math>
 
where the notation <math> \mathcal{L} = p_1 \partial_x + p_2 \partial_y </math> is used.
 
Without loss of generality, <math>
a_{20}\ne 0,
</math> i.e. <math> p_1\ne 0, </math>  and  it can be taken as 1, <math>
p_1 = 1. </math>  Now solution of the system of 6 equations on the variables
:<math> p_2, </math> <math>...  </math> <math> p_6 </math>
can be found in '''three steps'''.  
 
'''At the first step''', the roots of a '''''quadratic polynomial''''' have to be found.  
 
'''At the second step''', a linear system of '''''two algebraic equations''''' has to be solved.
 
'''At the third step''', '''''one algebraic condition'''''  has to be checked.
 
'''Step 1.'''
Variables 
:<math> p_2,</math>  <math>  p_4, </math>  <math> p_5
</math>
can be found from the first three equations,
 
:<math> a_{20} =  p_1p_4, </math>
 
:<math>  a_{11} =  p_2p_4+p_1p_5, </math>
 
:<math>  a_{02} =  p_2p_5. </math>
 
The (possible) solutions are then the functions of the roots of a quadratic polynomial:
 
:<math>  
\mathcal{P}_2(-p_2) =  a_{20}(- p_2)^2 +a_{11}(- p_2) +a_{02} = 0
</math>
 
Let  <math> \omega </math> be a root of the polynomial <math>
\mathcal{P}_2,
</math>
then
 
:<math>  p_1=1, </math>
:<math>  p_2=-\omega, </math>
:<math>  p_4=a_{20},</math>
:<math>  p_5=a_{20} \omega +a_{11},</math>
 
'''Step 2.'''
Substitution of the results obtained at the first step, into the next two equations
 
:<math>  a_{10} =  \mathcal{L}(p_4) + p_3p_4+p_1p_6, </math>
 
:<math>  a_{01} =  \mathcal{L}(p_5) + p_3p_5+p_2p_6, </math>
 
yields linear system of two algebraic equations:
 
:<math>  a_{10} =  \mathcal{L} a_{20} +p_3 a_{20} +p_6, </math>
:<math>  a_{01} =  \mathcal{L}(a_{11}+a_{20} \omega)+p_3( a_{11} + a_{20}\omega)-
  \omega p_6., </math>
 
'''In particularly''', if the root <math>\omega</math> is simple,
i.e.
 
:<math> \mathcal{P}_2'(\omega)=2a_{20}\omega+a_{11}\ne 0,</math> then these
equations have the unique solution:
 
:<math>  p_3 =  \frac{\omega a_{10}+a_{01} -\omega\mathcal{L}a_{20}- \mathcal{L}(a_{20} \omega+a_{11})}
{2a_{20}\omega+a_{11}},</math>
:<math>  p_6 =\frac{ (a_{20}\omega+a_{11})(a_{10}-\mathcal{L}a_{20})-a_{20}(a_{01}
-\mathcal{L}(a_{20}\omega+a_{11}))}{2a_{20}\omega+a_{11}}.</math>
 
At this step, for each
root of the polynomial <math> \mathcal{P}_2 </math> a corresponding set of coefficients  <math> p_j </math> is computed.  
 
'''Step 3.'''
Check factorization condition (which is the last of the initial 6 equations)
 
:<math> a_{00} =  \mathcal{L}(p_6)+p_3p_6, </math>
written in the known variables <math> p_j </math> and <math> \omega </math>):
 
:<math>
a_{00} = \mathcal{L} \left\{
  \frac{\omega a_{10}+a_{01} - \mathcal{L}(2a_{20} \omega+a_{11})}
{2a_{20}\omega+a_{11}}\right\}+ \frac{\omega a_{10}+a_{01} -
\mathcal{L}(2a_{20} \omega+a_{11})}
{2a_{20}\omega+a_{11}}\times
\frac{ a_{20}(a_{01}-\mathcal{L}(a_{20}\omega+a_{11}))+
(a_{20}\omega+a_{11})(a_{10}-\mathcal{L}a_{20})}{2a_{20}\omega+a_{11}}
</math>
 
If
 
:<math>
l_2= a_{00} - \mathcal{L} \left\{
\frac{\omega a_{10}+a_{01} - \mathcal{L}(2a_{20} \omega+a_{11})}
{2a_{20}\omega+a_{11}}\right\}+ \frac{\omega a_{10}+a_{01} -
\mathcal{L}(2a_{20} \omega+a_{11})}
{2a_{20}\omega+a_{11}}\times
\frac{ a_{20}(a_{01}-\mathcal{L}(a_{20}\omega+a_{11}))+
(a_{20}\omega+a_{11})(a_{10}-\mathcal{L}a_{20})}{2a_{20}\omega+a_{11}} =0,
</math>
 
the operator <math> \mathcal{A}_2</math> is factorizable and explicit form for the factorization coefficients <math> p_j</math>  is given above.
 
===Operator of order 3===
Consider an operator
 
:<math>
\mathcal{A}_3=\sum_{j+k\le3}a_{jk}\partial_x^j\partial_y^k =a_{30}\partial_x^3 +
a_{21}\partial_x^2 \partial_y + a_{12}\partial_x \partial_y^2 +a_{03}\partial_y^3 +
a_{20}\partial_x^2+a_{11}\partial_x\partial_y+a_{02}\partial_y^2+a_{10}\partial_x+a_{01}\partial_y+a_{00}.
</math>
 
with smooth coefficients and look for a factorization
 
:<math>
\mathcal{A}_3=(p_1\partial_x+p_2\partial_y+p_3)(p_4 \partial_x^2 +p_5 \partial_x\partial_y  + p_6 \partial_y^2 + p_7
\partial_x + p_8 \partial_y + p_9).
</math>
 
Similar to the case of the operator <math> \mathcal{A}_2, </math>  the conditions of factorization are described by the following system:
:<math>  a_{30} =  p_1p_4,</math>
 
:<math>  a_{21} =  p_2p_4+p_1p_5,</math>
 
:<math>  a_{12} =  p_2p_5+p_1p_6,</math>
 
:<math>  a_{03} =  p_2p_6,</math>
 
:<math>  a_{20} =  \mathcal{L}(p_4)+p_3p_4+p_1p_7,</math>
 
:<math>  a_{11} =  \mathcal{L}(p_5)+p_3p_5+p_2p_7+p_1p_8,</math>
 
:<math>  a_{02} =  \mathcal{L}(p_6)+p_3p_6+p_2p_8,</math>
 
:<math>  a_{10} =  \mathcal{L}(p_7)+p_3p_7+p_1p_9,</math>
 
:<math>  a_{01} =  \mathcal{L}(p_8)+p_3p_8+p_2p_9,</math>
 
:<math>  a_{00} =  \mathcal{L}(p_9)+p_3p_9,
</math>
with <math>\mathcal{L} = p_1 \partial_x + p_2 \partial_y,</math> and again <math>
a_{30}\ne 0,
</math> i.e. <math>  p_1=1, </math>  and three-step procedure yields:
 
'''At the first step''', the roots of a '''''cubic polynomial'''''
 
:<math>  \mathcal{P}_3(-p_2):=  a_{30}(-p_2)^3 +a_{21}(- p_2)^2 +
a_{12}(-p_2)+a_{03}=0.
</math>
 
have to be found. Again <math> \omega  </math> denotes a root and first four coefficients are
 
:<math> p_1=1,  </math>
:<math>p_2=-\omega, </math>
:<math>p_4=a_{30}, </math>
:<math>p_5=a_{30} \omega+a_{21}, </math>
:<math>p_6=a_{30}\omega^2+a_{21}\omega+a_{12}.
</math>
 
'''At the second step''', a linear system of '''''three algebraic equations''''' has to be solved:
 
:<math>  a_{20}-\mathcal{L} a_{30} = p_3 a_{30} +p_7,</math>
:<math>  a_{11}-\mathcal{L}(a_{30} \omega + a_{21}) = p_3(a_{30}\omega+a_{21})- \omega p_7+p_8,</math>
:<math>  a_{02}-\mathcal{L}(a_{30}\omega^2+a_{21}\omega+a_{12})= p_3 (a_{30}\omega^2+a_{21}\omega+a_{12})-\omega p_8.</math>
 
'''At the third step''', '''''two algebraic conditions'''''  have to be checked.
 
===Operator of order <math>n</math>===
==Invariant Formulation==
 
'''Definition''' The operators <math> \mathcal{A} </math>, <math> \tilde{\mathcal{A}}  </math> are called
equivalent if there is a gauge transformation that takes one to the
other:
:<math>
\tilde{\mathcal{A}} g= e^{-\varphi}\mathcal{A} (e^{\varphi}g).
</math>
BK-factorization is then pure algebraic procedure which allows to
construct explicitly a factorization of an arbitrary order LPDO <math> \tilde{\mathcal{A}}  </math>
in the form
:<math>
\mathcal{A}=\sum_{j+k\le n}a_{jk}\partial_x^j\partial_y^k=\mathcal{L}\circ
\sum_{j+k\le (n-1)}p_{jk}\partial_x^j\partial_y^k
</math>
with first-order operator <math> \mathcal{L}=\partial_x-\omega\partial_y+p</math> where <math> \omega</math> is '''an arbitrary  simple root'''  of the characteristic polynomial
:<math>
\mathcal{P}(t)=\sum^n_{k=0}a_{n-k,k}t^{n-k}, \quad
\mathcal{P}(\omega)=0.</math>
Factorization is possible then for each simple root <math>\tilde{\omega}</math>  '''iff'''
 
for <math>n=2 \ \ \rightarrow l_2=0,</math>
 
for <math>n=3 \ \ \rightarrow l_3=0, l_{31}=0,</math>
 
for <math>n=4 \ \ \rightarrow l_4=0,  l_{41}=0, l_{42}=0,</math>
 
and so on. All functions <math>l_2,  l_3,  l_{31},  l_4,
l_{41}, \ \ l_{42}, ...</math> are known functions, for instance,
 
:<math>  l_2= a_{00} - \mathcal{L}(p_6)+p_3p_6, </math>
 
:<math>  l_3= a_{00} - \mathcal{L}(p_9)+p_3p_9, </math>
 
:<math>  l_{31} = a_{01} -  \mathcal{L}(p_8)+p_3p_8+p_2p_9,</math>
 
and so on.
 
'''Theorem'''  All functions
:<math>l_2= a_{00} - \mathcal{L}(p_6)+p_3p_6, 
l_3= a_{00} - \mathcal{L}(p_9)+p_3p_9,
l_{31}, ....</math>
are '''invariants''' under gauge transformations.
 
'''Definition''' Invariants <math>l_2= a_{00} - \mathcal{L}(p_6)+p_3p_6, 
l_3= a_{00} - \mathcal{L}(p_9)+p_3p_9,
l_{31}, .... .</math> are
called '''generalized invariants''' of a bivariate operator of arbitrary
order.
 
In particular case of the bivariate hyperbolic operator  its generalized
invariants '''coincide with Laplace invariants''' (see [[Laplace invariant]]).
 
'''Corollary''' If an operator <math> \tilde{\mathcal{A}}  </math> is factorizable, then all
operators  equivalent to it, are also factorizable.
 
Equivalent operators are easy to compute:
:<math> e^{-\varphi} \partial_x e^{\varphi}= \partial_x+\varphi_x, \quad e^{-\varphi}\partial_y e^{\varphi}=
\partial_y+\varphi_y,</math>
 
:<math> e^{-\varphi} \partial_x \partial_y e^{\varphi}= e^{-\varphi} \partial_x e^{\varphi}
e^{-\varphi} \partial_y e^{\varphi}=(\partial_x+\varphi_x) \circ (\partial_y+\varphi_y)</math>
and so on. Some example are given below:
 
:<math> A_1=\partial_x \partial_y + x\partial_x + 1= \partial_x(\partial_y+x), \quad
l_2(A_1)=1-1-0=0;</math>
 
:<math>A_2=\partial_x \partial_y + x\partial_x + \partial_y +x + 1, \quad
A_2=e^{-x}A_1e^{x};\quad l_2(A_2)=(x+1)-1-x=0;</math>
 
:<math>A_3=\partial_x \partial_y + 2x\partial_x + (y+1)\partial_y +2(xy +x+1), \quad
A_3=e^{-xy}A_2e^{xy}; \quad l_2(A_3)=2(x+1+xy)-2-2x(y+1)=0;</math>
 
:<math>A_4=\partial_x \partial_y +x\partial_x + (\cos x +1) \partial_y + x \cos x +x +1, \quad
A_4=e^{-\sin x}A_2e^{\sin x}; \quad l_2(A_4)=0.</math>
 
==Transpose==
 
Factorization of an operator is the first step on the way of solving corresponding equation. But for solution we need '''right''' factors and BK-factorization constructs  '''left''' factors which are easy to construct. On the other hand, the existence of a certain right factor of a LPDO is equivalent to the existence of a corresponding left factor of the '''transpose''' of that operator.
 
'''Definition'''
The transpose <math> \mathcal{A}^t</math> of an operator
<math>
\mathcal{A}=\sum a_{\alpha}\partial^{\alpha},\qquad \partial^{\alpha}=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}.
</math>
is defined as
<math>
\mathcal{A}^t u = \sum (-1)^{|\alpha|}\partial^\alpha(a_\alpha u).
</math>
and the identity
<math>
\partial^\gamma(uv)=\sum \binom\gamma\alpha \partial^\alpha u,\partial^{\gamma-\alpha}v
</math>
implies that
<math>
\mathcal{A}^t=\sum (-1)^{|\alpha+\beta|}\binom{\alpha+\beta}\alpha (\partial^\beta a_{\alpha+\beta})\partial^\alpha.
</math>
 
Now the coefficients are
 
<math> \mathcal{A}^t=\sum \tilde{a}_{\alpha} \partial^{\alpha},</math>
<math> \tilde{a}_{\alpha}=\sum (-1)^{|\alpha+\beta|}
\binom{\alpha+\beta}{\alpha}\partial^\beta(a_{\alpha+\beta}).
</math>
 
with a standard convention for binomial coefficients in several
variables (see [[Binomial coefficient]]), e.g. in two variables
:<math>
\binom\alpha\beta=\binom{(\alpha_1,\alpha_2)}{(\beta_1,\beta_2)}=\binom{\alpha_1}{\beta_1}\,\binom{\alpha_2}{\beta_2}.
</math>
In particular, for the operator <math> \mathcal{A}_2 </math> the coefficients are
<math> \tilde{a}_{jk}=a_{jk},\quad j+k=2; \tilde{a}_{10}=-a_{10}+2\partial_x a_{20}+\partial_y
a_{11}, \tilde{a}_{01}=-a_{01}+\partial_x a_{11}+2\partial_y a_{02},</math>
:<math>
\tilde{a}_{00}=a_{00}-\partial_x a_{10}-\partial_y a_{01}+\partial_x^2 a_{20}+\partial_x \partial_x
a_{11}+\partial_y^2 a_{02}.
</math>
For instance,  the operator
:<math> \partial_{xx}-\partial_{yy}+y\partial_x+x\partial_y+\frac{1}{4}(y^2-x^2)-1 </math>
is factorizable as
:<math> \big[\partial_x+\partial_y+\tfrac12(y-x)\big]\,\big[...\big]</math>
and its transpose <math> \mathcal{A}_1^t </math>  is factorizable then as
<math> \big[...\big]\,\big[\partial_x-\partial_y+\tfrac12(y+x)\big].</math>
 
==See also==
* [[Partial derivative]]
* [[Invariant (mathematics)]]
* [[Invariant theory]]
* [[Characteristic polynomial]]
 
== Notes ==
<references/>
 
== References ==
* J. Weiss. Bäcklund transformation and the Painlevé property. [http://www2.appmath.com:8080/site/few/few.html] J. Math. Phys. '''27''', 1293-1305 (1986).
* R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. [http://www.springerlink.com/content/yx664142514k0217/ Theor. Math. Phys. '''145'''(2), pp. 1510-1523 (2005)] 
*  E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. [http://www.springerlink.com/content/lp81238030114354/ Theor. Math. Phys. '''147'''(3), pp. 839-846 (2006)]
*  E. Kartashova, O. Rudenko.  Invariant Form of BK-factorization and its Applications. Proc. GIFT-2006, pp.225-241, Eds.: J. Calmet, R. W. Tucker,  Karlsruhe University Press  (2006); [http://arxiv.org/abs/math-ph/0607040/ arXiv]
 
[[Category:Multivariable calculus]]
[[Category:Differential operators]]
[[Category:Partial differential equations]]

Latest revision as of 12:12, 25 November 2014

Today, there are several other types of web development and blogging software available to design and host your website blogs online and that too in minutes, if not hours. What I advise you do next is save the backup data file to a remote place like a CD-ROM, external disk drive if you have one or a provider such as Dropbox. The effect is to promote older posts by moving them back onto the front page and into the rss feed. If you're using Wordpress and want to make your blog a "dofollow" blog, meaning that links from your blog pass on the benefits of Google pagerank, you can install one of the many dofollow plugins available. If you are happy with your new look then click "Activate 'New Theme'" in the top right corner.

Always remember that an effective linkwheel strategy strives to answer all the demands of popular search engines while reacting to the latest marketing number trends. Some of the Wordpress development services offered by us are:. It sorts the results of a search according to category, tags and comments. So if you want to create blogs or have a website for your business or for personal reasons, you can take advantage of free Word - Press installation to get started. As soon as you start developing your Word - Press MLM website you'll see how straightforward and simple it is to create an online presence for you and the products and services you offer.

It is very easy to install Word - Press blog or website. Word - Press has ensured the users of this open source blogging platform do not have to troubleshoot on their own, or seek outside help. Those who cannot conceive with donor eggs due to some problems can also opt for surrogacy option using the services of surrogate mother. You or your web designer can customize it as per your specific needs. If you have any issues with regards to exactly where and how to use wordpress backup plugin, you can get hold of us at our own web site. For any web design and development assignment, this is definitely one of the key concerns, specifically for online retail outlets as well as e-commerce websites.

The disadvantage is it requires a considerable amount of time to set every thing up. * Robust CRM to control and connect with your subscribers. Thus it is difficult to outrank any one of these because of their different usages. If you are looking for Hire Wordpress Developer then just get in touch with him. Look for experience: When you are searching for a Word - Press developer you should always look at their experience level.

As a open source platform Wordpress offers distinctive ready to use themes for free along with custom theme support and easy customization. Visit our website to learn more about how you can benefit. In simple words, this step can be interpreted as the planning phase of entire PSD to wordpress conversion process. This is because of the customization that works as a keystone for a SEO friendly blogging portal website. Customers within a few seconds after visiting a site form their opinion about the site.