Invariant factorization of LPDOs

Introduction

Factorization of linear ordinary differential operators (LODOs) is known to be unique and in general, it finally reduces to the solution of a Riccati equation, i.e. factorization of LODOs is not a constructive procedure.

On the other hand, factorization of linear partial differential operators (LPDOs) though not even unique, can be performed constructively using Beals-Kartashova factorization procedure (BK-factorization). BK-factorization is an explicit algorithm for absolute factorization of a bivariate LPDO of arbitrary order $n$ into linear factors. The word absolute means that the coefficient field is not fixed from the very beginning and that the only demand on the coefficients is that they be smooth, i.e. they belong to an appropriate differential field.

BK-factorization is the procedure to find a first order left factor (when possible) in contrast to the use of right factorization, which is common in the papers of last few decades. Of course 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 (see below), so in principle nothing is lost by considering left factorization. Moreover taking transposes is trivial algebraically, so there is also nothing lost from the point of view of algorithmic computation.

BK-factorization allows also to factorize simultaneously the families of the LPDOs equivalent under the action of the gauge transformations and to construct the families of the LPDOs corresponding to integrable LPDEs.

Beals-Kartashova Factorization

Operator of order 2

Consider an operator

${\mathcal {A}}_{2}=a_{20}\partial _{x}^{2}+a_{11}\partial _{x}\partial _{y}+a_{02}\partial _{y}^{2}+a_{10}\partial _{x}+a_{01}\partial _{y}+a_{00}.$

with smooth coefficients and look for a factorization

${\mathcal {A}}_{2}=(p_{1}\partial _{x}+p_{2}\partial _{y}+p_{3})(p_{4}\partial _{x}+p_{5}\partial _{y}+p_{6}).$

Let us write down the equations on $p_{i}$ explicitly, keeping in mind the rule of left composition, i.e. that

$\partial _{x}(\alpha \partial _{y})=\partial _{x}(\alpha )\partial _{y}+\alpha \partial _{xy}.$

Then in all cases

$a_{20}=p_{1}p_{4},$

$a_{11}=p_{2}p_{4}+p_{1}p_{5},$

$a_{02}=p_{2}p_{5},$

$a_{10}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{6},$

$a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},$

$a_{00}={\mathcal {L}}(p_{6})+p_{3}p_{6},$

where the notation ${\mathcal {L}}=p_{1}\partial _{x}+p_{2}\partial _{y}$ is used.

Without loss of generality, $a_{20}\neq 0,$ i.e. $p_{1}\neq 0,$ and it can be taken as 1, $p_{1}=1.$ Now solution of the system of 6 equations on the variables

$p_{2},$ $...$ $p_{6}$

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

$p_{2},$ $p_{4},$ $p_{5}$

can be found from the first three equations,

$a_{20}=p_{1}p_{4},$

$a_{11}=p_{2}p_{4}+p_{1}p_{5},$

$a_{02}=p_{2}p_{5}.$

The (possible) solutions are then the functions of the roots of a quadratic polynomial:

${\mathcal {P}}_{2}(-p_{2})=a_{20}(-p_{2})^{2}+a_{11}(-p_{2})+a_{02}=0$

Let $\omega$ be a root of the polynomial ${\mathcal {P}}_{2},$ then

$p_{1}=1,$

$p_{2}=-\omega ,$

$p_{4}=a_{20},$

$p_{5}=a_{20}\omega +a_{11},$

Step 2. Substitution of the results obtained at the first step, into the next two equations

$a_{10}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{6},$

$a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},$

yields linear system of two algebraic equations:

$a_{10}={\mathcal {L}}a_{20}+p_{3}a_{20}+p_{6},$

$a_{01}={\mathcal {L}}(a_{11}+a_{20}\omega )+p_{3}(a_{11}+a_{20}\omega )-\omega p_{6}.,$

In particularly, if the root $\omega$ is simple, i.e.

${\mathcal {P}}_{2}'(\omega )=2a_{20}\omega +a_{11}\neq 0,$ then these

equations have the unique solution:

$p_{3}={\frac {\omega a_{10}+a_{01}-\omega {\mathcal {L}}a_{20}-{\mathcal {L}}(a_{20}\omega +a_{11})}{2a_{20}\omega +a_{11}}},$

$p_{6}={\frac {(a_{20}\omega +a_{11})(a_{10}-{\mathcal {L}}a_{20})-a_{20}(a_{01}-{\mathcal {L}}(a_{20}\omega +a_{11}))}{2a_{20}\omega +a_{11}}}.$

At this step, for each root of the polynomial ${\mathcal {P}}_{2}$ a corresponding set of coefficients $p_{j}$ is computed computed.

Step 3. Check factorization condition (which is the last of the initial 6 equations

$a_{00}={\mathcal {L}}(p_{6})+p_{3}p_{6},$

written in the known variables $p_{j}$ and $\omega$):

$a_{00}={\mathcal {L}}\left\{{\frac {\omega a_{10}+a_{01}-{\mathcal {L}}(2a_{20}\omega +a_{11})}{2a_{20}\omega +a_{11}}}\right\}+{\frac {\omega a_{10}+a_{01}-{\mathcal {L}}(2a_{20}\omega +a_{11})}{2a_{20}\omega +a_{11}}}\times {\frac {a_{20}(a_{01}-{\mathcal {L}}(a_{20}\omega +a_{11}))+(a_{20}\omega +a_{11})(a_{10}-{\mathcal {L}}a_{20})}{2a_{20}\omega +a_{11}}}$

If

$l_{2}=a_{00}-{\mathcal {L}}\left\{{\frac {\omega a_{10}+a_{01}-{\mathcal {L}}(2a_{20}\omega +a_{11})}{2a_{20}\omega +a_{11}}}\right\}+{\frac {\omega a_{10}+a_{01}-{\mathcal {L}}(2a_{20}\omega +a_{11})}{2a_{20}\omega +a_{11}}}\times {\frac {a_{20}(a_{01}-{\mathcal {L}}(a_{20}\omega +a_{11}))+(a_{20}\omega +a_{11})(a_{10}-{\mathcal {L}}a_{20})}{2a_{20}\omega +a_{11}}}=0,$

the operator ${\mathcal {A}}_{2}$ is factorizable and explicit form for the factorization coefficients $p_{j}$ is given above.

Operator of order 3

Consider an operator

${\mathcal {A}}_{3}=\sum _{j+k\leq 3}a_{jk}\partial _{x}^{j}\partial _{y}^{k}=a_{30}\partial _{x}^{3}+a_{21}\partial _{x}^{2}\partial _{y}+a_{12}\partial _{x}\partial _{y}^{2}+a_{03}\partial _{y}^{3}+a_{20}\partial _{x}^{2}+a_{11}\partial _{x}\partial _{y}+a_{02}\partial _{y}^{2}+a_{10}\partial _{x}+a_{01}\partial _{y}+a_{00}.$

with smooth coefficients and look for a factorization

${\mathcal {A}}_{3}=(p_{1}\partial _{x}+p_{2}\partial _{y}+p_{3})(p_{4}\partial _{x}^{2}+p_{5}\partial _{x}\partial _{y}+p_{6}\partial _{y}^{2}+p_{7}\partial _{x}+p_{8}\partial _{y}+p_{9}).$

Similar to the case of the operator ${\mathcal {A}}_{2},$ the conditions of factorization are described by the following system:

$a_{30}=p_{1}p_{4},$

$a_{21}=p_{2}p_{4}+p_{1}p_{5},$

$a_{12}=p_{2}p_{5}+p_{1}p_{6},$

$a_{03}=p_{2}p_{6},$

$a_{20}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{7},$

$a_{11}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{7}+p_{1}p_{8},$

$a_{02}={\mathcal {L}}(p_{6})+p_{3}p_{6}+p_{2}p_{8},$

$a_{10}={\mathcal {L}}(p_{7})+p_{3}p_{7}+p_{1}p_{9},$

$a_{01}={\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},$

$a_{00}={\mathcal {L}}(p_{9})+p_{3}p_{9},$

with ${\mathcal {L}}=p_{1}\partial _{x}+p_{2}\partial _{y}.$

Invariant Formulation

Definition The operators ${\mathcal {A}}$, ${\tilde {\mathcal {A}}}$ are called equivalent if there is a gauge transformation that takes one to the other:

${\tilde {\mathcal {A}}}g=e^{-\varphi }{\mathcal {A}}(e^{\varphi }g).$

BK-factorization is then pure algebraic procedure which allows to to construct explicitly a factorization of an arbitrary order LPDO ${\tilde {\mathcal {A}}}$ in the form

${\mathcal {A}}=\sum _{j+k\leq n}a_{jk}\partial _{x}^{j}\partial _{y}^{k}={\mathcal {L}}\circ \sum _{j+k\leq (n-1)}p_{jk}\partial _{x}^{j}\partial _{y}^{k}$

with first-order operator ${\mathcal {L}}=\partial _{x}-\omega \partial _{y}+p$ where $\omega$ is an arbitrary simple root of the characteristic polynomial (also called symbol of operator)

${\mathcal {P}}(t)=\sum _{k=0}^{n}a_{n-k,k}t^{n-k},\quad {\mathcal {P}}(\omega )=0.$

Factorization is possible then for each simple root ${\tilde {\omega }}$ iff

for $n=2\ \ \rightarrow l_{2}=0,$

for $n=3\ \ \rightarrow l_{3}=0,l_{31}=0,$

for $n=4\ \ \rightarrow l_{4}=0,l_{41}=0,l_{42}=0,$

and so on. All functions $l_{2},l_{3},l_{31},l_{4},l_{41},\ \ l_{42},...$ are explicit functions of $a_{ij}$ and ${\tilde {\omega }}$.

Theorem All $l_{2},l_{3},l_{31},....$ are nvariants under gauge transformations.

Definition Invariants $l_{2},l_{3},l_{31},....$ 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 ${\tilde {\mathcal {A}}}$ is factorizable, then all operators equivalent to it, are also factorizable.

Equivalent operators are easy to compute:

$e^{-\varphi }\partial _{x}e^{\varphi }=\partial _{x}+\varphi _{x},\quad e^{-\varphi }\partial _{y}e^{\varphi }=\partial _{y}+\varphi _{y},$

$e^{-\varphi }\partial _{x}\partial _{y}e^{\varphi }=e^{-\varphi }\partial _{x}e^{\varphi }e^{-\varphi }\partial _{y}e^{\varphi }=(\partial _{x}+\varphi _{x})\circ (\partial _{y}+\varphi _{y})$

and so on. Some example are given below:

$A_{1}=\partial _{x}\partial _{y}+x\partial _{x}+1=\partial _{x}(\partial _{y}+x),\quad l_{2}(A_{1})=1-1-0=0;$

$A_{2}=\partial _{x}\partial _{y}+x\partial _{x}+\partial _{y}+x+1,\quad A_{2}=e^{-x}A_{1}e^{x};\quad l_{2}(A_{2})=(x+1)-1-x=0;$

$A_{3}=\partial _{x}\partial _{y}+2x\partial _{x}+(y+1)\partial _{y}+2(xy+x+1),\quad A_{3}=e^{-xy}A_{2}e^{xy};\quad l_{2}(A_{3})=2(x+1+xy)-2-2x(y+1)=0;$

$A_{4}=\partial _{x}\partial _{y}+x\partial _{x}+(\cos x+1)\partial _{y}+x\cos x+x+1,\quad A_{4}=e^{-\sin x}A_{2}e^{\sin x};\quad l_{2}(A_{4})=0.$

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 ${\mathcal {A}}^{t}$ of an operator ${\mathcal {A}}=\sum a_{\alpha }\partial ^{\alpha },\qquad \partial ^{\alpha }=\partial _{1}^{\alpha _{1}}\cdots \partial _{n}^{\alpha _{n}}.$ is defined as ${\mathcal {A}}^{t}u=\sum (-1)^{|\alpha |}\partial ^{\alpha }(a_{\alpha }u).$ and the identity $\partial ^{\gamma }(uv)=\sum {\binom {\gamma }{\alpha }}\partial ^{\alpha }u,\partial ^{\gamma -\alpha }v$ implies that ${\mathcal {A}}^{t}=\sum (-1)^{|\alpha +\beta |}{\binom {\alpha +\beta }{\alpha }}(\partial ^{\beta }a_{\alpha +\beta })\partial ^{\alpha }.$

Now the coefficients are

${\mathcal {A}}^{t}=\sum {\tilde {a}}_{\alpha }\partial ^{\alpha },$ ${\tilde {a}}_{\alpha }=\sum (-1)^{|\alpha +\beta |}{\binom {\alpha +\beta }{\alpha }}\partial ^{\beta }(a_{\alpha +\beta }).$

with a standard convention for binomial coefficients in several variables (see Binomial coefficient), e.g. in two variables

${\binom {\alpha }{\beta }}={\binom {(\alpha _{1},\alpha _{2})}{(\beta _{1},\beta _{2})}}={\binom {\alpha _{1}}{\beta _{1}}}\,{\binom {\alpha _{2}}{\beta _{2}}}.$

In particular, for the operator ${\mathcal {A}}_{2}$ the coefficients are ${\tilde {a}}_{jk}=a_{jk},\quad j+k=2;{\tilde {a}}_{10}=-a_{10}+2\partial _{x}a_{20}+\partial _{y}a_{11},{\tilde {a}}_{01}=-a_{01}+\partial _{x}a_{11}+2\partial _{y}a_{02},$

${\tilde {a}}_{00}=a_{00}-\partial _{x}a_{10}-\partial _{y}a_{01}+\partial _{x}^{2}a_{20}+\partial _{x}\partial _{x}a_{11}+\partial _{y}^{2}a_{02}.$

For instance, the operator

$\partial _{xx}-\partial _{yy}+y\partial _{x}+x\partial _{y}+{\frac {1}{4}}(y^{2}-x^{2})-1$

is factorizable as

${\big [}\partial _{x}+\partial _{y}+{\tfrac {1}{2}}(y-x){\big ]}\,{\big [}...{\big ]}$

and its transpose ${\mathcal {A}}_{1}^{t}$ is factorizable then as ${\big [}...{\big ]}\,{\big [}\partial _{x}-\partial _{y}+{\tfrac {1}{2}}(y+x){\big ]}.$

References

• R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. Theor. Math. Phys. 145(2), pp. 1510-1523 (2005)

• E. Kartashova. BK-factorization and Darboux-Laplace transformations. Proc. CSC'05, pp. 144-150, Ed.: H. R. Arabnia, CSREA Press (2005) / RISC publications

• E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. 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); arXiv

See also

• Partial derivative
• Invariant (mathematics)
• Invariant theory
• Laplace invariant