Invariant factorization of LPDOs

The factorization of a linear partial differential operator (LPDO) is an important issue in the theory of integrability, due to the Laplace-Darboux transformations,[1] which allow construction of integrable LPDEs. Laplace solved the 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 operators of the second order. The factorization procedure is purely algebraic, the number of possible factorizations depending 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 arbitrary form, of order 2 and 3. Explicit factorization formulas for an operator of the order ${\displaystyle n}$ can be found in[2] General invariants are defined in[3] and invariant formulation of the Beals-Kartashova factorization is given in[4]

Beals-Kartashova Factorization

Operator of order 2

Consider an operator

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

with smooth coefficients and look for a factorization

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

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

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

Then in all cases

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

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

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

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

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

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

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

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

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

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

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

can be found from the first three equations,

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

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

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

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

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

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

${\displaystyle p_{1}=1,}$

${\displaystyle p_{2}=-\omega ,}$

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

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

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

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

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

yields linear system of two algebraic equations:

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

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

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

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

equations have the unique solution:

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

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

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

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

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

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

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

If

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

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

Operator of order 3

Consider an operator

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

with smooth coefficients and look for a factorization

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

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

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

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

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

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

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

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

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

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

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

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

with ${\displaystyle {\mathcal {L}}=p_{1}\partial _{x}+p_{2}\partial _{y},}$ and again ${\displaystyle a_{30}\neq 0,}$ i.e. ${\displaystyle p_{1}=1,}$ and three-step procedure yields:

At the first step, the roots of a cubic polynomial

${\displaystyle {\mathcal {P}}_{3}(-p_{2}):=a_{30}(-p_{2})^{3}+a_{21}(-p_{2})^{2}+a_{12}(-p_{2})+a_{03}=0.}$

have to be found. Again ${\displaystyle \omega }$ denotes a root and first four coefficients are

${\displaystyle p_{1}=1,}$

${\displaystyle p_{2}=-\omega ,}$

${\displaystyle p_{4}=a_{30},}$

${\displaystyle p_{5}=a_{30}\omega +a_{21},}$

${\displaystyle p_{6}=a_{30}\omega ^{2}+a_{21}\omega +a_{12}.}$

At the second step, a linear system of three algebraic equations has to be solved:

${\displaystyle a_{20}-{\mathcal {L}}a_{30}=p_{3}a_{30}+p_{7},}$

${\displaystyle a_{11}-{\mathcal {L}}(a_{30}\omega +a_{21})=p_{3}(a_{30}\omega +a_{21})-\omega p_{7}+p_{8},}$

${\displaystyle a_{02}-{\mathcal {L}}(a_{30}\omega ^{2}+a_{21}\omega +a_{12})=p_{3}(a_{30}\omega ^{2}+a_{21}\omega +a_{12})-\omega p_{8}.}$

At the third step, two algebraic conditions have to be checked.

Operator of order ${\displaystyle n}$

Invariant Formulation

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

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

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

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

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

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

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

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

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

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

and so on. All functions ${\displaystyle l_{2},l_{3},l_{31},l_{4},l_{41},\ \ l_{42},...}$ are known functions, for instance,

${\displaystyle l_{2}=a_{00}-{\mathcal {L}}(p_{6})+p_{3}p_{6},}$

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

${\displaystyle l_{31}=a_{01}-{\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},}$

and so on.

Theorem All functions

${\displaystyle l_{2}=a_{00}-{\mathcal {L}}(p_{6})+p_{3}p_{6},l_{3}=a_{00}-{\mathcal {L}}(p_{9})+p_{3}p_{9},l_{31},....}$

are invariants under gauge transformations.

Definition Invariants ${\displaystyle l_{2}=a_{00}-{\mathcal {L}}(p_{6})+p_{3}p_{6},l_{3}=a_{00}-{\mathcal {L}}(p_{9})+p_{3}p_{9},l_{31},.....}$ 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 ${\displaystyle {\tilde {\mathcal {A}}}}$ is factorizable, then all operators equivalent to it, are also factorizable.

Equivalent operators are easy to compute:

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

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

and so on. Some example are given below:

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

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

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

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

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

Now the coefficients are

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

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

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

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

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

For instance, the operator

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

is factorizable as

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

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

See also

• Partial derivative
• Invariant (mathematics)
• Invariant theory
• Characteristic polynomial

Notes

1. Weiss (1986)
2. R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. Theor. Math. Phys. 145(2), pp. 1510-1523 (2005)
3. E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. Theor. Math. Phys. 147(3), pp. 839-846 (2006)
4. 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

References

• J. Weiss. Bäcklund transformation and the Painlevé property. J. Math. Phys. 27, 1293-1305 (1986).
• R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. Theor. Math. Phys. 145(2), pp. 1510-1523 (2005)
• 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