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
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
[edit]
Operator of order 2
[edit]
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}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/0db6ef3814497239a93b87f63f063f0544a90f51)
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}).}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/57483d34de607e7a362b68e7bff1262c46002899)
Let us write down the equations on
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}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/12bea1c82ec321cbf7f716196eac3c7f00b9ea85)
Then in all cases
![{\displaystyle a_{20}=p_{1}p_{4},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e79b61c16badb424b4e6a644f48810eb02137c66)
![{\displaystyle a_{11}=p_{2}p_{4}+p_{1}p_{5},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/c8cbdef625be60f3990e88f0d09e89e8956933bf)
![{\displaystyle a_{02}=p_{2}p_{5},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/86048235e1687bda07af8537e9210b752f6d0627)
![{\displaystyle a_{10}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d7c9695e4d815be757f926041505414c17b6fe86)
![{\displaystyle a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d3c0d83a8e78b34608e420518bf2e0a33f7c324e)
![{\displaystyle a_{00}={\mathcal {L}}(p_{6})+p_{3}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/1d064a201b56cb22c9912d758a42894fd5b76eb4)
where the notation
is used.
Without loss of generality,
i.e.
and it can be taken as 1,
Now solution of the system of 6 equations on the variables
![{\displaystyle p_{6}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/a27354590cf8a617f5c84fca30bd5792ed6fd757)
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_{5}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/29cfe585b7eadbfb6d654fa8aea0e260db6e4a82)
can be found from the first three equations,
![{\displaystyle a_{20}=p_{1}p_{4},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e79b61c16badb424b4e6a644f48810eb02137c66)
![{\displaystyle a_{11}=p_{2}p_{4}+p_{1}p_{5},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/c8cbdef625be60f3990e88f0d09e89e8956933bf)
![{\displaystyle a_{02}=p_{2}p_{5}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/03669745b0869389a9f6d115d719d4ea6a6dc4b5)
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}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/709114b8dde19750f50059ba7052ec4ae37bdfad)
Let
be a root of the polynomial
then
![{\displaystyle p_{1}=1,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/b743118f6e692b1225612048bde610bf438cbf69)
![{\displaystyle p_{2}=-\omega ,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/6fa0fa4b758e65b99c6cdbea44455ca0e3ea7efb)
![{\displaystyle p_{4}=a_{20},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d00125680724367982849a59f56a976da1737e1b)
![{\displaystyle p_{5}=a_{20}\omega +a_{11},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/c7125acac12fdd6d91d6771d4dfe8554a65c33e5)
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},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d7c9695e4d815be757f926041505414c17b6fe86)
![{\displaystyle a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d3c0d83a8e78b34608e420518bf2e0a33f7c324e)
yields linear system of two algebraic equations:
![{\displaystyle a_{10}={\mathcal {L}}a_{20}+p_{3}a_{20}+p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/cd1a266ac5722d44a5bf05b49341b196c78b8ec2)
![{\displaystyle a_{01}={\mathcal {L}}(a_{11}+a_{20}\omega )+p_{3}(a_{11}+a_{20}\omega )-\omega p_{6}.,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/df584eeccc4a71ab55a9c28941299ef9387ea68c)
In particularly, if the root
is simple,
i.e.
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}}},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/3ae2aa7741b20a1dbc478e40502de6df27ba4c0e)
![{\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}}}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/8a58d1bbc4cb394cf29c76fac3159bbeea632452)
At this step, for each
root of the polynomial
a corresponding set of coefficients
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},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/1d064a201b56cb22c9912d758a42894fd5b76eb4)
written in the known variables
and
):
![{\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}}}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d86d5be9c3c120513c1a670fd7bc21210171ea27)
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,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/6cb573d20aa7e6da4696b00be2df2ec7797c2720)
the operator
is factorizable and explicit form for the factorization coefficients
is given above.
Operator of order 3
[edit]
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}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d235f4d5e1141707e3f879dcf8692eb77becdb86)
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}).}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/37c9043985eb02a0e26f7b006663b98ebfcde071)
Similar to the case of the operator
the conditions of factorization are described by the following system:
![{\displaystyle a_{30}=p_{1}p_{4},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/706157fafa17c11e602331551f78c78513464034)
![{\displaystyle a_{21}=p_{2}p_{4}+p_{1}p_{5},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/38ae20f5521af39737a13eb2729b012950a942f7)
![{\displaystyle a_{12}=p_{2}p_{5}+p_{1}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/508a657d36bfcf2f0820774521c807878aedce57)
![{\displaystyle a_{03}=p_{2}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/fb706c5c8a3600b54f0d2fc240c21bcaeb1ed764)
![{\displaystyle a_{20}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{7},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/63ea80f0aea4adfd9b8c30b2eafcdf14ca387f78)
![{\displaystyle a_{11}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{7}+p_{1}p_{8},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e994acdcef6aef6d2b2f3ea00e45e0ab5846f8a6)
![{\displaystyle a_{02}={\mathcal {L}}(p_{6})+p_{3}p_{6}+p_{2}p_{8},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/2a9eae11d34e1547f06b016af42bbed1bb51f959)
![{\displaystyle a_{10}={\mathcal {L}}(p_{7})+p_{3}p_{7}+p_{1}p_{9},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/7cbba556c025e083227492c37af8230a9fef0e06)
![{\displaystyle a_{01}={\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d80c2b5e535c49caa2cc5392f59f74a84e555839)
![{\displaystyle a_{00}={\mathcal {L}}(p_{9})+p_{3}p_{9},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/979ca303e7866da1c45edb6cfafee5db97a1ec05)
with
and again
i.e.
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.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/b054bb6100a55f97c38a2da6e25261092ee418a6)
have to be found. Again
denotes a root and first four coefficients are
![{\displaystyle p_{1}=1,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/b743118f6e692b1225612048bde610bf438cbf69)
![{\displaystyle p_{2}=-\omega ,}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/6fa0fa4b758e65b99c6cdbea44455ca0e3ea7efb)
![{\displaystyle p_{4}=a_{30},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e2205d519da331358e40889e3f92d2c6817ed344)
![{\displaystyle p_{5}=a_{30}\omega +a_{21},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e6977bf7632cc9c123eb41462d2b9109b6c73e4f)
![{\displaystyle p_{6}=a_{30}\omega ^{2}+a_{21}\omega +a_{12}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/2ccb3e84d6d4a7b556156a9a4f22aff29624a226)
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},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/05e10aa07843984c3e61afca2e0517db26ac81d6)
![{\displaystyle a_{11}-{\mathcal {L}}(a_{30}\omega +a_{21})=p_{3}(a_{30}\omega +a_{21})-\omega p_{7}+p_{8},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/9f45e1524c04ae9d0b110737655693adde518922)
![{\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}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/a73e53354fa3b26b22cf2fe853c2bbe9e0c00f12)
At the third step, two algebraic conditions have to be checked.
Definition The operators
,
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).}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/aede82eabaf032239ad84a866a5cf80664051a72)
BK-factorization is then pure algebraic procedure which allows to
construct explicitly a factorization of an arbitrary order LPDO
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}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d5065ff8cb2b50522e69097e0a8ec68f413e58d6)
with first-order operator
where
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.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/dd040b3e882baa1cadac9bf04014c695e26823c8)
Factorization is possible then for each simple root
iff
for
for
for
and so on. All functions
are known functions, for instance,
![{\displaystyle l_{2}=a_{00}-{\mathcal {L}}(p_{6})+p_{3}p_{6},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d7653150ea038055abf9e6eba326254c75d258c0)
![{\displaystyle l_{3}=a_{00}-{\mathcal {L}}(p_{9})+p_{3}p_{9},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/065880800bf37a73fadcac97e6f508ea3cf4b40f)
![{\displaystyle l_{31}=a_{01}-{\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/776225f5a4b86c2387dca911990b917e97b19eda)
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},....}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/01872b2df257523469a421b4d1b4603fc4e5e43b)
are invariants under gauge transformations.
Definition Invariants
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
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},}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/e84288f923cd0c7df75c67c3b1d9ea818bd7f5ff)
![{\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})}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/072d381c78282d2d505e62c1f77ebabf4d034983)
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;}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/dc73a1a39fb6e800986608cceba4a82e9f966c9f)
![{\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;}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/2247dd2c21b1bfd6c4bc8323d306103d7cf857cb)
![{\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;}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/de703e7792b45fe96d0293e27b8f820215815560)
![{\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.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/55335b9ed7b2713125c5357e34f1f498b0361bb7)
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
of an operator
is defined as
and the identity
implies that
Now the coefficients are
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}}}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/72b31cc9804ad02c1f406186e85bcbd7dcd6b544)
In particular, for the operator
the coefficients are
![{\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}.}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/978a0e33ef35e5ae5395e372cf6c1012bc7c541e)
For instance, the operator
![{\displaystyle \partial _{xx}-\partial _{yy}+y\partial _{x}+x\partial _{y}+{\frac {1}{4}}(y^{2}-x^{2})-1}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/1ce3bd83f6a97da178f656785f54d73b1b17b9f1)
is factorizable as
![{\displaystyle {\big [}\partial _{x}+\partial _{y}+{\tfrac {1}{2}}(y-x){\big ]}\,{\big [}...{\big ]}}](https://backend.710302.xyz/https/wikimedia.org/api/rest_v1/media/math/render/svg/d255ce024a21156fe87b8c4ec1dbf90e66adb7f3)
and its transpose
is factorizable then as
- J. Weiss. Bäcklund transformation and the Painlevé property. [1] 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