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Invariant factorization of LPDOs
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<h2 id="beals-kartashova-factorization">Beals-Kartashova Factorization</h2> <h3>Operator of order 2</h3> <p>Consider an operator </p> A 2 = a 20 ∂ x 2 + a 11 ∂ x ∂ y + a 02 ∂ y 2 + a 10 ∂ x + a 01 ∂ y + a 00 . {\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}.} <p>with smooth coefficients and look for a factorization </p> A 2 = ( p 1 ∂ x + p 2 ∂ y + p 3 ) ( p 4 ∂ x + p 5 ∂ y + p 6 ) . {\displaystyle {\mathcal {A}}_{2}=(p_{1}\partial _{x}+p_{2}\partial _{y}+p_{3})(p_{4}\partial _{x}+p_{5}\partial _{y}+p_{6}).} <p>Let us write down the equations on p i {\displaystyle p_{i}} explicitly, keeping in mind the rule of left composition, i.e. that </p> ∂ x ( α ∂ y ) = ∂ x ( α ) ∂ y + α ∂ x y . {\displaystyle \partial _{x}(\alpha \partial _{y})=\partial _{x}(\alpha )\partial _{y}+\alpha \partial _{xy}.} <p>Then in all cases </p> a 20 = p 1 p 4 , {\displaystyle a_{20}=p_{1}p_{4},} a 11 = p 2 p 4 + p 1 p 5 , {\displaystyle a_{11}=p_{2}p_{4}+p_{1}p_{5},} a 02 = p 2 p 5 , {\displaystyle a_{02}=p_{2}p_{5},} a 10 = L ( p 4 ) + p 3 p 4 + p 1 p 6 , {\displaystyle a_{10}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{6},} a 01 = L ( p 5 ) + p 3 p 5 + p 2 p 6 , {\displaystyle a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},} a 00 = L ( p 6 ) + p 3 p 6 , {\displaystyle a_{00}={\mathcal {L}}(p_{6})+p_{3}p_{6},} <p>where the notation L = p 1 ∂ x + p 2 ∂ y {\displaystyle {\mathcal {L}}=p_{1}\partial _{x}+p_{2}\partial _{y}} is used. </p><p>Without loss of generality, a 20 ≠ 0 , {\displaystyle a_{20}\neq 0,} i.e. p 1 ≠ 0 , {\displaystyle p_{1}\neq 0,} and it can be taken as 1, p 1 = 1. {\displaystyle p_{1}=1.} Now solution of the system of 6 equations on the variables </p> p 2 , {\displaystyle p_{2},} . . . {\displaystyle ...} p 6 {\displaystyle p_{6}} <p>can be found in three steps. </p><p>At the first step, the roots of a <i>quadratic polynomial</i> have to be found. </p><p>At the second step, a linear system of <i>two algebraic equations</i> has to be solved. </p><p>At the third step, <i>one algebraic condition</i> has to be checked. </p><p>Step 1. Variables </p> p 2 , {\displaystyle p_{2},} p 4 , {\displaystyle p_{4},} p 5 {\displaystyle p_{5}} <p>can be found from the first three equations, </p> a 20 = p 1 p 4 , {\displaystyle a_{20}=p_{1}p_{4},} a 11 = p 2 p 4 + p 1 p 5 , {\displaystyle a_{11}=p_{2}p_{4}+p_{1}p_{5},} a 02 = p 2 p 5 . {\displaystyle a_{02}=p_{2}p_{5}.} <p>The (possible) solutions are then the functions of the roots of a quadratic polynomial: </p> P 2 ( − p 2 ) = a 20 ( − p 2 ) 2 + a 11 ( − p 2 ) + a 02 = 0 {\displaystyle {\mathcal {P}}_{2}(-p_{2})=a_{20}(-p_{2})^{2}+a_{11}(-p_{2})+a_{02}=0} <p>Let ω {\displaystyle \omega } be a root of the polynomial P 2 , {\displaystyle {\mathcal {P}}_{2},} then </p> p 1 = 1 , {\displaystyle p_{1}=1,} p 2 = − ω , {\displaystyle p_{2}=-\omega ,} p 4 = a 20 , {\displaystyle p_{4}=a_{20},} p 5 = a 20 ω + a 11 , {\displaystyle p_{5}=a_{20}\omega +a_{11},} <p>Step 2. Substitution of the results obtained at the first step, into the next two equations </p> a 10 = L ( p 4 ) + p 3 p 4 + p 1 p 6 , {\displaystyle a_{10}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{6},} a 01 = L ( p 5 ) + p 3 p 5 + p 2 p 6 , {\displaystyle a_{01}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{6},} <p>yields linear system of two algebraic equations: </p> a 10 = L a 20 + p 3 a 20 + p 6 , {\displaystyle a_{10}={\mathcal {L}}a_{20}+p_{3}a_{20}+p_{6},} a 01 = L ( a 11 + a 20 ω ) + p 3 ( a 11 + a 20 ω ) − ω p 6 . , {\displaystyle a_{01}={\mathcal {L}}(a_{11}+a_{20}\omega )+p_{3}(a_{11}+a_{20}\omega )-\omega p_{6}.,} <p>In particularly, if the root ω {\displaystyle \omega } is simple, i.e. </p> P 2 ′ ( ω ) = 2 a 20 ω + a 11 ≠ 0 , {\displaystyle {\mathcal {P}}_{2}'(\omega )=2a_{20}\omega +a_{11}\neq 0,} then these <p>equations have the unique solution: </p> p 3 = ω a 10 + a 01 − ω L a 20 − L ( a 20 ω + a 11 ) 2 a 20 ω + a 11 , {\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}}},} p 6 = ( a 20 ω + a 11 ) ( a 10 − L a 20 ) − a 20 ( a 01 − L ( a 20 ω + a 11 ) ) 2 a 20 ω + 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}}}.} <p>At this step, for each root of the polynomial P 2 {\displaystyle {\mathcal {P}}_{2}} a corresponding set of coefficients p j {\displaystyle p_{j}} is computed. </p><p>Step 3. Check factorization condition (which is the last of the initial 6 equations) </p> a 00 = L ( p 6 ) + p 3 p 6 , {\displaystyle a_{00}={\mathcal {L}}(p_{6})+p_{3}p_{6},} <p>written in the known variables p j {\displaystyle p_{j}} and ω {\displaystyle \omega } ): </p> a 00 = L { ω a 10 + a 01 − L ( 2 a 20 ω + a 11 ) 2 a 20 ω + a 11 } + ω a 10 + a 01 − L ( 2 a 20 ω + a 11 ) 2 a 20 ω + a 11 × a 20 ( a 01 − L ( a 20 ω + a 11 ) ) + ( a 20 ω + a 11 ) ( a 10 − L a 20 ) 2 a 20 ω + a 11 {\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}}}} <p>If </p> l 2 = a 00 − L { ω a 10 + a 01 − L ( 2 a 20 ω + a 11 ) 2 a 20 ω + a 11 } + ω a 10 + a 01 − L ( 2 a 20 ω + a 11 ) 2 a 20 ω + a 11 × a 20 ( a 01 − L ( a 20 ω + a 11 ) ) + ( a 20 ω + a 11 ) ( a 10 − L a 20 ) 2 a 20 ω + a 11 = 0 , {\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,} <p>the operator A 2 {\displaystyle {\mathcal {A}}_{2}} is factorizable and explicit form for the factorization coefficients p j {\displaystyle p_{j}} is given above. </p> <h3>Operator of order 3</h3> <p>Consider an operator </p> A 3 = ∑ j + k ≤ 3 a j k ∂ x j ∂ y k = a 30 ∂ x 3 + a 21 ∂ x 2 ∂ y + a 12 ∂ x ∂ y 2 + a 03 ∂ y 3 + a 20 ∂ x 2 + a 11 ∂ x ∂ y + a 02 ∂ y 2 + a 10 ∂ x + a 01 ∂ y + a 00 . {\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}.} <p>with smooth coefficients and look for a factorization </p> A 3 = ( p 1 ∂ x + p 2 ∂ y + p 3 ) ( p 4 ∂ x 2 + p 5 ∂ x ∂ y + p 6 ∂ y 2 + p 7 ∂ x + p 8 ∂ y + p 9 ) . {\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}).} <p>Similar to the case of the operator A 2 , {\displaystyle {\mathcal {A}}_{2},} the conditions of factorization are described by the following system: </p> a 30 = p 1 p 4 , {\displaystyle a_{30}=p_{1}p_{4},} a 21 = p 2 p 4 + p 1 p 5 , {\displaystyle a_{21}=p_{2}p_{4}+p_{1}p_{5},} a 12 = p 2 p 5 + p 1 p 6 , {\displaystyle a_{12}=p_{2}p_{5}+p_{1}p_{6},} a 03 = p 2 p 6 , {\displaystyle a_{03}=p_{2}p_{6},} a 20 = L ( p 4 ) + p 3 p 4 + p 1 p 7 , {\displaystyle a_{20}={\mathcal {L}}(p_{4})+p_{3}p_{4}+p_{1}p_{7},} a 11 = L ( p 5 ) + p 3 p 5 + p 2 p 7 + p 1 p 8 , {\displaystyle a_{11}={\mathcal {L}}(p_{5})+p_{3}p_{5}+p_{2}p_{7}+p_{1}p_{8},} a 02 = L ( p 6 ) + p 3 p 6 + p 2 p 8 , {\displaystyle a_{02}={\mathcal {L}}(p_{6})+p_{3}p_{6}+p_{2}p_{8},} a 10 = L ( p 7 ) + p 3 p 7 + p 1 p 9 , {\displaystyle a_{10}={\mathcal {L}}(p_{7})+p_{3}p_{7}+p_{1}p_{9},} a 01 = L ( p 8 ) + p 3 p 8 + p 2 p 9 , {\displaystyle a_{01}={\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},} a 00 = L ( p 9 ) + p 3 p 9 , {\displaystyle a_{00}={\mathcal {L}}(p_{9})+p_{3}p_{9},} <p>with L = p 1 ∂ x + p 2 ∂ y , {\displaystyle {\mathcal {L}}=p_{1}\partial _{x}+p_{2}\partial _{y},} and again a 30 ≠ 0 , {\displaystyle a_{30}\neq 0,} i.e. p 1 = 1 , {\displaystyle p_{1}=1,} and three-step procedure yields: </p><p>At the first step, the roots of a <i>cubic polynomial</i> </p> P 3 ( − p 2 ) := a 30 ( − p 2 ) 3 + a 21 ( − p 2 ) 2 + a 12 ( − p 2 ) + a 03 = 0. {\displaystyle {\mathcal {P}}_{3}(-p_{2}):=a_{30}(-p_{2})^{3}+a_{21}(-p_{2})^{2}+a_{12}(-p_{2})+a_{03}=0.} <p>have to be found. Again ω {\displaystyle \omega } denotes a root and first four coefficients are </p> p 1 = 1 , {\displaystyle p_{1}=1,} p 2 = − ω , {\displaystyle p_{2}=-\omega ,} p 4 = a 30 , {\displaystyle p_{4}=a_{30},} p 5 = a 30 ω + a 21 , {\displaystyle p_{5}=a_{30}\omega +a_{21},} p 6 = a 30 ω 2 + a 21 ω + a 12 . {\displaystyle p_{6}=a_{30}\omega ^{2}+a_{21}\omega +a_{12}.} <p>At the second step, a linear system of <i>three algebraic equations</i> has to be solved: </p> a 20 − L a 30 = p 3 a 30 + p 7 , {\displaystyle a_{20}-{\mathcal {L}}a_{30}=p_{3}a_{30}+p_{7},} a 11 − L ( a 30 ω + a 21 ) = p 3 ( a 30 ω + a 21 ) − ω p 7 + p 8 , {\displaystyle a_{11}-{\mathcal {L}}(a_{30}\omega +a_{21})=p_{3}(a_{30}\omega +a_{21})-\omega p_{7}+p_{8},} a 02 − L ( a 30 ω 2 + a 21 ω + a 12 ) = p 3 ( a 30 ω 2 + a 21 ω + a 12 ) − ω 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}.} <p>At the third step, <i>two algebraic conditions</i> have to be checked. </p> <h2 id="invariant-formulation">Invariant Formulation</h2> <p>Definition The operators A {\displaystyle {\mathcal {A}}} , A ~ {\displaystyle {\tilde {\mathcal {A}}}} are called equivalent if there is a gauge transformation that takes one to the other: </p> A ~ g = e − φ A ( e φ g ) . {\displaystyle {\tilde {\mathcal {A}}}g=e^{-\varphi }{\mathcal {A}}(e^{\varphi }g).} <p>BK-factorization is then pure algebraic procedure which allows to construct explicitly a factorization of an arbitrary order LPDO A ~ {\displaystyle {\tilde {\mathcal {A}}}} in the form </p> A = ∑ j + k ≤ n a j k ∂ x j ∂ y k = L ∘ ∑ j + k ≤ ( n − 1 ) p j k ∂ x j ∂ y k {\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}} <p>with first-order operator L = ∂ x − ω ∂ y + p {\displaystyle {\mathcal {L}}=\partial _{x}-\omega \partial _{y}+p} where ω {\displaystyle \omega } is an arbitrary simple root of the characteristic polynomial </p> P ( t ) = ∑ k = 0 n a n − k , k t n − k , P ( ω ) = 0. {\displaystyle {\mathcal {P}}(t)=\sum _{k=0}^{n}a_{n-k,k}t^{n-k},\quad {\mathcal {P}}(\omega )=0.} <p>Factorization is possible then for each simple root ω ~ {\displaystyle {\tilde {\omega }}} iff </p><p>for n = 2 → l 2 = 0 , {\displaystyle n=2\ \ \rightarrow l_{2}=0,} </p><p>for n = 3 → l 3 = 0 , l 31 = 0 , {\displaystyle n=3\ \ \rightarrow l_{3}=0,l_{31}=0,} </p><p>for n = 4 → l 4 = 0 , l 41 = 0 , l 42 = 0 , {\displaystyle n=4\ \ \rightarrow l_{4}=0,l_{41}=0,l_{42}=0,} </p><p>and so on. All functions l 2 , l 3 , l 31 , l 4 , l 41 , l 42 , . . . {\displaystyle l_{2},l_{3},l_{31},l_{4},l_{41},\ \ l_{42},...} are known functions, for instance, </p> l 2 = a 00 − L ( p 6 ) + p 3 p 6 , {\displaystyle l_{2}=a_{00}-{\mathcal {L}}(p_{6})+p_{3}p_{6},} l 3 = a 00 − L ( p 9 ) + p 3 p 9 , {\displaystyle l_{3}=a_{00}-{\mathcal {L}}(p_{9})+p_{3}p_{9},} l 31 = a 01 − L ( p 8 ) + p 3 p 8 + p 2 p 9 , {\displaystyle l_{31}=a_{01}-{\mathcal {L}}(p_{8})+p_{3}p_{8}+p_{2}p_{9},} <p>and so on. </p><p>Theorem All functions </p> l 2 = a 00 − L ( p 6 ) + p 3 p 6 , l 3 = a 00 − L ( p 9 ) + p 3 p 9 , l 31 , . . . . {\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},....} <p>are invariants under gauge transformations. </p><p>Definition Invariants l 2 = a 00 − L ( p 6 ) + p 3 p 6 , l 3 = a 00 − L ( p 9 ) + p 3 p 9 , l 31 , . . . . . {\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. </p><p>In particular case of the bivariate hyperbolic operator its generalized invariants coincide with Laplace invariants (see <a href="/facts/Laplace_invariant/JQ3blrF9">Laplace invariant</a>). </p><p>Corollary If an operator A ~ {\displaystyle {\tilde {\mathcal {A}}}} is factorizable, then all operators equivalent to it, are also factorizable. </p><p>Equivalent operators are easy to compute: </p> e − φ ∂ x e φ = ∂ x + φ x , e − φ ∂ y e φ = ∂ y + φ y , {\displaystyle e^{-\varphi }\partial _{x}e^{\varphi }=\partial _{x}+\varphi _{x},\quad e^{-\varphi }\partial _{y}e^{\varphi }=\partial _{y}+\varphi _{y},} e − φ ∂ x ∂ y e φ = e − φ ∂ x e φ e − φ ∂ y e φ = ( ∂ x + φ x ) ∘ ( ∂ y + φ 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})} <p>and so on. Some example are given below: </p> A 1 = ∂ x ∂ y + x ∂ x + 1 = ∂ x ( ∂ y + x ) , l 2 ( A 1 ) = 1 − 1 − 0 = 0 ; {\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;} A 2 = ∂ x ∂ y + x ∂ x + ∂ y + x + 1 , A 2 = e − x A 1 e x ; l 2 ( A 2 ) = ( x + 1 ) − 1 − x = 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;} A 3 = ∂ x ∂ y + 2 x ∂ x + ( y + 1 ) ∂ y + 2 ( x y + x + 1 ) , A 3 = e − x y A 2 e x y ; l 2 ( A 3 ) = 2 ( x + 1 + x y ) − 2 − 2 x ( y + 1 ) = 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;} A 4 = ∂ x ∂ y + x ∂ x + ( cos x + 1 ) ∂ y + x cos x + x + 1 , A 4 = e − sin x A 2 e sin x ; l 2 ( A 4 ) = 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.} <h2 id="transpose">Transpose</h2> <p>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. </p><p>Definition The transpose A t {\displaystyle {\mathcal {A}}^{t}} of an operator A = ∑ a α ∂ α , ∂ α = ∂ 1 α 1 ⋯ ∂ n α n . {\displaystyle {\mathcal {A}}=\sum a_{\alpha }\partial ^{\alpha },\qquad \partial ^{\alpha }=\partial _{1}^{\alpha _{1}}\cdots \partial _{n}^{\alpha _{n}}.} is defined as A t u = ∑ ( − 1 ) | α | ∂ α ( a α u ) . {\displaystyle {\mathcal {A}}^{t}u=\sum (-1)^{|\alpha |}\partial ^{\alpha }(a_{\alpha }u).} and the identity ∂ γ ( u v ) = ∑ ( γ α ) ∂ α u , ∂ γ − α v {\displaystyle \partial ^{\gamma }(uv)=\sum {\binom {\gamma }{\alpha }}\partial ^{\alpha }u,\partial ^{\gamma -\alpha }v} implies that A t = ∑ ( − 1 ) | α + β | ( α + β α ) ( ∂ β a α + β ) ∂ α . {\displaystyle {\mathcal {A}}^{t}=\sum (-1)^{|\alpha +\beta |}{\binom {\alpha +\beta }{\alpha }}(\partial ^{\beta }a_{\alpha +\beta })\partial ^{\alpha }.} </p><p>Now the coefficients are </p><p> A t = ∑ a ~ α ∂ α , {\displaystyle {\mathcal {A}}^{t}=\sum {\tilde {a}}_{\alpha }\partial ^{\alpha },} a ~ α = ∑ ( − 1 ) | α + β | ( α + β α ) ∂ β ( a α + β ) . {\displaystyle {\tilde {a}}_{\alpha }=\sum (-1)^{|\alpha +\beta |}{\binom {\alpha +\beta }{\alpha }}\partial ^{\beta }(a_{\alpha +\beta }).} </p><p>with a standard convention for binomial coefficients in several variables (see <a href="/facts/Binomial_coefficient/Tsx7eY5h">Binomial coefficient</a>), e.g. in two variables </p> ( α β ) = ( ( α 1 , α 2 ) ( β 1 , β 2 ) ) = ( α 1 β 1 ) ( α 2 β 2 ) . {\displaystyle {\binom {\alpha }{\beta }}={\binom {(\alpha _{1},\alpha _{2})}{(\beta _{1},\beta _{2})}}={\binom {\alpha _{1}}{\beta _{1}}}\,{\binom {\alpha _{2}}{\beta _{2}}}.} <p>In particular, for the operator A 2 {\displaystyle {\mathcal {A}}_{2}} the coefficients are a ~ j k = a j k , j + k = 2 ; a ~ 10 = − a 10 + 2 ∂ x a 20 + ∂ y a 11 , a ~ 01 = − a 01 + ∂ x a 11 + 2 ∂ y a 02 , {\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},} </p> a ~ 00 = a 00 − ∂ x a 10 − ∂ y a 01 + ∂ x 2 a 20 + ∂ x ∂ x a 11 + ∂ y 2 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}.} <p>For instance, the operator </p> ∂ x x − ∂ y y + y ∂ x + x ∂ y + 1 4 ( y 2 − x 2 ) − 1 {\displaystyle \partial _{xx}-\partial _{yy}+y\partial _{x}+x\partial _{y}+{\frac {1}{4}}(y^{2}-x^{2})-1} <p>is factorizable as </p> [ ∂ x + ∂ y + 1 2 ( y − x ) ] [ . . . ] {\displaystyle {\big [}\partial _{x}+\partial _{y}+{\tfrac {1}{2}}(y-x){\big ]}\,{\big [}...{\big ]}} <p>and its transpose A 1 t {\displaystyle {\mathcal {A}}_{1}^{t}} is factorizable then as [ . . . ] [ ∂ x − ∂ y + 1 2 ( y + x ) ] . {\displaystyle {\big [}...{\big ]}\,{\big [}\partial _{x}-\partial _{y}+{\tfrac {1}{2}}(y+x){\big ]}.} </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Partial_derivative/JWHsBkAu">Partial derivative</a></li> <li><a href="/facts/Invariant_(mathematics)/9TeNN2ed">Invariant (mathematics)</a></li> <li><a href="/facts/Invariant_theory/5fIUh9h3">Invariant theory</a></li> <li><a href="/facts/Characteristic_polynomial/7m8roDqo">Characteristic polynomial</a></li></ul> <h2 id="notes">Notes</h2> <ul><li>J. Weiss. Bäcklund transformation and the Painlevé property. <a href="https://archive.today/20020906093934/http://www2.appmath.com:8080/site/few/few.html">[1]</a> J. Math. Phys. 27, 1293-1305 (1986).</li> <li>R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. <a href="https://doi.org/10.1007%2Fs11232-005-0178-7">Theor. Math. Phys. 145(2), pp. 1510-1523 (2005)</a></li> <li>E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. <a href="https://doi.org/10.1007%2Fs11232-006-0079-4">Theor. Math. Phys. 147(3), pp. 839-846 (2006)</a></li> <li>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); <a href="https://arxiv.org/abs/math-ph/0607040/">arXiv</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Weiss (1986) <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>R. Beals, E. Kartashova. Constructively factoring linear partial differential operators in two variables. Theor. Math. Phys. 145(2), pp. 1510-1523 (2005) <a href="https://doi.org/10.1007%2Fs11232-005-0178-7" target="_blank">https://doi.org/10.1007%2Fs11232-005-0178-7</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>E. Kartashova. A Hierarchy of Generalized Invariants for Linear Partial Differential Operators. Theor. Math. Phys. 147(3), pp. 839-846 (2006) <a href="https://doi.org/10.1007%2Fs11232-006-0079-4" target="_blank">https://doi.org/10.1007%2Fs11232-006-0079-4</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>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 <a href="https://arxiv.org/abs/math-ph/0607040/" target="_blank">https://arxiv.org/abs/math-ph/0607040/</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> </ol>