The Riemann-Hilbert problem for matrix-valued orthogonal polynomials

The Riemann-Hilbert problem for matrix-valued orthogonal polynomials1 Manuel Domínguez de la Iglesia Department of Mathematics, K. U. Leuven Seminar classical analysis Leuven, November 5, 2008 1joint work with Andrei Martínez Finkelshtein Preliminaries The RH problem for OMP An example Outline 1Preliminaries 2The RH problem for OMP 3An example Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Outline 1Preliminaries 2The RH problem for OMP 3An example Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Let Wbe a N×Naweight matrix such that dW (x) = W(x)dx. We can construct a family of OMP such that ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m⩾0 Pn(x) = γn(xn+an,n−1xn−1+···) = γnb Pn(x) The matrix-valued polynomials of the second kind, defined by Qn(x) = ZR Pn(t)W(t) t−xdt,n⩾0 (Pn)nand (Qn)nsatisfy a three term recurrence relation tPn(t) = An+1Pn+1(t) + BnPn(t) + A∗ nPn−1(t),n⩾0 det(An+1)6=0,Bn=B∗ n Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Let Wbe a N×Naweight matrix such that dW (x) = W(x)dx. We can construct a family of OMP such that ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m⩾0 Pn(x) = γn(xn+an,n−1xn−1+···) = γnb Pn(x) The matrix-valued polynomials of the second kind, defined by Qn(x) = ZR Pn(t)W(t) t−xdt,n⩾0 (Pn)nand (Qn)nsatisfy a three term recurrence relation tPn(t) = An+1Pn+1(t) + BnPn(t) + A∗ nPn−1(t),n⩾0 det(An+1)6=0,Bn=B∗ n Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Let Wbe a N×Naweight matrix such that dW (x) = W(x)dx. We can construct a family of OMP such that ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m⩾0 Pn(x) = γn(xn+an,n−1xn−1+···) = γnb Pn(x) The matrix-valued polynomials of the second kind, defined by Qn(x) = ZR Pn(t)W(t) t−xdt,n⩾0 (Pn)nand (Qn)nsatisfy a three term recurrence relation tPn(t) = An+1Pn+1(t) + BnPn(t) + A∗ nPn−1(t),n⩾0 det(An+1)6=0,Bn=B∗ n Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Let Wbe a N×Naweight matrix such that dW (x) = W(x)dx. We can construct a family of OMP such that ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m⩾0 Pn(x) = γn(xn+an,n−1xn−1+···) = γnb Pn(x) The matrix-valued polynomials of the second kind, defined by Qn(x) = ZR Pn(t)W(t) t−xdt,n⩾0 (Pn)nand (Qn)nsatisfy a three term recurrence relation tPn(t) = An+1Pn+1(t) + BnPn(t) + A∗ nPn−1(t),n⩾0 det(An+1)6=0,Bn=B∗ n Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The coefficients of the TTRR satisfy An=γn−1γ−1 n,Bn=γn(an,n−1−an+1,n)γ−1 n The TTRR for monic OMP xb Pn(x) = b Pn+1(x) + αnb Pn(x) + βnb Pn−1(x),n⩾0 αn=an,n−1−an+1,n,βn= (γ∗ nγn)−1(γ∗ n−1γn−1) Second-order differential equations of hypergeometric type P00 n(x)F2(x) + P0 n(x)F1(x) + Pn(x)F0(x) = ΛnPn(x),n⩾0 deg Fi⩽i,ΛnHermitian Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The coefficients of the TTRR satisfy An=γn−1γ−1 n,Bn=γn(an,n−1−an+1,n)γ−1 n The TTRR for monic OMP xb Pn(x) = b Pn+1(x) + αnb Pn(x) + βnb Pn−1(x),n⩾0 αn=an,n−1−an+1,n,βn= (γ∗ nγn)−1(γ∗ n−1γn−1) Second-order differential equations of hypergeometric type P00 n(x)F2(x) + P0 n(x)F1(x) + Pn(x)F0(x) = ΛnPn(x),n⩾0 deg Fi⩽i,ΛnHermitian Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Also we find a solution of the inverse (Yn)−1=  −ZR W(t)b P∗ n−1(t) t−zdtγ∗ n−1γn−1−1 2πiZR W(t)b P∗ n(t) t−zdt 2πib P∗ n−1(z)γ∗ n−1γn−1b P∗ n(z)   The Liouville-Ostrogradski formula Qn(z)P∗ n−1(z) − Pn(z)Q∗ n−1(z) = A−1 n The Hermitian property Qn(z)P∗ n(z) = Pn(z)Q∗ n(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The three-term recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→∞ Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) γ∗ n−1γn−1= − 1 2πiYn 121 = − 1 2πiYn−1 1−1 12 αn=Yn 111 −Yn+1 111,βn=Yn 112Yn 121 Bn=γnYn 111 −Yn+1 111γ−1 n,A∗ n=γnYn 112Yn 121γ−1 n−1 Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 =0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The three-term recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→∞ Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) γ∗ n−1γn−1= − 1 2πiYn 121 = − 1 2πiYn−1 1−1 12 αn=Yn 111 −Yn+1 111,βn=Yn 112Yn 121 Bn=γnYn 111 −Yn+1 111γ−1 n,A∗ n=γnYn 112Yn 121γ−1 n−1 Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 =0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The three-term recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→∞ Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) γ∗ n−1γn−1= − 1 2πiYn 121 = − 1 2πiYn−1 1−1 12 αn=Yn 111 −Yn+1 111,βn=Yn 112Yn 121 Bn=γnYn 111 −Yn+1 111γ−1 n,A∗ n=γnYn 112Yn 121γ−1 n−1 Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 =0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The three-term recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→∞ Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) γ∗ n−1γn−1= − 1 2πiYn 121 = − 1 2πiYn−1 1−1 12 αn=Yn 111 −Yn+1 111,βn=Yn 112Yn 121 Bn=γnYn 111 −Yn+1 111γ−1 n,A∗ n=γnYn 112Yn 121γ−1 n−1 Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 =0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The three-term recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→∞ Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) γ∗ n−1γn−1= − 1 2πiYn 121 = − 1 2πiYn−1 1−1 12 αn=Yn 111 −Yn+1 111,βn=Yn 112Yn 121 Bn=γnYn 111 −Yn+1 111γ−1 n,A∗ n=γnYn 112Yn 121γ−1 n−1 Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 =0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The kernel Let (Pn)nan orthonormal family. Then we have Kn(x,y) = n−1 X j=0 P∗ j(y)Pj(x) = P∗ n−1(y)AnPn(x) − P∗ n(y)A∗ nPn−1(x) x−y This kernel has the following properties 1Kn(x,y) = K∗ n(y,x) 2Kn(x,y) = RRKn(s,y)W(s)Kn(x,s)ds We also have that Kn(x,y) = 1 2πi(x−y)0I(Yn)−1 +(y)(Yn)+(x)I 0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The kernel Let (Pn)nan orthonormal family. Then we have Kn(x,y) = n−1 X j=0 P∗ j(y)Pj(x) = P∗ n−1(y)AnPn(x) − P∗ n(y)A∗ nPn−1(x) x−y This kernel has the following properties 1Kn(x,y) = K∗ n(y,x) 2Kn(x,y) = RRKn(s,y)W(s)Kn(x,s)ds We also have that Kn(x,y) = 1 2πi(x−y)0I(Yn)−1 +(y)(Yn)+(x)I 0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The kernel Let (Pn)nan orthonormal family. Then we have Kn(x,y) = n−1 X j=0 P∗ j(y)Pj(x) = P∗ n−1(y)AnPn(x) − P∗ n(y)A∗ nPn−1(x) x−y This kernel has the following properties 1Kn(x,y) = K∗ n(y,x) 2Kn(x,y) = RRKn(s,y)W(s)Kn(x,s)ds We also have that Kn(x,y) = 1 2πi(x−y)0I(Yn)−1 +(y)(Yn)+(x)I 0 Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example A differential equation We consider weight matrices of the form W(x) = ρ(x)T(x)T∗(x), where Tsatisfies T0(x) = G(x)T(x). Consider Xn(z) = Yn(z)J(z) = Yn(z)ρ(z)1/2T(z)0 0ρ(z)−1/2(T(z))−∗ Then d dz Xn(z)Xn(z)−1is entire and near infinity it behaves like I+Y1 z+O(z−2) 1 2 ρ0(z) ρ(z)+G(z)0 0−1 2 ρ0(z) ρ(z)−G(z)∗!I−Y1 z+O(z−2) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The Lax pair Let Ynbe the solution of the RH for Wand consider Xn(z) = Yn(z)e−z2/2eAz 0 0ez2/2e−A∗z Xn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Xn(z) d dz Xn(z) = −zI +A2Yn 112 −2Yn 121 zI −A∗Xn(z) Compatibility conditions 2(βn+1−βn) = Aαn−αnA+I αn=1 2(A+ (γ∗ nγn)−1A∗(γ∗ nγn)) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example The Lax pair Let Ynbe the solution of the RH for Wand consider Xn(z) = Yn(z)e−z2/2eAz 0 0ez2/2e−A∗z Xn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Xn(z) d dz Xn(z) = −zI +A2Yn 112 −2Yn 121 zI −A∗Xn(z) Compatibility conditions 2(βn+1−βn) = Aαn−αnA+I αn=1 2(A+ (γ∗ nγn)−1A∗(γ∗ nγn)) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Ladder operators Lowering operator b P0 n(z) = Ab Pn(z) − b Pn(z)A+2βnb Pn−1(z) Therefore βn=1 2(nI +an,n−1A−Aan,n−1) Raising operator b P0 n(z)=−2b Pn+1(z) + 2zb Pn(z) + Ab Pn(z) − b Pn(z)A−2αnb Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Ladder operators Lowering operator b P0 n(z) = Ab Pn(z) − b Pn(z)A+2βnb Pn−1(z) Therefore βn=1 2(nI +an,n−1A−Aan,n−1) Raising operator b P0 n(z)=−2b Pn+1(z) + 2zb Pn(z) + Ab Pn(z) − b Pn(z)A−2αnb Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Second-order differential equations Introduce the following differential/difference operators R1=∂1+∂0A,L1=2βnE−1+AE0 R2=∂1+∂0(A−2zI),L2= −2E1+ (A−2αn)E0 ∂k=dk dzk,Ekf(n) = f(n+k) The ladder operators are equivalent to b Pn(z)R1=L1b Pn(z),and b Pn(z)R2=L2b Pn(z) Therefore, the OMP b Pnsatisfy two second-order differential equations b Pn(z)R1R2=L1b Pn(z)R2=L1L2b Pn(z), b Pn(z)R2R1=L2b Pn(z)R1=L2L1b Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Second-order differential equations Introduce the following differential/difference operators R1=∂1+∂0A,L1=2βnE−1+AE0 R2=∂1+∂0(A−2zI),L2= −2E1+ (A−2αn)E0 ∂k=dk dzk,Ekf(n) = f(n+k) The ladder operators are equivalent to b Pn(z)R1=L1b Pn(z),and b Pn(z)R2=L2b Pn(z) Therefore, the OMP b Pnsatisfy two second-order differential equations b Pn(z)R1R2=L1b Pn(z)R2=L1L2b Pn(z), b Pn(z)R2R1=L2b Pn(z)R1=L2L1b Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example Second-order differential equations Introduce the following differential/difference operators R1=∂1+∂0A,L1=2βnE−1+AE0 R2=∂1+∂0(A−2zI),L2= −2E1+ (A−2αn)E0 ∂k=dk dzk,Ekf(n) = f(n+k) The ladder operators are equivalent to b Pn(z)R1=L1b Pn(z),and b Pn(z)R2=L2b Pn(z) Therefore, the OMP b Pnsatisfy two second-order differential equations b Pn(z)R1R2=L1b Pn(z)R2=L1L2b Pn(z), b Pn(z)R2R1=L2b Pn(z)R1=L2L1b Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example We will proof the following: Both equations are equivalent They are also equivalent to the second-order differential equation of hypergeometric type b P00 n(z) + b P0 n(z)(2A−2zI) + b Pn(z)(A2−2J) = (−2nI +A2−2J)b Pn(z) The first one is b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)A(A−2zI) = −4βnb Pn(z) + 2βn(A−2αn−1)b Pn−1(z) − 2Ab Pn+1(z) + A(A−2αn)b Pn(z) And the second b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)(A−2zI)A−2b Pn(z) = −4βn+1b Pn(z) + 2(A−2αn)βnb Pn−1(z) − 2Ab Pn+1(z) + (A−2αn)Ab Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example We will proof the following: Both equations are equivalent They are also equivalent to the second-order differential equation of hypergeometric type b P00 n(z) + b P0 n(z)(2A−2zI) + b Pn(z)(A2−2J) = (−2nI +A2−2J)b Pn(z) The first one is b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)A(A−2zI) = −4βnb Pn(z) + 2βn(A−2αn−1)b Pn−1(z) − 2Ab Pn+1(z) + A(A−2αn)b Pn(z) And the second b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)(A−2zI)A−2b Pn(z) = −4βn+1b Pn(z) + 2(A−2αn)βnb Pn−1(z) − 2Ab Pn+1(z) + (A−2αn)Ab Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP Preliminaries The RH problem for OMP An example We will proof the following: Both equations are equivalent They are also equivalent to the second-order differential equation of hypergeometric type b P00 n(z) + b P0 n(z)(2A−2zI) + b Pn(z)(A2−2J) = (−2nI +A2−2J)b Pn(z) The first one is b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)A(A−2zI) = −4βnb Pn(z) + 2βn(A−2αn−1)b Pn−1(z) − 2Ab Pn+1(z) + A(A−2αn)b Pn(z) And the second b P00 n(z) + 2b P0 n(z)(A−zI) + b Pn(z)(A−2zI)A−2b Pn(z) = −4βn+1b Pn(z) + 2(A−2αn)βnb Pn−1(z) − 2Ab Pn+1(z) + (A−2αn)Ab Pn(z) Manuel Domínguez de la Iglesia The RH problem of OMP