The inverse Berreman problem Bill Lionheart and Chris Newton School - PowerPoint PPT Presentation

Advances in the determination of liquid crystal director profiles The inverse Berreman problem Bill Lionheart and Chris Newton School of Mathematics University of Manchester and HP Labs Bristol Advances in the determination of liquid crystal director profiles – p. 1/13

Overview In stratified anisotropic media Maxwell’s equations can be replaced by Berreman’s ODE. • We want to solve for the unknown permittivity tensor in Berreman using data from a range of • incidence angles. Is such data sufficient to uniquely determine the parameters? • How ill-conditioned is the inverse problem and why? • Advances in the determination of liquid crystal director profiles – p. 2/13

Berreman’s formulation Berreman field vector X = ( E x , H y , E y , −H x ) T ∂z X = − iω ∂ c MX (1) where the matrix M is 2 3 µ 0 c ǫ 33 − ξ 2 − ǫ 13 − ǫ 23 ǫ 33 ξ ǫ 33 ξ 0 “ ” “ ” ǫ 33 6 7 ǫ 11 − ǫ 132 6 7 − ǫ 13 ǫ 12 − ǫ 13 ǫ 23 ǫ o c ǫ 33 ξ ǫ 0 c 0 6 7 ǫ 33 ǫ 33 6 7 M = (2) 6 7 0 0 0 µ 0 c 6 7 „ « “ ” 4 5 ǫ 22 − ǫ 2 ǫ 12 − ǫ 13 ǫ 23 − ǫ 23 ǫ 33 − ξ 2 ǫ 0 c ǫ 33 ξ ǫ 0 c 23 0 ǫ 33 The final value problem for this linear ODE can be solved to give the linear relationship between ‘initial data’ X (0) and X ( z ) = P ( z ) X (0) , where P ( z ) is a propagation matrix. We will write P ( d ) as simply P . Advances in the determination of liquid crystal director profiles – p. 3/13

Perturbation formula The inverse problem is to determine ǫ from P for a range of incident angles ξ , a kind of generalized inverse spectral problem for an ODE. We consider a perturbation δM in the material properties from some initial guess. The Berreman matrix changes to M + δM . We need to know how the transmitted light depends on the perturbation in M to first order. The final data can be obtained by solving this ODE for its final value δX ( d ) , or from the integral Z d δX ( d ) = G ( d, z ) δM ( z ) X ( z ) dz (3) 0 where G is the Green function of the ODE. Advances in the determination of liquid crystal director profiles – p. 4/13

Perturbation formula, constant background For the case of a constant background M we can solve explicitly, and while an over simplification it gives us some insight 0 1 Z z X ( z ) = − iω c exp( iω exp( − iω c Mz ′ ) h ( z ′ ) dz ′ + X (0) @ A c Mz ) (4) 0 or G ( z, z ′ ) = exp( − iω c M ( z − z ′ )) so that (3) becomes Z d δX ( d ) = iω c exp( − iω exp( iω c Mz ) δM ( z ) exp( − i ω c Md ) c Mz ) X (0) dz (5) 0 as for non-lossy materials M ξ is real, we can consider (5) as a generalized Fourier transform of δM . Advances in the determination of liquid crystal director profiles – p. 5/13

Fourier transform of the unknowns – sort of! Eigenvalues q i of M , distinct, real and q 1 = − q 2 , q 3 = − q 4 . U matrix of eigenvectors of M , Q = diag( q i ) . Equation (5) becomes Z d U exp( − iQz ) U − 1 δM ( z ) U exp( iQz ) U − 1 X (0) dz δX ( d ) = ( iω/c ) exp(( iω/c ) Md ) 0 (6) Set A = U − 1 δM ( z ) U . Note exp( − iQz ) A exp( iQz ) has elements e i ( q i − q j ) z A ij . As we know all possible pairs of initial and final data X (0) , X ( d ) , we know the linear response T defined by δX ( d ) = TX (0) hence we know the matrix Z d − i ( c/ω ) exp( − ( iω/c ) Md ) U − 1 TU = Y = exp( − iQz ) A exp( iQz ) dz with (7) 0 Z d e i ( q i − q j ) z A ij dz = d Y ij = A ij ( q i − q j ) (8) 0 where the b denotes the Fourier transform. Advances in the determination of liquid crystal director profiles – p. 6/13

Fourier transform of the unknowns, continued The q i are functions of ξ and in general varying ξ over an interval will vary q i − q j over an interval, for i � = j . This data over a variety of incident angles gives us some information about the deviation of the permittivity tensor from our initial constant assumption in terms of the deviation of the measurements from those we could calculate for the constant case. Advances in the determination of liquid crystal director profiles – p. 7/13

Orthorhombic example We can best illustrate this by an example that is tractable analytically, the orthorhombic case ǫ ij = 0 , i � = j . M reduces to 2 3 ξ 2 0 µ 0 c − 0 0 cǫ 0 ǫ 33 6 7 6 7 ǫ 0 cǫ 11 0 0 0 6 7 6 7 6 7 0 0 0 µ 0 c 4 5 ǫ 0 c ` ǫ 22 − ξ 2 ´ 0 0 0 The eigenvalues are s s 1 − ξ 2 1 − ξ 2 q 1 = √ ǫ 11 , q 2 = − q 1 , q 3 = √ ǫ 22 , q 4 = − q 3 ǫ 33 ǫ 22 Advances in the determination of liquid crystal director profiles – p. 8/13

Orthorhombic example, δM We consider a completely general perturbation δ ij from this orthorhombic case, and try to see if this can be detected by data from a range of ξ . The eigenvectors can be calculated explicitly, and of course are constant in z although they vary with ξ . We then calculate 2 3 ξ 2 δ 33 − ξ δ 13 − ξ δ 23 0 cǫ 0 ǫ 2 ǫ 33 ǫ 33 6 7 33 6 7 − ξ δ 1 , 3 cǫ 0 δ 11 cǫ 0 δ 12 0 6 7 ǫ 33 δM = 6 7 6 7 0 0 0 0 4 5 − ξ δ 23 cǫ 0 δ 12 cǫ 0 δ 22 0 ǫ 33 Advances in the determination of liquid crystal director profiles – p. 9/13

Orthorhombic example, what can we find? Look at coeff of δ ij in U − 1 δMU . We will be able to obtain Fourier data from off-diagonal terms. 2 3 − 1 1 0 0 6 7 6 7 − 1 1 0 0 q 3 6 7 coefficient of δ 11 = 6 7 6 7 2 cǫ 0 ǫ 11 0 0 0 0 4 5 0 0 0 0 2 3 − 1 1 0 0 q 3 q 3 6 7 6 7 − 1 1 0 0 1 6 7 q 3 q 3 coefficient of δ 12 = 6 7 6 7 − q 1 q 1 2 cǫ 0 0 0 4 5 2 ǫ 11 2 ǫ 11 − q 1 q 1 0 0 2 ǫ 11 2 ǫ 11 Advances in the determination of liquid crystal director profiles – p. 10/13

Orthorhombic example, what can we find? cont 2 3 1 0 0 0 6 7 6 7 0 1 0 0 ξ 6 7 coefficient of δ 13 = − 6 7 6 7 cǫ 0 ǫ 33 0 0 0 0 4 5 0 0 0 0 2 3 0 0 0 0 6 7 coefficient of δ 22 = cµ 0 6 7 0 0 − 1 1 4 5 2 q 3 0 0 − 1 1 Advances in the determination of liquid crystal director profiles – p. 11/13

Orthorhombic example, what can we find? cont 2 3 ξ √ ǫ 11 ξ √ ǫ 11 0 0 − 2 √ ǫ 33 q 1 q 3 2 √ ǫ 33 q 1 q 3 6 7 ξ √ ǫ 11 ξ √ ǫ 11 6 7 0 0 − 6 7 2 √ ǫ 33 q 1 q 3 2 √ ǫ 33 q 1 q 3 coefficient of δ 23 = 6 7 6 7 ξ ξ − − 0 0 4 5 2 ǫ 33 2 ǫ 33 ξ ξ − − 0 0 2 ǫ 33 2 ǫ 33 2 3 − 1 − 1 0 0 6 7 6 7 1 1 0 0 ξǫ 11 6 7 coefficient of δ 33 = 6 7 6 7 2 ǫ 33 q 1 0 0 0 0 4 5 0 0 0 0 We see that for all except δ 13 there are non-trivial off diagonal components so that for data using incident light in this plane for an interval of incident angles the Fourier transforms of all other δ ij are determined for some interval of frequencies, hence by analytic continuation the perturbation is determined uniquely everywhere. Although of course their recovery in the presence of noise will be unstable. Note now that by rotating the plane of incidence through a right angle, the x and y axis swap roles and we can with the data for these two planes recover all the perturbed coefficients. Advances in the determination of liquid crystal director profiles – p. 12/13

Conclusions The linearization or Jacobian can be calculated by solving an ODE. • Boundary data for a range of incident angles contains information about a depth dependent • permittivity tensor. The illposedness is similar to analytic continuation so is severe, but the problem is better • conditioned when the range of incident angles produces a wider separation of eigenvalues of the Berreman matrix. Sufficiency of data shown for perturbation to orthorhomic, where data is full optical transfer • matrix. Gives some insight in to real problem. Advances in the determination of liquid crystal director profiles – p. 13/13