Twist solution

The objective function of this optimization problem is the evaluation of the twist solution, being given by Eq. (10.19) in conjunction with Eqs. (10.17) and (10.18), at the free end of the beam. For the constitutive coefficients appearing in the respective ratios, the relative layer thickness h of Eq. (10.27) is required besides the material properties and the particular actuation scheme configuration from Table 10.2. After introduction of the geometry constraints, specified in Eqs. (10.22) and (10.24), the tip twist may be found for any combination of the variables oi, 02, and u ... 

Experimental observations of nematic Uquid crystals frequently report uniform alignment of the director, corresponding to the above twist solution with r = 0, resulting in a constant solution of the form in (i) above. This type of alignment occurs in samples between two parallel plates when the alignment of the director on both boimdary surfaces is identical. This set-up can be taken one stage further for example, if the director is fixed parallel to both plates and one plate were then to be rotated about its normal, it is anticipated that a twist orientation similar to the above twist solution with r 0 would be induced. This will be discussed further in the context of the twisted nematic device in Section 3.7. 

The energy Wp increases as (f>o increases. This observation led LesUe [167] to look at the possibility of what are called non-planar twist solutions. These solutions not only twist, as above, but also tilt out of the x -plane, despite the boundary tilt angle 0 being set to zero on both bormdaries. It will be of interest to compare the resulting energy Wnp for the non-planar twist solution with Wp for the planar twist solution. The boundary conditions (3.33) and (3.34) will be retained for this problem and solutions will be sought which satisfy the symmetries... 

A comparison will now be made between the energy Wp in equation (3.36) for the planar twist solution (3.35) and the energy for the non-planar twist solution provided by equations (3.50) and (3.51), the solution with the lesser energy being interpreted as the physically preferred solution. The total energy Wnp per unit area of the plates is obtained by integrating the energy w in (3.16) from z = 0 to z = d ... 

Hence the energy difference criterion (3.57) holds when Ki > 0 and therefore we have proved that when K2 = Kz and K2 > Ki > 0 then the non-planar twist solution is preferred to the planar twist solution. 

In this special case it follows that if the overall twist 24>o equals tt then the non-planar twist solutions for 0 and (j> in equations (3.50) and (3.51) edways exist for all 0 < < f. Surprisingly, as pointed out by Leslie [167], in this case... 

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