charge-free solutions 16 were found to be an excellent ap- proximation for ideal dielectric materials. 4 The condition ∇ · P = 0 implies that the polarization vector variation is restricted to a plane perpendicular to s (the trajectory is constrained to P$_{s}$ = P$_{s}$ ( ±∞ ) plane). In fact, the polarization trajectories representing the T180 and T90 walls calculated under ∇· P = 0 constraint were found to be the straight lines connecting the bound- ary values. 1,2,4,16 This greatly simplifies the algebra and the solution of the variational problem can be found an- alytically. Therefore, we decided to impose the condition of a direct, straight polarization trajectory for the varia- tional problem of all domain wall species of BaTiO$_{3}$. This condition, further referred as straight polarization path (SPP) approximation, implies that both P$_{s}$ and P$_{t}$ po- larization components are constant across the wall. We shall come back to the meaning and possible drawbacks of this approximation in Section VI. Polarization and strain in the mechanically compat- ible and electrically neutral SPP walls can be cast in the form: P = [ P$_{r}$ ( s ) , P$_{s}$ ( ±∞ ) , P$_{t}$ ( ±∞ )] and e = [ e$_{rr}$ ( ±∞ ) , e$_{ss}$ ( s ) , e$_{tt}$ ( ±∞ ) , 2 e$_{st}$ ( s ) , 2 e$_{rt}$ ( ) , 2 e$_{rs}$ ( s )], re- spectively. The s -dependent strain components are cal- culated from the mechanical equilibrium condition Boundary conditions for stress and the fact that all quan- tities vary only along direction s imply for ij ∈ { ss , st , rs } with the integration constant C to be determined from boundary conditions. The Euler-Lagrange equation for polarization reduces to where the strain components from Eqn. (14) were substi- tuted into f . Thus, the elastic field was eliminated. It turns out that the resulting Euler-Lagrange equation for P$_{r}$ is possible to rewrite in the form where p ( s ) stands for P$_{r}$ ( s ), and where the coefficients g, a$_{1}$, a$_{11}$ and a$_{111}$ , different for each domain type, depend only on the material tensors. The boundary values are − P$_{r}$ ( ∞ ) = P$_{r}$ ( −∞ ) = p$_{∞}$ . Solution of the Euler-Lagrange equation (16) is well known. 1,16–19 We shall follow the procedure of Ref. 20. Integrating Eqn. (16) one can obtain the equation: with U being the energy barrier between the domain states: The domain wall thickness (Fig. 3) is defined as Quantity A determines deviation of the profile (20) from the tanh profile, which occurs for the 4 $^{th}$-order potential (i.e., a$_{111}$ = 0 , A = 1). where and The differential Eqn. (16) has the analytical solution The function f$_{EL}$ ( p ) is a double-well ”Euler-Lagrange” potential with two minima ± p$_{∞}$ , where where (see Ref. 4) FIG. 3: Profile of the ”reversed” polarization component P$_{r}$ (Eqn. 20) for a domain wall with the shape factor A = 1 (solid) and A = 3 (dashed). Both profiles have the same derivative for s/ 2 ξ = 0 (dotted) and therefore also the same thickness (indicated by vertical lines) according to the defini- tion in Eqn. (23). 5 3 ∑ j =1 ∂σ$_{ij}$ ∂x$_{j}$ = 3 ∑ j =1 ∂ ∂x$_{j}$ ( ∂f ∂e$_{ij}$ ) = 0 . (13) f$_{EL}$ ( p ) = a$_{1}$p ( s ) 2 + a$_{11}$p ( s ) 4 + a$_{111}$p ( s ) 6 . (18) p 2 ∞ = − a$_{11}$ + √ a 2 11 − 3 a$_{111}$a$_{1}$ 3 a$_{111}$ . (19) ∂f$_{Cq}$ ∂e$_{ij}$ + C = 0 (14) p ( s ) = p$_{∞}$ sinh( s/ξ $^{′}$) √ A + sinh $^{2}$( s/ξ $^{′}$) , (20) A = 3 a$_{111}$p 2 ∞ + a$_{11}$ 2 a$_{111}$p 2 ∞ + a$_{11}$ . (21) ξ ′ = ξ √$_{A}$ . (22) ∂ ∂s ∂f ∂P$_{r}$$_{,}$$_{s}$ − ∂f ∂P$_{r}$ = 0 . (15) g d $^{2}$p ( s ) ds 2 = 2 a$_{1}$p ( s ) + 4 a$_{11}$p $^{3}$( s ) + 6 a$_{111}$p $^{5}$( s ) , (16) 2 ξ = p$_{∞}$ √ 2 g U . (23) U = f$_{EL}$ (0) − f$_{EL}$ ( p$_{∞}$ ) = 2 a$_{111}$p 6 ∞ + a$_{11}$p 4 ∞ . (24) g 2 ( ∂p ∂s ) 2 = f$_{EL}$ ( p ) (17)