The problem is discribed firstly, and a possoble strategy which might work (yet don't know how exactly) is suggested.
How to gain an analytical solution? Suggestion on numerical method is equally welcomed.
Thank you SO much ~
1. Problem Discription
This is a nonhomogenous BVP. The domain is rectangular on xoz plane with vertexes of $(0,-h/2)$ and $(L,h/2)$. The dependent functions are components of displacement $u(x,z),w(x,z)$.
Two coupled PDEs are derived from 2D stress problem of transversely isotropic materials, i.e. $$ (c_{11}\partial_{xx}+c_{44}\partial_{zz})u+(c_{13}+c_{44})\partial_{xz}w=0 \tag{1} $$ $$ (c_{33}\partial_{zz}+c_{44}\partial_{xx})w+(c_{13}+c_{44})\partial_{xz}u=0 \tag{2} $$ where, the only known relationship among the coefficients is $c_{13}=c_{11}\nu_t=c_{33}\nu_p$.
To prescribe the bcs, let's give the derivation of the components of stress, i.e., $$ \sigma_x=c_{11}\partial_xu+c_{13}\partial_zw $$ $$ \sigma_z=c_{33}\partial_zw+c_{13}\partial_xu $$ $$ \tau_{zx}=c_{44}(\partial_xw+\partial_zu) $$ Then, the bcs are given as $$ u(0,z)=u(L,z)=w(0,z)=w(L,z)=0 $$ $$ \sigma_z(x,-h/2)=-q $$ $$ \tau_{zx}(x,-h/2)=\sigma_z(x,h/2)=\tau_{zx}(x,h/2)=0 $$
2. Possible Strategy
Following the strategy for isotropic materials with more constraints among the coefficients, another function $\Phi(x,z)$ is introduced to represent $u$ and $w$ as follows, $$ u=\partial_{xz}\Phi $$ $$ w=-\frac{c_{11}}{c_{13}+c_{44}}\partial_{xx}\Phi-\frac{c_{44}}{c_{13}+c_{44}}\partial_{zz}\Phi $$
Then, eq.(1) is automatically satified, and eq.(2) becomes $$ c_{11}c_{44}\partial_{xxxx}\Phi+(c_{11}c_{33}-c_{13}(c_{13}+2c_{44}))\partial_{xxzz}\Phi+c_{44}c_{33}\partial_{zzzz}\Phi=0 $$
It would be a Biharmonic equation for isotropic materials, which is not true for this case considered transversely isotropic materials.