For an arbitrary strain field $\textstyle {\boldsymbol {\varepsilon }}$, the strain-displacement relation $\textstyle {\boldsymbol {\varepsilon }}={\cfrac {1}{2}}({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})$ is the partial differential equation that can be solved to obtain the displacement $\textstyle \mathbf {u}$. The solution to this equation must exist and must be unique.

If two displacement fields $\textstyle \mathbf {u}$ and $\textstyle \mathbf {u} ^{'}$ correspond to the same strain field, then $\textstyle \mathbf {u} =\mathbf {u} ^{'}+\mathbf {w}$, where $\textstyle \mathbf {w}$ is a rigid displacement field.

The strain field $\textstyle {\boldsymbol {\varepsilon }}$ corresponding to a $\textstyle C^{2}$ continuous displacement field satisfies the compatibility equation

The compatibility condition also implies the following relationship between the infinitesimal strain tensor and the axial vector corresponding to the infinitesimal rotation tensor:

Show that the compatibility relation for plane stress is satisfied
by unrestrained thermal expansion ($\varepsilon _{11}=\varepsilon _{22}=\alpha T$, $\varepsilon _{12}=0$), where $\alpha$ is the coefficient of thermal expansion and $T$ is the temperature, provided that the temperature is a two-dimensional harmonic function, i.e.,

Hence deduce that, subject to certain restrictions which you should
explicitly specify, no thermal stresses will be induced in a thin body with a steady-state, two-dimensional temperature distribution and no surface tractions.

The above equation is the steady-state heat conduction equation without any internal sources.

If there are no surface tractions, the state $\sigma _{11}=\sigma _{22}=\sigma _{12}$ satisfies the BCs. Since the steady-state heat conduction equation is also the compatibility equation, compatibility is automatically satisfied by the above stress state. Therefore, no thermal stresses are induced in this situation. However, extra conditions need to be applied if the body is multiply-connected.