2-dimensions

For two-dimensional, plane strain problems the strain-displacement relations are

${\displaystyle \varepsilon _{11}={\cfrac {\partial u_{1}}{\partial x_{1}}}~;~~\varepsilon _{12}={\cfrac {1}{2}}\left[{\cfrac {\partial u_{1}}{\partial x_{2}}}+{\cfrac {\partial u_{2}}{\partial x_{1}}}\right]~;~~\varepsilon _{22}={\cfrac {\partial u_{2}}{\partial x_{2}}}}$

Repeated differentiation of these relations, in order to remove the displacements ${\displaystyle u_{1}}$ and ${\displaystyle u_{2}}$, gives us the two-dimensional compatibility condition for strains

${\displaystyle {\cfrac {\partial ^{2}\varepsilon _{11}}{\partial x_{2}^{2}}}-2{\cfrac {\partial ^{2}\varepsilon _{12}}{\partial x_{1}\partial x_{2}}}+{\cfrac {\partial ^{2}\varepsilon _{22}}{\partial x_{1}^{2}}}=0}$

The only displacement field that is allowed by a compatible plane strain field is a plane displacement field, i.e., ${\displaystyle \mathbf {u} =\mathbf {u} (x_{1},x_{2})}$.

3-dimensions

In three dimensions, in addition to two more equations of the form seen for two dimensions, there are three more equations of the form

${\displaystyle {\cfrac {\partial ^{2}\varepsilon _{33}}{\partial x_{1}\partial x_{2}}}={\cfrac {\partial }{\partial x_{3}}}\left[{\cfrac {\partial \varepsilon _{23}}{\partial x_{1}}}+{\cfrac {\partial \varepsilon _{31}}{\partial x_{2}}}-{\cfrac {\partial \varepsilon _{12}}{\partial x_{3}}}\right]}$

Therefore, there are 3⁴=81 partial differential equations, however due to symmetry conditions, this number reduces to six different compatibility conditions. We can write these conditions in index notation as^[4]

${\displaystyle e_{ikr}~e_{jls}~\varepsilon _{ij,kl}=0}$

where ${\displaystyle e_{ijk}}$ is the permutation symbol. In direct tensor notation

${\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }})^{T}={\boldsymbol {0}}}$

where the curl operator can be expressed in an orthonormal coordinate system as ${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }}=e_{ijk}\varepsilon _{rj,i}\mathbf {e} _{k}\otimes \mathbf {e} _{r}}$.

The second-order tensor

${\displaystyle {\boldsymbol {R}}:={\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\times {\boldsymbol {\varepsilon }})^{T}~;~~R_{rs}:=e_{ikr}~e_{jls}~\varepsilon _{ij,kl}}$

is known as the incompatibility tensor, and is equivalent to the Saint-Venant compatibility tensor

Compatibility condition for the right Cauchy-Green deformation tensor

The compatibility condition for the right Cauchy-Green deformation tensor can be expressed as

${\displaystyle R_{\alpha \beta \rho }^{\gamma }:={\frac {\partial }{\partial X^{\rho }}}[\Gamma _{\alpha \beta }^{\gamma }]-{\frac {\partial }{\partial X^{\beta }}}[\Gamma _{\alpha \rho }^{\gamma }]+\Gamma _{\mu \rho }^{\gamma }~\Gamma _{\alpha \beta }^{\mu }-\Gamma _{\mu \beta }^{\gamma }~\Gamma _{\alpha \rho }^{\mu }=0}$

where ${\displaystyle \Gamma _{ij}^{k}}$ is the Christoffel symbol of the second kind. The quantity ${\displaystyle R_{ijk}^{m}}$ represents the mixed components of the Riemann-Christoffel curvature tensor.

Necessary conditions

For the necessary conditions we assume that the field ${\displaystyle \mathbf {v} }$ exists and satisfies ${\displaystyle v_{i,j}=A_{ij}}$. Then

${\displaystyle v_{i,jk}=A_{ij,k}~;~~v_{i,kj}=A_{ik,j}}$

Since changing the order of differentiation does not affect the result we have

${\displaystyle v_{i,jk}=v_{i,kj}}$

Hence

${\displaystyle A_{ij,k}=A_{ik,j}}$

From the well known identity for the curl of a tensor we get the necessary condition

${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {A}}={\boldsymbol {0}}}$

Sufficient conditions

Figure 2. Integration paths used in proving the sufficiency conditions for compatibility.

To prove that this condition is sufficient to guarantee existence of a compatible second-order tensor field, we start with the assumption that a field ${\displaystyle {\boldsymbol {A}}}$ exists such that ${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {A}}={\boldsymbol {0}}}$. We will integrate this field to find the vector field ${\displaystyle \mathbf {v} }$ along a line between points ${\displaystyle A}$ and ${\displaystyle B}$ (see Figure 2), i.e.,

${\displaystyle \mathbf {v} (\mathbf {X} _{B})-\mathbf {v} (\mathbf {X} _{A})=\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}{\boldsymbol {\nabla }}\mathbf {v} \cdot ~d\mathbf {X} =\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}{\boldsymbol {A}}(\mathbf {X} )\cdot d\mathbf {X} }$

If the vector field ${\displaystyle \mathbf {v} }$ is to be single-valued then the value of the integral should be independent of the path taken to go from ${\displaystyle A}$ to ${\displaystyle B}$.

From Stokes' theorem, the integral of a second order tensor along a closed path is given by

${\displaystyle \oint _{\partial \Omega }{\boldsymbol {A}}\cdot d\mathbf {s} =\int _{\Omega }\mathbf {n} \cdot ({\boldsymbol {\nabla }}\times {\boldsymbol {A}})~da}$

Using the assumption that the curl of ${\displaystyle {\boldsymbol {A}}}$ is zero, we get

${\displaystyle \oint _{\partial \Omega }{\boldsymbol {A}}\cdot d\mathbf {s} =0\quad \implies \quad \int _{AB}{\boldsymbol {A}}\cdot d\mathbf {X} +\int _{BA}{\boldsymbol {A}}\cdot d\mathbf {X} =0}$

Hence the integral is path independent and the compatibility condition is sufficient to ensure a unique ${\displaystyle \mathbf {v} }$ field, provided that the body is simply connected.

Necessary conditions

Suppose that there exists ${\displaystyle \mathbf {u} }$ such that the expression for ${\displaystyle {\boldsymbol {\epsilon }}}$ holds. Now

${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} ={\boldsymbol {\epsilon }}+{\boldsymbol {\omega }}}$

where

${\displaystyle {\boldsymbol {\omega }}:={\frac {1}{2}}[{\boldsymbol {\nabla }}\mathbf {u} -({\boldsymbol {\nabla }}\mathbf {u} )^{T}]}$

Therefore, in index notation,

${\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {\omega }}\equiv \omega _{ij,k}={\frac {1}{2}}(u_{i,jk}-u_{j,ik})={\frac {1}{2}}(u_{i,jk}+u_{k,ji}-u_{j,ik}-u_{k,ji})=\varepsilon _{ik,j}-\varepsilon _{jk,i}}$

If ${\displaystyle {\boldsymbol {\omega }}}$ is continuously differentiable we have ${\displaystyle \omega _{ij,kl}=\omega _{ij,lk}}$. Hence,

${\displaystyle \varepsilon _{ik,jl}-\varepsilon _{jk,il}-\varepsilon _{il,jk}+\varepsilon _{jl,ik}=0}$

In direct tensor notation

${\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})^{T}={\boldsymbol {0}}}$

The above are necessary conditions. If ${\displaystyle \mathbf {w} }$ is the infinitesimal rotation vector then ${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }}={\boldsymbol {\nabla }}\mathbf {w} +{\boldsymbol {\nabla }}\mathbf {w} ^{T}}$. Hence the necessary condition may also be written as ${\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\mathbf {w} +{\boldsymbol {\nabla }}\mathbf {w} ^{T})^{T}={\boldsymbol {0}}}$.

Sufficient conditions

Let us now assume that the condition ${\displaystyle {\boldsymbol {\nabla }}\times ({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})^{T}={\boldsymbol {0}}}$ is satisfied in a portion of a body. Is this condition sufficient to guarantee the existence of a continuous, single-valued displacement field ${\displaystyle \mathbf {u} }$?

The first step in the process is to show that this condition implies that the infinitesimal rotation tensor ${\displaystyle {\boldsymbol {\omega }}}$ is uniquely defined. To do that we integrate ${\displaystyle {\boldsymbol {\nabla }}\mathbf {w} }$ along the path ${\displaystyle \mathbf {X} _{A}}$ to ${\displaystyle \mathbf {X} _{B}}$, i.e.,

${\displaystyle \mathbf {w} (\mathbf {X} _{B})-\mathbf {w} (\mathbf {X} _{A})=\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}{\boldsymbol {\nabla }}\mathbf {w} \cdot d\mathbf {X} =\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})\cdot d\mathbf {X} }$

Note that we need to know a reference ${\displaystyle \mathbf {w} (\mathbf {X} _{A})}$ to fix the rigid body rotation. The field ${\displaystyle \mathbf {w} (\mathbf {X} )}$ is uniquely determined only if the contour integral along a closed contour between ${\displaystyle \mathbf {X} _{A}}$ and ${\displaystyle \mathbf {X} _{b}}$ is zero, i.e.,

${\displaystyle \oint _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})\cdot d\mathbf {X} ={\boldsymbol {0}}}$

But from Stokes' theorem for a simply-connected body and the necessary condition for compatibility

${\displaystyle \oint _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})\cdot d\mathbf {X} =\int _{\Omega _{AB}}\mathbf {n} \cdot ({\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }})~da={\boldsymbol {0}}}$

Therefore, the field ${\displaystyle \mathbf {w} }$ is uniquely defined which implies that the infinitesimal rotation tensor ${\displaystyle {\boldsymbol {\omega }}}$ is also uniquely defined, provided the body is simply connected.

In the next step of the process we will consider the uniqueness of the displacement field ${\displaystyle \mathbf {u} }$. As before we integrate the displacement gradient

${\displaystyle \mathbf {u} (\mathbf {X} _{B})-\mathbf {u} (\mathbf {X} _{A})=\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}{\boldsymbol {\nabla }}\mathbf {u} \cdot d\mathbf {X} =\int _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}({\boldsymbol {\epsilon }}+{\boldsymbol {\omega }})\cdot d\mathbf {X} }$

From Stokes' theorem and using the relations ${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }}={\boldsymbol {\nabla }}\mathbf {w} =-{\boldsymbol {\nabla }}\times \omega }$ we have

${\displaystyle \oint _{\mathbf {X} _{A}}^{\mathbf {X} _{B}}({\boldsymbol {\epsilon }}+{\boldsymbol {\omega }})\cdot d\mathbf {X} =\int _{\Omega _{AB}}\mathbf {n} \cdot ({\boldsymbol {\nabla }}\times {\boldsymbol {\epsilon }}+{\boldsymbol {\nabla }}\times {\boldsymbol {\omega }})~da={\boldsymbol {0}}}$

Hence the displacement field ${\displaystyle \mathbf {u} }$ is also determined uniquely. Hence the compatibility conditions are sufficient to guarantee the existence of a unique displacement field ${\displaystyle \mathbf {u} }$ in a simply-connected body.

Necessary conditions

Suppose that a field ${\displaystyle \mathbf {x} (\mathbf {X} )}$ exists that satisfies condition (1). In terms of components with respect to a rectangular Cartesian basis

${\displaystyle {\frac {\partial x^{i}}{\partial X^{\alpha }}}{\frac {\partial x^{i}}{\partial X^{\beta }}}=C_{\alpha \beta }}$

From finite strain theory we know that ${\displaystyle C_{\alpha \beta }=g_{\alpha \beta }}$. Hence we can write

${\displaystyle \delta _{ij}~{\frac {\partial x^{i}}{\partial X^{\alpha }}}~{\frac {\partial x^{j}}{\partial X^{\beta }}}=g_{\alpha \beta }}$

For two symmetric second-order tensor field that are mapped one-to-one we also have the relation

${\displaystyle G_{ij}={\frac {\partial X^{\alpha }}{\partial x^{i}}}~{\frac {\partial X^{\beta }}{\partial x^{j}}}~g_{\alpha \beta }}$

From the relation between of ${\displaystyle G_{ij}}$ and ${\displaystyle g_{\alpha \beta }}$ that ${\displaystyle \delta _{ij}=G_{ij}}$, we have

${\displaystyle _{(x)}\Gamma _{ij}^{k}=0}$

Then, from the relation

${\displaystyle {\frac {\partial ^{2}x^{m}}{\partial X^{\alpha }\partial X^{\beta }}}={\frac {\partial x^{m}}{\partial X^{\mu }}}\,_{(X)}\Gamma _{\alpha \beta }^{\mu }-{\frac {\partial x^{i}}{\partial X^{\alpha }}}~{\frac {\partial x^{j}}{\partial X^{\beta }}}\,_{(x)}\Gamma _{ij}^{m}}$

we have

${\displaystyle {\frac {\partial F_{~\alpha }^{m}}{\partial X^{\beta }}}=F_{~\mu }^{m}\,_{(X)}\Gamma _{\alpha \beta }^{\mu }\qquad$ ;~~F_{~\alpha }^{i}:={\frac {\partial x^{i}}{\partial X^{\alpha }}}}

From finite strain theory we also have

${\displaystyle _{(X)}\Gamma _{\alpha \beta \gamma }={\frac {1}{2}}\left({\frac {\partial g_{\alpha \gamma }}{\partial X^{\beta }}}+{\frac {\partial g_{\beta \gamma }}{\partial X^{\alpha }}}-{\frac {\partial g_{\alpha \beta }}{\partial X^{\gamma }}}\right)~;~~_{(X)}\Gamma _{\alpha \beta }^{\nu }=g^{\nu \gamma }\,_{(X)}\Gamma _{\alpha \beta \gamma }~;~~g_{\alpha \beta }=C_{\alpha \beta }~;~~g^{\alpha \beta }=C^{\alpha \beta }}$

Therefore,

${\displaystyle \,_{(X)}\Gamma _{\alpha \beta }^{\mu }={\cfrac {C^{\mu \gamma }}{2}}\left({\frac {\partial C_{\alpha \gamma }}{\partial X^{\beta }}}+{\frac {\partial C_{\beta \gamma }}{\partial X^{\alpha }}}-{\frac {\partial C_{\alpha \beta }}{\partial X^{\gamma }}}\right)}$

and we have

${\displaystyle {\frac {\partial F_{~\alpha }^{m}}{\partial X^{\beta }}}=F_{~\mu }^{m}~{\cfrac {C^{\mu \gamma }}{2}}\left({\frac {\partial C_{\alpha \gamma }}{\partial X^{\beta }}}+{\frac {\partial C_{\beta \gamma }}{\partial X^{\alpha }}}-{\frac {\partial C_{\alpha \beta }}{\partial X^{\gamma }}}\right)}$

Again, using the commutative nature of the order of differentiation, we have

${\displaystyle {\frac {\partial ^{2}F_{~\alpha }^{m}}{\partial X^{\beta }\partial X^{\rho }}}={\frac {\partial ^{2}F_{~\alpha }^{m}}{\partial X^{\rho }\partial X^{\beta }}}\implies {\frac {\partial F_{~\mu }^{m}}{\partial X^{\rho }}}\,_{(X)}\Gamma _{\alpha \beta }^{\mu }+F_{~\mu }^{m}~{\frac {\partial }{\partial X^{\rho }}}[\,_{(X)}\Gamma _{\alpha \beta }^{\mu }]={\frac {\partial F_{~\mu }^{m}}{\partial X^{\beta }}}\,_{(X)}\Gamma _{\alpha \rho }^{\mu }+F_{~\mu }^{m}~{\frac {\partial }{\partial X^{\beta }}}[\,_{(X)}\Gamma _{\alpha \rho }^{\mu }]}$

or

${\displaystyle F_{~\gamma }^{m}\,_{(X)}\Gamma _{\mu \rho }^{\gamma }\,_{(X)}\Gamma _{\alpha \beta }^{\mu }+F_{~\mu }^{m}~{\frac {\partial }{\partial X^{\rho }}}[\,_{(X)}\Gamma _{\alpha \beta }^{\mu }]=F_{~\gamma }^{m}\,_{(X)}\Gamma _{\mu \beta }^{\gamma }\,_{(X)}\Gamma _{\alpha \rho }^{\mu }+F_{~\mu }^{m}~{\frac {\partial }{\partial X^{\beta }}}[\,_{(X)}\Gamma _{\alpha \rho }^{\mu }]}$

After collecting terms we get

${\displaystyle F_{~\gamma }^{m}\left(\,_{(X)}\Gamma _{\mu \rho }^{\gamma }\,_{(X)}\Gamma _{\alpha \beta }^{\mu }+{\frac {\partial }{\partial X^{\rho }}}[\,_{(X)}\Gamma _{\alpha \beta }^{\gamma }]-\,_{(X)}\Gamma _{\mu \beta }^{\gamma }\,_{(X)}\Gamma _{\alpha \rho }^{\mu }-{\frac {\partial }{\partial X^{\beta }}}[\,_{(X)}\Gamma _{\alpha \rho }^{\gamma }]\right)=0}$

From the definition of ${\displaystyle F_{\gamma }^{m}}$ we observe that it is invertible and hence cannot be zero. Therefore,

${\displaystyle R_{\alpha \beta \rho }^{\gamma }:={\frac {\partial }{\partial X^{\rho }}}[\,_{(X)}\Gamma _{\alpha \beta }^{\gamma }]-{\frac {\partial }{\partial X^{\beta }}}[\,_{(X)}\Gamma _{\alpha \rho }^{\gamma }]+\,_{(X)}\Gamma _{\mu \rho }^{\gamma }\,_{(X)}\Gamma _{\alpha \beta }^{\mu }-\,_{(X)}\Gamma _{\mu \beta }^{\gamma }\,_{(X)}\Gamma _{\alpha \rho }^{\mu }=0}$

We can show these are the mixed components of the Riemann-Christoffel curvature tensor. Therefore, the necessary conditions for ${\displaystyle {\boldsymbol {C}}}$-compatibility are that the Riemann-Christoffel curvature of the deformation is zero.

Sufficient conditions

The proof of sufficiency is a bit more involved.^[5][6] We start with the assumption that

${\displaystyle R_{\alpha \beta \rho }^{\gamma }=0~;~~g_{\alpha \beta }=C_{\alpha \beta }}$

We have to show that there exist ${\displaystyle \mathbf {x} }$ and ${\displaystyle \mathbf {X} }$ such that

${\displaystyle {\frac {\partial x^{i}}{\partial X^{\alpha }}}{\frac {\partial x^{i}}{\partial X^{\beta }}}=C_{\alpha \beta }}$

From a theorem by T.Y.Thomas ^[7] we know that the system of equations

${\displaystyle {\frac {\partial F_{~\alpha }^{i}}{\partial X^{\beta }}}=F_{~\gamma }^{i}~\,_{(X)}\Gamma _{\alpha \beta }^{\gamma }}$

has unique solutions ${\displaystyle F_{~\alpha }^{i}}$ over simply connected domains if

${\displaystyle _{(X)}\Gamma _{\alpha \beta }^{\gamma }=_{(X)}\Gamma _{\beta \alpha }^{\gamma }~;~~R_{\alpha \beta \rho }^{\gamma }=0}$

The first of these is true from the defining of ${\displaystyle \Gamma _{jk}^{i}}$ and the second is assumed. Hence the assumed condition gives us a unique ${\displaystyle F_{~\alpha }^{i}}$ that is ${\displaystyle C^{2}}$ continuous.

Next consider the system of equations

${\displaystyle {\frac {\partial x^{i}}{\partial X^{\alpha }}}=F_{~\alpha }^{i}}$

Since ${\displaystyle F_{~\alpha }^{i}}$ is ${\displaystyle C^{2}}$ and the body is simply connected there exists some solution ${\displaystyle x^{i}(X^{\alpha })}$ to the above equations. We can show that the ${\displaystyle x^{i}}$ also satisfy the property that

${\displaystyle \det \left|{\frac {\partial x^{i}}{\partial X^{\alpha }}}\right|\neq 0}$

We can also show that the relation

${\displaystyle {\frac {\partial x^{i}}{\partial X^{\alpha }}}~g^{\alpha \beta }~{\frac {\partial x^{j}}{\partial X^{\beta }}}=\delta ^{ij}}$

implies that

${\displaystyle g_{\alpha \beta }=C_{\alpha \beta }={\frac {\partial x^{k}}{\partial X^{\alpha }}}~{\frac {\partial x^{k}}{\partial X^{\beta }}}}$

If we associate these quantities with tensor fields we can show that ${\displaystyle {\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}}$ is invertible and the constructed tensor field satisfies the expression for ${\displaystyle {\boldsymbol {C}}}$.