The compatibility conditions in linear elasticity are obtained by observing that there are six strain-displacement relations that are functions of only three unknown displacements. This suggests that the three displacements may be removed from the system of equations without loss of information. The resulting expressions in terms of only the strains provide constraints on the possible forms of a strain field.

2-dimensions edit

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 edit

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

For solids in which the deformations are not required to be small, the compatibility conditions take the form

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

where ${\displaystyle {\boldsymbol {F}}}$  is the deformation gradient. In terms of components with respect to a Cartesian coordinate system we can write these compatibility relations as

${\displaystyle e_{ABC}~{\cfrac {\partial F_{iB}}{\partial X_{A}}}=0}$ 

This condition is necessary if the deformation is to be continuous and derived from the mapping ${\displaystyle \mathbf {x} ={\boldsymbol {\chi }}(\mathbf {X} ,t)}$  (see Finite strain theory). The same condition is also sufficient to ensure compatibility in a simply connected body.

Compatibility condition for the right Cauchy-Green deformation tensor edit

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.

The problem of compatibility in continuum mechanics involves the determination of allowable single-valued continuous fields on simply connected bodies. More precisely, the problem may be stated in the following manner.^[5]

Consider the deformation of a body shown in Figure 1. If we express all vectors in terms of the reference coordinate system ${\displaystyle \{(\mathbf {E} _{1},\mathbf {E} _{2},\mathbf {E} _{3}),O\}}$ , the displacement of a point in the body is given by

${\displaystyle \mathbf {u} =\mathbf {x} -\mathbf {X} ~;~~u_{i}=x_{i}-X_{i}}$ 

Also

${\displaystyle {\boldsymbol {\nabla }}\mathbf {u} ={\frac {\partial \mathbf {u} }{\partial \mathbf {X} }}~;~~{\boldsymbol {\nabla }}\mathbf {x} ={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}}$ 

What conditions on a given second-order tensor field ${\displaystyle {\boldsymbol {A}}(\mathbf {X} )}$  on a body are necessary and sufficient so that there exists a unique vector field ${\displaystyle \mathbf {v} (\mathbf {X} )}$  that satisfies

${\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {A}}\quad \equiv \quad v_{i,j}=A_{ij}}$ 

Necessary conditions edit

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 edit

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.

The compatibility condition for the deformation gradient is obtained directly from the above proof by observing that

${\displaystyle {\boldsymbol {F}}={\cfrac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}\mathbf {x} }$ 

Then the necessary and sufficient conditions for the existence of a compatible ${\displaystyle {\boldsymbol {F}}}$  field over a simply connected body are

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

The compatibility problem for small strains can be stated as follows.

Given a symmetric second order tensor field ${\displaystyle {\boldsymbol {\epsilon }}}$  when is it possible to construct a vector field ${\displaystyle \mathbf {u} }$  such that

${\displaystyle {\boldsymbol {\epsilon }}={\frac {1}{2}}[{\boldsymbol {\nabla }}\mathbf {u} +({\boldsymbol {\nabla }}\mathbf {u} )^{T}]}$ 

Necessary conditions edit

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 edit

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.

The compatibility problem for the Right Cauchy-Green deformation field can be posed as follows.

Problem: Let ${\displaystyle {\boldsymbol {C}}(\mathbf {X} )}$  be a positive definite symmetric tensor field defined on the reference configuration. Under what conditions on ${\displaystyle {\boldsymbol {C}}}$  does there exist a deformed configuration marked by the position field ${\displaystyle \mathbf {x} (\mathbf {X} )}$  such that

${\displaystyle (1)\quad \left({\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}\right)^{T}\left({\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}\right)={\boldsymbol {C}}}$ 

Necessary conditions edit

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 edit

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 ${\displaystyle {\boldsymbol {C}}}$ .

• Saint-Venant's compatibility condition
• Linear elasticity
• Deformation (mechanics)
• Infinitesimal strain theory
• Finite strain theory
• Tensor derivative (continuum mechanics)
• Curvilinear coordinates

• Amit Acharya's notes on compatibility on iMechanica
• Plasticity by J. Lubliner, sec. 1.2.4 p. 35