A static problem in linear elasticity is typically written as the following boundary value problem:

find $\boldsymbol u$ and $\boldsymbol \sigma$ such that:

$\text{div} \boldsymbol \sigma + \boldsymbol f = \boldsymbol 0$ in $\Omega$,

$\boldsymbol \sigma^T = \boldsymbol \sigma$ in $\Omega$,

$\boldsymbol \sigma = 2\mu \boldsymbol \epsilon + \lambda \text{tr}\boldsymbol \epsilon \, \boldsymbol I$ in $\Omega$,

$\boldsymbol \epsilon = \frac{1}{2}( \nabla \boldsymbol u + \nabla^T \boldsymbol u )$ in $\Omega$,

$\boldsymbol \sigma \cdot \boldsymbol n = \boldsymbol T^d$ on $\partial\Omega_T$,

$\boldsymbol u = \boldsymbol u^d$ on $\partial\Omega_u$.

And it can be proved that the solution is unique both on displacement field and stress field.

I wonder if we have unicity for the boundary value problem:

find $\boldsymbol \sigma$ such that:

$\text{div} \boldsymbol \sigma + \boldsymbol f = \boldsymbol 0$ in $\Omega$,

$\boldsymbol \sigma^T = \boldsymbol \sigma$ in $\Omega$,

$\boldsymbol \sigma \cdot \boldsymbol n = \boldsymbol T^d$ on $\partial\Omega$.

We have three equations, three boundary conditions and three independent component fields (in case the coordinate system is chosen so that basis vectors correspond to eigen vectors at each point of $\Omega$). I am aware about the indertermination of the displacement field due to a rigid body motion.

Do you have knowledge, references that treat this, I would say intermediate, problem?

Thank you in advance for sharing it.

  • $\begingroup$ I might be misunderstanding something, but I'm pretty sure the answer is no. Consider the buckling of a rod under load. The flat rod is always a solution but there are also buckled solutions if the applied stress is sufficiently large. $\endgroup$
    – Bort
    Dec 17, 2015 at 17:53

1 Answer 1


The aswer is negative. You may, for instance, construct a vector field $v$ whose integral lines are circles centered on the $z$ axis and with $\mbox{div}\: {\bf v}=0$. (Think of an incompressible fluid rotating around the $z$ axis). You may confine this field to a torus $\Omega$. The tensor field $\sigma = {\bf v}\otimes {\bf v}$ satisfies all your requirements with $f=0$, ${\bf T}^d=0$. But also ${\bf v}=0$ does.

This result implies that, on that $\Omega$ the problem with generic $f$ and ${\bf T}^d$ does not admit unique solution. Indeed, if $\sigma$ is a solution with given $f$ and ${\bf T}^d$, $\sigma + a{\bf v}\otimes {\bf v}$ satisfies the same problem for every $a \in \mathbb R$ and ${\bf v}$ defined as above.

  • $\begingroup$ In the b.v.p. we can suppose the existence of two solutions and see that the difference leads to the same b.v.p. but with $\boldsymbol f = \boldsymbol 0$ and $\boldsymbol T^d = \boldsymbol 0$, then your counterexample gives the non unicity for this specific b.v.p. and finally for all others. $\endgroup$ Dec 18, 2015 at 14:19
  • $\begingroup$ This does not constitute a solution, not in the usual sense of linear elasticity at least. For a solution, stress needs to derive from a gradient, i.e., strains need to be compatible. In general, in terms of displacement, the given bvp has a unique solution up to a rigid body motion. In terms of stress, the solution is indeed unique. In nonlinear elasticity, say with possible buckling, this no longer holds and all hell breaks loose. $\endgroup$
    – Hussein
    Jun 23, 2018 at 23:25

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