1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '15 at 17:53
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ – KevMoriarty Dec 18 '15 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 '18 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.