10 The case of a single mode
In this chapter, we discuss the case \(m = 1\); all tensors considered previously (\(\dot{E}_{ij}\), \(E_{ijk}\), \(E_{ijkl}\), etc) then reduce to simple scalars. To avoid ambiguity, indices are kept: \(\dot{E}_{11}\), \(E_{111}\), \(E_{1111}\). Since \(\dot{\E}_2\) is negative definite over \(V\) (see note in Section 2.2), we have \(\dot{E}_{11} < 0\).
It is first observed that the following conditions are necessary to ensure stability of the critical point \[ E_{111} = 0 \quad \text{and} \quad E_{1111} \geq 0, \] which shows that asymmetric critical points are always unstable.
10.1 Asymmetric bifurcation
We first consider the case \(E_{111} \neq 0\). Owing to the discussion above, the bifurcation point is unstable. It follows from Eq. (8.1) that \[ \order[1]{\lambda} = - \frac{E_{111} \, \order[1]{\xi}_1}{\dot{E}_{11}}. \]
Then, Eq. (8.2) delivers, with \(\order[1]{u} = \order[1]{\xi}_1 \, v_1\) \[ \bigl( \order[1]{\xi}_1 \bigr)^2 \E_{, uu}[u(\lambda), \lambda; v_1, v_1] = -\bigl( \lambda - \lambda_0 \bigr) \, \bigl( \order[1]{\xi}_1 \bigr)^2 \, \dot{\E}_2(v_1, v_1) + o(\lambda - \lambda_0) \] or \[ \E_{, uu}[u(\lambda), \lambda; v_1, v_1] = -\bigl( \lambda - \lambda_0 \bigr) \, \dot{E}_{11} + o(\lambda - \lambda_0). \]
Since \(\dot{E}_{11} < 0\) we conclude that asymmetric bifurcations branches are unstable for \(\lambda \leq \lambda_0\) and stable for \(\lambda > \lambda_0\) (stability switch).
10.2 Symmetric bifurcation
We now consider the case \(E_{111}=0\). From the general discussion of Chapter 3, the critical point is stable if \(E_{1111} > 0\) and unstable if \(E_{1111} < 0\).
Is this really true? Even for \(m = 1\) the proof that conditions (3.4) are sufficient to ensure stability does not seem so obvious to me.
The bifurcation equation (9.1) reduces to \[ \tfrac{1}{3}E_{1111} \, \bigl( \order[1]{\xi_1} \bigr)^2 + \order[2]\lambda \dot{E}_{11} = 0, \] which in particular shows that \(\order[2]{\lambda}\) has the same sign as \(E_{1111}\). Since the expansion of \(\lambda\) reads: \(\lambda = \lambda_0 + \order[2]\lambda \eta^2 / 2 + o(\eta^2)\), the bifurcation branch exists only for loads above the critical load (\(\lambda \geq \lambda_0\)) if \(E_{1111} > 0\) and only for loads below the critical load (\(\lambda \leq \lambda_0\)) if \(E_{1111} < 0\).
From Eq. (6.2), the hessian of the energy along the bifurcated branch reads \[ \begin{aligned}[b] \E_{, uu}[u(\eta), \lambda(\eta); v_1, v_1] ={}& \tfrac{1}{2} \eta^2 \, \bigl[ \bigl( \order[1]{\xi}_1 \bigr) ^2 \, E_{1111} + \order[2]{\lambda} \, \dot{E}_{11} \bigr] + o(\eta^2)\\ ={}& - \eta^2 \, \order[2]{\lambda} \, \dot{E}_{11} + o(\eta^2), \end{aligned} \] which has the sign of \(\order[2]{\lambda}\). Therefore the Hessian is positive (resp. negative) definite if \(E_{1111} > 0\) (resp \(< 0\)).
To sum up, if \(E_{1111} > 0\), then the bifurcation branch (including the critical point) is stable and exists only for loads greater than the critical load. Conversely, if \(E_{1111} < 0\), then the bifurcation branch (including the critical point) is unstable and exists only for loads lower than the critical load.