Linearisation derivation

Start with

$$\begin{array}{rl}\dot{X}& ={\mathcal{D}}_{u}N\\ {\mathcal{D}}_t^{\ast }S& ={\mathcal{D}}_u^{\ast }Q-\Gamma N\end{array}(1)$$

Set ${\mathcal{D}}_t^{\ast }S=0$ so that $N={\Gamma }^{-1}{\mathcal{D}}_u^{\ast }Q$. Substituting, you get

$$\dot{X}=F(X,X^{\prime },X^{\prime \prime })={\mathcal{D}}_u({\Gamma }^{-1}{\mathcal{D}}_u^{\ast }Q)$$

Now $X_f:[0,L]\to \mathfrak{g}$ satsifies $F(X_f,X_f^{\prime },X_f^{\prime \prime })=0$ for all $u$. That is, $X_f$ is a fixed point of $F$. Recall that the stress itself is of the form $Q=\mathrm{K}(X-X_0)$. Where $X_0=(100000{)}^T$.

We write $Q$ as

$$Q=\mathrm{K}(\delta X+X_f-X_0)$$

where $\delta X=X-X_f$. In general $\mathrm{K}$ would be $u$-dependent, but we will assume it is not. If we consider small perturbations $|\delta X|\ll 1$, then any term of order $\delta X_i\delta X_j$ can be discarded.

Now recall that ${\mathcal{D}}_u={\partial }_u+{\text{ad}}_X$. So we have that

$${\mathcal{D}}_u={\partial }_u+{\text{ad}}_{\delta X}+{\text{ad}}_{X_f}$$

Similarly, we have that

$${\mathcal{D}}_u^{\ast }={\partial }_u+{\text{ad}}_{\delta X}^{\ast }+{\text{ad}}_{X_f}^{\ast }$$

Note: Actually, I’m not sure if ${\text{ad}}_{A+B}^{\ast }={\text{ad}}_A^{\ast }+{\text{ad}}_B^{\ast }$. Can you check this? Perhaps your index formula for the dual adjoint will help to do this.

Inserting all of this, we get

$$\delta \dot{X}=({\partial }_u+{\text{ad}}_{\delta X}+{\text{ad}}_{X_f}){\Gamma }^{-1}({\partial }_u+{\text{ad}}_{\delta X}^{\ast }+{\text{ad}}_{X_f}^{\ast })\mathrm{K}\delta X$$

We can clearly see that no terms involving ${\text{ad}}_{\delta X}$ can survive, since ${\text{ad}}_{\delta X}\delta X\approx 0$, so we have

$$\delta \dot{X}=({\partial }_u+{\text{ad}}_{X_f}){\Gamma }^{-1}({\partial }_u+{\text{ad}}_{X_f}^{\ast })\mathrm{K}\delta X$$

After having computed all terms, you’ll get something that looks like

$$\begin{array}{rl}\delta \dot{X}(u,t)& =C(u)+A_0(u)\delta X(u,t)+A_1(u){\partial }_u\delta X(u,t)+A_2(u){\partial }_u^2\delta X(u,t)\end{array}$$

However, the $C(u)$ term should vanish, if we impose that $F(X_f,X_f^{\prime },X_f^{\prime \prime })=0$.

We can then neatly write the equation as

$$\delta \dot{X}={\mathcal{A}}^{(g)}\delta X(2)$$

where ${\mathcal{A}}^{(g)}$ is a $u$-dependent 2nd-order differential operator

$${\mathcal{A}}^{(g)}=A_0(u)+A_1(u){\partial }_u+A_2(u){\partial }_u^2$$

and where the superscript signifies the dependence on the amplitude of the follower force.

The key thing now is to find the spectrum

$${\mathcal{A}}^{(g)}{\psi }_k^{(g)}(u)={\lambda }_k^{(g)}{\psi }_k^{(g)}(u)(3)$$

Hopefully, since (3) is linear, this should be fairly straightforward. You will probably have to do some massaging though.

To simplify the problem, you may for now assume that $\mathrm{K}$ is diagonal. This should simplify equation (2) and (3) quite a lot.

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In a previous note, we found that

$$X_f^{(g)}(u)=({\theta }_1^{(g)}(u)00000{)}^T(4)$$

where ${\partial }_u^2{\theta }_1^{(g)}=0$, with boundary conditions ${\theta }_1^{(g)}(0)=g/{\mathrm{K}}_{11}+1$ and ${\theta }_1^{(g)}(L)=1$.

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If it is more convenient, you may also repeat the above derivation but explicitly for the translational and rotational parts of the Cosserat rod. So start with

$$\begin{array}{rl}{D}_t\mathbf{\theta }& =D_u\mathbf{V}\\ \dot{\mathbf{\pi }}& =D_u\mathbf{\Omega }\\ D_t\mathbf{P}& =D_u\mathbf{F}-\gamma \mathbf{V}\\ D_t\mathbf{L}& =D_u\mathbf{M}+\mathbf{\theta }\times \mathbf{F}-{\gamma }^R\mathbf{\Omega }\end{array}$$

Following the same steps as above, setting $D_t\mathbf{P}=0$ and $D_t\mathbf{L}=0$, should leave you with a set of equations analogous to (2) but for the translational and rotational parts separately.

Note, we have

$$\begin{array}{rl}{\mathbf{\theta }}_f& =({\theta }_1^{(g)}00{)}^T\\ {\mathbf{\pi }}_f& =(000{)}^T\end{array}$$

and $\delta \mathbf{\theta }=\mathbf{\theta }-{\mathbf{\theta }}_f$ and $\delta \mathbf{\pi }=\mathbf{\pi }$.

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