The Complementing and Agmon’s Conditions in Finite Elasticity, Three Dimensions

Abstract

We consider the complementing condition and Agmon’s condition for linearized elasticity in three-dimensions. With an elasticity tensor $\mathsf{C}$ derived from a compressible, isotropic stored energy $W$, linearized about a homogeneous deformation $\mathbf{f}_{0}$, we apply the complementing and Agmon’s conditions to a traction portion of the surface of a body with unit normal $\mathbf{n}$. We examine three problems including a general and a neo-Hookean $W$.

Appendix A

We summarize the calculations in Sect. 3.1. As in the paragraph following (3.16), the set $S_{\alpha } = \mathrm{span} \{\mathbf{z} _{\alpha }^{(1)}, \mathbf{z}_{\alpha }^{(2)},\mathbf{z}_{\alpha }^{(3)} \}$ of bounded solutions of (2.12a) is given in three cases analogous to Simpson [21], Theorem 4.3 (or following (4.38) in Sect. 4.2 there). In each case we let ${\mathbf{z}}_{\alpha }^{(1)}=e^{r_{1}t}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$, $r_{1} = - \sqrt{\frac{\beta _{3\alpha }}{\beta _{1}}}$. Then

Case 1.$\kappa _{1} \ne 0$, $r_{2} \ne r_{3}$.

$$ \mathbf{a}^{(i)} = \left [ \textstyle\begin{array}{c} 0 \\ -\kappa _{1} r_{i} \\ i(\beta _{1} r_{i}^{2} - \tau _{1\alpha }) \end{array}\displaystyle \right ], \quad \mathbf{z}^{(i)}(t) = e^{r_{i} t} \mathbf{a}^{(i)}, \ i = 2,3. $$

Case 2.$\kappa _{1} = 0$, $r_{2}^{2} = \frac{\tau _{1\alpha }}{ \beta _{1}}$, $r_{3}^{2} = \frac{\beta _{1\alpha }}{\tau _{3}}$.

$$ \mathbf{a}^{(2)} = \left [ \textstyle\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\displaystyle \right ], \qquad \mathbf{a}^{(3)} = \left [ \textstyle\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\displaystyle \right ], \quad \mathbf{z}^{(i)}(t) \mbox{ as in Case 1}, \ i = 2,3. $$

Case 3.$\kappa \ne 0$, $r_{2} = r_{3}$. Defining

$$ \mathbf{a}_{\alpha }(r) = \left [ \textstyle\begin{array}{c} 0 \\ -\kappa _{1} r \\ i(\beta _{1} r^{2} - \tau _{1\alpha }) \end{array}\displaystyle \right ], \quad \mbox{then} \ \mathbf{z}^{(2)}(t) = e^{r_{2} t} \mathbf{a} _{\alpha }(r_{2}), \ \mathbf{z}^{(3)}(t) = \left . \frac{\partial }{ \partial r}\bigl[e^{r t} \mathbf{a}_{\alpha }(r) \bigr] \right \vert _{r=r_{2}}. $$

(in fact, $[A(r_{i}) - \alpha I]\mathbf{a}^{(i)} = \mathbf{0}$ in Cases 1, 2 and in Case 3, $[A(r)-\alpha I]{\mathbf{a}}_{\alpha }(r)=i({\beta }_1{\tau }_3{r}^4-{\lambda }_{\alpha }{r}^2+{\beta }_{1\alpha }{\tau }_{1\alpha })\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=i{\beta }_1{\tau }_3{(r^2-r_2^2)}^2\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ so the proof in Simpson [21], Theorem 4.3 applies).

Using these solutions we obtain $C_{\xi ,\alpha }^{(2)}$ in (3.17) in the three cases. With $p_{\alpha }^{(1)}(r) = \alpha _{1} \beta _{1} r ^{2} + \nu _{1} \tau _{1\alpha }$, $p_{\alpha }^{(2)}(r) = i [(\tau _{3} \tau _{1\alpha } - \alpha _{1} \kappa _{1}) r - \beta _{1} \tau _{3} r^{3}]$

Case 1.$C_{\xi ,\alpha }^{(2)}=\left[\begin{array}{cc}{p}_{\alpha }^{(1)}(r_2)& p_{\alpha }^{(1)}(r_3)\\ p_{\alpha }^{(2)}(r_2)& p_{\alpha }^{(2)}(r_3)\end{array}\right]$,

Case 2.$C_{\xi ,\alpha }^{(2)}=\left[\begin{array}{cc}-{\beta }_1{r}_2& i{\nu }_1\\ i{\alpha }_1& -{\tau }_3{r}_3\end{array}\right]$,

Case 3.$C_{\xi ,\alpha }^{(2)}=\left[\begin{array}{cc}{p}_{\alpha }^{(1)}(r_2)& \frac{d{p}_{\alpha }^{(1)}}{dr}(r_2)\\ p_{\alpha }^{(2)}(r_2)& \frac{d{p}_{\alpha }^{(2)}}{dr}(r_2)\end{array}\right]$.

We used (2.19), (3.13), (3.14) and $\mathcal{B}_{\xi }[\mathbf{z}^{(i)}] = B(r_{i})\mathbf{a}^{(i)}$ in Cases 1, 2 and the reasoning following Simpson [21], (4.42) in Case 3.

Finally we evaluate $\det C_{\xi ,\alpha }^{(2)}$. Note $\hat{A}_{ \alpha }$ in (3.18).

$$\begin{aligned} \textbf{Case\ 1.} \quad & \det C_{\xi ,\alpha }^{(2)} = -i \kappa _{1} \sqrt{\frac{ \tau _{1\alpha }}{\tau _{3}}} \hat{A}_{\alpha } (r_{3} - r_{2}), \end{aligned}$$

(A.1)

$$\begin{aligned} \textbf{Case\ 2.} \quad & \det C_{\xi ,\alpha }^{(2)} = (\sqrt{\beta _{1} \beta _{1\alpha }} + \sqrt{\tau _{3} \tau _{1\alpha }})^{-1} \hat{A}_{\alpha }, \end{aligned}$$

(A.2)

$$\begin{aligned} \textbf{Case\ 3.} \quad &\det C_{\xi ,\alpha }^{(2)} = -i \kappa _{1} \sqrt{\frac{ \tau _{1\alpha }}{\tau _{3}}} \hat{A}_{\alpha }. \end{aligned}$$

(A.3)

In fact, in Case 1 we have

$$\begin{aligned} \det C_{\xi ,\alpha }^{(2)} &= -i \bigl\{ \alpha _{1} \tau _{3} \beta _{1}^{2} r_{2}^{2} r_{3}^{2} + \tau _{3} \tau _{1\alpha } \nu _{1} \beta _{1} \bigl(r _{2}^{2} + r_{3}^{2} \bigr) + (- \alpha _{1} \beta _{1} \delta _{\alpha } + \tau _{3} \tau _{1\alpha } \nu _{1} \beta _{1})r_{2} r_{3} + \nu _{1} \tau _{1\alpha } \delta _{\alpha } \bigr\} \\ &\quad\times{} (r_{3} - r_{2}) \end{aligned}$$

(A.4)

where $\delta _{\alpha } = \alpha _{1} \kappa _{1} - \tau _{3} \tau _{1 \alpha }$. From (3.16) we have

$$ r_{2}^{2} + r_{3}^{2} = \frac{\lambda _{\alpha }}{\beta _{1} \tau _{3}}, \qquad r_{2}^{2} r_{3}^{2} = \frac{\beta _{1\alpha } \tau _{1\alpha }}{\beta _{1} \tau _{3}}, \qquad r_{2} r_{3} = \sqrt{ \frac{\beta _{1\alpha } \tau _{1\alpha }}{\beta _{1} \tau _{3}}} $$

(A.5)

and substituting these into (A.4) yields (A.1). In Case 2 we get $\det C_{\xi ,\alpha }^{(2)} = \sqrt{\beta _{1} \beta _{1\alpha } \tau _{3} \tau _{1\alpha }} - \alpha _{1}^{2}$ using $0 = \kappa _{1} = \alpha _{1} + \nu _{1}$. This equals the right side of (A.2). In Case 3, $\det C_{\xi ,\alpha }^{(2)} = \frac{d}{dr} \Delta (r) \vert_{r = r_{2}}$ where

$$ \Delta (r) = \det \left [ \textstyle\begin{array}{c@{\quad }c} p_{\alpha }^{(1)}(r_{2}) & p_{\alpha }^{(1)}(r) \\ p_{\alpha }^{(2)}(r_{2}) & p_{\alpha }^{(2)}(r) \end{array}\displaystyle \right ]. $$

But $\Delta (r)$ equals the right side of (A.4) with $r_{3}$ replaced $r$. Thus $\frac{d}{dr} \Delta (r_{2})$ equals the right side of (A.4) with the factor $(r_{3} - r_{2})$ deleted. Then using (A.5) yields (A.3).