Asymptotic behavior of the eigenfrequencies of a thin elastic rod with non-uniform cross-section of extremely oblate shape

Abstract

We consider the eigenvalue problem for the Lamé system, which describes the oscillation of the homogeneous isotropic elastic bodies in $\mathbb {R}^3$. In this paper, we deal with the elastic body which is supposed to be like a rod with non-uniform cross-section. In that case, we consider the asymptotic behavior of the low frequency eigenvalue, when the cross-section of this rod is not geometrically isotopic, and obtain the characterization formula for the limit value.

Appendix

To begin with, we introduce the sketch of the proof of Remark 4.1.

Proof of Remark 4.1

By (4.3) and (4.6), we see that

$$\begin{aligned}&\int _{\Omega _1}\{(\Phi _{1}^{(k)})^2 +(\Phi _{2}^{(k)})^2+(\Phi _{3}^{(k)})^2\}\,dy=1, \end{aligned}$$

(4.42)

$$\begin{aligned}&\int _{\Omega _1} \Phi _{1}^{(k_1)}\Phi _{1}^{(k_2)}\,dy=0 \end{aligned}$$

(4.43)

for any $k,k_1,k_2\in \mathbb {N}$ with $k_1\ne k_2$. Then, it follows from Lemma 4.1 and (4.43) that

$$\begin{aligned} \int _{0}^{l}\eta _{1}^{(k_1)}(\tau )\eta _{1}^{(k_2)}(\tau ) M(\tau )\,d\tau =0 \end{aligned}$$

for all $k_1,k_2\in \mathbb {N}$ with $k_1\ne k_2$. We next prove that $\eta _{1}^{(k)}\not \equiv 0$ for any $k\in \mathbb {N}$ by a contradiction argument. For that purpose, we note that the matrix defined by for any $\tau \in (0,l)$,

$$\begin{aligned} \mathfrak {M}(\tau )=\begin{pmatrix} M_{11}(\tau ) &{} M_{12}(\tau ) &{} -M_{1}(\tau )\\ M_{21}(\tau ) &{} M_{22}(\tau ) &{} -M_{2}(\tau )\\ -M_{1}(\tau ) &{} -M_{2}(\tau ) &{} M(\tau ) \end{pmatrix} \end{aligned}$$

is positive definite, where $\{M_{ij}\}_{i,j=1,2}$, $\{M_{i}\}_{i=1,2}$ and M are as in (1.2). Indeed, by a direct calculation, for any $\tau \in (0,l)$ and for any vector $(w_1,w_2,w_3)\in \mathbb {R}^3$, we have that

$$\begin{aligned} (\mathfrak {M}(\tau )w,w)_{\mathbb {R}^3}&=\sum _{i,j=1}^{2}M_{ij}(\tau )w_i w_j -2\sum _{i=1}^{2}M_{i}(\tau )w_i w_3 +M(\tau ) w_3^2 \nonumber \\&=\int _{\hat{\Omega }(\tau )} \left( \sum _{i,j=1}^{2} y_i y_j w_i w_j -2\sum _{i=1}^{2}y_{i}w_i w_3 +w_3^2 \right) \,dy^\prime \nonumber \\&=\int _{\hat{\Omega }(\tau )} (y_1w_1+y_2 w_2-w_3)^2\,dy^{\prime }. \end{aligned}$$

(4.44)

Furthermore, noting that $\{y_{1},y_{2},-1\}$ is a linearly independent set of $L^{2}(\hat{\Omega }(\tau ))$, there exists a constant $\delta >0$ (independent of $\tau$) such that for any $\tau \in (0,l)$

$$\begin{aligned} \int _{\hat{\Omega }(\tau )} (y_1 w_1+y_2 w_2-w_3)^2\,dy^{\prime } \ge \delta (w_1^2+w_2^2+w_3^2) \end{aligned}$$

for any vector $(w_1,w_2,w_3) \in \mathbb {R}^3$, and so we see by (4.44) that for any $\tau \in (0,l)$,

$$\begin{aligned} (\mathfrak {M}(\tau )w,w)_{\mathbb {R}^3}\ge \delta (w_1^2+w_2^2+w_3^2) \end{aligned}$$

(4.45)

for any vector $(w_1,w_2,w_3) \in \mathbb {R}^3$, which yields the positivity of the matrix $\mathfrak {M}$. Suppose that there exists $k_{0} \in \mathbb {N}$ such that $\eta _{1}^{(k_{0})}\equiv 0$, and we take $(w_1,w_2,w_3)=(0, \frac{d^2 \eta ^{(k_{0})}_{2}}{d \tau ^2}, \frac{d \eta ^{(k_{0})}_{3}}{d \tau })$ in (4.45). Then by (4.22), it holds that $(\mathfrak {M}(\tau )w,w)_{\mathbb {R}^3}=0$ for any $\tau \in (0,l)$, and it follows from (4.45) that

$$\begin{aligned} 0=(\mathfrak {M}(\tau )w,w)_{\mathbb {R}^3}\ge \delta \left\{ \left( \frac{d^2 \eta ^{(k_{0})}_{2}}{d \tau ^2} \right) ^2 + \left( \frac{d \eta ^{(k_{0})}_{3}}{d \tau } \right) ^2 \right\} \end{aligned}$$

for any $\tau \in (0,l)$. Therefore we have by the boundary condition of (4.22) that $\eta ^{(k_{0})}_{2}=\eta ^{(k_{0})}_{3}=0$ on (0, l), and so we obtain $\Phi ^{(k_{0})}=0$. This fact is contrary to (4.42). $\square$

Finally, we shall sketch the proof of Proposition 4.3 via the Rayleigh quotient, which is a often used theory of the eigenvalue problem of the second order elliptic operator like Laplacian.

Proof of Proposition 4.3

We first should remark that the numerator $\hat{B}$ defined by (4.32) is coercive, More precisely, it follows from (4.45) that for some constant $\delta >0$,

$$\begin{aligned} \hat{B}(\xi ) \ge \delta \int _{0}^{l} \left\{ \left( \frac{d^2\xi _1}{d\tau ^2}\right) ^2 + \left( \frac{d^2\xi _2}{d\tau ^2}\right) ^2 + \left( \frac{d\xi _3}{d\tau } \right) ^2 \right\} \, d\tau \end{aligned}$$

(4.46)

for any $\xi =(\xi _1,\xi _{2},\xi _3)$ satisfying (4.12).

Moreover, it is easy to verify that the variational equation (Euler-Lagrange equation) of $\hat{R}$ as in (4.31) is given by (4.30), and so $\Lambda$ appears as the Euler-Lagrange multiplier. Therefore, we can deal with the eigenvalue problem (4.30) through the variational problem $\hat{R}$.

Define $\Lambda _{1}$ by

$$\begin{aligned} \Lambda _1:=\inf \left\{ \hat{R}(\xi )\,;\, \xi =(\xi _1,\xi _{2},\xi _3) ~\text {satisfies}~ (4.12),~ \xi \ne 0 \right\} . \end{aligned}$$

By the coercivity of the numerator of $\hat{R}$ as (4.46), the minimizing sequence $\{\xi ^{(p)}\}_{p=1}^{\infty }=\{(\xi _{1}^{(p)},\xi _{2}^{(p)},\xi _{3}^{(p)})\}_{p=1}^{\infty }$ is relatively compact in $L^{2}((0,l))\times L^{2}((0,l))\times L^{2}((0,l))$ provided that $\Vert \xi _1^{(p)}\Vert _{L^{2}(M,(0,l))}=1$. We can get a minimizer $\eta ^{(1)}=(\eta _{1}^{(1)},\eta _{2}^{(1)},\eta _{3}^{(1)})$ which satisfies (4.12).

We next define $\Lambda _2$ by

$$\begin{aligned} \Lambda _2:=\inf \left\{ \hat{R}(\xi )\,;\, \xi ~\text {satisfies}~ (4.12),~ (\xi ,\eta ^{(1)})_{L^2(M,(0,l))}=0,~ \xi \ne 0 \right\} , \end{aligned}$$

which satisfies $\Lambda _{1}\le \Lambda _2$. The similar argument shows the existence of the minimizer $\eta ^{(2)}=(\eta _{1}^{(2)},\eta _{2}^{(2)},\eta _{3}^{(2)})$ with the property as (4.12). We thus inductively have a sequence $\{\Lambda _k \}_{k=1}^{\infty }$ and $\{\eta ^{(k)}\}_{k=1}^{\infty }$.

We next prove $\lim _{k\rightarrow \infty }\Lambda _{k}=\infty$ by a contradiction argument. Assume that there exists a constant $C>0$ such that $0< \Lambda _{k}\le C$ for any $k\in \mathbb {N}$. By making use of (4.46), we see that $\{\eta ^{(k)}\}_{k=1}^{\infty } =\{(\eta ^{(k)}_1,\eta ^{(k)}_2,\eta ^{(k)}_3)\}_{k=1}^{\infty }$ is bounded with respect to k in $H^2((0,l))\times H^2((0,l))\times H^1((0,l))$, and $\{\eta ^{(k)}\}_{k=1}^{\infty }$ is relatively compact in $L^2((0,l))\times L^2((0,l))\times L^2((0,l))$. However, since $\{\eta _{1}^{(k)}\}_{k=1}^{\infty }$ is an orthogonal system in $L^2(M,(0,l))$, it holds that for $k_1<k_2$, $\Vert \eta _{1}^{(k_1)}-\eta _{1}^{(k_2)}\Vert _{L^{2}(M,(0,l))}\ge \sqrt{2}$, which means that any subsequence of $\{\eta _{1}^{(k)}\}_{k=1}^{\infty }$ dose not converge in $L^{2}(M,(0,l))$. This causes a contradiction. We thus have that $\lim _{k\rightarrow \infty }\Lambda _{k}=\infty$.

Finally, we shall prove that $\{\eta ^{(k)}_{1}\}_{k=1}^{\infty }$ is a complete orthogonal system in $L^{2}(M,(0,l))$ by a contradiction argument. Assume that

$$\begin{aligned} L^2(M,(0,l))\ne \overline{L.H.[\{ \eta _{1}^{(k)}\}_{k\ge 1}]}=:Z. \end{aligned}$$

Take any $\xi \in L^{2}(M,(0,l))$ satisfying $(\xi ,\eta _{1}^{(k)})_{L^2(M,(0,l))}=0$ for all $k\in \mathbb {N}$ with $\Vert \xi \Vert _{L^2(M,(0,l))}=1$, and for any $\delta >0$ we may choose $\tilde{\xi }\in H^2((0,l))$ satisfying (4.12) such that $\Vert \tilde{\xi }-\xi \Vert _{L^2(M,(0,l))}<\delta$. Furthermore, since $\widetilde{\Upsilon }=\widetilde{\xi } -\sum _{p=1}^k c_p \eta _1^{(p)}$, $c_p=(\widetilde{\xi },\eta _1^{(p)})_{L^2(M,(0,\ell ))}/\Vert \eta _{1}^{(p)}\Vert ^2_{L^2(M,(0,l))}$ satisfies (4.12), we have that

$$\begin{aligned} \widehat{R}(\Upsilon )\geqq \Lambda _{k+1}, \end{aligned}$$

where $\Upsilon =(\widetilde{\Upsilon },-\sum _{p=1}^{k}c_p \eta _{2}^{(p)}, -\sum _{p=1}^{k}c_p \eta _{3}^{(p)})$. From the above inequality, we have

$$\begin{aligned} \widehat{B}(\Upsilon )\geqq \Lambda _{k+1}\Vert \widetilde{\Upsilon } \Vert _{L^2(M,(0,\ell ))}^2. \end{aligned}$$

(4.47)

Dividing $\Upsilon =\Upsilon _1-\Upsilon _2$ by $\Upsilon _1 =\left( \tilde{\xi }, 0, 0 \right)$, $\Upsilon _2 =\sum _{p=1}^{k}c_p \left( \eta _{1}^{(p)}, \eta _{2}^{(p)}, \eta _{3}^{(p)} \right)$, since by a direct calculation, it holds that $\widehat{B}(\Upsilon _2)=\sum _{p=1}^k \Lambda _p c_p^2$, we see that the left hand side of the above inequality (4.47) agrees to

$$\begin{aligned} \widehat{B}(\Upsilon _1)-\sum _{p=1}^k \Lambda _p c_p^2 \end{aligned}$$

which is not greater than $\widehat{B}(\Upsilon _1)$. On the other hand, denoting the orthogonal projection

$$\begin{aligned} \mathcal {P}(\widetilde{\xi }) =\sum _{p=1}^k c_p \eta _1^{(p)}, \end{aligned}$$

we consider the right hand side (4.47) (from below). Using $\mathcal {P}(\xi )=0$ and take $\delta =3/4$, we have

$$\begin{aligned}&\Vert \widetilde{\Upsilon } \Vert _{L^2(M,(0,\ell ))}= \Vert \widetilde{\xi }-\mathcal {P}(\widetilde{\xi })\Vert _{L^2(M,(0,\ell ))} =\Vert \xi +(I-\mathcal {P})(\widetilde{\xi }-\xi )\Vert _{L^2(M,(0,\ell ))} \\&\quad \ge \Vert \xi \Vert _{L^2(M,(0,\ell ))} -\Vert \widetilde{\xi }-\xi \Vert _{L^2(M,(0,\ell ))}\ge 1/4. \end{aligned}$$

Accordingly we get $\widehat{B}(\Upsilon _1)\ge \Lambda _{k+1}/16$ and by taking $k\rightarrow \infty$, we get a contradiction. $\square$