Null Lagrangians in linear theories of micropolar type and few other generalizations of elasticity

Abstract

In the context of linear theories of generalized elasticity including those for homogeneous micropolar media, quasicrystals, and piezoelectric and piezomagnetic media, we explore the concept of null Lagrangians. For obtaining the family of null Lagrangians, we employ the sufficient conditions of H. Rund. In some cases, a nonzero null Lagrangian is found and the stored energy admits a split into a null Lagrangian and a remainder. However, the null Lagrangian vanishes whenever the relevant elasticity tensor obeys certain symmetry conditions which can be construed as an analogue of the Cauchy relations.

Appendix A: Characterization of null Lagrangians

If the right-hand side of (1.33) is to vanish for all values of $\frac{{\partial }^2y^j}{{\partial } x^{{\beta }}{\partial } x^{{\gamma }}}$, the coefficient of these quantities which is symmetric in ${{\beta }}$ and ${{\gamma }}$ must be skew-symmetric in these indices, i.e.

$$\begin{aligned} \frac{{\partial }^2\Psi }{{\partial } y^j_{{\beta }}{\partial } y^k_{{\gamma }}}=-\frac{{\partial }^2\Psi }{{\partial } y^j_{{\gamma }}{\partial } y^k_{{\beta }}}. \end{aligned}$$

Following the analysis of [17], in general, $\Psi$ is a polynomial in its dependence on the derivative of y. In the class of Lagrangians which are interesting in this document, it is sufficient to consider $M=2.$ Thus, $\Psi$ can be written as

$$\begin{aligned}&\displaystyle \Psi =\Psi ^{(2)}+\Psi ^{(1)}+\Phi , \end{aligned}$$

(A.1)

$$\begin{aligned}&\displaystyle \text {where } \Psi ^{(1)}:=A^{\alpha _1}_{i_1}y^{i_1}_{\alpha _1},\qquad \Psi ^{(2)}:=\frac{1}{2}A^{\alpha _1\alpha _2}_{i_1i_2}y^{i_1}_{\alpha _1}y^{i_2}_{\alpha _2}. \end{aligned}$$

(A.2)

Here, $\Phi ,\mathbf {A},{{\mathbb {A}}}$ are functions of $\varvec{x}$ and $\varvec{y}$ of class $\mathcal {C}^2$. Clearly $A^{\alpha _1\alpha _2}_{i_1i_2}=A^{\alpha _2\alpha _1}_{i_2i_1}.$ Then (1.34) becomes

$$\begin{aligned} \sum _{n=1}^2{\mathscr {E}}_k(\Psi ^{(n)})+{\mathscr {E}}_k(\Phi )\equiv 0. \end{aligned}$$

(A.3)

On substituting from (1.33), we see that this represents a set of PDEs for $\mathbf {A}$ and ${{\mathbb {A}}}$, whose precise form we have to determine.

From (A.2), we have $\frac{{\partial }\Psi ^{(2)}}{{\partial } y^k_{{\gamma }}}=A^{{{\gamma }}\alpha _2}_{ki_2}y^{i_2}_{\alpha _2}\qquad \frac{{\partial }\Psi ^{(1)}}{{\partial } y^k_{{\gamma }}}=A^{{\gamma }}_k.$ Therefore,

$$\begin{aligned} {\mathscr {E}}_k(\Psi ^{(2)})&=\frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{{\gamma }}\alpha _2}_{ki_2}\right) y^{i_2}_{\alpha _2}+\left[ \frac{\partial }{{\partial } y^{i_1}}\left( A^{\alpha _1\alpha _2}_{ki_2}\right) -\frac{1}{2}\frac{\partial }{{\partial } y^k}\left( A^{\alpha _1\alpha _2}_{i_1i_2}\right) \right] y^{i_1}_{\alpha _1}y^{i_2}_{\alpha _2}+A^{{{\gamma }}{{\beta }}}_{kj}\frac{{\partial }^2y^j}{{\partial } x^{{\beta }}{\partial } x^{{\gamma }}},\\ {\mathscr {E}}_k(\Psi ^{(1)})&=\frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{\gamma }}_k\right) +\left[ \frac{\partial }{{\partial } y^{i_1}}\left( A^{\alpha _1}_k\right) -\frac{\partial }{{\partial } y^k}\left( A^{\alpha _1}_{i_1}\right) \right] y^{i_1}_{\alpha _1}. \end{aligned}$$

From (A.3), we get

$$\begin{aligned}&\left[ \frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{{\gamma }}\alpha _2}_{ki_2}\right) +\frac{\partial }{{\partial } y^{i_2}}\left( A^{\alpha _2}_k\right) -\frac{\partial }{{\partial } y^k}\left( A^{\alpha _2}_{i_2}\right) \right] y^{i_2}_{\alpha _2}+\left[ \frac{\partial }{{\partial } y^{i_1}}\left( A^{\alpha _1\alpha _2}_{ki_2}\right) -\frac{1}{2}\frac{\partial }{{\partial } y^k}\left( A^{\alpha _1\alpha _2}_{i_1i_2}\right) \right] y^{i_1}_{\alpha _1}y^{i_2}_{\alpha _2}\nonumber \\&\quad +A^{{{\gamma }}{{\beta }}}_{kj}\frac{{\partial }^2y^j}{{\partial } x^{{\beta }}{\partial } x^{{\gamma }}}+\left[ \frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{\gamma }}_k\right) -\frac{{\partial }\Phi }{{\partial } y^k}\right] =0. \end{aligned}$$

(A.4)

This holds for all values of $y^i_\alpha$ and $\frac{{\partial }^2y^j}{{\partial } x^{{\beta }}{\partial } x^{{\gamma }}}$. So every term is zero separately. In particular,

$$\begin{aligned} \left[ \frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{{\gamma }}\alpha _2}_{ki_2}\right) +\frac{\partial }{{\partial } y^{i_2}}\left( A^{\alpha _2}_k\right) -\frac{\partial }{{\partial } y^k}\left( A^{\alpha _2}_{i_2}\right) \right] =0,\qquad \left[ \frac{\partial }{{\partial } x^{{\gamma }}}\left( A^{{\gamma }}_k\right) -\frac{{\partial }\Phi }{{\partial } y^k}\right] =0. \end{aligned}$$

(A.5)

for $k=1,\dots ,N.$

Let us consider 3 functions (of $\mathcal {C}^2$ smoothness) $\{S^\alpha (\varvec{x},\varvec{y})\}_{\alpha =1,2,3}$ and define

$$\begin{aligned} S^\alpha _i:=\frac{{\partial } S^\alpha }{{\partial } y^i}; \qquad S^\alpha _{|\beta }:=\frac{{\partial } S^\alpha }{{\partial } x^\beta }. \end{aligned}$$

Recall $\mathfrak {D}$s as defined by (1.36). Therefore,

$$\begin{aligned} \frac{{\partial } \mathfrak {D}(\alpha _1;i_1)}{{\partial } x^{\alpha _1}}&= \begin{vmatrix} S^{\alpha _1}_{i_1|\alpha _1}&S^{\alpha _2}_{i_1|\alpha _1}\\ S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2} \end{vmatrix}+ \begin{vmatrix} S^{\alpha _1}_{i_1}&S^{\alpha _2}_{i_1}\\ S^{\alpha _1}_{|\alpha _1\alpha _2}&S^{\alpha _2}_{|\alpha _1\alpha _2} \end{vmatrix}\nonumber \\&\quad (\alpha _1 and \alpha _2\text { both being dummy indices, the latter determinant is }0)\end{aligned}$$

(A.6)

$$\begin{aligned}&=\frac{\partial }{{\partial } y^{i_1}} \begin{vmatrix} S^{\alpha _1}_{|\alpha _1}&S^{\alpha _2}_{|\alpha _1}\\ S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2} \end{vmatrix} -\begin{vmatrix} S^{\alpha _1}_{|\alpha _1}&S^{\alpha _2}_{|\alpha _1}\\ S^{\alpha _1}_{i_1|\alpha _2}&S^{\alpha _2}_{i_1|\alpha _2} \end{vmatrix}=\frac{{\partial } \mathfrak {D}(0;0)}{{\partial } y^{i_1}}-\frac{{\partial } \mathfrak {D}(\alpha _1;i_1)}{{\partial } x^{\alpha _1}}\nonumber \\ \implies 2\frac{{\partial } \mathfrak {D}(\alpha _1;i_1)}{{\partial } x^{\alpha _1}}&=\frac{{\partial } \mathfrak {D}(0;0)}{{\partial } y^{i_1}}. \end{aligned}$$

(A.7)

Also,

$$\begin{aligned} \frac{{\partial } \mathfrak {D}(\alpha _1,\alpha _2;i_1,i_2)}{{\partial } x^{\alpha _2}}&= \begin{vmatrix} S^{\alpha _1}_{i_1|\alpha _2}&S^{\alpha _2}_{i_1|\alpha _2}\\ S^{\alpha _1}_{i_2}&S^{\alpha _2}_{i_2} \end{vmatrix} + \begin{vmatrix} S^{\alpha _1}_{i_1}&S^{\alpha _2}_{i_1}\\ S^{\alpha _1}_{i_2|\alpha _2}&S^{\alpha _2}_{i_2|\alpha _2} \end{vmatrix}\\&=\frac{\partial }{{\partial } y^{i_1}} \begin{vmatrix} S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2}\\ S^{\alpha _1}_{i_2}&S^{\alpha _2}_{i_2} \end{vmatrix} +\frac{\partial }{{\partial } y^{i_2}} \begin{vmatrix} S^{\alpha _1}_{i_1}&S^{\alpha _2}_{i_1}\\ S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2} \end{vmatrix} - \begin{vmatrix} S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2}\\ S^{\alpha _1}_{i_1i_2}&S^{\alpha _2}_{i_1i_2} \end{vmatrix} - \begin{vmatrix} S^{\alpha _1}_{i_1i_2}&S^{\alpha _2}_{i_1i_2}\\ S^{\alpha _1}_{|\alpha _2}&S^{\alpha _2}_{|\alpha _2} \end{vmatrix}\\&=-\frac{{\partial } \mathfrak {D}(\alpha _1;i_2)}{{\partial } y^{i_1}}+\frac{{\partial } \mathfrak {D}(\alpha _1;i_1)}{{\partial } y^{i_2}}\\&\quad \text {(the remaining two determinants cancel each other)}. \end{aligned}$$

Comparing this with (A.5)${}_1$, we get

$$\begin{aligned} A^{\alpha _1\alpha _2}_{i_1i_2}=\mathfrak {D}(\alpha _1,\alpha _2;i_1,i_2),\qquad A^{\alpha _1}_{i_1}=\mathfrak {D}(\alpha _1;i_1). \end{aligned}$$

(A.8)

From (A.7), $\frac{1}{2}\frac{{\partial } \mathfrak {D}(0;0)}{{\partial } y^{i_1}}=\frac{{\partial } A^{\alpha _1}_{i_1}}{{\partial } x^{\alpha _1}}=\frac{{\partial }\Phi }{{\partial } y^{i_1}}$ by (A.5)${}_2$, which upon integrating yields

$$\begin{aligned} \Phi =\frac{1}{2}\mathfrak {D}(0;0). \end{aligned}$$

Therefore, reverting back to (A.1) we get (1.35).

Using (A.1), (A.2), (A.4) and (A.8), we can summarize the result as:

Theorem 1

([17], pg 257-258) Any function of the form (1.35) satisfies Euler–Lagrange equation (1.34) identically, where $\mathcal {C}^2$ functions $S^\alpha (\varvec{x},\varvec{y})$ are entirely arbitrary.