Stationary viscoelastic wave fields generated by scalar wave functions

TL;DR

This paper introduces a refined Helmholtz decomposition for vector fields in viscoelastic media, showing that divergence-free parts depend on only two scalar functions, which simplifies solving boundary value problems in physics.

Contribution

It presents a new decomposition of divergence-free vector fields into polarized components and demonstrates the completeness of this representation for stationary wave fields in viscoelastic media.

Findings

Each scalar function satisfies a Helmholtz equation.

The decomposition simplifies boundary value problem solutions.

Application to torsional deformation of cylindrical viscoelastic media.

Abstract

The usual Helmholtz decomposition gives a decomposition of any vector valued function into a sum of gradient of a scalar function and rotation of a vector valued function under some mild condition. In this paper we show that the vector valued function of the second term i.e. the divergence free part of this decomposition can be further decomposed into a sum of a vector valued function polarized in one component and the rotation of a vector valued function also polarized in the same component. Hence the divergence free part only depends on two scalar functions. Further we show the so called completeness of representation associated to this decomposition for the stationary wave field of a homogeneous, isotropic viscoelastic medium. That is by applying this decomposition to this wave field, we can show that each of these three scalar functions satisfies a Helmholtz equation. Our completeness of representation is useful for solving boundary value problem in a cylindrical domain for several partial differential equations of systems in mathematical physics such as stationary isotropic homogeneous elastic/viscoelastic equations of system and stationary isotropic homogeneous Maxwell equations of system. As an example, by using this completeness of representation, we give the solution formula for torsional deformation of a pendulum of cylindrical shaped homogeneous isotropic viscoelastic medium.

Equations

$${\mu }_R,{\mu }_I>0,3{\lambda }_R+2{\mu }_R,3{\lambda }_I+2{\mu }_I>0$$

$$L_{\lambda ,\mu }u:=(\lambda +\mu )\text{grad}\text{div}u+\mu \Delta u+\rho {\omega }^{2}u.$$

$$u=\nabla \phi +\nabla \times (\psi e)+\nabla \times \nabla \times (\chi e)$$

$$\left\{\begin{array}[]{l}(\Delta+\frac{\rho\omega^{2}}{\lambda+2\mu})\varphi=0,\\ (\Delta+\frac{\rho\omega^{2}}{\mu})\psi=0,\\ (\Delta+\frac{\rho\omega^{2}}{\mu})\chi=0,\end{array}\right.$$

$$v\in W_{p,loc}^{l,m}({\mathbb{R}}^3)⟺\sum _{|\alpha |+j\le l,j\le m}{\int }_{{\mathbb{R}}^3}|{\partial }_{x^{\prime }}^{\alpha }{\partial }_z^j(\eta v)(x^{\prime },z){|}^{p}d{x}^{\prime }dz<\infty$$

$${\partial }_{x^{\prime }}^{\beta }{\partial }_z^{k}v\in W_{p,loc}^{l-|\beta |-k,\min \{m-k,l-|\beta |-k\}}({\mathbb{R}}^3)$$

$$\left\{\begin{array}[]{lll}|\alpha|+|\beta|+j+k\leq l,\\ j+k\leq m.\\ \end{array}\right.\Longleftrightarrow\left\{\begin{array}[]{lll}|\alpha|+j\leq l-|\beta|-k,\\ j\leq m-k.\\ \end{array}\right.$$

$$u=\nabla \phi +\nabla \times (\psi e_z)+\nabla \times \nabla \times (\chi e_z),$$

$$u=\nabla \phi +\nabla \times A\text{in}{\mathbb{R}}^3\text{with}$$

$$\begin{array}[]{l}\varphi(x)=-(4\pi)^{-1}\int\frac{\nabla_{y}\cdot u(y)}{|y-x|}dy\in W^{7}_{p,loc}({\mathbb{R}}^{3}),\\ A(x)=(4\pi)^{-1}\int\frac{\nabla_{y}\times u(y)}{|y-x|}dy\in H^{7}_{p,loc}({\mathbb{R}}^{3}).\end{array}$$

$${\Delta }^{\prime }\chi =-e_z\cdot (\nabla \times A):=f\in W_{p,loc}^6({\mathbb{R}}^3)\text{in}{\mathbb{R}}^2,$$

$$\begin{array}[]{ll}(\nabla\times A(x),b)&=(4\pi)^{-1}\int(\nabla_{y}\times u(y),\nabla_{x}\times(|y-x|^{-1}b)\,dy\\ &=-(4\pi)^{-1}\int\big{(}u(y),\nabla_{y}\times\nabla_{x}\times(|y-x|^{-1}b)\big{)}\,dy,\end{array}$$

$${\partial }_{x^{\prime }}^{\alpha }(\nabla \times A(x^{\prime },z))=O((1+|x^{\prime }|{)}^{-3-|\alpha |})(|x^{\prime }|\to \infty )$$

$$p>2,-\frac{2}{p}-1<\delta <-\frac{2}{p}\text{except}\delta +\frac{2}{p}\text{or}-(\delta +\frac{2}{p})\in {\mathbb{Z}}_{+},$$

$$\sum _{|\alpha |\le 2}\Vert (1+|x^{\prime }{|}^2{)}^{(\delta +|\alpha |)/2}{\partial }_{x^{\prime }}^{\alpha }h{\Vert }_{L^p({\mathbb{R}}^2)}$$

$$O\{(1+|x^{\prime }|{)}^{\delta +2}(e_z\cdot (\nabla \times A(x^{\prime },z)))\}=O((1+|x^{\prime }|{)}^{-1+\delta })$$

$${\Delta }^{\prime }:{\dot{M}}_{p,\delta }\to L_{\delta +2}^p$$

$$\mathbb{R}\ni z\to e_z\cdot (\nabla \times A(\cdot ,z))\in L_{\delta +2}^p$$

$${\partial }_{x^{\prime }}\chi (x^{\prime },z)=\frac{1}{2\pi }{\int }_{{\mathbb{R}}^2}\log (|x^{\prime }-y^{\prime }|){\partial }_{y^{\prime }}f(y^{\prime },z)d{y}^{\prime }.$$

$${\Delta }^{\prime }\psi =-e_z\cdot (\nabla \times \nabla \times A(x)):=g\in W_{p,loc}^5({\mathbb{R}}^3)\text{in}{\mathbb{R}}^2$$

$$V=\nabla \times A-\nabla \times \nabla \times (\chi e_z)-\nabla \times (\psi e_z).$$

$$({\Delta }^{\prime }V)(x^{\prime },z)=0\text{in}{\mathbb{R}}_{x^{\prime }}^2\text{for any z∈R}.$$

$$\left\{\begin{array}[]{l}\text{div}\,V=0,\\ e_{z}\cdot V=e_{z}\cdot(\nabla\times A)-e_{z}\cdot\big{(}\nabla\times\nabla\times(\chi e_{z})\big{)}=e_{z}\cdot(\nabla\times A)+\Delta^{\prime}\chi=0,\\ e_{z}\cdot\big{(}\nabla\times V\big{)}=e_{z}\cdot\big{(}\nabla\times\nabla\times A\big{)}-e_{z}\cdot\big{(}\nabla\times\nabla\times(\psi e_{z})\big{)}=e_{z}\cdot\big{(}\nabla\times\nabla\times A\big{)}+\Delta^{\prime}\psi=0.\end{array}\right.$$

$$-\frac{2}{p}<{\delta }^{\prime }<-\frac{2}{p}+1\text{except}{\delta }^{\prime }+\frac{2}{p}\text{or}-({\delta }^{\prime }+\frac{2}{p})\in {\mathbb{Z}}_{+},$$

$${\Delta }^{\prime }(b\cdot \nabla \times \nabla \times (\chi e_z))=b\cdot \nabla \times \nabla \times ((-e_z\cdot \nabla \times A)e_z)=O((1+|x^{\prime }|{)}^{-5}).$$

$${\Delta }^{\prime }(b\cdot \nabla \times (\psi e_z))=b\cdot \nabla \times ((-e_z\cdot \nabla \times \nabla \times A)e_z)=O((1+|x^{\prime }|{)}^{-5})$$

$$\left\{\begin{array}[]{lll}(\Delta+\frac{\rho\omega^{2}}{\lambda+2\mu})\varphi_{1}&=&0,\\ (\Delta+\frac{\rho\omega^{2}}{\mu})\psi_{1}&=&0,\\ (\Delta+\frac{\rho\omega^{2}}{\mu})\chi_{1}&=&0\end{array}\right.$$

$$(\lambda +2\mu )(\nabla {\square }_1\phi )+\mu \nabla \times ({\square }_2\psi e_z)+\mu \nabla \times \nabla \times ({\square }_2\chi e_z)=0\text{ in }\Omega$$

$${\square }_2{\Psi }_1=0,$$

$${\Psi }_1$$

Full text

1 Introduction

The Helmholtz decomposition has played an important role to solve boundary value problems in a rectangle paralellpiped, sphere and cylinder for several partial differential equations of systems in mathematical physics such as stationary isotropic homogeneous elastic/viscoelastic equations of system and stationary isotropic Maxwell equations of system.

The aim of this paper is to provide a special Helmholtz decomposition for solutions of an isotropic homogeneous elastic/viscoelastic equations of system in a finite cylindrical domain. The special Helmholtz decomposition enables to represent solutions of this homogeneous viscoelastic system in a finite cylindrical shaped domain by scalar wave functions via a special Helmholtz decomposition associated to the cylindrical coordinates. This is called the completeness of representation. More precisely let $\Omega$ be a homogeneous isotropic viscoelastic medium in ${\mathbb{R}}^{3}$ with complexified Lamé moduli $\lambda,\,\mu$ such that their real parts $\lambda_{R},\,\mu_{R}$ and imaginary parts $\lambda_{I},\,\mu_{I}$ satisfy the strong convexity condition:

[TABLE]

(see [9]). The shape of $\Omega$ could be either a ball or cylinder. Define the stationary homogeneous isotropic viscoelastic operator $L_{\lambda,\mu}$ with density $\rho>0$ and angular frequency $\omega>0$ by

[TABLE]

Then the completeness of representation is given as follows. That is for $u$ satisfying $L_{\lambda,\mu}u=0$ in $\Omega$, there exists scalar functions $\varphi,\,\psi,\,\chi$ such that $u$ has a special Helmholtz decomposition given as

[TABLE]

and $\varphi,\,\psi,\,\chi$ satisfy the following Helmoholtz type equations:

[TABLE]

where $e$ is a unit vector in the axial direction and direction of $z$ axis for the case $\Omega$ is a ball and cylinder, respectively. If there is a global orthogonal curvilinear coordinates $(x^{1},x^{2},x^{3})$ in a neighborhood of $\overline{\Omega}$, then there is a question about the completeness of representation when $e$ is replaced by $w(x^{3})e_{3}$ with unit vector $e_{3}$ say in the $x^{3}$ direction and some auxilary function $w(x^{3})$ for the seperation of variables.

This completeness of representation was rigorously proved in [3] when a medium has a shape of ball. After that in [5], a more general discussion was made for the completeness of representation. It included the case that the medium has a shape of cylinder. However it did not give any rigorous proof for this case. To the best of our knowledge, it seems that the proof for this case has been left opened for more than 40 years.

In this paper our main aim is to give a rigorous proof for the completeness of representation for the case $\Omega$ is a finite cylinder. Hence in the rest of the paper, $\Omega$ denotes an open finite cylinder.

Needless to say that this completeness of representation (1.3) is very useful for giving a solution formula to solve given boundary value problem. As an example, we apply the torque boundary condition(see (4.2)) and find the solution which satisfies our representation (1.3). In fact we used the solution formula to identified the complex shared modulus of the sample using a digitized commercial base pendulum-type viscoelastic spectrometer (PVS) (see [6] for its details).

The rest of this paper is organized as follows. In Section 2, we give a special type of Helmholtz decomposition. Then we give the completeness of representation for this decomposition in Section 3. The last section is devoted to an application of the completeness of representation. That is we apply it to solve a boundary value problem for a cantilever homogeneous isotropic viscoelastic cylinder with a given torque at its bottom by using the completeness of representation.

2 Special Helmholtz Decomposition

For $l,m\in{\mathbb{Z}_{+}}:={\mathbb{N}}\cup\{0\}$, $l\geq m$ and $p\geq 2$, we define a function space

[TABLE]

for all $\eta\in C_{0}^{\infty}({\mathbb{R}}^{3})$, where we used $x=(x,y,z)\in{\mathbb{R}}^{3}$, $x^{\prime}=(x,y)$. We also define $W_{p}^{l,m}(\Omega)$ by deleting $\eta$ and replacing ${\mathbb{R}}^{3}$ by $\Omega$ in the above definition. For $p=2$, we use the notations $H_{loc}^{l,m}({\mathbb{R}}^{3})$, $H^{l,m}(\Omega)$. Also, we can infer that

[TABLE]

which follows from

[TABLE]

Theorem 2.1**.**

Assume that $u\in W_{p}^{6}({\mathbb{R}}^{3})$, $p\geq 2$ and it has a compact support. Then $u$ admits the representation

[TABLE]

where $\varphi\in W^{7}_{p,loc}({\mathbb{R}}^{3})$, $\psi\in W_{p,loc}^{7,4}(({\mathbb{R}}^{3})$, $\chi\in W_{p,loc}^{8,5}(({\mathbb{R}}^{3})$. Note that the regularities of $\psi,\,\chi$ given above imply $\nabla\times(\psi e_{z}),\,\,\nabla\times\nabla\times(\chi e_{z})\in W_{p,loc}^{6,4}({\mathbb{R}}^{3})$.

Proof. First of all, it is well known that for any given $u\in W_{p}^{6}({\mathbb{R}}^{3})$ with compact support, we can have the following Helmholtz decomposition [4]:

[TABLE]

where $\varphi$ and $A$ are given by

[TABLE]

where the integration are take over ${\mathbb{R}}^{3}$.

Next in order to decompose $\nabla\times A$ into two potentials, we prepare the solvability of the equation for $\chi$ which satisfies

[TABLE]

where $\Delta^{\prime}$ is the Laplacian with respect to $x^{\prime}=(x_{1},x_{2})=(x,y)$. Since the inner product $(A(x),b)$ of $A(x)$ with any constant vector $b$ is given as

[TABLE]

we have $\partial_{x^{\prime}}^{\alpha}(\nabla\times A(x))=O((1+|x|)^{-3-|\alpha|})\,\,(|x|\rightarrow\infty)$ for $|\alpha|\leq 1$. Hence fixing $z\in{\mathbb{R}}$, we have

[TABLE]

for $|\alpha|\leq 1$. In the rest of this proof we will omit $(|x|\rightarrow\infty)$ and $(|x^{\prime}|\rightarrow\infty)$ for simplicity.

Now we will apply the following result in [7] for the case $n=2$ in which we are interested. That is, if $p,\,\delta$ satisfies the condition

[TABLE]

then $\Delta^{\prime}:M_{p,\delta}^{2}\rightarrow L_{\delta+2}^{2}$ is surjective and the kernel of $\Delta^{\prime}$ is ${\mathbb{C}}$. Here $L_{\delta+2}^{p}=L_{\delta+2}^{p}({\mathbb{R}}^{2})$ and $M_{p,\delta}^{2}$ is the completion of $C_{0}^{\infty}({\mathbb{R}}^{2})$ in the norm

[TABLE]

with $x=(x^{\prime},z)=(x,y,z)$, $\partial_{x^{\prime}}^{\alpha}=\partial_{x_{1}}^{i}\partial_{x_{2}}^{j}$, $i+j=\alpha$ for nonnegative $i,j$ and $L_{\delta+2}^{p}$ is the $L^{2}({\mathbb{R}}^{2})$ with weight $(1+|x^{\prime}|^{2})^{(\delta+2)/2}$. Let $\delta$ satisfy (2.7), then we have $e_{z}\cdot(\nabla\times A(x))\in L_{\delta+2}^{p}$ because

[TABLE]

and $p(-1+\delta)<-2$ since $\delta<-2/p$. Therefore, the solution $\chi\in M_{p,\delta}^{2}$ of (2.4) exists.

Let $\dot{M}_{p,\delta}=\{\zeta\in M_{p,\delta}^{2}:\int_{|x^{\prime}|\leq R}\zeta(x^{\prime})\,dx^{\prime}=0\}$ with a large enough fixed $R>0$. Then, since the kernel of $\Delta^{\prime}$ is ${\mathbb{C}}$,

[TABLE]

is a bi-continuous isomorphism by Banach’s closed graph theorem and open mapping theorem. Since the mapping

[TABLE]

is locally continuous, the solution $\chi=\chi(\cdot,z)\in\dot{M}_{p,\delta}$ depends continously on $z\in{\mathbb{R}}$.

Furthermore, we can show $\partial_{x^{\prime}}\chi\in W_{p,loc}^{7,5}({\mathbb{R}}^{3})$. By using the fundamental solution for $\Delta^{{}^{\prime}}$, we have following representation [8], that

[TABLE]

Above representation holds because $\partial_{x^{\prime}}f(x^{\prime},z)\in L_{\delta+1}^{p}({\mathbb{R}}^{3})\subset L^{p}({\mathbb{R}}^{3})$ when we take $\delta+1>0$. So, since $\partial_{x^{\prime}}f(x^{\prime},z)\in W^{5}_{p,loc}({\mathbb{R}}^{3})$, we have $\partial_{x^{\prime}}\chi\in W_{p,loc}^{7,5}({\mathbb{R}}^{3})$. Therefore, we have $\chi\in W_{p,loc}^{8,5}({\mathbb{R}}^{3})$.

In a similar way, we can easily check that the existence of a solution $\psi\in\dot{M}_{2,\delta}^{2}$ of

[TABLE]

and its properties which implies $\partial_{x^{\prime}}\psi\in W_{p,loc}^{6,4}({\mathbb{R}}^{3})$, where $p,\delta$ satisfy (2.7) and $\delta+1>0$. So, we have $\psi\in W_{p,loc}^{7,4}({\mathbb{R}}^{3})$.

Now define $V$ by

[TABLE]

Then, we have the following lemma.

Lemma 2.2**.**

[TABLE]

Proof. Observe that for $V=V_{1}e_{x}+V_{2}e_{y}+V_{3}e_{z}$, we have

[TABLE]

Then the proof can be finished by just observing that the first and third equations give the Cauchy Riemann system of equations for $V-(e_{z}\cdot V)e_{z}=V_{1}e_{x}+V_{2}e_{y}$. This completes the proof of Lemma2.2.

To proceed further, we recall the following result in [7]. That is if $\delta^{\prime}$ satisfies the condition

[TABLE]

then $\Delta^{\prime}:M_{p,\delta^{\prime}}^{2}\rightarrow L_{\delta^{\prime}+2}^{p}$ is injective. We will consider $\delta^{\prime}=\delta+1$ for $\delta$ satisfying (2.7).

Now, we show that $V\in M_{2,\delta^{\prime}}^{2}$ for each fixed $z\in{\mathbb{R}}$. Since $\nabla\times A=O((1+|x^{\prime}|)^{-3})$, we need $p(\delta^{\prime}-3)<-2$. It holds from the condition $\delta^{\prime}<0$ . Hence $\nabla\times A\in M_{2,\delta^{\prime}}^{2}$.

To examine the second term of $V$, we need a following lemma in [7].

Lemma 2.3**.**

If $u\in L_{\delta^{\prime}}^{p}$, $\Delta^{\prime}u\in L_{\delta^{\prime}+2}^{p}$, then $u\in M_{p,\delta^{\prime}}^{2}$.

For any constant vector $b$, we have from (2.4) that

[TABLE]

We can easily check that $p(-5+\delta^{\prime}+2)<-2$. Hence $\Delta^{\prime}(b\cdot\nabla\times\nabla\times(\chi e_{z}))\in L_{\delta^{\prime}+2}^{p}$. Also, for the potential $\chi$ of (2.4), we have $\chi\in M_{p,\delta}^{2}$ and by the definition of $M_{p,\delta}^{2}$, we have $b\cdot\nabla\times\nabla\times(\chi e_{z})\in L_{\delta+2}^{p}\subset L_{\delta^{\prime}}^{2}$ with the $\delta$ of (2.7). Therefore by Lemma 2.3, we have $\nabla\times\nabla\times(\chi e_{z})\in M_{p,\delta^{\prime}}^{2}$.

As we did in the second term, for the third term of $V$, we have

[TABLE]

and hence $\Delta^{\prime}(b\cdot\nabla\times(\psi e_{z}))\in L_{\delta^{\prime}+2}^{p}$. Also, from $\psi\in M_{p,\delta}^{2}$ with $\delta$ of (2.7), we have $\nabla\times(\psi e_{z})\in L_{\delta^{\prime}}^{p}$. Hence by Lemma 2.2, we have $\nabla\times(\psi e_{z})\in M_{p,\delta^{\prime}}^{2}$.

Therefore we have shown $V=\nabla\times A-\nabla\times\nabla\times(\psi e_{z})-\nabla\times(\chi e_{z})\in M_{p,\delta^{\prime}}^{2}$ which together with Lemma 2.2 immediately gives $V=0$. This completes the proof of Theorem 2.1.

3 Completeness of representation

In this section we will show the completeness of representation based on the special Helmholtz decomposition. From now on for our convenience, we will use the convention embedding $L^{p}$- spaces into $L^{2}$- spaces for $p>2$ and bounded domains. For example, $W_{p}^{7}(\Omega)\subset H^{7}(\Omega)$, $W_{p}^{7,5}(\Omega)\subset H^{7,5}(\Omega)$. If we don’t use this convention, it will be stated.

We first preare a Boggio’s type result as lemma for further arguments.

Lemma 3.1**.**

Let $\varphi\in H^{m+2}(\Omega),\,m\geq 0$ be a solution for $\nabla^{2}(\Box_{1}\varphi)=0$ in $\Omega$ where $\Box_{1}=\nabla^{2}+c_{1}^{2}$, $c$ is constant. Then $\varphi$ can be decomposed into $\varphi=\varphi_{1}+\varphi_{2}$ with $\varphi_{1},\varphi_{2}\in H^{m}(\Omega)$ such that $\nabla^{2}\varphi_{1}=0$ and $\Box_{1}\varphi_{2}=0$ in $\Omega$.

Proof. Let $\varphi=\varphi_{1}+(\varphi-\varphi_{1})$ such that $\varphi_{1}=\frac{1}{c_{1}^{2}}\Box_{1}\varphi\in H^{m}(\Omega)$. Then we can easily check that $\nabla^{2}\varphi_{1}=0$ and $\Box_{1}(\varphi-\varphi_{1})=\Box_{1}(-\frac{1}{c_{1}^{2}}\nabla^{2}\varphi)=0$ in $\Omega$. Hence we can finish the proof by setting $\varphi_{2}=\varphi-\varphi_{1}\in H^{m}(\Omega)$. Thus we have completed the proof of Lemma 3.1.

Now using the lemma given above and Theorem 2.1, we have the following completeness result for homogeneouse viscoelastic system. Let $\Omega\subset{\mathbb{R}}^{3}$ be the cylinder given by $\Omega=\{(r,\theta,z):\,0\leq r\leq a,0\leq\theta\leq 2\pi,0\leq z\leq h\}$ in terms of cylindrical coordinates $(r,\theta,z)$, where $a,\,h$ are positive constants.

Theorem 3.2**.**

If $u\in W_{p}^{6}(\Omega)$ is a solution of $L_{\lambda,\mu}u=0$ in $\Omega$, then $u=\nabla\varphi_{1}+\nabla\times(\psi_{1}e_{z})+\nabla\times\nabla\times(\chi_{1}e_{z})$, where $e_{z}$ is the unit vector in the $z$ direction, and $\varphi_{1}$, $\psi_{1}$, $\chi_{1}$ are solutions of

[TABLE]

in $\Omega$, where $\varphi_{1}\in H^{5}(\Omega)$ and $\psi_{1}\in H^{5,2}(\Omega)$, $\chi_{1}\in H^{6,3}(\Omega)$.

Proof. Since the solution $u$ is in $W_{p}^{6}(\Omega)$, we have $u=\nabla\varphi+\nabla\times(\psi e_{z})+\nabla\times\nabla\times(\chi e_{z})$ in $\Omega$ with $\varphi\in H_{loc}^{7}({\mathbb{R}}^{3})$, $\psi\in H_{loc}^{7,4}({\mathbb{R}}^{3})$, $\chi\in H_{loc}^{8,5}({\mathbb{R}}^{3})$ by extending $u$ to $W_{p}^{6}({\mathbb{R}}^{3})$ with compact support (see [10] about this extension) and Theorem 2.1. Then $L_{\lambda,\mu}u=0$ in $\Omega$ has the form

[TABLE]

where $\square_{1}=\Delta+\frac{\rho\omega^{2}}{\lambda+2\mu}$, $\square_{2}=\Delta+\frac{\rho\omega^{2}}{\mu}$. In the rest of the proof we will omit ”in $\Omega$” for simplicity. Taking the divergence of (3.2), we have $(\lambda+2\mu)(\nabla^{2}\square_{1}\varphi)=0$. Here we decompose $\varphi$ using Lemma 3.1 into $\varphi=\varphi_{1}+\varphi_{2}$ with $\varphi_{1},\,\varphi_{2}\in H^{5}(\Omega)$ such that $\square_{1}\varphi_{1}=0$, $\nabla^{2}\varphi_{2}=0$.

On the other hand, taking curl of (3.2), we have $\mu\nabla\times\nabla\times(\square_{2}\psi e_{z}+\nabla\times(\square_{2}\chi e_{z})=0$. Taking one more curl there, we have $-\mu\nabla\times\nabla^{2}(\square_{2}\psi e_{z}+\nabla\times(\square_{2}\chi e_{z}))=0$ from the identity $\nabla\times\nabla\times\nabla\times=-\nabla\times\nabla^{2}$. Let $\Psi:=\nabla\times(\psi e_{z})+\nabla\times\nabla\times(\chi e_{z})$. Then $\Psi\in H^{6,4}(\Omega)$ and $\nabla^{2}\square_{2}\Psi=0$. Now we decompose $\Psi$ using Lemma 3.1 into $\Psi=\Psi_{1}+\Psi_{2}$ with $\Psi_{1},\,\Psi_{2}\in H^{4,2}(\Omega)$ such that

[TABLE]

and $\Psi_{1}=-\mu(\rho\omega^{2})^{-1}\nabla^{2}\Psi$, $\Psi_{2}=\mu(\rho\omega^{2})^{-1}\square_{2}\Psi$. That is $\Psi_{1},\,\Psi_{2}$ are given by

[TABLE]

Hence by defining

[TABLE]

we clearly have

[TABLE]

Put $\Xi:=\psi_{1}e_{z}+\nabla\times(\chi_{1}e_{z})\in H_{loc}^{5,2}({\mathbb{R}}^{3})$. Then from (3.6), we have $\nabla\times\square_{2}\Xi=0$ and hence $\square_{2}\Xi=\nabla{h}$ for some scalar potential $h\in H^{4,1}({\mathbb{R}}^{3})$. Here we consider the solvability for the equation $\square_{2}\Xi_{2}=\nabla{h}\in H_{loc}^{3,0}({\mathbb{R}}^{3})$. We can show that $\Xi_{2}:=\square_{2}^{-1}(\nabla h)\in H^{5,2}({\mathbb{R}}^{3})$ (see [12]). Define $\Xi_{1}:=\Xi-\Xi_{2}\in H^{5,2}(\mathbb{R}^{3})$. Then, we can have a Boggio type decomposition for $\Xi$ as $\Xi=\Xi_{1}+\Xi_{2}$ with $\square_{2}\Xi_{1}=0$ and $\nabla\times\Xi_{2}=0$.

Taking curl on $\Xi_{1}$, we have

[TABLE]

with

[TABLE]

Taking $\square_{2}$ on (3.9), we have

[TABLE]

Here, we will use the uniqueness for $\Delta^{\prime}$ in Theorem 2.1. To do so, we have to show $\square_{2}\psi_{1}$, $\square_{2}\chi_{1}\in M_{2,\delta}({\mathbb{R}}^{2})$ where $-1<\delta<0$ as in the condition (2.13). From (3.4), we have $\square_{2}\psi_{1}\in H_{loc}^{3,0}({\mathbb{R}}^{3})\subset L^{2}({\mathbb{R}}^{3})$ and $\square_{2}\chi_{1}\in H_{loc}^{4,1}({\mathbb{R}}^{3})\in L^{2}({\mathbb{R}}^{3})$. Since $\delta<0$, we have $L^{2}({\mathbb{R}}^{3})\subset L_{\delta}^{2}({\mathbb{R}}^{3})$. So, we have for fixed $z\in{\mathbb{R}}$, $\square_{2}\psi_{1}(x^{\prime},z),\square_{2}\chi_{1}(x^{\prime},z)\in L_{\delta}^{2}({\mathbb{R}}^{2})$. Obviously, we have $\Delta{{}^{\prime}}\square_{2}\psi_{1}=\Delta{{}^{\prime}}\square_{2}\chi_{1}=0\in L_{\delta+2}^{2}({\mathbb{R}}^{2})$. By applying the lemma 2.3, for any fixed $z\in{\mathbb{R}}$, we have $\square_{2}\psi_{1}(x^{\prime},z),\square_{2}\chi_{1}(x^{\prime},z)\in M_{2,\delta}({\mathbb{R}}^{2})$. So, the uniqueness for $\Delta^{{}^{\prime}}$ give us that

[TABLE]

Then it follows from the decompositions on $\varphi,\psi,\chi$ and (3.2) that we have

[TABLE]

And using $\nabla^{2}\varphi_{2}=0$ and (3.7), we have

[TABLE]

It follows that our solution $u$ has the form

[TABLE]

where the three scalar functions $\varphi_{1}$, $\psi_{1}$ and $\chi_{1}$ satisfy (3.1) in $\Omega$. Thus we have completed the proof of Theorem 3.2.

4 Solution formula to torsional deformation

In this section we will give the solution for a torsional deformation of cylinder. For that we consider again the viscoelastic equations of system

[TABLE]

where $\Omega\subset{\mathbb{R}}^{3}$ is the cylinder given by $\Omega=\{(r,\theta,z):\,0\leq r\leq a,0\leq\theta\leq 2\pi,0\leq z\leq h\}$. Decompose the boundary $\partial\Omega$ of $\Omega$ into $D=\{(r,\theta)~{}|~{}\,0\leq r<a,\ 0\leq\theta\leq 2\pi\}$, a$D_{0}=D\times\{0\}$ and $D_{h}=D\times\{h\}$ which are the bottom and top parts of the boundary $\partial\Omega$, respectively. Then the mixed type boundary condition for torsional deformation is given as

[TABLE]

Here we used the notations $t_{ij}$’s to denote tractions in cylindrical coordinates whose definitions can be found in [1, p. 74-75] (see the the next page for precise definitions). The function $f$ represents the torque given at the bottom surface of the cylinder. We assume that it is smooth enough and depends only on the radial variable.

The aim of this section is to give the form of solution $u$ of (4.1) satisfying the mixed type boundary type boundary condition (4.2) by using Theorem 3.2. To be precise, we can see from (4.21) given later that the solution $u$ has a form of

[TABLE]

for some scalar function $\psi(r,z)$ satisfying (3.1). This type is well known as torsional wave which involves a displacement in the circumferential direction only (see [2]).

We recall the definitions of $t_{ij}$’s. For a vector valued function $u=(u_{1},u_{2},u_{3})$, we define $u_{r}$, $u_{\theta}$ and $u_{z}$ by

[TABLE]

Then the components of the traction, denoted by $t_{ij}$ in (4.2), are given by the following formulae:

[TABLE]

Now we define a basis function for series solution of (1.2) satisfying (4.2). Let $J_{1}$ be the Bessel function of order 1, i.e., the solution to

[TABLE]

and let $0<k_{2}<k_{3}<\cdots$ be positive numbers such that

[TABLE]

Define $\varphi_{n}$ by

[TABLE]

where $c_{n}$ is a normalization constant such that

[TABLE]

We note that $\varphi_{n}$ ($n\geq 2$) satisfies

[TABLE]

and

[TABLE]

It is known that $\{\varphi_{n}\}_{n=1}^{\infty}$ forms a complete orthonormal system in $L^{2}((0,1),rdr)$ (see [11]).

Let $a$ be the radius of the cylinder, and define

[TABLE]

Then $\{\varphi_{n}^{a}\}_{n=1}^{\infty}$ is a complete orthonormal system in $L^{2}((0,a),rdr)$. So, the torque $f$ can be expanded as

[TABLE]

Moreover, one can see from (4.9) that the following holds:

[TABLE]

Let $k^{2}=\rho\omega^{2}/\mu$ and define $\gamma_{n}$ by

[TABLE]

Also let

[TABLE]

Then we have the following lemma and theorem (see [6, Section2] for their proofs).

Lemma 4.1**.**

For $n=1,2,\cdots$ let

[TABLE]

and

[TABLE]

where ${\bf e}_{1}=(1,0,0)$, ${\bf e}_{2}=(0,1,0)$. Then $w_{n}(r,\theta,z):=v_{n}(r,z)\Theta(\theta)$ is the solution of mixed type boundary value problem (4.1), (4.2) with $f$ replace by $f_{n}\varphi_{n}^{a}(r)$.

Theorem 4.2**.**

Let $f=\sum_{n=1}^{\infty}f_{n}\varphi_{n}^{a}(r)$ satisfy

[TABLE]

or equivalently $f(r,\theta):=f(r)\Theta(\theta)$ belong to $H^{-1/2}(D_{0})$. Then $u$ defined by

[TABLE]

is the unique solution of the mixed type boundary value problem (4.1), (4.2) in $H^{1}(\Omega)$. Further $u$ satisfies the estimate

[TABLE]

with some constant $C$ independent of $f$.

We remark on Theorem 4.2 that if we define $\psi$ by

[TABLE]

then $u$ can be expressed as

[TABLE]

where $\psi$ satifies (3.1). This is really giving the formula of solution based on the completeness of representation.

Acknowledgement

The second author was partially supported by grant-in-aid for Scientific Research (15K21766 and 15H05740) of the Japan Society for the Promotion of Science doing the research of this paper.