Static and Dynamic Green’s Functions in Peridynamics

Abstract

We derive the static and dynamic Green’s functions for one-, two- and three-dimensional infinite domains within the formalism of peridynamics, making use of Fourier transforms and Laplace transforms. Noting that the one-dimensional and three-dimensional cases have been previously studied by other researchers, in this paper, we develop a method to obtain convergent solutions from the divergent integrals, so that the Green’s functions can be uniformly expressed as conventional solutions plus Dirac functions, and convergent nonlocal integrals. Thus, the Green’s functions for the two-dimensional domain are newly obtained, and those for the one and three dimensions are expressed in forms different from the previous expressions in the literature. We also prove that the peridynamic Green’s functions always degenerate into the corresponding classical counterparts of linear elasticity as the nonlocal length tends to zero. The static solutions for a single point load and the dynamic solutions for a time-dependent point load are analyzed. It is analytically shown that for static loading, the nonlocal effect is limited to the neighborhood of the loading point, and the displacement field far away from the loading point approaches the classical solution. For dynamic loading, due to peridynamic nonlinear dispersion relations, the propagation of waves given by the peridynamic solutions is dispersive. The Green’s functions may be used to solve other more complicated problems, and applied to systems that have long-range interactions between material points.

Appendices

Appendix A: Peridynamic Material Parameters

Peridynamic material parameters can be determined by equating the strain energy density from the peridynamic theory to that of classical continuum theory [32]. By this approach (following [32]), the peridynamic parameters for one-dimensional and two-dimensional domains are derived as follows.

1.1 A.1 PD Material Parameters for One-Dimensional Deformation

The peridynamic material parameters can be determined by considering an infinitely long bar subjected to a uniform deformation of $s=\zeta$. This uniform deformation can be a stretch, a torsion or a shear. The micropotential for this loading is

$$ w=\frac{1}{2}C(\xi )\xi^{2}\zeta^{2}. $$

(A.1)

Thus, the peridynamic strain energy density can be expressed as

$$ W=\frac{1}{2} \int_{\mathcal {H}} \biggl( \frac{1}{2}C(\xi )\xi^{2} \zeta ^{2} \biggr) \mathrm{d}\xi = \frac{1}{4}\zeta^{2} \int_{\mathcal {H}}C( \xi )\xi^{2}\mathrm{d}\xi . $$

(A.2)

For the same deformation condition, the strain energy density based on the classical linear theory is

$$ W=\frac{1}{2}\varLambda \zeta^{2}, $$

(A.3)

where $\varLambda$ is the classical material parameter as listed in Table 1. By equating the strain density Eq. (A.2) to Eq. (A.3), the peridynamic material parameters can be determined by the following equation:

$$ \varLambda =\frac{1}{2} \int_{\mathcal {H}}C ( \xi ) \xi^{2} \mathrm{d}\xi . $$

(A.4)

1.2 A.2 PD Material Parameters for Two-Dimensional Deformation

In an infinite domain, the displacements of material points $\boldsymbol{x}'$ and $\boldsymbol{x}$ are represented by $\boldsymbol{u}'$ and $\boldsymbol{u}$, respectively. The relative position vector is $\boldsymbol{\xi }=\boldsymbol{x}'- \boldsymbol{x}$ and the relative displacement vector is $\boldsymbol{\eta }=\boldsymbol{u}'- \boldsymbol{u}$. Then the micropotential can be evaluated as

$$ w=\frac{1}{2}\boldsymbol{\eta }\cdot \boldsymbol{C} ( \boldsymbol{\xi } ) \cdot \boldsymbol{\eta }. $$

(A.5)

The PD material parameters for two-dimensional plane-strain or plane-stress deformation can be determined by considering two different loading conditions. We take the derivation of the parameters for the plane-stress deformation as an example.

In the first loading condition, an isotropic expansion is applied to an infinite domain. In other words, all PD bonds is subjected to a uniform stretch $s=\zeta$, that is, $\boldsymbol{\eta }=\zeta \boldsymbol{\xi }$ for all the PD bonds. Using Eq. (A.5) and Eq. (35), the corresponding micropotential is equal to

$$ w=\frac{1}{2}\lambda ( \xi ) \xi^{4} \zeta^{2}. $$

(A.6)

The peridynamic strain energy density for this deformation can be evaluated as

$$ W=\frac{1}{2} \int_{\mathcal {H}} \biggl( \frac{1}{2}\lambda ( \xi ) \xi^{4}\zeta^{2} \biggr) \mathrm{d}A_{\boldsymbol{\xi }} = \frac{\pi }{2}\zeta ^{2}\int_{0}^{l}\lambda ( \xi ) \xi^{5}\mathrm{d}\xi . $$

(A.7)

Based on linear elasticity, the strain energy density becomes

$$ W=\frac{E}{1-\nu }\zeta^{2}. $$

(A.8)

The equality of the two quantities leads to the following relation

$$ \frac{E}{1-\nu }=\frac{\pi }{2}\int_{0}^{l} \lambda ( \xi ) \xi^{5}\mathrm{d}\xi . $$

(A.9)

In the second loading condition, an infinite domain is subjected to a pure shear loading of $\gamma_{xy}=\zeta /2$, that is, all PD bonds are subjected to stretches of $\varepsilon_{xx}=\zeta$, $\varepsilon_{yy}=- \zeta$. For material points $\boldsymbol{x}'$ and $\boldsymbol{x}$, the relative displacement under this loading is $u'_{x}-u_{x}=\zeta \xi_{x}$ and $u'_{y}-u_{y}=-\zeta \xi_{y}$. Assume that the angle between the relative position vector $\boldsymbol{\xi }$ and the $x$-axis is $\theta$. The relative displacement vector can be expressed as

$$ \boldsymbol{\eta }= \bigl[ \bigl( u'_{x}-u_{x} \bigr) \cos \theta + \bigl( u'_{y}-u _{y} \bigr) \sin \theta \bigr] \boldsymbol{\xi }. $$

(A.10)

Due to $\xi_{x}=\xi \cos \theta$ and $\xi_{y}=\xi \sin \theta$, Eq. (A.10) becomes

$$ \boldsymbol{\eta }=\zeta \bigl( \cos^{2}\theta - \sin^{2}\theta \bigr) \boldsymbol{\xi }. $$

(A.11)

By substituting Eq. (A.10) into Eq. (A.5), the micropotential for this pure shear deformation is

$$ w=\frac{1}{2}\lambda ( \xi ) \xi^{4} \zeta^{2} \bigl( \cos^{2} \theta -\sin^{2}\theta \bigr) ^{2}. $$

(A.12)

Then the peridynamic strain energy density is

$$ W=\frac{1}{2} \int_{\mathcal {H}} \biggl[ \frac{1}{2}\lambda ( \xi ) \xi^{4}\zeta^{2} \bigl( \cos^{2}\theta - \sin^{2}\theta \bigr) ^{2} \biggr] \mathrm{d}A_{\boldsymbol{\xi }} =\frac{\pi }{4}\zeta^{2}\int_{0}^{l} \lambda ( \xi ) \xi^{5}\mathrm{d}\xi . $$

(A.13)

The strain energy density of classical theory is

$$ W=\frac{E}{1+\nu }\zeta^{2}. $$

(A.14)

Equating Eq. (A.13) to Eq. (A.14) leads to

$$ \frac{E}{1+\nu }=\frac{\pi }{4}\int_{0}^{l} \lambda ( \xi ) \xi^{5}\mathrm{d}\xi . $$

(A.15)

Combining Eq. (A.9) with Eq. (A.15) results in $\nu =\frac{1}{3}$ and

$$ E=\frac{\pi }{3}\int_{0}^{l}\lambda ( \xi ) \xi^{5} \mathrm{d}\xi . $$

(A.16)

As for plane-strain deformation, the PD parameters can be obtained with the same derivation process through replacing $\nu$ with $\frac{ \nu }{1-\nu }$ and $E$ with $\frac{E}{1-\nu^{2}}$. Thus, the parameters for the plane-strain deformation satisfy $\nu =\frac{1}{4}$ and

$$ E=\frac{5\pi }{16}\int_{0}^{l}\lambda ( \xi ) \xi^{5} \mathrm{d}\xi . $$

(A.17)

Appendix B: Calculation of Tensor $M( \boldsymbol{k})$

We need to calculate the following two tensors:

$$ \boldsymbol{M} ( \boldsymbol{k} ) = \int _{\mathcal {H}}\boldsymbol{C} ( \boldsymbol{\xi } ) ( 1-\cos \boldsymbol{k}\cdot \boldsymbol{\xi } ) \mathrm{d}A_{\boldsymbol{\xi }} $$

(B.1)

and

$$ \boldsymbol{M} ( \boldsymbol{k} ) = \int _{\mathcal {H}}\boldsymbol{C} ( \boldsymbol{\xi } ) ( 1-\cos \boldsymbol{k}\cdot \boldsymbol{\xi } ) \mathrm{d}V_{\boldsymbol{\xi }}. $$

(B.2)

For the second tensor, Silling [6] first gave the solution in the special case of $\boldsymbol{k}=\boldsymbol{e}_{1}$, and the solution for the general case has also been obtained by Weckner et al. [27]. The approach of Weckner et al. [27] is not suitable for the first tensor, so it has not been calculated. Now we adopt a new method to solve the second tensor in spherical coordinates. Then the first tensor can be solved in polar coordinates in the same way.

Inspired by the solutions of Silling [6], we find that the tensor $M ( \boldsymbol{k} )$ is easily obtained when the direction of vector $\boldsymbol{k}$ is parallel to any of the base vectors of an orthonormal basis $\{ \boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3} \}$. Therefore, by coordinate transformation, we establish a new orthonormal basis $\{ \boldsymbol{e}'_{1},\boldsymbol{e}'_{2},\boldsymbol{e}'_{3} \}$ and make the new base vector $\boldsymbol{e}'_{1}$ parallel to the vector $\boldsymbol{k}$. The new orthonormal basis can be expressed as

$$ \left [ \textstyle\begin{array}{c} \boldsymbol{e}'_{1} \\ \boldsymbol{e}'_{2} \\ \boldsymbol{e}'_{3} \\ \end{array}\displaystyle \right ] =\boldsymbol{D}\left [ \textstyle\begin{array}{c} \boldsymbol{e}_{1} \\ \boldsymbol{e}_{2} \\ \boldsymbol{e}_{3} \\ \end{array}\displaystyle \right ] , $$

(B.3)

where

$$ \boldsymbol{D} =\left [ \textstyle\begin{array}{c@{\quad }c@{\quad }c} \frac{k_{1}}{k} & \frac{k_{2}}{k} & \frac{k_{3}}{k} \\ \frac{k_{2}}{\sqrt{k_{1}^{2}+k_{2}^{2}}} & \frac{-k_{1}}{\sqrt{k _{1}^{2}+k_{2}^{2}}} & 0 \\ \frac{k_{1}k_{3}}{k\sqrt{k_{1}^{2}+k_{2}^{2}}} & \frac{k_{2}k_{3}}{k\sqrt{k _{1}^{2}+k_{2}^{2}}} & \frac{-\sqrt{k_{1}^{2}+k_{2}^{2}}}{k} \\ \end{array}\displaystyle \right ] . $$

(B.4)

The components of the vector $\boldsymbol{\xi }$ in the two sets of orthonormal bases satisfy the following relations:

$$ \left [ \textstyle\begin{array}{c} \xi_{1} \\ \xi_{2} \\ \xi_{3} \\ \end{array}\displaystyle \right ] =\boldsymbol{D}^{\mathrm{T}}\left [ \textstyle\begin{array}{c} \xi '_{1} \\ \xi '_{2} \\ \xi '_{3} \\ \end{array}\displaystyle \right ] . $$

(B.5)

We first calculate the component $M_{11}$ as follows

$$ \begin{aligned}[b] M_{11}&= \int_{0}^{l} \int_{0}^{\pi }\int_{0}^{2\pi }\lambda ( \xi ) \xi_{1}^{2} \bigl( 1-\cos ( k\xi \cos \theta ) \bigr) \xi^{2}\sin \theta \mathrm{d}\phi \mathrm{d}\theta \mathrm{d}\xi \\ &= \frac{k_{1}^{2}}{k^{2}}M_{\Vert }^{\mathrm{III}}+ \biggl( 1- \frac{k _{1}^{2}}{k^{2}} \biggr) M_{\bot }^{\mathrm{III}} , \end{aligned} $$

(B.6)

where

$$ M_{\Vert }^{\mathrm{III}}=\int_{0}^{l} \int_{0}^{\pi }\int_{0} ^{2\pi }\lambda ( \xi ) ( \xi \cos \theta ) ^{2} \bigl( 1-\cos ( k\xi \cos \theta ) \bigr) \xi^{2}\sin \theta \mathrm{d}\phi \mathrm{d}\theta \mathrm{d}\xi $$

(B.7)

and

$$ M_{\bot }^{\mathrm{III}}=\int_{0}^{l} \int_{0}^{\pi }\int_{0} ^{2\pi }\lambda ( \xi ) ( \xi \sin \theta \cos \phi ) ^{2} \bigl( 1-\cos ( k\xi \cos \theta ) \bigr) \xi^{2}\sin \theta \mathrm{d}\phi \mathrm{d}\theta \mathrm{d}\xi . $$

(B.8)

Eq. (B.5) has been used in the above derivation. $M_{\Vert } ^{\mathrm{III}}$ and $M_{\bot }^{\mathrm{III}}$ can be easily calculated, as shown by Silling [6]. Similarly, the other components are calculated and the component form of tensor $\boldsymbol{M} ( \boldsymbol{k} )$ is finally expressed as

$$ M_{im}=M_{\bot }^{\mathrm{III}} \delta_{im}+\frac{k_{i}k_{m}}{k^{2}} \bigl( M_{\Vert }^{\mathrm{III}}-M_{\bot }^{\mathrm{III}} \bigr) , $$

(B.9)

for which

$$\begin{aligned} M_{\Vert }^{\mathrm{III}}&=4\pi \int_{0}^{l} \lambda ( \xi ) \xi^{4}A_{1}^{\mathrm{III}} ( k\xi ) \mathrm{d}\xi , \end{aligned}$$

(B.10)

$$\begin{aligned} M_{\bot }^{\mathrm{III}}&=4\pi \int_{0}^{l} \lambda ( \xi ) \xi^{4}A_{2}^{\mathrm{III}} ( k\xi ) \mathrm{d}\xi , \end{aligned}$$

(B.11)

$$\begin{aligned} A_{1}^{\mathrm{III}} ( k\xi ) &=\frac{1}{3}- \frac{\sin ( k \xi ) }{k\xi }-\frac{2\cos ( k\xi ) }{ ( k\xi ) ^{2}} +\frac{2\sin ( k\xi ) }{ ( k\xi ) ^{3}} , \end{aligned}$$

(B.12)

$$\begin{aligned} A_{2}^{\mathrm{III}} ( k\xi ) &=\frac{1}{3}+ \frac{\cos ( k \xi ) }{ ( k\xi ) ^{2}} -\frac{\sin ( k\xi ) }{ ( k\xi ) ^{3}}. \end{aligned}$$

(B.13)

For the first tensor in (B.1), the above method is adopted and a new orthonormal basis $\{ \boldsymbol{e}'_{1},\boldsymbol{e}'_{2} )$ is established with $\boldsymbol{e}'_{1}$ parallel to the vector $\boldsymbol{k}$. The component form of the tensor in (B.1) is finally given as

$$ M_{im}=M_{\bot }^{\mathrm{II}} \delta_{im}+\frac{k_{i}k_{m}}{k^{2}} \bigl( M_{\Vert }^{\mathrm{II}}-M_{\bot }^{\mathrm{II}} \bigr) , $$

(B.14)

for which

$$\begin{aligned} & \begin{aligned}[b] M_{\Vert }^{\mathrm{II}}&= \int_{0}^{l}\int_{0}^{2\pi }\lambda ( \xi ) ( \xi \cos \theta ) ^{2} \bigl( 1-\cos ( k \xi \cos \theta ) \bigr) \xi \mathrm{d}\theta \mathrm{d}\xi \\ &= 2\pi \int_{0}^{l}\lambda ( \xi ) \xi^{3}A_{1} ( k \xi ) \mathrm{d}\xi , \end{aligned} \end{aligned}$$

(B.15)

$$\begin{aligned} & \begin{aligned}[b] M_{\bot }^{\mathrm{II}}&= \int_{0}^{l}\int_{0}^{2\pi }\lambda ( \xi ) ( \xi \sin \theta ) ^{2} \bigl( 1-\cos ( k \xi \cos \theta ) \bigr) \xi \mathrm{d}\theta \mathrm{d}\xi \\ &= 2\pi \int_{0}^{l}\lambda ( \xi ) \xi^{3}A_{2} ( k \xi ) \mathrm{d}\xi , \end{aligned} \end{aligned}$$

(B.16)

$$\begin{aligned} & A_{1}^{\mathrm{II}} ( k\xi ) =\frac{1}{2}- \mathrm{J}_{0} ( k \xi ) +\frac{\mathrm{J}_{1} ( k\xi ) }{k\xi } , \end{aligned}$$

(B.17)

$$\begin{aligned} & A_{2}^{\mathrm{II}} ( k\xi ) =\frac{1}{2}- \frac{\mathrm{J} _{1} ( k\xi ) }{k\xi }. \end{aligned}$$

(B.18)

Appendix C: Betti’s Reciprocal Theorem, Somigliana Formula and Boundary Integral Equation in Peridynamics

By the definitions of a nonlocal divergence operator and a nonlocal gradient operator in [33], and the concept of peridynamic stress tensor in [34], we can obtain

$$ \int_{\varOmega } ( \boldsymbol{u}\cdot \mathcal{L}\boldsymbol{v}-\boldsymbol{v}\cdot \mathcal{L}\boldsymbol{u} ) \mathrm{d}V= \int_{\partial \varOmega } \bigl( \boldsymbol{u} \cdot \boldsymbol{\tau }(\boldsymbol{v})-\boldsymbol{v}\cdot \boldsymbol{\tau }(\boldsymbol{u}) \bigr) \mathrm{d}S , $$

(C.1)

in which $\boldsymbol{u}$ and $\boldsymbol{v}$ are any vector fields; $\tau$ is the surface traction due to the vector field $\boldsymbol{u}$ or $\boldsymbol{v}$. $\varOmega$ is the reference configuration of a closely bounded body with the boundary $\partial \varOmega$. ℒ represents the peridynamic linear operator

$$ \mathcal{L}\boldsymbol{u}\equiv \int _{\mathcal {H}}\boldsymbol{C} ( \boldsymbol{\xi } )\cdot \bigl[ \boldsymbol{u} ( \boldsymbol{x}+\boldsymbol{\xi },t ) -\boldsymbol{u} ( \boldsymbol{x},t ) \bigr] {\mathrm{d}}V_{\boldsymbol{\xi }} . $$

(C.2)

Eq. (C.1) is the Betti’s reciprocal theorem in peridynamics. The derivation is as follows.

The nonlocal divergence operator $\mathcal{D}$ on $\boldsymbol{\beta}$ is defined as [33]

$$ {\mathcal{D}}(\boldsymbol{\beta }) (\boldsymbol{x})\equiv \int_{\mathcal{H}} \bigl(\boldsymbol{\beta } ( \boldsymbol{x},\boldsymbol{x}' )+ \boldsymbol{\beta } ( \boldsymbol{x}',\boldsymbol{x} )\bigr)\cdot \boldsymbol{\alpha} (\boldsymbol{x},\boldsymbol{x}') \mathrm{d} V_{\boldsymbol{x}'} , $$

(C.3)

in which $\boldsymbol{\beta}(\boldsymbol{x},\boldsymbol{x}')$ is a tensor, and $\boldsymbol{\alpha}$ is an antisymmetric vector, i.e., $\boldsymbol{\alpha}(\boldsymbol{x}',\boldsymbol{x}) = -\boldsymbol{\alpha}(\boldsymbol{x}, \boldsymbol{x}')$. The adjoint operator corresponding to $\mathcal{D}$ is defined as [33]

$$ {\mathcal{D}}^{*}(\boldsymbol{u}) (\boldsymbol{x},\boldsymbol{x}')\equiv -\bigl( \boldsymbol{u}(\boldsymbol{x}')-\boldsymbol{u}(\boldsymbol{x}) \bigr) \otimes \boldsymbol{\alpha} ( \boldsymbol{x}, \boldsymbol{x}' ). $$

(C.4)

If $\boldsymbol{\beta}$ is a second-order tensor, based on (C.3) and (C.4), there is

$$ \mathcal{D}(\boldsymbol{u}\cdot \boldsymbol{\beta}) = \boldsymbol{u}\cdot \mathcal{D}(\boldsymbol{\beta})- \int_{\mathcal{H}}\mathcal{D}^{*}(\boldsymbol{u}): \boldsymbol{\beta} \bigl(\boldsymbol{x}',\boldsymbol{x}\bigr) \mathrm{d}V_{\boldsymbol{x}'}. $$

(C.5)

For linear elasticity in peridynamics, $\boldsymbol{\beta} (\boldsymbol{x}, \boldsymbol{x}')$ is expressed as $\boldsymbol{\beta} (\boldsymbol{x}, \boldsymbol{x}') = \boldsymbol{\varTheta}: \mathcal{D}^{*}(\boldsymbol{u})$. $\boldsymbol{\varTheta}$ is a fourth-order tensor satisfying $\boldsymbol{\varTheta} (\boldsymbol{x}, \boldsymbol{x}') = \boldsymbol{\varTheta} (\boldsymbol{x}', \boldsymbol{x})$ with the symmetry ${\varTheta}_{ijkl} = {\varTheta}_{klij} = {\varTheta}_{jikl} = {\varTheta}_{ijlk}$, which is similar to the classical stiffness tensor. Then

$$\begin{aligned} \mathcal{D}\bigl(\boldsymbol{\varTheta}: \mathcal{D}^{*}(\boldsymbol{u})\bigr) &=-2 \int_{\mathcal{H}} \bigl\{ \boldsymbol{\varTheta}: \bigl[\bigl(\boldsymbol{u}(\boldsymbol{x}')-\boldsymbol{u} (\boldsymbol{x})\bigr) \otimes \boldsymbol{\alpha}(\boldsymbol{x},\boldsymbol{x}')\bigr]\bigr\} \cdot \boldsymbol{\alpha}(\boldsymbol{x},\boldsymbol{x}')\mathrm{d}V_{\boldsymbol{x}'} \\ &=-2\int_{\mathcal{H}} \bigl[\boldsymbol{\alpha}\bigl(\boldsymbol{x},\boldsymbol{x}'\bigr) \cdot\boldsymbol{\varTheta}\cdot\boldsymbol{\alpha} \bigl(\boldsymbol{x},\boldsymbol{x}'\bigr)\bigr]\cdot \bigl(\boldsymbol{u}\bigl(\boldsymbol{x}'\bigr)-\boldsymbol{u}(\boldsymbol{x})\bigr) \mathrm{d}V_{\boldsymbol{x}'}, \end{aligned}$$

(C.6)

in which the symmetry of $\boldsymbol{\varTheta}$ has been used. Comparing Eq. (C.6) with Eq. (C.2), we can get

$$ \mathcal{L}\boldsymbol{u}= -\mathcal{D}\bigl(\boldsymbol{\varTheta}:\mathcal{D}^{*}(\boldsymbol{u})\bigr), $$

(C.7)

where

$$ \boldsymbol{C} (\boldsymbol{\xi}) =2\boldsymbol{\alpha } ( \boldsymbol{x}, \boldsymbol{x}') \cdot \boldsymbol{\varTheta } \cdot\boldsymbol{\alpha} ( \boldsymbol{x},\boldsymbol{x}' ), $$

(C.8)

with $\boldsymbol{\xi}= \boldsymbol{x}' - \boldsymbol{x}$. Thus, the linear elastic micromodulus Eq. (75) is retrieved by choosing

$$ \begin{aligned} & \boldsymbol{\alpha } ( \boldsymbol{x},\boldsymbol{x}' ) = \frac{\boldsymbol{\xi}}{ \vert \boldsymbol{\xi} \vert ^{2} }, \\ &\boldsymbol{\varTheta }=\frac{1}{2}\lambda \bigl(\vert\boldsymbol{\xi}\vert\bigr) (\boldsymbol{\xi}\otimes \boldsymbol{\xi}\otimes\boldsymbol{\xi}\otimes\boldsymbol{\xi}) . \end{aligned} $$

(C.9)

The peridynamic force flux vector or surface traction at a point $\boldsymbol{x}$ is given by [34]

$$ \boldsymbol{\tau}(\boldsymbol{x},\boldsymbol{n})= \boldsymbol{\nu}(\boldsymbol{x})\cdot \boldsymbol{n}, $$

(C.10)

in which $\boldsymbol{\nu}(\boldsymbol{x})$ is the peridynamic stress tensor defined in [34], and $\boldsymbol{n}$ denotes the unit normal vector of a plane. It is proved in [34] that

$$ \mathcal{L}\boldsymbol{u}=\nabla\cdot \boldsymbol{\nu}(\boldsymbol{u}), $$

(C.11)

where $\boldsymbol{\nu}(\boldsymbol{u})$ represents the stress field corresponding to the displacement field $\boldsymbol{u}$. Combining Eqs. (C.7) and (C.11), we can obtain the following relation by the Gauss theorem:

$$ \int_{{\varOmega}} \mathcal{L}\boldsymbol{u}\mathrm{d}V_{\boldsymbol{x}} = -\int_{{\varOmega}} \mathcal{D}\bigl(\boldsymbol{\varTheta}: \mathcal{D}^{*}(\boldsymbol{u})\bigr)\mathrm{d}V_{\boldsymbol{x}} =\int_{{\varOmega}}\nabla\cdot \boldsymbol{\nu}(\boldsymbol{u})\mathrm{d}V_{\boldsymbol{x}} =\int_{\partial{\varOmega}} \boldsymbol{\nu}\cdot \boldsymbol{n} \mathrm{d}S_{\boldsymbol{x}} =\int_{\partial{\varOmega}}\boldsymbol{\tau}(\boldsymbol{u})\mathrm{d}S_{\boldsymbol{x}}. $$

(C.12)

For any displacement vector filed $\boldsymbol{v}$, we can get

$$\begin{aligned} \int_{\varOmega} \boldsymbol{v}\cdot \mathcal{L}\boldsymbol{u} \mathrm{d}V_{\boldsymbol{x}} &=-\int_{\varOmega} \boldsymbol{v}\cdot \mathcal{D} \bigl(\boldsymbol{\varTheta}: \mathcal{D}^{*} (\boldsymbol{u})\bigr)\mathrm{d}V_{\boldsymbol{x}} \\ &=\int_{\partial\varOmega} \boldsymbol{v}\cdot\boldsymbol{\tau}(\boldsymbol{u})\mathrm{d}S_{\boldsymbol{x}} -\int_{\varOmega}\int_{\mathcal{H}}\mathcal{D}^{*} (\boldsymbol{v}):\bigl(\boldsymbol{\varTheta}: \mathcal{D}^{*} (\boldsymbol{u})\bigr) \mathrm{d}V_{\boldsymbol{x}'}\mathrm{d}V_{\boldsymbol{x}}, \end{aligned}$$

(C.13)

where Eqs. (C.5), (C.7) and (C.12) have been used. Exchanging $\boldsymbol{v}$ and $\boldsymbol{u}$ leads to

$$ \int_{\varOmega} \boldsymbol{u}\cdot \mathcal{L}\boldsymbol{v} \mathrm{d}V_{\boldsymbol{x}} =\int_{\partial\varOmega} \boldsymbol{u}\cdot\boldsymbol{\tau} (\boldsymbol{v}) \mathrm{d}S_{\boldsymbol{x}} - \int_{\varOmega}\int_{\mathcal{H}} \mathcal{D}^{*} (\boldsymbol{u}): \bigl(\boldsymbol{\varTheta}: \mathcal{D}^{*} (\boldsymbol{u})\bigr) \mathrm{d}V_{\boldsymbol{x}'}\mathrm{d}V_{\boldsymbol{x}}. $$

(C.14)

Due to the symmetry of the last term of Eq. (C.13) or Eq. (C.14) with respect to $\boldsymbol{u}$ and $\boldsymbol{v}$, subtracting Eq. (C.13) from Eq. (C.14) leads to the Betti’s reciprocal theorem in Eq. (C.1).

In the Betti’s reciprocal theorem Eq. (C.1), let $\boldsymbol{u}$ be the Green’s function of the infinite domain, that is,

$$ \bigl(\mathcal{L}\boldsymbol{u}^{K}\bigr)_{i} =\delta_{iK}\delta (\boldsymbol{x}-\boldsymbol{x}_{0}). $$

(C.15)

The superscript $K$ denotes the direction of the point force. Substituting Eq. (C.15) into Eq. (C.1) and changing $\boldsymbol{x}_{0}$ to $\boldsymbol{x}$ lead to the Somigliana formula of peridynamics,

$$ \boldsymbol{u}(\boldsymbol{x}) \cdot \boldsymbol{e}^{K} = \int_{\partial\varOmega} \bigl[\boldsymbol{u}^{K} (\boldsymbol{\zeta}-\boldsymbol{x})\cdot \boldsymbol{\tau}(\boldsymbol{x})- \boldsymbol{u}(\boldsymbol{\zeta}) \cdot \boldsymbol{\tau}^{K} (\boldsymbol{\zeta}-\boldsymbol{x})\bigr] \mathrm{d}S_{\boldsymbol{\zeta}} + \int_{\varOmega} \bigl(\boldsymbol{u}^{K} (\boldsymbol{\zeta}-\boldsymbol{x}) \cdot \boldsymbol{b}(\boldsymbol{\zeta})\bigr) \mathrm{d}V_{\boldsymbol{\zeta}}, $$

(C.16)

in which $\boldsymbol{e}^{K}$ is the base vector in the direction of the point force, and the static governing equation $\mathcal{L}\boldsymbol{u}+\boldsymbol{b} = 0$ has been used.

If the point $\boldsymbol{x}$ is on the smooth boundary $\partial\varOmega$, the integral $\int_{\varOmega} (\boldsymbol{u}\cdot \mathcal{L}\boldsymbol{v})\mathrm{d}V$ is equal to half of that when the point is not on the boundary. Thus the boundary integral equation in peridynamics is

$$ \frac{1}{2} \boldsymbol{u}(\boldsymbol{x})\cdot \boldsymbol{e}^{K} = \int_{\partial\varOmega} \bigl[\boldsymbol{u}^{K} (\boldsymbol{\zeta}-\boldsymbol{x})\cdot \boldsymbol{\tau}(\boldsymbol{x})-\boldsymbol{u}(\boldsymbol{\zeta}) \cdot \boldsymbol{\tau}^{K} (\boldsymbol{\zeta}-\boldsymbol{x})\bigr] \mathrm{d}S_{\boldsymbol{\zeta}} + \int_{\varOmega} \bigl(\boldsymbol{u}^{K} (\boldsymbol{\zeta}-\boldsymbol{x})\cdot \boldsymbol{b}(\boldsymbol{\zeta})\bigr) \mathrm{d}V_{\boldsymbol{\zeta}}. $$

(C.17)

From Eq. (C.17), the unknown displacement $\boldsymbol{u}$ on the boundary can be calculated from the given traction $\boldsymbol{\tau}$ on the boundary; if the traction $\boldsymbol{\tau}$ is unknown, it can be calculated from the given displacement $\boldsymbol{u}$. Then, using the determined boundary displacement and traction, one can obtain the displacement field in the region $\varOmega$ by the Somigliana formula (C.16).