Analysis of the Volume-Constrained Peridynamic Navier Equation of Linear Elasticity

Abstract

Well-posedness results for the state-based peridynamic nonlocal continuum model of solid mechanics are established with the help of a nonlocal vector calculus. The peridynamic strain energy density for an elastic constitutively linear anisotropic heterogeneous solid is expressed in terms of the field operators of that calculus, after which a variational principle for the equilibrium state is defined. The peridynamic Navier equilibrium equation is then derived as the first-order necessary conditions and are shown to reduce, for the case of homogeneous materials, to the classical Navier equation as the extent of nonlocal interactions vanishes. Then, for certain peridynamic constitutive relations, the peridynamic energy space is shown to be equivalent to the space of square-integrable functions; this result leads to well-posedness results for volume-constrained problems of both the Dirichlet and Neumann types. Using standard results, well-posedness is also established for the time-dependent peridynamic equation of motion.

Appendix A

We provide proofs of Lemma 1 and the space equivalence results that lead to the proofs of Theorems 2 and 3. For the sake of brevity, for the latter result, we only treat the case corresponding to a homogeneous Dirichlet volume constraint.

1.1 A.1 Proof of Lemma 1

Let $\mathbf{v}\in U_{0}({\varOmega\cup\varOmega_{\mathcal{I}}})$. The variation of the potential energy E(u;b,g) is calculated as

The relation (8b) implies that

$$ \operatorname{Tr} \bigl(\mathcal{D}_\omega^\ast \mathbf{u} \bigr) (\mathbf{x}) = \int_{{\varOmega\cup\varOmega_{\mathcal{I}}}} \operatorname{Tr} \bigl(\mathcal{D}^\ast\mathbf{u} \bigr) (\mathbf{x},\mathbf{y}) \omega ( \mathbf{x},\mathbf{y}) \,d\mathbf{y}\quad \mbox{for}\ \mathbf{x}\in\varOmega $$

(26a)

and the identity (a⊗b)c=(a⊗c)b=(b⋅c)a leads to the nonlocal Green’s identities [9] for the trace of tensor divergences:

(26b)

(26c)

Together with relations (4) and (10), we then have

so that Lemma 1 is established.

1.2 A.2 Space Equivalences

Under the interior cone condition on the domain stated in Assumption 2, for any $\mathbf{x}\in{\bar{\varOmega}}_{\mathcal{I}}$ we may assume that a cone $B_{r_{0}}^{\theta_{0}}(\mathbf{x})$ centered at x with radius r ₀ and angle θ ₀ is contained in the domain ${\varOmega\cup \varOmega_{\mathcal{I}}}$. For brevity, we hereby only present the case associated with the homogeneous Dirichlet type volume constrained condition. We now show that, under Assumptions 1–4, the energy space $U({\varOmega\cup\varOmega_{\mathcal{I}}})$ is in fact equivalent to $L^{2}({\varOmega\cup\varOmega_{\mathcal{I}}})$. First, we note that ϕ(x) and τ(x), defined in (10) and (13b), respectively, have the following upper and lower bounds.

Lemma 3

If the domain and coefficients in the peridynamic model (16a) satisfy Assumptions 1–4, we have that

where

Proof

Let $\pi(\mathbf{x})=\frac{1}{3}\int_{B_{r_{0}}^{\theta_{0}}(\mathbf {x})}|\mathbf{y}-\mathbf{x} |^{2}\varpi(\mathbf{x},\mathbf{y})\,d\mathbf{y}$. By (10), we estimate

By the interior cone condition on the domain ${\varOmega\cup\varOmega _{\mathcal{I}}}$, we have ϕ(x)≥π(x). On the other hand, because |y−x|² ϖ(x,y)>0 on any set of nonzero measure in $B_{r_{0}}^{\theta_{0}}(\mathbf{x})$ and by the non-degeneracy condition, we have π(x)>0 for any x in ${\bar{\varOmega}}_{\mathcal{I}}$. Given the integrability condition on ϖ(x,y), we have that π(x) is continuous in ${\bar{\varOmega}}_{\mathcal{I}}$ so that

$$\phi(\mathbf{x}) \geq\min_{\mathbf{x}\in{\overline{\varOmega \cup\varOmega}_{\mathcal{I}}}}\pi(\mathbf{x}) = \pi_0(r_0, \theta_0) > 0. $$

Then, one readily sees that

$$\bigl|\tau(\mathbf{x})\bigr| = \frac{|\sigma(\mathbf{x})|}{\phi^2(\mathbf {x})} < \frac{\kappa_1 + \frac {1}{3}\eta_1 \pi_1({\varOmega\cup\varOmega_{\mathcal{I}}})}{\pi_0(r_0,\theta_0)} $$

and the desired conclusion now follows. □

The following result expresses $\mathcal{L}$ in a form suitable for analysis.

Lemma 4

Let the operator $\mathcal{L}$ and the functions σ and τ be given defined by (12) and (13a), (13b) respectively. Then, for $\mathbf{u}\in U({\varOmega\cup\varOmega_{\mathcal {I}}})$, we have the equivalent expression

$$ \mathcal{L}(\mathbf{u}) (\mathbf{x}) = \int _{{\varOmega\cup \varOmega_{\mathcal{I}}}}\boldsymbol{\varPhi}(\mathbf{x},\mathbf {y}) \bigl( \mathbf{u}(\mathbf{y})-\mathbf{u}(\mathbf{x}) \bigr)\,d\mathbf{y}, $$

(27)

where Φ(x,y)=Φ ₁(x,y)+Φ ₂(x,y) with

(28a)

(28b)

Proof

We first rewrite the first term of (12) as

and the second term of (12) as

By interchanging the order of integration and relabeling the variables, we have

and

Therefore,

$$\mathcal{D}_{\omega} \bigl(\sigma \operatorname{Tr} \bigl( \mathcal{D}_\omega^\ast(\mathbf{u}) \bigr)\mathbf {I} \bigr) ( \mathbf{x}) = \int_{{\varOmega\cup\varOmega_{\mathcal {I}}}}\boldsymbol{\varPhi}_2( \mathbf{x},\mathbf{y}) \bigl(\mathbf {u}(\mathbf{x})-\mathbf{u}(\mathbf{y}) \bigr) \,d\mathbf{y} $$

and the conclusion of the lemma now follows. □

We now define the tensor field

$$ \mathbf{P}(\mathbf{x}) = \int_{\varOmega\cup\varOmega_{\mathcal{I}}}\boldsymbol{ \varPhi }(\mathbf{x},\mathbf{y})\,d\mathbf{y} = \int_{{\varOmega\cup\varOmega_{\mathcal{I}}}} \bigl(\boldsymbol {\varPhi}_1(\mathbf{x},\mathbf{y}) + \boldsymbol{ \varPhi }_2(\mathbf{x},\mathbf{y}) \bigr)\,d\mathbf{y}, $$

(29)

where Φ ₁ and Φ ₂ are defined in (28a) and (28b), respectively. We next show that the tensor field P(x) is uniformly positive definite and uniformly bounded.

Lemma 5

If the domain and coefficients in the peridynamic model (16a) satisfy Assumptions 1–4, we have that P=P(x) is uniformly bounded from above and is uniformly positive definite, i.e., P(x)≥P ₀>0 in ${\varOmega\cup\varOmega_{\mathcal{I}}}$, where

$$ \mathbf{P}_0 = \eta_0 \min_{\mathbf {x}\in{\overline{\varOmega\cup\varOmega}_{\mathcal{I}}} }\int _{B_{r_0}^{\theta_0}(\mathbf{x})}\frac{\varpi(\mathbf {x},\mathbf{y})}{|\mathbf{y}-\mathbf{x}|^2}(\mathbf{y} -\mathbf{x})\otimes( \mathbf{y}-\mathbf{x})\,d\mathbf{y} $$

(30)

and $B_{r_{0}, \theta_{0}}(\mathbf{x})$ represents the intersection of a ball with radius r ₀ centered at x and a cone with vertex x and angle θ ₀.

Proof

By Assumption 4, it is easy to see that P=P(x) is uniformly bounded from above in ${\varOmega\cup\varOmega_{\mathcal{I}}}$. Next, for any unit vector e∈ℝ³, we may use a similar argument as in the proof of the Lemma 3 to obtain that

$$\mathbf{e}\cdot \biggl[\int_{B_{r_0}^{\theta_0}(\mathbf{x})}\frac{\varpi(\mathbf {x},\mathbf{y})}{|\mathbf{y}-\mathbf{x} |^2}(\mathbf{y}- \mathbf{x})\otimes(\mathbf{y}-\mathbf{x})\, d\mathbf{y} \biggr] \mathbf{e} = \int _{B_{r_0}^{\theta_0}(\mathbf{x})}\frac{\varpi(\mathbf {x},\mathbf{y})}{|\mathbf{y}-\mathbf{x}|^2}\bigl[\mathbf{e} \cdot(\mathbf{y}- \mathbf{x})\bigr]^2\,d\mathbf{y} $$

is uniformly bounded below by a positive constant, independent of $\mathbf{x}\in{\varOmega\cup \varOmega_{\mathcal{I}}}$. Taking the minimum over all e on the unit sphere in ℝ³, we see that P ₀ is a positive definite matrix.

From the definitions (13b) and (28b), we obtain

By changing the order of integrals, we obtain

By Assumption 3, κ(x)≥κ ₀ and by Lemma 3, ϕ(x)≥π ₀(r ₀,θ ₀)>0 so that A ₁≥0. By the definition of Φ ₁ in (28a), we obtain

$$\int_{\varOmega\cup\varOmega_{\mathcal{I}}}\boldsymbol{\varPhi }_1(\mathbf{x},\mathbf{y}) = 2\eta(\mathbf{x})\int_{\varOmega \cup\varOmega_{\mathcal{I}}}\frac{\varpi(\mathbf{x} ,\mathbf{y})}{|\mathbf{y}-\mathbf{x}|^2}(\mathbf{y}-\mathbf {x})\otimes(\mathbf{y}-\mathbf{x})\,d\mathbf{y}= 2\mathbf{A}_3. $$

Because of the interior cone condition on the domain ${\varOmega\cup \varOmega_{\mathcal{I}}}$, we have

$$\mathbf{A}_3 = \eta(\mathbf{x})\int_{\varOmega\cup\varOmega _{\mathcal{I}}}\frac{\varpi(\mathbf{x},\mathbf{y})}{|\mathbf {y}-\mathbf{x} |^2}(\mathbf{y}-\mathbf{x})\otimes(\mathbf{y}-\mathbf{x})\, d\mathbf{y}\geq\mathbf{P}_0. $$

We can rewrite P(x) as P(x)=2A ₃+A ₁−A ₂=A ₃+A ₁+(A ₃−A ₂). Because A ₃≥P ₀>0 and A ₁≥0, we only need to show that A ₃−A ₂ is semi-positive definite, i.e., A ₃−A ₂≥0, which is equivalent to demonstrating m(A ₃−A ₂)≥0 so that a factor of 3 in the second term cancels. Let e denote an arbitrary nonzero vector. By the definition of ϕ(x) (3), we have

By the Cauchy–Schwarz inequality, we have

which implies that e⋅m(A ₃−A ₂)e≥0 for any nonzero e, i.e., m(A ₃−A ₂)≥0. The conclusion of the Lemma now follows. □

We then have the following nonlocal Korn inequality for the peridynamic Navier equation.

Lemma 6

(Nonlocal Korn inequality)

Let the domain and coefficients in the bilinear form (18) satisfy Assumptions 1–4. Then, there exists a constant c ₁>0 such that, provided that $\varOmega_{\mathcal{I}}$ has a non-empty interior, for $\mathbf {u}\in U_{0}({\varOmega\cup\varOmega_{\mathcal{I}}})$ we have

(31)

This inequality also holds for $\mathbf{u}\in U({\varOmega\cup \varOmega_{\mathcal{I}}})/Z({\varOmega\cup\varOmega_{\mathcal{I}}})$.

Proof

We show Φ ₁(x,y) and Φ ₂(x,y) are square integrable which in turn implies that Φ ₁(x,y)+Φ ₂(x,y) is square integrable. First,

Similarly,

By a change of variables, we have

where $|\mathrm{diam}({\varOmega\cup\varOmega_{\mathcal{I}}})|$ denotes the diameter of the domain ${\varOmega\cup\varOmega_{\mathcal{I}}}$. The final inequality holds by the integrability condition on ϖ(y,x). By Lemma 3, we obtain $\int_{\varOmega\cup\varOmega_{\mathcal{I}}}\int_{\varOmega\cup \varOmega_{\mathcal{I}}}|\boldsymbol{\varPhi}_{2}(\mathbf{x},\mathbf {y})|^{2}\,\allowbreak\,d\mathbf{x}\,d\mathbf{y} <\infty$, so that

$$ \int_{\varOmega\cup\varOmega_{\mathcal{I}}}\int_{\varOmega\cup \varOmega_{\mathcal{I}}}\bigl| \boldsymbol{\varPhi}(\mathbf{x},\mathbf {y})\bigr|^2\, d\mathbf{x}\,d\mathbf{y}<\infty. $$

(32)

For simplicity, we separate the operator $\mathcal{L}$ as

$$-\mathcal{L}\mathbf{u}= -\int_{\varOmega\cup\varOmega_{\mathcal {I}}}\boldsymbol{\varPhi}( \mathbf{x},\mathbf{y})\mathbf{u}(\mathbf {y})\,d\mathbf{y}+ \biggl(\int _{\varOmega\cup\varOmega_{\mathcal{I}}}\boldsymbol{\varPhi }(\mathbf{x},\mathbf{y})\,d\mathbf{y} \biggr)\mathbf{u}(\mathbf {x}) = \mathcal{A}\mathbf{u}+ \mathbf {P}\mathbf{u}, $$

where P=P(x) is defined as in (29); Lemma 5 implies that P is a positive-definite operator uniformly bounded from above and with an uniformly bounded inverse. Combining with (32), we have

$$\int_{\varOmega\cup\varOmega_{\mathcal{I}}}\int_{\varOmega\cup \varOmega_{\mathcal{I}}}\bigl| \mathbf{P}^{-1/2}( \mathbf{x})\boldsymbol {\varPhi}(\mathbf{x},\mathbf{y})\mathbf {P}^{-1/2}( \mathbf{y})\bigr|^2\, d\mathbf{x}\,d\mathbf{y}<\infty. $$

We can then invoke the theory on Hilbert-Schmidt integral operators [20] to conclude that the operator $\mathbf{P}^{-1/2}\mathcal{A}\mathbf{P}^{-1/2}$ is a compact operator with at most a sequence of real eigenvalues λ _i with only 0 as the possible accumulation point of the spectrum. Thus, we see that $\mathbf{P}^{-1/2}\mathcal{A}\mathbf{P}^{-1/2}+\mathbf{I}$ can have at most a sequence of real eigenvalues λ _i with only 1 as the possible accumulation point of the spectrum. Because the operator $-\mathcal{L}= \mathcal{A} + \mathbf{P}$ is non-negative and, under the Assumptions 1–4, the kernel of ⦀⋅⦀ only contains the zero element in $U_{0}({\varOmega\cup \varOmega_{\mathcal{I}}})$ or $U({\varOmega\cup\varOmega_{\mathcal{I}}})/Z({\varOmega\cup \varOmega_{\mathcal{I}}})$, we have that $\mathbf{P}^{-1/2}\mathcal {A}\mathbf {P}^{-1/2}+\mathbf{I}$ is non-negative as well with 0 not in the spectrum. This implies that $\mathbf {P}^{-1/2}\mathcal{A}\mathbf{P}^{-1/2}+\mathbf{I}$ and thus $-\mathcal{L}= \mathcal{A} + \mathbf{P}$ are in fact positive-definite operators with positive eigenvalues. This establishes the desired estimate (31). □

We also provide an upper bound for the energy norm.

Lemma 7

Under the hypotheses of Lemma 6, we have

(33)

where c ₂ is a constant.

Proof

We establish the bound (33) by showing that both

$$ \int_{\varOmega\cup\varOmega_{\mathcal{I}}}\int_{\varOmega\cup \varOmega_{\mathcal{I}}}\eta\varpi(\mathbf{x}, \mathbf{y}) \operatorname{Tr}\bigl(\mathcal{D}^\ast\mathbf{u}\bigr) \operatorname{Tr}\bigl(\mathcal{D}^\ast \mathbf{u}\bigr)\,\,d\mathbf{y} \,d\mathbf{x}\quad \mbox{and} \quad\int_{\varOmega\cup\varOmega_{\mathcal {I}}}|\sigma| \operatorname{Tr}\bigl(\mathcal{D}_\omega^\ast\mathbf {u}\bigr) \operatorname{Tr}\bigl(\mathcal{D}_\omega^\ast\mathbf{u}\bigr)\,d \mathbf{x} $$

are bounded from above by $c\|\mathbf{u}\|_{L^{2}}^{2}$ for a generic constant c. The upper bound for first term follows from the estimates

The final inequality above holds because

$$\int_{\varOmega\cup\varOmega_{\mathcal{I}}}\varpi(\mathbf {x},\mathbf{y})\,d\mathbf{y}< |{ \varOmega\cup\varOmega_{\mathcal {I}}}|^{1/2} \biggl(\int _{\varOmega\cup\varOmega_{\mathcal{I}}} \varpi(\mathbf{x},\mathbf{y})^2\,d\mathbf{y} \biggr)^{1/2} < |{\varOmega\cup\varOmega_{\mathcal{I}}}|^{1/2} M^{1/2}. $$

Because $|\sigma(\mathbf{x})| \leq\kappa_{1} + \eta_{1} \pi _{1}({\varOmega\cup\varOmega_{\mathcal{I}}})/3$, we have that

where $\hat{\kappa}_{1}=\kappa_{1} + \eta_{1} \pi_{1}({\varOmega\cup \varOmega_{\mathcal{I}}})/3$. Combining results leads to the conclusion of the lemma. □

Together, Lemmas 6 and 7 demonstrate that the energy space is a Hilbert Space and is, in fact, $L^{2}({\varOmega\cup\varOmega_{\mathcal{I}}})$ and therefore Theorems 2 and 3 are established.