A new phase field model for material fatigue in oscillating elastoplastic beam

We pursue the study of fatigue accumulation in an oscillating elastoplastic beam under the additional hypothesis that the material can partially recover by the effect of melting. The full system consists of the momentum and energy balance equations, an evolution equation for the fatigue rate, and a differential inclusion for the phase dynamics. The main result consists in proving the existence and uniqueness of a strong solution.


Introduction
It was shown in [16] that the Kirchhoff-Love method of reducing the 3D problem of transversal oscillations of a solid elastoplastic beam with the single yield von Mises plasticity law leads to the beam equation with a multiyield hysteresis Prandtl-Ishlinskii constitutive operator. The present authors have used in [7] (see also [5], [11]) the Prandtl-Ishlinskii formalism to propose a model for the cyclic fatigue accumulation in an oscillating beam and to study its properties; results have been obtained correspondingly also for the plate, see [6], [8], [1]. Here, we extend the model by taking into account the possibility of partial fatigue recovery by the effect of melting when a solid-liquid phase transition takes place.
The fatigue accumulation law is still based on the observation that there exists a proportionality between accumulated fatigue and dissipated energy, see [2,9]. Unlike in [7] and similarly as in [9], we assume that out of all dissipative components in the energy balance, only the purely plastic dissipation produces damage. This makes the mathematical problem easier: the system of equations then does not develop singularities in finite time and a unique regular solution is proved to exist on every bounded time interval. On the other hand we consider here an additional difficulty -we assume that the weight function ϕ in the definition of the Prandtl-Ishlinskii operator depends also on the fatigue parameter m; this has been considered also in [8] and [1].
The unknowns of the problem are the transversal displacement w ∈ R of the beam, the absolute temperature θ > 0, the fatigue variable m ≥ 0, and the phase variable χ ∈ [0, 1]. The full system of equations consists of the momentum balance equation (the beam equation with a fatigue dependent hysteresis operator), the energy balance equation, the fatigue accumulation equation and the phase transition equation. The model is derived in detail in Section 1.
The problem is rigorously stated in Section 2, where we also check the thermodynamic consistency of the system and collect some preliminary material in Section 3. In Section 4 we carry out formally the a priori estimates that allow us to construct the solution of the full system. In Section 5, we apply these ideas to a spatially discrete scheme and derive estimates that are sufficient for proving that the space discrete approximations converge to a solution of the original problem in appropriate function spaces. The main existence and uniqueness Theorem 2.2 is proved in Section 6.
1 The model 1

.1 Governing equations
We consider a transversally inhomogeneous beam of length 1, and denote by x ∈ [0, 1] the longitudinal variable, by t ∈ [0, T ] the time variable, by w(x, t) the transversal displacement of the point x at time t, by ε(x, t) = w xx (x, t) the linearized curvature, and by σ(x, t) the bending moment. We assume a thermo-visco-elasto-plastic scalar constitutive law in the form where B > 0 is a constant hardening modulus, m ≥ 0 is a scalar time and space dependent parameter describing the accumulation of fatigue, where m = 0 corresponds to zero fatigue, P [m, ε] is a fatigue dependent Prandtl-Ishlinskii constitutive operator of elastoplasticity defined below in Subsection 1.2, ν is the viscosity coefficient, β is the thermal bending coefficient related to a layered structure of the beam, θ > 0 is the absolute temperature, and θ ref is a fixed referential temperature (more specifically, the melting temperature). Following [16], Newton's law of motion is formally written as where α = ρl 2 /12 and l > 0 is the thickness of the beam, ρ the mass density and F is the external load.
With the constitutive law (1.1), we associate the free energy operator where V [m, ε] is the fatigue dependent Prandtl-Ishlinskii potential (1.19), c (the specific heat capacity) and L (the latent heat) are given constants, and I [0,1] is the indicator function of the interval [0, 1]. The entropy operator S and internal energy operator U then read We consider the first and the second principles of thermodynamics in the form where q = −κθ x is the heat flux with a constant heat conductivity κ > 0, and g is the heat source density. Note that (1.6) is the energy conservation law, (1.7) is the Clausius-Duhem inequality.
The evolution of the phase variable χ is governed by the inclusion −γχ t ∈ ∂ χ F , that is, . The analysis of the so-called rainflow method of cyclic fatigue accumulation in elastoplastic materials carried out in [2] has shown a close relation between accumulated fatigue and dissipated energy, similarly as in [9]. Here, we assume in addition that partial recovery of the damaged material is possible under strong local melting. Mathematically, this is expressed in terms of the evolution equation for the fatigue variable m where h is a nonnegative nondecreasing function, λ is a nonnegative smooth function with (small) compact support and D[m, ε] is the fatigue dependent dissipation operator, (1.20). The subdifferential ∂I [0,∞) of the indicator function I [0,∞) ensures that the fatigue parameter remains nonnegative.
The meaning of (1.9) is simple. If no phase transition takes place or if the material solidifies, that is, χ t ≤ 0, then fatigue at a point x increases proportionally to the energy dissipated in a neighborhood of the point x. On the other hand, under strong melting if χ grows faster than the plastic dissipation rate, the fatigue may decrease until it possibly reaches the unperturbed state m = 0.

Hysteresis operators
Let us first recall the definition of the stop. Definition 1.1. Let u ∈ W 1,1 (0, T ) and a closed connected set Z ⊂ R be given. The variational inequality defines the stop and play operators s Z and p Z by the formula For a canonical choice of Z = [−r, r] with some r > 0 and for the initial condition z(0) = Q r (u(0)), where Q r is the projection of R onto the interval [−r, r], we simply write (1.12) A simple proof of the following easy properties of the play and stop can be found e.g. in [13].
The variational inequality (1.10) can be equivalently written as the inclusionż(t) + ∂I Z (z(t)) ∋ u(t). This enables us to rewrite the differential inclusions (1.8) and (1.9) for the phase variable χ and fatigue variable m with a choice χ 0 (x) ∈ [0, 1], m 0 (x) ≥ 0 of initial conditions in the form where The advantage of this representation is that now, χ and m are defined by equations involving, by virtue of Proposition 1.2, only operators that are Lipschitz continuous in C[0, T ] and in W 1,1 (0, T ).
The variational inequality (1.10) is also used to model single-yield elastoplasticity. In this case, the constraint Z = [−r, r] is the admissible stress domain, the input u = ε is the strain, and the output z = σ r := s r [ε] is the stress. We can rewrite (1.10) equivalently in "energetic" forṁ Indeed,ε(t)σ r (t) is the power supplied to the system, part of it is used for the increase of the potential 1 2 σ 2 r (t), and the rest r|ξ(t)| is dissipated. The Prandtl-Ishlinskii model is constructed as a linear combination of stops with all possible yield points r > 0. Here, given a measurable function ϕ : [0, ∞) × (0, ∞) → [0, ∞) satisfying Hypothesis 2.1 (i) below, we define the fatigue dependent Prandtl-Ishlinskii operator P : (W 1,1 (0, T )) 2 → W 1,1 (0, T ) by the integral ( we can write the Prandtl-Ishlinskii energy balance in the forṁ

Statement of the problem
For any T > 0, we denote Ω T : for unknown functions u, w, θ, m, χ with initial and boundary conditions The zero initial conditions for w and m are motivated by the fact that it is difficult to determine the initial degree of fatigue for a material with unknown loading history, and the most transparent hypothesis consists in assuming that no deformation (and therefore no fatigue) has taken place prior to the time t = 0.
The data are required to fulfill the following hypotheses: , locally Lipschitz continuous in the first variable, and there existφ, (iv) f ∈ L 2 (Ω T ) is a given function for some fixed T > 0, such that f tt , f xt ∈ L 2 (Ω T ).
The assumption that ϕ(m, r) decreases with increasing fatigue m corresponds to the observation that the stiffness of the material decreases with increasing fatigue. Also the assumption that g 0 (x, t) ≥ 0 makes sense. Note that g is the heat source density, so that at zero temperature, we cannot remove heat from the system.
By Hypothesis 2.1 (i) and (2.5), hence, by Hypothesis 2.1 (vi) provided we check that the absolute temperature θ stays positive. In Subsection 5.1, we will find a positive lower bound for the discrete approximations of the temperature, which is independent of the discretization parameter, and therefore is preserved in the limit and implies the positivity of the temperature.
The main result of this paper reads as follows.
Theorem 2.2. Let Hypothesis 2.1 hold. Then there exists a unique solution to the system ∈ Ω T , and with the regularity
The Gagliardo-Nirenberg inequality states that there exists a constant C > 0 such that for every v ∈ W 1,p (0, 1) we have In fact, (3.1) is straightforward: If we introduce an auxiliary parameter r = 1 + s(1 − 1 p ) and use the chain rule d dx |v(x)| r ≤ r|v(x)| r−1 |v ′ (x)| a.e., we obtain from Hölder's inequality the estimate

be a vector, and let us denote
The discrete counterpart of (3.1) reads and can be easily derived from (3.1) by defining v e.g. as equidistant piecewise linear interpolations of v k .

Formal estimates
In order to explain the estimation technique, we first proceed formally, assuming that the positivity of temperature is already established. For the sake of simplicity we set from now on Due to the fact that, by Definition 1.1 we have |s r [w xx ](t)| ≤ r and from Hypothesis 2.
Finally, due to Hypothesis 2.1 (i) and (iii) and (1.22), we have We will denote in the sequel by C any constant possibly depending on the constants in Hypothesis 2.1 and on the initial data of the problem.

The energy estimate
We multiply (2.1) by w xxt , obtaining we differentiate (2.2) in time and multiply by w t , getting and finally we sum up (4.5), (4.6) and (2.3), all integrated in space. The first term in (4.5) simplifies with the third term in (4.6) due to integration by parts and our choice of boundary conditions; moreover the viscosity terms cancel out and also a term β θ. Concerning the term with hysteresis, using (1.21) we deduce and thus in the sum of (4.5), (4.6) and (2.3) what remains is just the term containing V . More precisely we have the energy balance (4.8) and Gronwall's argument together with Hypothesis 2.1 (iv) and (vii) gives the estimate (4.9)

The Dafermos estimate
We test the equation for the temperature (2.3) by θ −1/3 and obtain, using notations (4.1) and (4.2), that We keep the terms where we have integrated by parts and used the boundary conditions (2.7). All the other terms will be estimated from below. First, In the identity we have T 3 ≥ 0 whenever m t ≥ 0. On the other hand, if m t < 0, then by (2.5), (4.2), and Hypothesis 2.1 (vi) we have Now the assumption m t < 0 implies that χ t > 0. Then, however, by (2.4), we have that (4.14) Combining the above inequalities with (4.3), we obtain for m t < 0 that We obviously have The term can be treated in a similar way as the term T 3 and using (4.14) for χ t = 0 we get Finally, we find a lower bound for T 8 by Hypothesis 2.1 (vii) as follows: Coming back to (4.10), integrating it in time, we deduce where we put The first two terms on the right hand side of (4.16) are bounded due to (4.9). The last term we estimate by Hölder inequality as follows and (4.16) yields that Now we apply (3.1) with v = θ 1/3 , s = 3, q = 5, p = 2, ρ = 4/25 and notice that due to (4.9). We therefore have Combining this last estimate with (4.17), we deduce Applying again the Gagliardo-Nirenberg inequality with the choices v = θ 1/3 , s = 3, q = 8, p = 2, ρ = 1/4, we obtain that and this after space integration, together with (4.9) and (4.18) brings the estimate To derive a further estimate, we sum again (4.5) and (4.6), and obtain d dt We estimate the first term on the right hand side using the inequality βθw xxt ≤ β 2 2ν θ 2 + ν 2 w 2 xxt and the previous estimate (4.19

Approximation
Here, we make rigorous the estimates derived formally in the previous section. From now on, the values of all physical constants are set to 1 for simplicity.
This is a system of ODEs for u k , w k , θ k . We proceed as follows: We claim that (5.1)-(5.6) admits a W 1,∞ solution in an interval [0, T n ]. First, denoting by w the vector (w 1 , . . . , w n−1 ), and ε = (ε 1 , . . . , ε n−1 ), we have, by (5.3), −ε = Sw with a positive definite matrix S , which has the form Hence, the left hand side of (5.2) reads (I + S)ẇ. By (5.2),ε is itself a Lipschitz continuous mapping of u = (u 1 , . . . , u n−1 ). Using Proposition 1.2 (ii) we see that (5.1)-(5.4) can be considered as an ODE system in u k , w k , θ k , with a locally Lipschitz continuous and locally bounded right hand side and the existence and uniqueness of a local solution in an interval [0, T n ] follows from the standard theory of ODEs, and the solution belongs to W 1,∞ (0, T n ).

Positivity of the temperature
In this subsection, we prove that θ k remain positive in the whole range of existence. As a first step, we test On the other hand, by (5.9) we also have −n Moreover, D k (t) ≥ 0 and g k (θ, t) ≥ 0 for θ ≤ 0 by Hypothesis 2.1 (vii), hence Now we deal with the phase term. We have thaṫ by virtue of (5.10). Summarizing the above computations, we get d dt 1 2n where we put K ε,n := max{|ε k (t)| : k = 1, . . . , n − 1, t ∈ [0, T n ]} and Gronwall's argument yields θ − k (t) = 0 for all k and t ∈ [0, T n ]. We now prove that in fact, θ k (t) are bounded away from 0 for all k and all t ∈ [0, T n ]. First of all we notice that ifχ k = 0 then On the other handṁ Using the estimates above together witḣ where we set Let p be the solution of the differential equatioṅ with θ * > 0 from Hypothesis 2.1 (v). It is easy to check that Testing (5.11) by (p − θ k ) + and using (5.9), we obtain hence, as ψ is nondecreasing for positive arguments, so that θ k (t) ≥ p(t) > 0 for all k and all t ∈ [0, T n ], which is the desired result.

The discrete Dafermos estimate
We test (5.4) by θ −1/3 k and we proceed similarly as in Subsection 4.2. The integration by parts is replaced by the elementary inequality with the choice x = θ k , y = θ k−1 . We obtain for all t ∈ [0, T ] after summing up from k = 1, ..., n − 1 and integrating in time the following counterpart of (4.16): The last two terms on the right hand side are bounded by virtue of (5.16). By (5.15),

Higher order discrete estimates
We define ε 0 , ε n for k = 0, k = n, as solutions to the differential equations Then (5.1) holds for all k = 0, . . . , n, and we havė for all k = 1, . . . n. By (5.9) we have hence, by (5.1)-(5.2) and (5.23), This yields, as a counterpart to (4.23), d dt As in (4.30), we have |P [m k , ε k ] t | ≤ C(1 + θ k + 1 n n−1 k=1 |ε j |), and this enables us to estimate the terms on the right hand side of (5.25) as follows: We estimate the initial conditions as in (4.33), and integrating (5.25) in time we conclude from the above considerations that 1 2n Gronwall's argument and (5.20) then yields the following counterpart to (4.36) 1 2n We now test (5.4) by θ k and obtain d dt where, by (5.20) we infer from (5.32) that and we rewrite (5.34) as (compare with (4.43)) We estimate the right hand side of (5.35) using (3.3) as follows: We have t 0 |ε(τ )| 2 2 dτ ≤ C by virtue of (5.21), hence Combining (5.35) with (5.36) yields Therefore there exist a constant C > 0 such that

Uniqueness
Let (u, w, θ, χ, m), (ũ,w,θ,χ,m) be two solutions of (2.1)-(2.7), with the same initial conditions and the same right hand sides. We integrate the difference of (2.3) for θ andθ in time, and estimate the terms on the right hand side as follows: We now test the resulting inequality by θ −θ and integrate in x. Taking into account the above estimates, we finally obtain In the next step, we test the difference of the time derivatives of (2.2) for w andw by w t −w t , the difference of (2.1) for u andũ by w xxt −w xxt , and sum up. Arguing as above, we obtain Gronwall's argument now yields that w =w, θ =θ, and the proof of Theorem 2.2 is complete.