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# An inverse problem related to an elasto-plastic beam

Laurent Bourgeois, Jean-François Mercier

► To cite this version:

Laurent Bourgeois, Jean-François Mercier. An inverse problem related to an elasto-plastic beam. Inverse Problems, 2025, 41 (10), 10.1088/1361-6420/ae0e49. hal-05036216v2

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https://inria.hal.science/hal-05036216v2

Submitted on 20 Aug 2025

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# An inverse problem related to an elasto-plastic beam

August 20, 2025

LAURENT BOURGEOIS AND JEAN-FRANÇOIS MERCIER

Laboratoire Poems, CNRS, INRIA, Ensta Paris, Institut Polytechnique de Paris,
828 Boulevard des Maréchaux, 91762 Palaiseau, France

## Abstract

We consider an elasto-plastic beam and address the following inverse problem: external forces have created some plastic strains in this beam, which therefore shows a residual observable deformation once the structure is load-free. Can we retrieve the loading history from this observation, or at least the plastic strains ? After proving the well-posedness of the forward problem, we show that the solution can be described in a semi-explicit way in the pure bending case, so that the forward problem amounts to a one dimensional non linear problem. Such problem is smooth enough for us to solve the inverse problem by using a classical least-square method, which is illustrated with the help of some numerical examples.

## 1 Introduction

This paper is dedicated to an inverse problem related to a two-dimensional elasto-plastic beam. The context is the Non Destructive Testing of a beam which has been submitted to a certain unknown loading history, this loading being sufficiently severe to have created plastic strains within the structure. Observing the deformation of the beam at the end of this loading history, that is once the structure is load-free again, the question is: are we able to reconstruct such loading history, or at least the plastic zone and the plastic strains ? This question was first raised in the PhD thesis [3] of the first author, the results of which were briefly presented by his PhD advisor in [22]. There are not many contributions in the field of inverse problems in elasto-plasticity. Most contributions concern the identification of the elasto-plastic parameters from measurements, like in [13, 12]. In [1], an ill-posed data completion problem for an elasto-plastic structure is addressed. A shape optimization problem for an elasto-plastic material is considered in [18] in the simpler non-incremental case and in [7] in the time domain. But the problem of retrieving the plastic strains from the residual deformation after an unknown loading history has received very little attention, probably because it is extremely ill-posed. Maybe the first contribution in this direction goes back to [9], but their approach
does not exploit the elasto-plasticity law, which leads to poor numerical results. Another
pioneering contribution is [10], but it considers a specific indentation test. In [25] a similar
study as in [3] is carried out in the case of pure bending plates, but does not provide
numerical experiments. If the displacements are measured at the end of an increasing
loading history, then the problem is simpler and the plastic zone can be detected by
using a level set method like in [2]. It should be noted that in the presence of large
deformations, the plastic strains (or analogously the residual stresses) affect the elastic
tensor itself, which implies that NDT procedures using waves could be used to identify the
modified elastic tensor in a first step, and then to identify the plastic strains themselves
in a second step. This method was formally proposed in [5] but has never been used so
far, to our best knowledge. However, a method in the same spirit was used to identify
an initial displacement in an hyperelastic structure in [6].

In the present paper, we first describe the forward evolution problem of an elasto-plastic
beam in the presence of a kinematic hardening. Well-posedness of evolution problems
in elasto-plasticity has been studied for a long time since the pioneering work of J.-J.
Moreau [21] and is pretty well known, see for example [15, 14, 24, 16]. Note that the
monograph [11] gathers a comprehensive list of well-posedness results in elasto-plasticity,
in particular with various assumptions on the type of hardening under consideration.
The elasto-plastic beam model that we consider in this article is very simple and rather
classical, relying in particular on the Bernoulli assumption. It is for example used in
[23] and [28]. In a first part, we show that we can recast the forward problem in an
abstract framework proposed in [11] and prove an existence/uniqueness result for the
displacement and the plastic strain fields in an appropriate functional space. In a second
part, we propose an alternative formulation to prove an existence/uniqueness result for
the thermodynamical force associated with the plastic strain. This formulation consists
of a differential inclusion that perfectly fits the general theory of multivalued maximal
monotonous operators exposed in [4]. In a third part, for the particular case of the pure
bending loading, we partially solve the forward problem explicitely (in the sense that the
dependence of fields in the thickness of the beam is explicit), showing that it amounts
to a one dimensional (non-linear) problem that can be easily solved numerically. This
one dimensional problem paves the way for the inverse problem, which is solved with the
help of a classical least-square method based on a gradient descent, such gradient being
obtained by using an adjoint state. This classical technique could not be used directly
starting from the former equations of the problem. Indeed, the adjoint variables result
from the differentiation of the equations with respect to the state variables, which seems
impossible in the presence of a differential inclusion. The fact that we partially solve
the problem in the transverse direction of the beam in order to derive a one-dimensional
problem enables us to circumvent this issue. Note that a related problem to ours is the
optimal control of elasto-plasticity, the theory of which is addressed for example in [19],
[20], [26] and [27]. The added value of our contribution compared to [22] is twofold: firstly
we provide a mathematical framework of our approach and the corresponding rigorous
justifications, both for the forward and the inverse problems, and secondly we illustrate
its efficiency with the help of numerical simulations.

Our article is organized as follows. In the second section we describe the simple elastic-
plastic model that we use. Our well-posedness result with the help of the abstract framework of [11] is established in section 3, the one obtained with the help of the abstract framework of [4] is established in section 4. The special case of pure traction or pure bending is addressed in section 5. A quasi-explicit solution to the forward problem in the particular pure bending case is established in section 6 while the inverse problem is addressed in section 7. Lastly, some numerical experiments are presented in section 8.

## 2 Description of the model

In this article, a generic point of the two dimensional beam is given by $(x, y) \in \Omega := (0, l) \times (-h, h)$, where $l, h > 0$ and $x$ is the (horizontal) coordinate in the axis of the beam while $y$ is the transverse (vertical) coordinate. The length of the beam, that is $l$, is supposed to be much larger than the thickness of the beam, that is $2h$. We restrict to a very simple model, that is the strain and the stress tensors are given by $\varepsilon e_x \otimes e_x$ and $\sigma e_x \otimes e_x$, respectively, where $e_x$ is the unit vector along the $x$-axis, and $\varepsilon, \sigma$ are scalar functions which depends on $(t, x, y)$, where $t$ is the time. In what follows, $g'$ and $\dot{g}$ will denote the derivative of $g$ with respect to $x$ and $t$, respectively. In addition, we assume that the strain field is of the Bernoulli form, that is

$$ \varepsilon(t, x, y) = u'(t, x) - y v''(t, x), \quad (1) $$

where $u, v$ denote the horizontal and vertical displacement of the neutral fiber of the beam defined by $(0, l) \times \{0\}$. We assume that the beam is simply supported at points $x = 0$ and $x = l$, which implies that for all $t$

$$ u(t, 0) = 0, \quad v(t, 0) = v(t, l) = 0. \quad (2) $$

Assuming that the beam is made of an elasto-plastic material, the plastic strain is of the form $\alpha(t, x, y) e_x \otimes e_x$, while the elastic strain is hence given by $(\varepsilon - \alpha)(t, x, y) e_x \otimes e_x$. We assume that the behaviour of the material is governed by the following energy

$$ W(\varepsilon, \alpha) := \frac{1}{2} E(\varepsilon - \alpha)^2 + \frac{1}{2} H \alpha^2, $$

which corresponds to an elasto-plastic material with a kinematical hardening, where $E > 0$ is the Young modulus and $H > 0$ is the hardening parameter. Note that the case $H = 0$ would correspond to perfect plasticity. The state laws provide

$$ \sigma := \frac{\partial W}{\partial \varepsilon} = E(\varepsilon - \alpha) \quad \text{and} \quad a := -\frac{\partial W}{\partial \alpha} = \sigma - H\alpha, \quad (3) $$

where the first relationship is the Hooke's law while the second one provides the thermodynamic force $a$ which must belong to the convex of plasticity given by the segment $[-k, k]$, where $k > 0$ is the plasticity threshold. More precisely, the plasticity flow rule is given by

$$ \dot{\alpha} \in \partial I_{[-k, k]}(a), \quad (4) $$
where $I_{[-k,k]}$ is the indicator function of the convex set $[-k, k]$ and $\partial I_{[-k,k]}(a)$ its sub-gradient at point $a$. From the definition of the subgradient, we obtain that

$$ \partial I_{[-k,k]}(a) = \begin{cases} (-\infty, 0] & \text{if } a = -k \\ \{0\} & \text{if } a \in (-k, k) \\ [0, +\infty) & \text{if } a = k. \end{cases} $$

We also assume that the beam is subjected to a distributed force $(n, -p)$, where $n(t, x)$ and $-p(t, x)$ are the longitudinal and transverse components of such force. The quasi-static equilibrium of the beam at each time $t$ is obtained by minimizing for fixed $\alpha$ the potential energy

$$ U(t) := \int_{0}^{l} \int_{-h}^{h} W(\varepsilon(t), \alpha(t)) dy dx - \int_{0}^{l} n(t) u(t) dx + \int_{0}^{l} p(t) v(t) dx $$

with respect to the functions $u(t)$ and $v(t)$ satisfying the kinematical conditions (2). Writing that the Fréchet derivatives of $U(t)$ with respect to $u(t)$ and $v(t)$ satisfying those conditions both vanish implies that

$$ \int_{0}^{l} \int_{-h}^{h} E(\varepsilon(t) - \alpha(t)) \hat{u}' dy dx - \int_{0}^{l} n(t) \hat{u} dx = 0, \quad \forall \hat{u} \text{ such that } \hat{u}(0) = 0, \quad (5) $$

and

$$ - \int_{0}^{l} \int_{-h}^{h} E(\varepsilon(t) - \alpha(t)) y \hat{v}'' dx dy + \int_{0}^{l} p(t) \hat{v} dx = 0, \quad \forall \hat{v} \text{ such that } \hat{v}(0) = \hat{v}(l) = 0. \quad (6) $$

From (5) and (6) we deduce that

$$ N(t, l) = 0, \quad M(t, 0) = M(t, l) = 0, \quad N' = -n, \quad M'' = p, \quad (7) $$

where we have defined

$$ N := \int_{-h}^{h} \sigma \, dy, \quad M := \int_{-h}^{h} y \sigma \, dy, \quad (8) $$

which are the normal force and the bending moment, respectively. Starting from the initial conditions

$$ u(0, x) = v(0, x) = 0, \quad \alpha(0, x, y) = 0 \quad (9) $$

and assuming a loading distribution $(n(t, x), -p(t, x))$ such that

$$ n(0, x) = p(0, x) = 0, \quad (10) $$

the forward problem consists in determining the values of $u(t, x)$, $v(t, x)$ and $\alpha(t, x, y)$ for $t > 0$ by solving the system of equations (1)-(4) and (7)-(10).
### 3 Well-posedness of the forward problem

In this section we adopt the approach proposed in the monograph [11], and solve the so-called primal variational problem of elastoplasticity (see chapter 7 of [11]). The following well-posedness result for an abstract variational inequality is proved in [11] (see Theorem 6.15).

**Theorem 3.1.** Let us consider the problem: find $w: [0, T] \to V$, $w(0) = 0$, such that for almost all $t \in (0, T)$, $\dot{w}(t) \in V$ and

$$b(w(t), \tilde{w} - \dot{w}(t)) + j(\tilde{w}) - j(\dot{w}(t)) \geq \langle \ell(t), \tilde{w} - \dot{w}(t) \rangle, \quad \forall \tilde{w} \in V. \quad (11)$$

Here, $V$ denotes a Hilbert space, $b: V \times V \to \mathbb{R}$ a symmetric, bounded and $V$-coercive bilinear form and $j: V \to \mathbb{R}$ a non-negative, convex, positively homogeneous and Lipschitz continuous function. We have also denoted $\ell \in H^1(0, T; V')$ with $\ell(0) = 0$, while $\langle \cdot, \cdot \rangle$ consists of the duality bracket between $V'$ and $V$, where $V'$ is the dual space of $V$. Such problem has a unique solution $w \in H^1(0, T; V)$.

Let us now specify the space $V$, the bilinear form $b$, the function $j$ as well as the source term $\ell$. In this view we need to introduce the dissipation potential $D$, defined as the Fenchel-Legendre conjugate function of the indicator function $I_{[-k,k]}$. We obtain

$$D(\dot{\alpha}) = k|\dot{\alpha}|$$

and from a classical result of convex analysis, the inclusion (4) is equivalent to the inclusion

$$a \in \partial D(\dot{\alpha}). \quad (12)$$

Let us introduce the spaces $H_{0,0}^1(0, l) := \{u \in H^1(0, l), u(0) = 0\}$, $H_0^1(0, l) := \{v \in H^1(0, l), v(0) = v(l) = 0\}$ and $V := H_{0,0}^1(0, l) \times (H^2(0, l) \cap H_0^1(0, l)) \times L^2(\Omega)$, which may be equipped with the norm $\|w\|_V$ defined for $w = (u, v, \alpha) \in V$ by

$$\|w\|_V^2 := \int_0^l (u')^2 dx + \int_0^l (v'')^2 dx + \int_{\Omega} \alpha^2 dxdy.$$

For $w = (u, v, \alpha) \in V$ and $\tilde{w} = (\tilde{u}, \tilde{v}, \tilde{\alpha}) \in V$, let us define the bilinear form

$$b(w, \tilde{w}) := \int_{\Omega} (E(u' - yv'' - \alpha)(\tilde{u}' - y\tilde{v}' - \tilde{\alpha}) + H\alpha\tilde{\alpha}) dxdy,$$

the function

$$j(\tilde{w}) := \int_{\Omega} D(\tilde{\alpha}) dxdy = k \int_{\Omega} |\tilde{\alpha}| dxdy$$

and the linear form

$$\langle \ell, \tilde{w} \rangle := \int_0^l n \tilde{u} dx - \langle p, \tilde{v} \rangle_{H^{-1}(0,l), H_0^1(0,l)},$$

where $n \in H^1(0, T; L^2(0, l))$ and $p \in H^1(0, T; H^{-1}(0, l))$ with $n(0) = p(0) = 0$. The second term above is well-defined since $H^{-1}(0, l) = (H_0^1(0, l))' \subset (H^2(0, l) \cap H_0^1(0, l))'$.
That the function $j$ and the source term $\ell$ fulfill the assumptions of Theorem 3.1 is straightforward. The symmetry property as well as the boundedness of the bilinear form $b$ are obvious. It remains to prove the $V$-coercivity of $b$, which is the subject of the following lemma, in which we make use of the notations

$$S := 2h, \quad I := \frac{2}{3}h^3, \quad \langle \cdot \rangle := \frac{1}{2h} \int_{-h}^{h} \cdot dy.$$

**Lemma 3.2.** For $H > 0$, there exists a constant $c > 0$ such that for all $w \in V$,

$$b(w, w) \geq c \|w\|_V^2.$$

*Proof.* For $w = (u, v, \alpha) \in V$,

$$ \begin{aligned} b(w, w) &= \int_{\Omega} E(u' - yv'' - \alpha)^2 dx dy + \int_{\Omega} H\alpha^2 dx dy \\ &= \int_{\Omega} E(\varepsilon - \alpha)^2 dx dy + \int_{\Omega} H\alpha^2 dx dy \\ &= \int_{\Omega} E\varepsilon^2 dx dy + \int_{\Omega} E\alpha^2 dx dy - 2 \int_{\Omega} E\varepsilon\alpha dx dy + \int_{\Omega} H\alpha^2 dx dy. \end{aligned} $$

For any $\eta > 0$, we have

$$-2 \int_{\Omega} E\varepsilon\alpha \,dx\,dy \geq -\eta \int_{\Omega} E\varepsilon^2 \,dx\,dy - \frac{1}{\eta} \int_{\Omega} E\alpha^2 \,dx\,dy.$$

We hence obtain that

$$b(w, w) \geq E(1 - \eta) \int_{\Omega} \varepsilon^2 \,dx\,dy + \left(E + H - \frac{E}{\eta}\right) \int_{\Omega} \alpha^2 \,dx\,dy.$$

Choosing $\eta \in (E/(E+H), 1)$, there exists $c > 0$ such that

$$b(w, w) \geq c \left( \int_{\Omega} \varepsilon^2 \,dx\,dy + \int_{\Omega} \alpha^2 \,dx\,dy \right).$$

Using that $\varepsilon = u' - yv''$, we observe that

$$\int_{\Omega} \varepsilon^2 \,dx\,dy = \int_{\Omega} ((u')^2 + y^2(v'')^2) \,dx\,dy = S \int_{0}^{l} (u')^2 \,dx + I \int_{0}^{l} (v'')^2 \,dx,$$

which completes the proof. $\square$

Now let us establish the variational inequation (11). From (3) and (12), we have that

$$D(\tilde{\alpha}) \geq D(\dot{\alpha}) + (E(\varepsilon - \alpha) - H\alpha)(\tilde{\alpha} - \dot{\alpha}), \quad \forall \tilde{\alpha} \in L^2(\Omega),$$

which implies after integration on $\Omega$ that

$$j(\tilde{w}) \geq j(\dot{w}) + \int_{\Omega} ((E(\varepsilon - \alpha) - H\alpha)(\tilde{\alpha} - \dot{\alpha})dx dy, \quad \forall \tilde{\alpha} \in L^2(\Omega). \quad (13)$$
Applying (5) with $\hat{u} = \tilde{u} - \dot{u}(t)$ implies that

$$ \int_{\Omega} E(\varepsilon - \alpha)(\tilde{u}' - \dot{u}') \, dxdy - \int_{0}^{l} n(t)(\tilde{u} - \dot{u}) \, dx = 0, \quad \forall \tilde{u} \in H_{0,0}^{1}(0, l). \quad (14) $$

Applying (6) with $\hat{v} = \tilde{v} - \dot{v}(t)$ implies that

$$ - \int_{\Omega} E(\varepsilon - \alpha)y(\tilde{v}'' - \dot{v}'') \, dxdy + \langle p(t), (\tilde{v} - \dot{v}) \rangle = 0, \quad \forall \tilde{v} \in H^{2}(0, l) \cap H_{0}^{1}(0, l). \quad (15) $$

Adding (13), (14) and (15), we obtain (11). From Theorem 3.1 and Lemma 3.2, we directly obtain the following proposition.

**Proposition 1.** Assume that $n \in H^1(0, T; L^2(0, l))$ and $p \in H^1(0, T; H^{-1}(0, l))$ with $n(0) = p(0) = 0$. If $H > 0$, the problem (1)-(4) and (7)-(10) has a unique solution $(u, v, \alpha) \in H^1(0, T; V)$.

**Remark 1.** In the statement of the Proposition 1, $p(t)$ is supposed to belong to $H^{-1}(0, l)$, which allows us to consider for example a finite linear combination of Dirac distributions, in other words pointwise forces.

# 4 A link with Brezis theory of multivalued maximal monotonous operators

In this section, we show that the forward problem may be alternatively written as a differential inclusion involving the thermodynamic force $a$ solely, such problem being well-posed in view of the theory of multivalued maximal monotonous operators developed in [4].

Using the Hooke's law in (3) and the Bernoulli assumption (1), we obtain

$$ N = ES(u' - \langle \alpha \rangle), \quad M = -EIv'' - ES \langle y \alpha \rangle. \quad (16) $$

From (1), (3) and (16), we derive an expression of $a$ directly in terms of $N$, $M$ and $\alpha$ which will be crucial in what follows:

$$ a = \frac{N}{S} + \frac{M}{I}y + E\langle\alpha\rangle + \frac{ES}{I}y\langle y\alpha\rangle - (E+H)\alpha. \quad (17) $$

Since the normal force $N$ and the bending moment $M$ are completely determined from $(n, -p)$ by using (7), the forward problem boils down to find $a$ and $\alpha$ satisfying (4) and (17). Let us set

$$ f := \frac{N}{S} + \frac{M}{I}y \quad (18) $$

and

$$ L\alpha := (E + H)\alpha - E\langle\alpha\rangle - \frac{ES}{I}y\langle y\alpha\rangle. \quad (19) $$
From (4) and (17), we obtain pointwise

$$ \dot{f} \in \dot{a} + (L \circ \partial I_{[-k,k]})(a), $$

with initial condition $a(0, x, y) = 0$.

Let us now recast such problem in an appropriate functional space. In this view we denote $V^T := L^2(0, T; L^2(\Omega))$ and the set $K \subset V^T$ given by

$$ K := \{a \in V^T, a(t, x, y) \in [-k, k] \text{ for almost all } (t, x, y) \in (0, T) \times \Omega\}. \quad (20) $$

We first need the following functional inequality.

**Lemma 4.1.** For all $\varphi \in L^2(-h, h)$,

$$ \int_{-h}^{h} \left( \langle \varphi \rangle^2 + \frac{S^2}{I^2} y^2 \langle y\varphi \rangle^2 \right) dy \le \int_{-h}^{h} \varphi^2 dy. $$

*Proof.* Assume firstly that $\varphi$ is an even function. We hence have $\langle y\varphi \rangle = 0$ and by using the Cauchy-Schwarz inequality,

$$ \langle \varphi \rangle^2 = \frac{1}{S^2} \left( \int_{-h}^{h} \varphi dy \right)^2 \le \frac{1}{S} \int_{-h}^{h} \varphi^2 dy. $$

Then

$$ \int_{-h}^{h} \langle \varphi \rangle^2 dy = S \langle \varphi \rangle^2 \le \int_{-h}^{h} \varphi^2 dy, $$

which is the result. Assume secondly that $\varphi$ is an odd function. We hence have $\langle y\varphi \rangle = 0$ and by using the Cauchy-Schwarz inequality,

$$ \langle y\varphi \rangle^2 = \frac{1}{S^2} \left( \int_{-h}^{h} y\varphi dy \right)^2 \le \frac{I}{S^2} \int_{-h}^{h} \varphi^2 dy. $$

Then

$$ \int_{-h}^{h} \frac{S^2}{I^2} y^2 \langle y\varphi \rangle^2 dy = \frac{S^2}{I} \langle y\varphi \rangle^2 \le \int_{-h}^{h} \varphi^2 dy, $$

which again is the result. In order to address the general case, we use the unique decomposition

$$ \varphi = \varphi_e + \varphi_o, $$

where $\varphi_e$ and $\varphi_o$ are even and odd functions respectively. Hence

$$ \int_{-h}^{h} \left( \langle \varphi \rangle^2 + \frac{S^2}{I^2} y^2 \langle y\varphi \rangle^2 \right) dy = \int_{-h}^{h} \left( \langle \varphi_e \rangle^2 + \frac{S^2}{I^2} y^2 \langle y\varphi_o \rangle^2 \right) dy. $$

By using the previous results for $\varphi_e$ and $\varphi_o$, we get

$$ \int_{-h}^{h} \left( \langle \varphi \rangle^2 + \frac{S^2}{I^2} y^2 \langle y\varphi \rangle^2 \right) dy \le \int_{-h}^{h} \varphi_e^2 dy + \int_{-h}^{h} \varphi_o^2 dy. $$
It suffices now to observe that

$$ \int_{-h}^{h} \varphi^2 dy = \int_{-h}^{h} \varphi_e^2 dy + \int_{-h}^{h} \varphi_o^2 dy + \int_{-h}^{h} 2\varphi_e \varphi_o dy = \int_{-h}^{h} \varphi_e^2 dy + \int_{-h}^{h} \varphi_o^2 dy, $$

which is the result.

□

We have the following lemma which specifies the properties of L.

**Lemma 4.2.** For $H > 0$, the operator $L: V^T \to V^T$ is a self-adjoint and positive isomorphism.

*Proof.* Let us first check that the operator $L$ defined by (19) is continuous. For $\alpha \in V^T$, for almost all $t \in (0, T)$,

$$ \int_{\Omega} \langle \alpha \rangle^2 dxdy = S \int_{0}^{l} \langle \alpha \rangle^2 dx = S \int_{0}^{l} \left( \frac{1}{S} \int_{-h}^{h} \alpha dy \right)^2 dx \\ \leq \frac{1}{S} \int_{0}^{l} \left( S \int_{-h}^{h} \alpha^2 dy \right) dx = \int_{\Omega} \alpha^2 dxdy. $$

Integrating the above inequality with respect to $t$, we obtain there exists a constant $C > 0$ such that

$$ \|\langle \alpha \rangle\|_{V^T} \leq C \|\alpha\|_{V^T}, \quad \forall \alpha \in V^T. $$

The same inequality is obtained for $y \langle y \alpha \rangle$ instead of $\langle \alpha \rangle$, which proves that $L$ is continuous.

Now let us set $\beta = L\alpha$. From (19), computing $\langle \beta \rangle$ and $\langle y\beta \rangle$ provides

$$ \langle \beta \rangle = (E + H)\langle \alpha \rangle - E\langle \alpha \rangle = H\langle \alpha \rangle $$

and

$$ \langle y \beta \rangle = (E + H)\langle y \alpha \rangle - E\langle y \alpha \rangle = H\langle y \alpha \rangle. $$

We hence obtain that

$$ \alpha = \frac{1}{E+H} \left( \frac{E}{H} \langle \beta \rangle + \frac{ES}{HI} y \langle y \beta \rangle + \beta \right), $$

from which we infer that $L$ is invertible. For $\alpha, \beta \in V^T$, for almost all $t \in (0, T)$, we have using Fubini's theorem,

$$ \int_{\Omega} \langle \alpha \rangle \beta \, dxdy = \frac{1}{S} \int_{0}^{l} \int_{-h}^{h} \left( \int_{-h}^{h} \alpha(x, z, t) \, dz \right) \beta(x, y, t) \, dxdy \\ = \frac{1}{S} \int_{0}^{l} \int_{-h}^{h} \alpha(x, z, t) \left( \int_{-h}^{h} \beta(x, y, t) \, dy \right) dxdz = \int_{\Omega} \alpha \langle \beta \rangle \, dxdy. $$

and

$$ \int_{\Omega} y \langle y \alpha \rangle \beta \, dxdy = \frac{1}{S} \int_{0}^{l} \int_{-h}^{h} y \left( \int_{-h}^{h} z \alpha(x, z, t) \, dz \right) \beta(x, y, t) \, dxdy $$
$$ = \frac{1}{S} \int_{0}^{l} \int_{-h}^{h} \alpha(x, z, t) z \left( \int_{-h}^{h} y \beta(x, y, t) dy \right) dxdz = \int_{\Omega} \alpha y \langle y \beta \rangle dxdy. $$

This shows that $L$ is self-adjoint. It remains to prove that $L$ is positive. For we have to compute $(L\alpha, \alpha)_{L^2(\Omega)}$ for almost all $t$. We have

$$ (L\alpha, \alpha)_{L^2(\Omega)} = (E + H) \int_{\Omega} \alpha^2 dxdy - E \int_{\Omega} \alpha \left( \langle \alpha \rangle + \frac{S}{I} y \langle y \alpha \rangle \right) dxdy. $$

Since

$$ \begin{aligned} - \int_{\Omega} \alpha \left( \langle \alpha \rangle + \frac{S}{I} y \langle y \alpha \rangle \right) dxdy &\geq -\frac{1}{2} \int_{\Omega} \alpha^2 dxdy - \frac{1}{2} \int_{\Omega} \left( \langle \alpha \rangle + \frac{S}{I} y \langle y \alpha \rangle \right)^2 dxdy \\ &= -\frac{1}{2} \int_{\Omega} \alpha^2 dxdy - \frac{1}{2} \int_{\Omega} \left( \langle \alpha \rangle^2 + \frac{S^2}{I^2} y^2 \langle y \alpha \rangle^2 \right) dxdy \\ &\geq -\frac{1}{2} \int_{\Omega} \alpha^2 dxdy - \frac{1}{2} \int_{\Omega} \alpha^2 dxdy = - \int_{\Omega} \alpha^2 dxdy, \end{aligned} $$

where we have used Lemma 4.1 to obtain the last inequality. We finally obtain that for $\alpha \in V^T$ and almost all $t$,

$$ (L\alpha, \alpha)_{L^2(\Omega)} \geq (E + H) \int_{\Omega} \alpha^2 dxdy - E \int_{\Omega} \alpha^2 dxdy = H \int_{\Omega} \alpha^2 dxdy, $$

which proves that the operator $L$ is positive if $H > 0$. $\square$

<figure><img src="image_3.png" /></figure>

We are now in a position to specify the correct framework to solve our differential inclusion in $V^T$:

$$ \left\{ \begin{array}{l} \dot{f} \in \dot{a} + (L \circ \partial I_K)(a) \\ a(0) = 0 \end{array} \right. \qquad (21) $$

with $f$ and $L$ given by (18) and (19), respectively. In this view we need the following theorem proved in [4].

**Theorem 4.3.** Let $H$ be a Hilbert space, and $R$ the multivalued operator of $H$ defined by $R := \partial\Phi$, where $\Phi$ is a convex, proper and lower semi-continuous function on $H$, with $\min \Phi = 0$.

For $F \in L^2(0, T; H)$ and $A_0 \in D(R)$, where $D(R)$ is the domain of $R$, the problem of finding $A(t) \in H$ for $t \in (0, T)$ such that

$$ \left\{ \begin{array}{l} F \in \frac{dA}{dt} + R(A) \\ A(0) = A_0 \end{array} \right. \qquad (22) $$

as a unique strong solution $A \in H^1(0, T; H)$, that is $A(t) \in D(R)$ and $F \in dA/dt + R(A)$ for almost all $t \in (0, T)$.

From theorem 4.3, we obtain the following corollary.
**Proposition 2.** Let us assume that $n \in H^1(0, T; L^2(0, l))$ and $p \in H^1(0, T; H^{-1}(0, l))$, with $n(0) = p(0) = 0$. Let us define $N$ and $M$ by (7), $f$ by (18) and $L$ by (19). The problem (21) has a unique solution $a \in H^1(0, T; L^2(\Omega))$ if $H > 0$.

*Proof.* The regularity assumptions for $n$ and $p$ imply that $\dot{n} \in L^2(0, T; L^2(0, l))$ and $\dot{p} \in L^2(0, T; H^{-1}(0, l))$. In view of (7), we have that $\dot{N} \in L^2(0, T; L^2(0, l))$ and $\dot{M} \in L^2(0, T; L^2(0, l))$, which from (18) implies that $\dot{f} \in V^T$. Assume that $H > 0$. From Lemma 4.2, we can define $L^{1/2}, L^{-1/2} : V^T \to V^T$, which implies that

$$L^{-1/2} \dot{f} \in L^{-1/2} \dot{a} + (L^{1/2} \circ \partial I_K)(a).$$

Setting $F = L^{-1/2} \dot{f} \in V^T$, $A = L^{-1/2} a \in V^T$ and $K = L^{-1/2} K \subset V^T$, we have

$$F \in \dot{A} + (L^{1/2} \circ \partial I_K)(L^{1/2} A) = \dot{A} + \partial(I_K \circ L^{1/2})(A) = \dot{A} + \partial I_K(A).$$

We have used the fact that $L^{1/2}$ is self-adjoint in the first equality. We also have $a = 0$ at $t = 0$, so that $A = 0$ at $t = 0$. The function $A$ is hence solution to the problem (22) with $H = L^2(\Omega)$, $\Phi = I_K$ and $A_0 = 0$. Besides, $I_K$ is a convex, proper and lower semi-continuous function on $H$, with $\min I_K = 0$. It is clear that in this case $D(R) = D(\partial I_K) = K$, so that $0 \in D(R)$, and $F \in L^2(0, T; L^2(\Omega))$. We are hence exactly in the framework of Theorem 4.3, which guarantees existence of a unique solution $A \in H^1(0, T; L^2(\Omega))$ such that $A \in K$, hence of a unique solution $a \in H^1(0, T; L^2(\Omega))$ such that $a \in K$. $\square$

**Remark 2.** It should be noted that the statements of Propositions 1 and 2 are perfectly consistent, both from the point of view of the assumptions and from the point of view of the conclusions.

# 5 The pure traction and pure bending cases

The pure traction case corresponds to $p = 0$ while the pure bending case corresponds to $n = 0$. From (7), we obtain that $M = 0$ in the pure traction case while $N = 0$ in the pure bending case. From (17) and (4), as long as the thermodynamical force $a$ lies in the plasticity convex set $(-k, k)$, we have $a = N/S$ in the pure traction case, that is $a$ is symmetric with respect to $y$ and $a = (M/I)y$ in the pure bending case, that is $a$ is antisymmetric with respect to $y$. In view of (4) again, we have that the plastic strain $\alpha$ is also symmetric with respect to $y$ in the pure traction case, antisymmetric with respect to $y$ in the pure bending case. Now, from (16) and (2), we deduce that $v = 0$ in the pure traction case while $u = 0$ in the pure bending case. In order to specify Proposition 1 in the pure traction/bending cases, let us define the spaces

$$L_0^2(\Omega) := \{\alpha \in L^2(\Omega), \alpha(x, -y) = -\alpha(x, y) \text{ for almost all } (x, y) \in \Omega\}$$

and

$$L_1^2(\Omega) := \{\alpha \in L^2(\Omega), \alpha(x, -y) = \alpha(x, y) \text{ for almost all } (x, y) \in \Omega\},$$
as well as

$$V_0 := (H^2(0, l) \cap H_0^1(0, l)) \times L_0^2(\Omega)$$

and

$$V_1 := H_{0,0}^1(0, l) \times L_1^2(\Omega).$$

A modified version of Lemma 3.2, the proof of which is similar, is the following.

**Lemma 5.1.** For $H > 0$, there exists a constant $c > 0$ such that for all $w_0 = (v, \alpha) \in V_0$,

$$b_0(w_0, w_0) \geq c \|w_0\|_{V_0}^2$$

and for all $w_1 = (u, \alpha) \in V_1$,

$$b_1(w_1, w_1) \geq c \|w_1\|_{V_1}^2,$$

where $b_0((v, \alpha), (\tilde{v}, \tilde{\alpha})) := b((0, v, \alpha), (0, \tilde{v}, \tilde{\alpha}))$ and $b_1((u, \alpha), (\tilde{u}, \tilde{\alpha})) := b((u, 0, \alpha), (\tilde{u}, 0, \tilde{\alpha}))$.

In view of Lemma 5.1, Theorem 3.1 provides the following adapted version of Proposition 1.

**Proposition 3.** Assume that $n \in H^1(0, T; L^2(0, l))$ and $p \in H^1(0, T; H^{-1}(0, l))$ with $n(0) = p(0) = 0$. If $n = 0$ and $H > 0$, the problem (1)-(4) and (7)-(10) has a unique solution $(v, \alpha) \in H^1(0, T; V_0)$ and $u = 0$. If $p = 0$ and $H > 0$, the problem (1)-(4) and (7)-(10) has a unique solution $(u, \alpha) \in H^1(0, T; V_1)$ and $v = 0$.

**Remark 3.** The assumption $H > 0$ seems crucial in the statement of all our well-posedness results concerning the forward problem, which excludes perfect plasticity, that is $H = 0$. However, we will mention this case in the sequel.

# 6 A quasi-explicit solution in the pure bending case

In this section, we consider the particular case

$$n = 0, \quad p(x, t) = \text{Hist}(t) p_m(x),$$

where $p_m \in H^{-1}(0, l)$, $p_m \geq 0$ in the distributional sense and $\text{Hist} \in H_0^1(0, T)$, in particular $\text{Hist}(0) = \text{Hist}(T) = 0$. We assume in addition that the function $\text{Hist}$ is increasing from $t = 0$ to some $t = t_m \in (0, T)$, then is decreasing from $t_m$ to $T$, with $\text{Hist}(t_m) = 1$. Note that $n = 0$ corresponds to a pure bending case, the beam being submitted to a pressure $p$ which is first increasing from 0 to a maximum $p_m$, secondly decreasing from $p_m$ to 0, with $p \in H^1(0, T; H^{-1}(0, l))$, $p(0) = p(T) = 0$ and $p \geq 0$ in the distributional sense. This type of loading history, that is the pressure is first increasing from 0 to a certain distribution then decreasing to 0, is supposed to mimick an accident (like a bump) which happened in the past and which couldn't be observed by anybody, except once it is over. This is a typical Non Destructive Testing situation. From Proposition 3, we obtain that the problem (1)-(4) and (7)-(10) has a unique solution $(v, \alpha) \in H^1(0, T; V_0)$ and $u = 0$ for any $H > 0$. If the maximum pressure $p_m$ is sufficiently large, then some
plastic strains $\alpha$ appear, which implies that the transverse displacement $v$ of the beam will not be zero at final time $T$.

We now show that the forward problem can be described in a more precise way in this particular case. In addition, it will be easily solved numerically since it amounts to a one dimensional (non-linear) problem.

## 6.1 The loading phase $t \in [0, t_m]$

Let us denote $M_m \in H_0^1(0, l)$ the solution to the problem

$$M_m(0) = M_m(l) = 0, \quad M_m'' = p_m. \qquad (23)$$

In view of the weak maximum principle and $p_m \ge 0$, we have that $M_m \le 0$ and $M_m$ is a convex function. Let us assume that the minimum of $M_m$, denoted $M_m^{\min}$, is reached on the segment $S_m \subset (0, l)$, which may be reduced to one point. Starting from $t = 0$, the beam has a linear behaviour as long as $|a(t, x, y)| < k$, a.e., which implies from (17) that

$$a(t, x, y) = \frac{M(t, x)}{I} y, \quad M(t, x) = \text{Hist}(t) M_m(x).$$

From the above expression, the plastification starts at points $(x, y) \in S_m \times (\{-h\} \cup \{h\})$ and for time $t_e \in (0, t_m)$ such that

$$\frac{-M(t_e, x)}{I} h = k,$$

which leads to

$$\text{Hist}(t_e) M_m^{\min} = M_0, \quad M_0 := -\frac{2}{3} k h^2. \qquad (24)$$

More precisely, for $x \in S_m$, $a(t_e, x, h) = -k$ while $a(t_e, x, -h) = k$.

For $t > t_e$, the plastic strain $\alpha$ is no longer zero. As already said, we have

$$\alpha(t, x, -y) = -\alpha(t, x, y), \qquad (25)$$

which implies that $\langle \alpha \rangle(x, t) = 0$ for almost all $(t, x) \in (0, t_m) \times (0, l)$. For a given $t > t_e$, let us define $x^{-}(t) \le x^{+}(t)$ such that

$$M(t, x^{-}(t)) = M(t, x^{+}(t)) = M_0. \qquad (26)$$

We divide the beam $\Omega$ into an elastic and a plastic zone, that is $\Omega = \Omega_e(t) \cup \Omega_p(t)$, so that $\alpha = 0$ in the elastic zone and $\alpha \ne 0$ in the plastic one. We have

$$\Omega_p(t) = \{(x, y) \in \Omega, |y| > q(t, x)\} \quad \text{and} \quad \Omega_e(t) = \Omega \setminus \Omega_p(t),$$

where $q: [t_e, T] \times [0, l] \to [0, h]$ is a continuous function such that $0 \le q(t, x) < h$ for $x \in (x^{-}(t), x^{+}(t))$ and $q(t, x) = h$ for $x \in [0, x^{-}(t)] \cup [x^{+}(t), l]$. The plastic and elastic zones are represented on Figure 1. The objective is now to compute $q$ and $\alpha$. The identity
<figure><img src="image_4.png" /><figcaption>Figure 1: Plastic zone is in red</figcaption></figure>

(17) becomes

$$a(t, x, y) = \frac{M}{I} y + 2 \frac{E}{I} y \int_{-h}^{-q(t, x)} y \alpha(t, x, y) dy - (E + H) \alpha(t, x, y). \quad (27)$$

In view of (27), the plastic strain in the plastic zone $\Omega_p(t)$ is necessarily of the form

$$\alpha(t, x, y) = \alpha_0(t, x) + y \alpha_1(t, x), \quad y < -q(t, x), \quad (28)$$

which determines $\alpha(t, x, y)$ for $y > q(x, t)$ by (25). Then (27) can be rewritten in the whole domain $\Omega$:

$$a(t, x, y) = \frac{M}{I} y - \frac{E}{I} \alpha_0 y (h^2 - q^2) + \frac{2}{3} \frac{E}{I} y (h^3 - q^3) \alpha_1 - (E + H) \alpha(t, x, y). \quad (29)$$

In the domain $\Omega_p(t)$, precisely $y < -q(t, x)$, we have that $a = k$, which implies from (28) and (29) that

$$\alpha_0 = -\frac{k}{E+H} \quad (30)$$

and

$$\frac{M}{I} + \frac{k}{I} \frac{E}{E+H} (h^2 - q^2) - \frac{2}{3} \frac{E}{I} q^3 \alpha_1 - H \alpha_1 = 0. \quad (31)$$

In the domain $\Omega_e(t)$, we have from (29) and $\alpha = 0$ that

$$a(t, x, y) = \frac{M}{I} y + \frac{k}{I} \frac{E}{E+H} y (h^2 - q^2) + \frac{2}{3} \frac{E}{I} y (h^3 - q^3) \alpha_1.$$

Taking $y = -q(t, x)$ and using the continuity of $a$ at the interface $y = -q(t, x)$ of the plastic zone $\Omega_p(t)$, in which $a = k$, and the elastic one $\Omega_e(t)$, implies that for $x \in (x^-(t), x^+(t))$

$$\frac{M}{I} + \frac{k}{I} \frac{E}{E+H} (h^2 - q^2) + \frac{2}{3} \frac{E}{I} (h^3 - q^3) \alpha_1 = -\frac{k}{q}. \quad (32)$$
By combining the equations (30), (31) and (32), we finally obtain $\alpha_1 = -k/(E + H)q$ and that the plastic strain is given in $\Omega_p(t)$ by

$$ \alpha(t, x, y) = \frac{k}{E + H} \left( \operatorname{sgn}(y) - \frac{y}{q(t, x)} \right), \quad (33) $$

where $q$ satisfies the following cubic equation for $x \in (x^-(t), x^+(t))$

$$ \frac{kH}{E+H} + \frac{M(t,x)}{I}q(t,x) + \frac{k}{I}\frac{E}{E+H}\left(h^2q(t,x) - \frac{1}{3}q^3(t,x)\right) = 0. \quad (34) $$

We have to check that for all $(t, x) \in (t_e, t_m) \times (x^-(t), x^+(t))$, the function $q(t, x)$ is well-defined by the equation (34). By studying the sign of the derivative of the function of $q$ given by the left-hand side of equation (34) and by using the intermediate value theorem, we obtain the following result.

**Proposition 4.** For all $(t, x) \in (t_e, t_m) \times (x^-(t), x^+(t))$, the equation (34) has a unique solution $q(t, x) \in (0, h)$.

To sum up, in $\Omega_p(t)$, we have $a = -\operatorname{sgn}(y)k$, $\alpha$ is given by (33) and $q < h$ whereas in $\Omega_e(t)$, $|a| < k$, $\alpha = 0$ and $q = h$. From (3), which implies that $E\varepsilon = a + (E + H)\alpha$, and from what precedes, is it easy to obtain that the strain $\varepsilon$ is given by

$$ \varepsilon(t, x, y) = \begin{cases} -\frac{k}{E} \frac{y}{q(x, t)} & \text{if } x \in (x^-(t), x^+(t)), \\ \frac{M(t, x)}{EI} y & \text{if } x \in [0, x^-(t)] \cup [x^+(t), l], \end{cases} $$

exhibiting a linear expression with respect to $y$. From (1), (2) and (16), we obtain that $u = 0$ and $v(t) \in H_0^1(0, l)$ satisfies $\varepsilon = -yv''$, thus

$$ v''(t, x) = \begin{cases} \frac{k}{E} \frac{1}{q(x, t)} & \text{if } x \in (x^-(t), x^+(t)), \\ -\frac{M(t, x)}{EI} & \text{if } x \in [0, x^-(t)] \cup [x^+(t), l], \end{cases} \quad (35) $$

which completely determines $v$ on $[0, l]$.

**Remark 4.** In the pure bending case, we have shown directly that the problem (1)-(4) and (7)-(10) has a solution, in accordance with Proposition 1.

**Remark 5.** In the case $H > 0$, from (34) the function $q$ never vanishes, which implies that the two components $|y| > q(x, t)$ of the plastic zone $\Omega_p(t)$ will always be separated. In the case $H = 0$ (perfect plasticity), the equation (34) can easily be solved by hand, we obtain that for $x \in (x^-(t), x^+(t))$,

$$ q(t, x) = \sqrt{3 \left( h^2 + \frac{M(t, x)}{k} \right)}. $$
The above computations are still valid for $H = 0$ as soon as $M(t_m, x) > -kh^2$ for all $x \in (0, l)$. If on the contrary there exists some $t \in (t_e, t_m)$ and $x \in (0, l)$ such that $M(t, x) = -kh^2$, we get $q(t, x) = 0$, which means that the two components $|y| > q(x, t)$ of the plastic zone $\Omega_p(t)$ meet each other, this corresponds to the beam failure. In the pure bending case, we have hence checked that for $H = 0$, Proposition 3 does not hold.

## 6.2 The unloading phase $t \in [t_m, T]$

At the end of the loading phase, in view of (35) the transverse displacement $v_m = v(t_m) \in H_0^1(0, l)$ satisfies the problem

$$v_m''(x) = \begin{cases} \frac{k}{E} \frac{1}{q_m(x)} & \text{if } x \in (x_m^-, x_m^+), \\ -\frac{M_m(x)}{EI} & \text{if } x \in [0, x_m^-] \cup [x_m^+, l], \end{cases} \quad (36)$$

where $M_m(x) = M(t_m, x)$, $q_m(x) := q(t_m, x)$ and $x_m^\pm := x^\pm(t_m)$.

The unloading phase is purely elastic, which means that if we denote $v_T := v(T) \in H_0^1(0, l)$ the final transverse displacement, the displacement step $(v_T - v_m) \in H_0^1(0, l)$ satisfies the problem

$$(v_T - v_m)''(x) = \frac{M_m(x)}{EI}. \quad (37)$$

Adding (36) and (37), we obtain that $v_T \in H_0^1(0, l)$ satisfies the problem

$$v_T''(x) = \begin{cases} \frac{k}{E} \frac{1}{q_m(x)} + \frac{M_m(x)}{EI} & \text{if } x \in (x_m^-, x_m^+), \\ 0 & \text{if } x \in [0, x_m^-] \cup [x_m^+, l]. \end{cases} \quad (38)$$

Having in mind that $q_m$ can be viewed as a function of $M_m$ by solving the cubic equation (34) and that $x \in (x_m^-, x_m^+)$ is equivalent to $M_m(x) < M_0$, where $M_0$ is defined in (24), the system (38) is equivalent to find $v_T \in H_0^1(0, l)$ such that

$$v_T''(x) = g(M_m)(x) \quad \forall x \in (0, l),$$

with

$$g(M) := \begin{cases} \frac{k}{E} \frac{1}{q(M)} + \frac{3}{2} \frac{M}{Eh^3} & \text{if } M < M_0, \\ 0 & \text{if } M \ge M_0. \end{cases} \quad (39)$$

**Remark 6.** As can be deduced from (39), if $M_m(x) \ge M_0$, for all $x \in [0, l]$, then $v_T = 0$, which means that the loading history is purely elastic and the beam recovers its initial shape at the end of such loading history.

**Remark 7.** It is remarkable that the final transverse displacement $v_T$ does not depend on the precise shape of the time profile Hist. Indeed, the quasi-static evolution of elasto-plastic structures does not depend on the physical time.
## 7 The inverse problem in the pure bending case

Having described and partially solved the forward problem in the previous section, we now tackle the following inverse problem: we wish to recover the loading history $p(t, x) = \text{Hist}(t) p_m(x)$ for $(t, x) \in (0, T) \times (0, l)$ from the measurement of the final transverse displacement $v_T(x)$ for $x \in (0, l)$. In view of Remark 7, it will not be possible to recover the time profile Hist, this is why we will stick to recover the space profile $p_m$ from the function $v_T$. In the sequel, to simplify notations we simply denote $p_m = p$, $M_m = M$, $q_m = q$ and $v_T = v$. Let us define the set

$$L_+^2(0, l) := \{p \in L^2(0, l), p(x) \ge 0 \text{ a.e.}\}.$$

Given some $v \in L^2(0, l)$, the inverse problem consists in finding $p \in L_+^2(0, l)$ such that

$$\left\{ \begin{array}{ll} M'' = p & M(0) = M(l) = 0 \\ v'' = g(M) & v(0) = v(l) = 0, \end{array} \right. \qquad (40)$$

in which the function $g$ is defined by (39).

We have the following lemma.

**Lemma 7.1.** The function $g$ defined by (39) is $C^1((-\infty, 0])$. Furthermore, $\frac{dg}{dM}(M) < 0$ for all $M < M_0$.

*Proof.* From the definition of $g$ given by (39), we have that for $M < M_0$,

$$g(M) = \frac{k}{E} \frac{1}{q(M)} + \frac{3}{2} \frac{M}{Eh^3},$$

where $q(M)$ is implicitly given by (34). Firstly, the function $g$ is continuous. Indeed, since $q$ is continuous with $q(M_0) = h$,

$$\lim_{M \to M_0, M < M_0} g(M) = \frac{k}{E} \frac{1}{h} + \frac{3 M_0}{2 E h^3} = 0 = \lim_{M \to M_0, M > M_0} g(M).$$

Let us secondly prove that $dg/dM$, as a function of $M$, is continuous as well. The only possible discontinuity is at point $M_0$. The cubic equation (34) also reads for $M < M_0$

$$q^3(M) - 3 \left( \frac{M}{k} \left( 1 + \frac{H}{E} \right) + h^2 \right) q(M) - 2 \frac{H}{E} h^3 = 0.$$

Let us compute $\lim_{M \to M_0, M < M_0} \frac{dg}{dM}(M)$. From the equation above, we have

$$\begin{aligned} \frac{dg}{dM}(M) &= \frac{q(M)}{k} \left(1 + \frac{H}{E}\right) \frac{1}{q^2(M) - \left(\frac{M}{k}\left(1 + \frac{H}{E}\right) + h^2\right)} \\ &= \frac{3 q^2(M)}{2k} \left(1 + \frac{H}{E}\right) \frac{1}{q^3(M) + \frac{H}{E}h^3}. \end{aligned}$$
<figure><img src="image_5.png" /><figcaption>Figure 2: Graph of the function $g(M)$</figcaption></figure>

If $M \to M_0$, $M < M_0$, then $q(M) \to h$, which implies

$$ \lim_{M \to M_0, M < M_0} \frac{dq}{dM}(M) = \frac{3}{2} \frac{1}{kh}. $$

From the definition of $g$, for $M < M_0$ we have

$$ \frac{dg}{dM}(M) = -\frac{k}{E} \frac{(dq/dM)(M)}{q^2(M)} + \frac{3}{2} \frac{1}{Eh^3} = -\frac{3}{2} \frac{1}{Eh^3} \frac{h^3 - q^3(M)}{q^3(M) + \frac{H}{E}h^3}. $$

We conclude on the one hand that

$$ \lim_{M \to M_0, M < M_0} \frac{dg}{dM}(M) = 0 = \lim_{M \to M_0, M > M_0} \frac{dg}{dM}(M), $$

on the other hand that $\frac{dg}{dM}(M) < 0$ for all $M < M_0$. A picture of the function $g$ is given on Figure 2. $\square$

**Remark 8.** As a consequence of Lemma 7.1, for any $p \in L_+^2(0, l)$, we have $v \in C^3([0, l])$, where $v$ is the solution to problem (40). Indeed, from a classical regularity result, we have that $M \in H^2(0, l)$, hence $M \in C^1([0, l])$, which implies that $g \circ M \in C^1([0, l])$, and lastly $v \in C^3([0, l])$.

We now establish a partial uniqueness result for the inverse problem.

**Proposition 5.** Let us denote $(M_1, v_1)$ and $(M_2, v_2)$ the solutions in $H^1(0, l)$ to the problem (40) associated with $p_1$ and $p_2$ in $L_+^2(0, l)$, respectively. Let us denote the corresponding functions $q_1$ and $q_2$ given by (34) for $t = t_m$ and $\alpha_1$ and $\alpha_2$ given by (33) for $t = t_m$. If $v_1 = v_2$, then $q_1 = q_2$ and $\alpha_1 = \alpha_2$.

*Proof.* From Lemma 7.1, since $g(M_0) = 0$, $\frac{dg}{dM}(M_0) = 0$ and $\frac{dg}{dM}(M) < 0$ for $M < M_0$, we get that for any $p \in L_+^2(0, l)$, we have $g(M) > 0$ for $M < M_0$ and thus

$$ \left\{ \begin{array}{ll} g(M)(x) > 0 & \text{if } x \in (x_m^-, x_m^+), \\ g(M)(x) = 0 & \text{if } x \in [0, x_m^-] \cup [x_m^+, l]. \end{array} \right. \qquad (41) $$
Let us assume that $v_1(x) = v_2(x)$ for all $x \in [0, l]$, then from (40) we obtain that $g(M_1)(x) = g(M_2)(x)$ for all $x \in [0, l]$. If the transition points associated with $p_1$ and $p_2$ are denoted $(x_{m1}^-, x_{m1}^+)$ and $(x_{m2}^-, x_{m2}^+)$, respectively, we have from (41) that $x_{m1}^- = x_{m2}^- =: x_m^-$ and $x_{m1}^+ = x_{m2}^+ =: x_m^+$. Lemma 7.1 implies in particular that $g$ is injective in $(-\infty, M_0]$, and since we have $g(M_1)(x) = g(M_2)(x)$ for all $x \in (x_m^-, x_m^+)$, this yields $M_1 = M_2$ in $(x_m^-, x_m^+)$. We conclude that $q_1 = q_2$ and $\alpha_1 = \alpha_2$ from (34) and (33), respectively. $\square$

**Remark 9.** The interpretation of Proposition 5 is that there are at most one plastic zone delimited by $q$ and one plastic strain field $\alpha$ which are compatible with a given final transverse displacement $v$. However, we note that $v_1 = v_2$ on $[0, l]$ does not imply $p_1 = p_2$ on $[0, l]$. In other words, uniqueness of $p$ given $v$ does not hold. But in the framework of Non Destructive Testing, the real goal is not to retrieve the loading $p$ but to identify the internal damages created by such loading, that is the plastic zone and the plastic strain within such plastic zone.

We now propose an effective inverse method to identify $p$ from some measurement $v_{mes}$ of $v$. In view of Remark 9, we only expect to retrieve local minima which will produce valuable solutions in terms of the plastic zone and the plastic strain. As it is classical, we formulate the inverse problems as an optimal control problem (in the terminology of [17]) or in other words as a Tikhonov regularization (in the terminology of [8]), that is the least-square problem

$$ \inf_{p \in L_+^2(0,l)} J_r(p), \qquad (42) $$

where the cost function $J_r$ is given by

$$ J_r(p) := \frac{1}{2} \int_0^l (v(p) - v_{mes})^2 dx + \frac{r}{2} \int_0^l p^2 dx, \qquad (43) $$

$r > 0$ being a regularization parameter, and where $v, M$ are the solutions in $H^1(0, l)$ of the system (40). We have the following existence result.

**Proposition 6.** The problem (42) has at least one solution.

*Proof.* Let us denote $(p_n)_{n \in \mathbb{N}} \in L_+^2(0, l)$ a minimizing sequence, that is $J_r(p_n) \to J_{inf}$, where $J_{inf}$ is the minimum for problem (42). Since $J_r(p_n) \ge (r/2)\|p_n\|_{L^2(0,l)}^2$, the sequence $(p_n)$ is bounded in $L^2(0, l)$ and there exists a subsequence, still denoted $(p_n)$, such that $p_n \to p$ in $L^2(0, l)$. The convex set $L_+^2(0, l)$ is closed, it is hence weakly closed and $p \in L_+^2(0, l)$. The mapping $p \in L^2(0, l) \mapsto M \in H^2(0, l)$, where $M$ is the solution to (40) associated with $p$, is continuous, hence $M_n \to M$ in $H^2(0, l)$, where $M_n$ corresponds to $p_n$. It follows that $M_n \to M$ in $H^1(0, l)$, thus $M_n \to M$ in $C^0([0, l])$. Since the sequence $(M_n)$ converges in $C^0([0, l])$, there exists $B > 0$, such that $-B \le M_n(x) \le 0$ for all $n \in \mathbb{N}$ and for all $x \in [0, l]$. In view of Lemma 7.1, $g \in C^1([-B, 0])$, which implies that there exists some $G > 0$ such that

$$ |(g \circ M_n)(x) - (g \circ M)(x)| \le G |M_n(x) - M(x)|, \quad \forall x \in [0, l], \forall n \in \mathbb{N}, $$
then

$$ \|g \circ M_n - g \circ M\|_{L^2(0,l)} \le G \|M_n - M\|_{L^2(0,l)}. $$

We hence have $g \circ M_n \to g \circ M$ in $L^2(0, l)$, and thus $v_n \to v \in H^2(0, l)$, where $v_n$ and $v$ are associated with $p_n$ and $p$ via (40), respectively. We conclude that $J_0(p_n) \to J_0(p)$. We also have

$$ \|p\|_{L^2(0,l)}^2 \le \liminf_{n \to +\infty} \|p_n\|_{L^2(0,l)}^2, $$

hence

$$ J_r(p) \le \liminf_{n \to +\infty} J_r(p_n) = J_{\inf}. $$

Finally, $J_r(p) = J_{\inf}$, that is $p$ is a solution to problem (42). □

**Remark 10.** We note that the cost function $J_r$ given by (43) is not convex, which prevents us from using classical existence results in minimization problems, and which does not ensure uniqueness of the minimizer.

**Remark 11.** If we consider the bounded set $L_b^2(0, l) = \{p \in L^2(0, l), p(x) \in [0, P] \text{ a.e.}\}$, where $P > 0$, then the problem

$$ \inf_{p \in L_b^2(0,l)} J_0(p) $$

has a solution. The proof of such result is the same as that of Proposition 6.

In order to solve the minimization problem (42), we need to compute the Fréchet derivative of the cost function $J_r$. In this view we make use of an adjoint state.

**Proposition 7.** For any $p \in U \subset L_+^2(0, l)$, where $U$ is an open set of $L^2(0, l)$, the function $J_r$ is Fréchet differentiable at point $p$ and the Fréchet derivative of $J_r$ is given by

$$ J'_r(p) = M_* + r p, $$

where $(M_*, v_*)$ are the solutions in $H^1(0, l)$ of the adjoint problem

$$ \left\{ \begin{array}{ll} v_{*}'' = v - v_{\text{mes}} & v_{*}(0) = v_{*}(l) = 0 \\ M_{*}'' = v_{*} \frac{dg}{dM}(M) & M_{*}(0) = M_{*}(l) = 0, \end{array} \right. \qquad (44) $$

and $(v, M)$ are the solutions in $H^1(0, l)$ of the problem (40).

*Proof.* Let us define the Lagrangian mapping $\mathcal{L}: U \times (H_0^1(0, l))^4 \to \mathbb{R}$ by

$$ \begin{aligned} \mathcal{L}(p, N, w, N_*, w_*) ={}& \frac{1}{2} \int_0^l (w - v_{\text{mes}})^2 dx + \frac{r}{2} \int_0^l p^2 dx \\ & + \int_0^l (N' N_*' + p N_*) dx + \int_0^l (w' w_*' + g(N) w_*) dx, \end{aligned} \qquad (45) $$

for all $(p, N, w, N_*, w_*) \in U \times (H_0^1(0, l))^4$. We observe that since $(v, M)$ is solution to the problem (40) associated with $p$, we have

$$ \mathcal{L}(p, M(p), v(p), N_*, w_*) = J_r(p), \quad \forall (w_*, N_*) \in (H_0^1(0, l))^2. $$
Using the chain rule, for all $\tilde{p} \in L^2(0, l)$,

$$ (J'_r(p), \tilde{p})_{L^2(0,l)} = (\mathcal{L}'_p(p, M(p), v(p), N_*, w_*), \tilde{p})_{L^2(0,l)} \\ + (\mathcal{L}'_N(p, M(p), v(p), N_*, w_*), M'(p)\tilde{p})_{H_0^1(0,l)} \\ + (\mathcal{L}'_w(p, M(p), v(p), N_*, w_*), v'(p)\tilde{p})_{H_0^1(0,l)}, $$

where $\mathcal{L}'_p, \mathcal{L}'_N, \mathcal{L}'_w$ are the partial Fréchet derivatives of $\mathcal{L}$ with respect to $p, N$ and $w$, respectively, while $M'(p)$ and $v'(p)$ are the Fréchet derivatives of the mappings $p \in U \mapsto M(p) \in H_0^1(0, l)$ and $p \in U \mapsto v(p) \in H_0^1(0, l)$, respectively.

Let us compute $\mathcal{L}'_p, \mathcal{L}'_N$ and $\mathcal{L}'_w$.

We have, for all $\tilde{p} \in L^2(0, l)$, $\tilde{N} \in H_0^1(0, l)$ and $\tilde{w} \in H_0^1(0, l)$,

$$ (\mathcal{L}'_p(p, N, w, N_*, w_*), \tilde{p})_{L^2(0,l)} = \int_0^l (N_* + r p) \tilde{p} \, dx, $$

$$ (\mathcal{L}'_N(p, N, w, N_*, w_*), \tilde{N})_{H_0^1(0,l)} = \int_0^l N'_* \tilde{N}' dx + \int_0^l \frac{dg}{dM}(N) w_* \tilde{N} dx, $$

$$ (\mathcal{L}'_w(p, N, w, N_*, w_*), \tilde{w})_{H_0^1(0,l)} = \int_0^l w'_* \tilde{w}' dx + \int_0^l (w - v_{\text{mes}}) \tilde{w} dx. $$

It remains to choose $N_*$ and $w_*$ such that

$$ \mathcal{L}'_N(p, M(p), v(p), N_*, w_*) = 0, \quad \mathcal{L}'_w(p, M(p), v(p), N_*, w_*) = 0, $$

that is $N_* = M_*$ and $w_* = v_*$, where $(M_*, v_*)$ are the solutions of the adjoint problem (44), which implies the result. $\square$

# 8 Numerical experiments

## 8.1 Validation of the model

In this section we test the validity of our simple elasto-plastic beam model with the help of a comparison between a numerical simulation based on such model and experimental data. These experimental data correspond to a classical 4-point bending test of a steel specimen (a scheme of the device is represented on Figure 3). Such test is consistent with the model presented in section 2 in the pure bending case $n = 0$ and for some distribution $p(t)$ given by

$$ p(t) = \frac{F_{\text{tot}}(t)}{2d}(\delta_{l_1} + \delta_{l-l_1}), $$

where $F_{\text{tot}}$ is the total applied strength, $d$ is the dimension of the specimen in the direction $z$ which is orthogonal to the plane of Figure 3 and where $\delta_x$ is a Dirac located at point $x$. In particular, the kinematical conditions of the experiment are such that the all the fields are independent of the coordinate $z$. This enables us to consider our two dimensional model. The actual physical parameters related to the experiment are

$$ E = 195 \text{ GPa}, \quad k = 450 \text{ Mpa}, $$
<figure><img src="image_6.png" /><figcaption>Figure 3: A scheme of the 4-point bending experiment.</figcaption></figure>

where $E$ and $k$ are identified thanks to a traction test and

$$l/2 = 5.47 \text{ cm}, \quad l_1 = 2.25 \text{ cm}, \quad h = 2.525 \text{ mm}, \quad d = 3.73 \text{ cm}.$$

During the experiment we measure the total strength $F_{tot}(t)$, which increases from 0 to a large value that creates plastic strains, as well as the corresponding transverse displacement at the middle of the beam, that is $v(t, l/2)$. Such experimental curve is compared to a simulated curve obtained by solving the problem (35), where $M(t, x) \in H_0^1(0, l)$ satisfies $M'' = p$, with the help of a simple one dimensional finite element method. Precisely, we use some P1 finite elements, given a regular subdivision of the interval $(0, l)$ into 100 subintervals. The state variables $M, v \in H_0^1(0, l)$ are therefore approached by piecewise linear and continuous functions in $[0, l]$ which vanish at 0 and $l$. The resolution of problem (35) requires to solve the cubic equation (34) at each time step, which is done with the help of a simple dichotomy method. A consistent value of the kinematical hardening $H$ is difficult to find in the literature, hence we deduce it from the experimental curve. A good choice seems to be $H = 0.02 E$. The comparison between the experimental curve and two numerical curves is presented in Figure 4, the first one is plotted choosing $H = 0$ and the second one choosing $H = 0.02 E$, this second curve being closer to the experimental curve. Figure 4 emphasizes the fact that the elasto-plastic behaviour of our steel specimen is not perfect and indeed involves hardening, which is a well-known fact.

## 8.2 Resolution of the inverse problem

In this section we apply the optimal control approach proposed in section 7. We use the previous simple one dimensional P1 finite element method for solving both the forward problem (40) and the adjoint problem (44). The size of the beam is given by $l = 10$ and $h = 1$, we set $E = 1$, $H = 0.02$ and $k = 1$. We solve the minimization problem (42) with $r = 0$ (see Remark 11), that is we minimize $J_0$, simply denoted $J$ from now on, by using a gradient method. Starting from an initial guess $p_0 \in L^2(0, l)$, for $p \in L^2(0, l)$ close to $p_0$ we have

$$J(p) \approx J(p_0) + (J'(p_0), p - p_0)_{L^2(0,l)},$$
<figure><img src="image_7.png" /><figcaption>Figure 4: The strength $F_{tot}(t)$ in kN as a function of the displacement $v(t, l/2)$ in mm. Continuous black line: experimental curve. Dashed red line: numerical curve with $H = 0.02 E$. Dashed blue line: numerical curve with $H = 0$.</figcaption></figure>

which classically leads us to choose

$$p - p_0 = -\lambda(p_0)J'(p_0),$$

where $\lambda(p_0) > 0$ is chosen such that

$$\frac{|J(p) - J(p_0)|}{J(p_0)} = \epsilon,$$

where $\epsilon$ is supposed to be small, which corresponds to

$$\lambda(p_0) = \epsilon \frac{J(p_0)}{\|J'(p_0)\|_{L^2(0,l)}^2}.$$

In conclusion, the iteration formula is given, for $n \in \mathbb{N}$, by

$$p_{n+1} = p_n - \epsilon_n \frac{J(p_n)}{\|J'(p_n)\|_{L^2(0,l)}^2} J'(p_n),$$

where we choose $\epsilon_n = \epsilon/\sqrt{n+1}$, and $\epsilon$ is a fixed value in $[0.1, 0.4]$, that is the value of $\epsilon_n$ decreases with $n$. The precise value of $\epsilon$ will be specified in each case.

**Remark 12.** We could of course have chosen a more refined gradient method, but since the computations for this one dimensional problem are very quick, no effort was spent to improve the convergence speed.

In the following numerical experiments, $a_{ref}$, $b_{ref}$, $c_{ref}$, $d_{ref}$ being real constants, we produce an artificial function $v_{mes}$ obtained by solving the forward problem (40) for

$$p = a_{ref} \sin(\pi x/l) + b_{ref} \sin(2\pi x/l) + c_{ref} \sin(3\pi x/l) \quad (46)$$
or

$$p = d_{\text{ref}} (\delta_{l/4} + \delta_{3l/4}). \qquad (47)$$

The parameter $b_{\text{ref}}$ introduces some asymmetry in the data. From the “exact” data $v_{\text{mes}}$, we build a random noisy data $v_{\text{mes}}^{\delta}$ such that

$$v_{\text{mes}}^{\delta} = v_{\text{mes}} + \rho \text{rand},$$

where $\text{rand}$ is a random function and $\rho > 0$ is calibrated such that

$$\frac{\|v_{\text{mes}}^{\delta} - v_{\text{mes}}\|_{L^2(0,l)}}{\|v_{\text{mes}}\|_{L^2(0,l)}} = \Delta,$$

where $\Delta$ is a prescribed relative amplitude of noise. Besides, we consider two amplitudes of noise: $\Delta = 0$ (exact data) and $\Delta = 10\%$.

The different cases we consider for the exact pressure $p$ to retrieve are the following:

* Case 1: $p$ is given by (46) for $a_{\text{ref}} = 0.09$, $b_{\text{ref}} = c_{\text{ref}} = 0$

* Case 2: $p$ is given by (46) for $a_{\text{ref}} = 0.09$, $b_{\text{ref}} = 0.02$, $c_{\text{ref}} = 0$

* Case 3: $p$ is given by (46) for $a_{\text{ref}} = 0.09$, $b_{\text{ref}} = 0.01$, $c_{\text{ref}} = 0.05$

* Case 4: $p$ is given by (47) for $d_{\text{ref}} = 0.32$

In all the following experiments, the initial guess $p_0$ is a constant function. On Figures 5, 6, 7 and 8, we have represented a comparison between the retrieved solution and the exact solution for uncontaminated data ($\Delta = 0$), for the cases 1, 2, 3 and 4, respectively. On the top left of each figure, the red curve corresponds to the initial guess $p_0$, the solid dark blue curve corresponds to the real pressure $p$, while the dashed light blue curve corresponds to the retrieved pressure $p$. On the top right of each figure, we have plotted the corresponding transverse displacement $v$, while on the bottom figure we have plotted the corresponding functions $\pm q$ delimiting the plastic zone. On Figures 9, 10, 11, we have represented a comparison between the retrieved solution and the exact solution for noisy data ($\Delta = 10\%$), for the cases 1, 2, 3, respectively. In Figure 12, we have represented the evolution of the cost function $J$ as a function of the iteration number $n$, in the case 1, both with uncontaminated and noisy data.

In the three cases 1, 2 and 3 (smooth pressure $p$), we observe that the transverse displacement is very well retrieved, that the plastic zone is very well retrieved in the symmetrical case 1 and quite well retrieved in the asymmetrical cases 2 and 3, which is a consequence of the uniqueness result of Proposition 5, while the pressure is badly retrieved, which is consistent with Remark 9 about the non-uniqueness of the inverse problem. We also observe that our approach is rather robust with respect to noise. Unsurprisingly, the case 4, which corresponds to a 4-point bending test, is more difficult. Indeed, the pressure to retrieve is not a $L^2(0,l)$ function, instead it is the sum of two pointwise forces. This explains why we obtain poor numerical results, even for the plastic zone. Figure 12 illustrates that the choice of a $1/\sqrt{n}$ law for $\varepsilon_n$ allows us to obtain a monotonous decay of the cost function $J_n$.
<figure><img src="image_8.png" /><figcaption>Figure 5: Case 1 (no noise). Red: initial guess. Solid dark blue: exact solution. Dashed light blue: retrieved solution ($\epsilon = 0.3$). Top left: pressure $p$. Top right: transverse displacement $v$. Bottom: functions $\pm q$ defining the plastic zone.</figcaption></figure>

## Acknowledgement

The authors are indebted to Fabien Szmytka, from ENSTA/UME, who carried out the 4-point bending test.

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<figure><img src="image_13.png" /><figcaption>Figure 10: Case 2 (10% noise). Red: initial guess. Solid dark blue: exact solution. Magenta: noisy data. Dashed light blue: retrieved solution ($\epsilon = 0.2$). Top left: pressure $p$. Top right: transverse displacement $v$. Bottom: functions $\pm q$ defining the plastic zone.</figcaption></figure>
<figure><img src="image_14.png" /><figcaption>Figure 11: Case 3 (10% noise). Red: initial guess. Solid dark blue: exact solution. Magenta: noisy data. Dashed light blue: retrieved solution ($\epsilon = 0.15$). Top left: pressure $p$. Top right: transverse displacement $v$. Bottom: functions $\pm q$ defining the plastic zone.</figcaption></figure>

<figure><img src="image_15.png" /><figcaption>Figure 12: Case 1. Log of the cost function $J$ after $n$ iterations. Left: no noise. Right: 10% noise.</figcaption></figure>