Biswajit Banerjee

Nonlinear programming and closest point return plasticity

Part 6 of the series on plasticity return algorithms

Introduction

In Part 5 I briefly hinted at the closest-point return algorithm. The ideas behind this were made rigorous in the mid-to-late 1980s by a group of researchers influenced by developments in convex optimization. Since then there has been a statis in the development of return algorithms for phenomenological plasticity. However, the recent surge in improvements in machine learning has the potential of leading to another leap forward in our understanding and implementation of nonlinear, history-dependent, constitutive models.

Let us explore the basic ideas behind closest-point algorithms.

Background

In nonlinear optimization, the method of Lagrange multipliers has been used since the mid 1800s to solve minimization problems with equality constraints. In 1950, this approach was generalized by Kuhn and Tucker to allow for inequality constraints. Later it was discovered that W. Karush from the University of Chicago had reached the same conclusions in his MSc thesis from 1939.

Primal form

The primal form of the optimization problem is

$$ \begin{align} & \text{minimize} & & f(\mathbf{x}) \\ & \text{subject to} & & g_i(\mathbf{x}) \le 0, \quad i = 1, \dots, m \\ & & & h_j(\mathbf{x}) = 0, \quad j = 1, \dots, p \end{align} $$

Note that there is no convexity requirement for this problem.

The Lagrangian

The Lagrangian () associated with the primal form is just the weighted sum of the objective function and the constraint functions and . Thus

$$ \mathcal{L}(\mathbf{x}, \boldsymbol{\lambda}, \boldsymbol{\nu}) = f(\mathbf{x}) + \boldsymbol{\lambda}\cdot\mathbf{g}(\mathbf{x}) + \boldsymbol{\nu}\cdot\mathbf{h}(\mathbf{x}) $$

where

$$ \boldsymbol{\lambda} = \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_m \end{bmatrix} ~,~~ \mathbf{g} = \begin{bmatrix} g_1 \\ g_2 \\ \vdots \\ g_m \end{bmatrix} ~,~~ \boldsymbol{\nu} = \begin{bmatrix} \nu_1 \\ \nu_2 \\ \vdots \\ \nu_p \end{bmatrix} ~,~~ \mathbf{h} = \begin{bmatrix} h_1 \\ h_2 \\ \vdots \\ h_p \end{bmatrix} \,. $$

The vectors and are called Lagrange multiplier vectors or, more frequently, the dual variables of the primal problem.

Dual function

The dual function () to the primal problem is defined as

$$ \mathcal{F}(\boldsymbol{\lambda},\boldsymbol{\nu}) = \inf_{\mathbf{x}} \mathcal{L}(\mathbf{x}, \boldsymbol{\lambda}, \boldsymbol{\nu}) = \inf_{\mathbf{x}} \left[f(\mathbf{x}) + \boldsymbol{\lambda}\cdot\mathbf{g}(\mathbf{x}) + \boldsymbol{\nu}\cdot\mathbf{h}(\mathbf{x})\right] $$

Note that the dual function is the minimum of a family of affine functions (linear + a constant term) in . This makes the dual problem concave. Note also that since the dual function is affine, it is bounded from below by when the value of is unbounded.

Simplified forms for can be found for many problems, including problems that can be expressed as quadratic forms.

Dual form

Since the dual function is the largest lower bound on the Lagrangian, the Lagrange dual form of the primal minimization can be expressed as

$$ \begin{align} & \text{maximize} & & \mathcal{F}(\boldsymbol{\lambda},\boldsymbol{\nu}) \\ & \text{subject to} & & \boldsymbol{\lambda} \succeq \mathbf{0} \end{align} $$

We don’t have any constraint on because .

Karush-Kuhn-Tucker optimality conditions

Let be the optimal solution for the primal problem and let be the optimal solution of the dual problem. When these two solutions lead to a zero duality gap, i.e.,

$$ f(\mathbf{x}^\star) = \mathcal{F}(\boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) $$

the Lagrangian at that optimal point is

$$ \mathcal{L}(\mathbf{x}^\star, \boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) = f(\mathbf{x}^\star) + \boldsymbol{\lambda}^\star\cdot\mathbf{g}(\mathbf{x}^\star) + \boldsymbol{\nu}^\star\cdot\mathbf{h}(\mathbf{x}^\star) $$

Also, since and ,

$$ f(\mathbf{x}^\star) = \mathcal{F}(\boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) = \inf_{\mathbf{x}} \mathcal{L}(\mathbf{x}, \boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) \le \mathcal{L}(\mathbf{x}^\star, \boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) \le f(\mathbf{x}^\star) $$

The only way for the above to be true is when

$$ \boldsymbol{\lambda}^\star\cdot\mathbf{g}(\mathbf{x}^\star) = 0 \quad \leftrightarrow \quad \lambda^\star_i g_i(\mathbf{x}^\star) = 0 \,. $$

Also, since minimizes the Lagrangian, its gradient is zero at that point:

$$ \frac{\partial}{\partial \mathbf{x}} \mathcal{L}(\mathbf{x}^\star, \boldsymbol{\lambda}^\star, \boldsymbol{\nu}^\star) = \mathbf{0} = \frac{\partial f(\mathbf{x}^\star)}{\partial\mathbf{x}} + \boldsymbol{\lambda}^\star\cdot\frac{\partial\mathbf{g}(\mathbf{x}^\star)}{\partial\mathbf{x}} + \boldsymbol{\nu}^\star\cdot\frac{\partial\mathbf{h}(\mathbf{x}^\star)}{\partial\mathbf{x}} $$

These results, along with the original constraints of the primal and dual problems, are collected together into the Karush-Kuhn-Tucker optimality conditions:

$$ \begin{align} & g_i(\mathbf{x}^\star) \le 0 & & h_j(\mathbf{x}^\star) = 0 \\ & \lambda_i^\star \ge 0 & & \lambda^\star_i g_i(\mathbf{x}^\star) = 0 \\ & \frac{\partial f(\mathbf{x}^\star)}{\partial\mathbf{x}} + \boldsymbol{\lambda}^\star\cdot\frac{\partial\mathbf{g}(\mathbf{x}^\star)}{\partial\mathbf{x}} + \boldsymbol{\nu}^\star\cdot\frac{\partial\mathbf{h}(\mathbf{x}^\star)}{\partial\mathbf{x}} = \mathbf{0} \end{align} $$
Similarity with plasticity

The plastic loading-unloading conditions are similar to the Karush-Kush-Tucker optimality conditions in that we have

$$ \begin{align} g(\boldsymbol{\sigma}) \le 0 ~,~~ \dot{\lambda} \ge 0 ~,~~ \dot{\lambda} g(\boldsymbol{\sigma}) = 0 \end{align} $$

where is the yield surface constraining the values of . We may also interpret the flow rule as the last Karush-Kuhn-Tucker condition:

$$ -\dot{\boldsymbol{\varepsilon}}^p + \dot{\lambda}\frac{\partial g}{\partial \boldsymbol{\sigma}} = 0 \quad \text{where} \quad -\dot{\boldsymbol{\varepsilon}}^p =: \frac{\partial f}{\partial \boldsymbol{\sigma}} $$

and is the quantity that is minimized in the primal problem. We can interpret as the negative of the maximum plastic dissipation, i.e.,

If we use a first-order update approach, the discretized equations for perfect plasticity are (see Part 5)

$$ \begin{align} \boldsymbol{\sigma}_{n+1} & = \mathbf{C}:(\boldsymbol{\varepsilon}_{n+1} - \boldsymbol{\varepsilon}_{n+1}^p) = \boldsymbol{\sigma}_{n+1}^{\text{trial}} - \mathbf{C}:(\boldsymbol{\varepsilon}_{n+1}^p - \boldsymbol{\varepsilon}_{n}^p)\\ \boldsymbol{\varepsilon}_{n+1}^p & = \boldsymbol{\varepsilon}_n^p + \Delta\lambda \left.\frac{\partial g}{\partial \boldsymbol{\sigma}}\right|_{\boldsymbol{\sigma}_{n}} \quad \text{or} \quad \boldsymbol{\varepsilon}_{n+1}^p = \boldsymbol{\varepsilon}_n^p + \Delta\lambda \left.\frac{\partial g}{\partial \boldsymbol{\sigma}}\right|_{\boldsymbol{\sigma}_{n+1}} \\ & g(\boldsymbol{\sigma}_{n+1}) \le 0 ~,~~ \Delta\lambda \ge 0 ~,~~ \Delta\lambda g(\boldsymbol{\sigma}_{n+1}) = 0 \end{align} $$

Note that if we interpret the flow rule as an optimality condition a backward Euler update is consistent with the Karush-Kuhn-Tucker conditions and a forward Euler update is ruled out.

Closest point return

Let be the trial stress and let be the value of the yield function at that state. Let be actual stress and let be the value of the yield function at the actual stress state.

Let us assume the actual stress state on the yield surface is at the closest distance from the trial stress. Then we can devise the primal minimization problem:

$$ \begin{align} & \text{minimize} & & f(\boldsymbol{\sigma}) = \lVert \boldsymbol{\sigma}^{\text{trial}} - \boldsymbol{\sigma}\rVert^2 \\ & \text{subject to} & & g(\boldsymbol{\sigma}) \le 0 \\ \end{align} $$

where

$$ \lVert \boldsymbol{\sigma} \rVert = \sqrt{\boldsymbol{\sigma}:\boldsymbol{\sigma}} $$

The Lagrangian for this problem is

$$ \mathcal{L}(\boldsymbol{\sigma},\lambda) = f(\boldsymbol{\sigma})+ \Delta\lambda g(\boldsymbol{\sigma}) = \lVert \boldsymbol{\sigma}^{\text{trial}} - \boldsymbol{\sigma}\rVert^2 + \Delta\lambda g(\boldsymbol{\sigma}) $$

The Karush-Kuhn-Tucker conditions for this problem at the optimum value are

$$ \begin{align} & g(\boldsymbol{\sigma}_{n+1}) \le 0 ~,~~ \Delta\lambda \ge 0 ~,~~ \Delta\lambda g(\boldsymbol{\sigma}_{n+1}) = 0 \\ & \frac{\partial f(\boldsymbol{\sigma}_{n+1})}{\partial\boldsymbol{\sigma}} + \Delta\lambda \frac{\partial g(\boldsymbol{\sigma}_{n+1})}{\partial \boldsymbol{\sigma}} = -2(\boldsymbol{\sigma}^{\text{trial}} - \boldsymbol{\sigma}_{n+1}) + \Delta\lambda \frac{\partial g(\boldsymbol{\sigma}_{n+1})}{\partial \boldsymbol{\sigma}} = \mathbf{0} \end{align} $$

From the last condition we see that the closest distance using this criterion leads to a stress value of

$$ \boldsymbol{\sigma}_{n+1} = \boldsymbol{\sigma}^{\text{trial}} - \tfrac{1}{2} \Delta\lambda \frac{\partial g(\boldsymbol{\sigma}_{n+1})}{\partial \boldsymbol{\sigma}} $$

But we have seen previously that the first-order stress update with backward Euler leads to

$$ \boldsymbol{\sigma}_{n+1} = \boldsymbol{\sigma}^{\text{trial}} - \Delta\lambda\mathbf{C}: \frac{\partial g(\boldsymbol{\sigma}_{n+1})}{\partial \boldsymbol{\sigma}} $$

The similarity between the two indicates that we are on the right track, i.e., the actual stress is at the closest distance from the trial stress to the yield surface. But the correct closest distance is not in the standard standard stress space, but in a space where the norm to be minimized is given by

$$ \lVert \boldsymbol{\sigma} \rVert_{\mathbf{C}^{-1}} = \sqrt{\boldsymbol{\sigma}:\mathbf{C}^{-1}:\boldsymbol{\sigma}} $$

This can be verified by repeating the above exercise with the new definition of the norm.

Remarks

Most methods used for finding using the closest-point projection idea use variations on the Newton method that require the computation of second-derivatives of the yield function. We will discuss a method that avoids those computations in the next part of this series.

If you have questions/comments/corrections, please contact banerjee at parresianz dot com dot zen (without the dot zen).

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