The proper generalized decomposition for advanced numerical simulations ch06 ex Many problems in scientific computing are intractable with classical numerical techniques. These fail, for example, in the solution of high-dimensional models due to the exponential increase of the number of degrees of freedom. Recently, the authors of this book and their collaborators have developed a novel technique, called Proper Generalized Decomposition (PGD) that has proven to be a significant step forward. The PGD builds by means of a successive enrichment strategy a numerical approximation of the unknown fields in a separated form. Although first introduced and successfully demonstrated in the context of high-dimensional problems, the PGD allows for a completely new approach for addressing more standard problems in science and engineering. Indeed, many challenging problems can be efficiently cast into a multi-dimensional framework, thus opening entirely new solution strategies in the PGD framework. For instance, the material parameters and boundary conditions appearing in a particular mathematical model can be regarded as extra-coordinates of the problem in addition to the usual coordinates such as space and time. In the PGD framework, this enriched model is solved only once to yield a parametric solution that includes all particular solutions for specific values of the parameters. The PGD has now attracted the attention of a large number of research groups worldwide. The present text is the first available book describing the PGD. It provides a very readable and practical introduction that allows the reader to quickly grasp the main features of the method. Throughout the book, the PGD is applied to problems of increasing complexity, and the methodology is illustrated by means of carefully selected numerical examples. Moreover, the reader has free access to the Matlab© software used to generate these examples.
Chapter 6: THE HR VARIATIONAL PRINCIPLE OF ELASTOSTATICS 6–16 Homework Exercises for Chapter Solutions EXERCISE 6.1 Not assigned EXERCISE 6.2 Integrate the N u term by parts: L HR = Nu − L N2 d x − f1 u − f2 u = − 2E A N2 dx − 2E A L u N d x + N u|0L − f u − f u (E6.1) The exact solution is constant N , which is assumed in the element Thus N = and the functional reduces to L HR =− N2 d x + N (u − u ) − f u − f u 2E A (E6.2) The internal u(x) has disappeared, and the axial displacements only come in through their end values u and u Therefore, it does not matter what is taken for the displacement field inside the element as long as a constant N is assumed This is a variational freak since it applies only to that specific example problem EXERCISE 6.3 Not assigned EXERCISE 6.4 The weak links are (cf Figure E6.1): eiuj − eiσj = uˆ i − u i = in V, on Su , (E6.3) where eiuj = 12 (u i, j + u j,i ) and eiσj = Ci jk σk Multiply the residuals (E6.3) by δσi j and δti = δσi j n j and integrate over V and Su , respectively: (eiuj − eiσj ) δσi j d V + V (uˆ i − u i ) δσi j n j d S = (E6.4) Su Apply the divergence theorem to the first term on the left: eiuj δσi j d V = V (u i, j + u j,i ) δσi j d V = − V u i δσi j, j d V + V u i δσi j n j d S, (E6.5) S in which the indicated term vanishes because the BE are strongly satisfied: δ(σi j, j + bi ) = δσi j, j = in V Replacing into (E6.4) gives δ TCPE = −Ci jk σk δσi j d V + V u i δσi j n j d S + S (uˆ i − u i ) δσi j n j d S = Su 6–16 (E6.6) 6–17 Exercises (uˆ i − u i ) δσi j n j d S = Su u^ u Ignorable b σi j, j + bi = in V (eiuj − eiσj ) δσi j d V = Master Slave V CE: eσ σ σi j = E i jk ek in V FBC: σi j n j = tˆi on St ^t Figure E6.1 Departure Weak Form to derive the TCPE functional Split the S integral over St and Su Over St we can set δσi j n j = δ(σi j n j ) = δ tˆi = because the FBCs are strongly satisfied The integrals of u i δσi j n j over Su cancel out and we are left with δ TCPE =− Ci jk σk δσi j d V + V uˆ i δσi j n j d S = (E6.7) Su This is the exact first variation of the complementary energy functional (E6.1) EXERCISE 6.5 The only weak connection is σi j = E i jk ek We begin as above, trying δ S = (σi j − E i jk ek ) δei j d V, (E6.8) V where σi j must be viewed as a data field.10 This is the exact variation of S (ei j ) = (σi j − 12 E i jk ek ) ei j d V (E6.9) V And this is the end No further progress can be made This is the only canonical functional of elasticity that contains no boundary integrals In the Exercise statement it was noted that functional S has limited practical value The reason is that all of the difficult field equations are taken as strong The stress field must satisfy both equilibrium equations and stress BC point by point a priori, while the strain field must be compatible with a displacement field that satisfies the displacement BC The only relaxation of the governing equations pertains to the constitutive equations Nevertheless the principle may be occasionally useful in material homogenization, as the following simple example illustrates Consider a bar of uniform cross section A and total length L = L + L + L made up 10 Why? Because the stresses must satisfy the volume equilibrium equations and the surface traction BC a priori Thus the stress field must be known at every point in V 6–17 6–18 Chapter 6: THE HR VARIATIONAL PRINCIPLE OF ELASTOSTATICS E1 , A E2 , A E3 , A σ σ L1 L2 L3 Figure E6.2 Application of the strain-only canonical functional to material homogenization of three materials with elastic moduli E , E and E , respectively Using the functional modulus E S, find an average To carry out the homogenization process, assume that the bar is under a constant axial stress field σ = P/A (see Figure E6.2), which obviously satisfies all stress equilibrium equations and surface traction boundary conditions The average strain e = E −1 σ is taken as the only unknown to be varied in S : S (e) The condition δ S = gives ∂ e= = ALσ e − 12 A(E L + E L + E L )e2 S /∂e (E6.10) = 0, from which L σ, E1 L + E2 L + E3 L E= σ E1 L + E2 L + E3 L = e L (E6.11) This technique essentially amounts to equating the strain energies absorbed by the actual and homogenized (fictitious) bars Note that the displacement field does not appear in this statically determinate problem 6–18 ... while the strain field must be compatible with a displacement field that satisfies the displacement BC The only relaxation of the governing equations pertains to the constitutive equations Nevertheless... δσi j d V + V uˆ i δσi j n j d S = (E6.7) Su This is the exact first variation of the complementary energy functional (E6.1) EXERCISE 6.5 The only weak connection is σi j = E i jk ek We begin... viewed as a data field.10 This is the exact variation of S (ei j ) = (σi j − 12 E i jk ek ) ei j d V (E6.9) V And this is the end No further progress can be made This is the only canonical functional