The proper generalized decomposition for advanced numerical simulations ch02 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.
2–16 Chapter 2: DECOMPOSITION OF POISSON PROBLEMS Homework Exercises for Chapter Solutions EXERCISE 2.1 Written as vectors: grad = ∂/∂ x1 ∂/∂ x2 ∂/∂ x3 , div = [ ∂/∂ x1 d dx EA ∂/∂ x2 (E2.4) ∂/∂ x3 ] hence div T = grad and grad T = div EXERCISE 2.2 (a) Yes Elimination of e and N gives du dx + q = 0, (E2.5) Thus the correspondence is k → E A, q → −s (or, if you prefer, k → −E A and q → s); u is the same, x1 ≡ x and ∂ becomes the ordinary differential d If E is constant, this could be further transformed to (d/d x)(A du/d x) = −q/E, although this does not buy much (b) The diagram is shown in Figure E2.3 u^0 u(0)=u^0 q u dN/dx+q=N'+q=0 e=du/dx=u' e N=EAe N ^ N(L)=NL Figure E2.3 Tonti diagram for bar problem Here prime denotes derivative respect to x 2–16 ^ NL 2–17 Solutions to Exercises EXERCISE 2.3 (a) ST consists of AD and BC, and Sq consists of AB and C D Mathematically: ST : AD ∪ BC, Sq : AB ∪ C D Because of geometry and B.C.s, x1 = is a symmetry plane That is, T (x1 , x2 ) = T (−x1 , x2 ) Hence the normal temperature gradient gn = ∂ T /∂n = ∂ T /∂ x1 vanishes there and so does the flux Consequently one can reduce the problem to one half by placing the boundary condition qˆ = on x1 = (b) Elimination of g and q yields ∇ · (k∇T ) = If k is constant it can be moved out as a factor: k∇ · ∇T = k∇ T = Hence ∇ T = and the temperature distribution satisfies Laplace’s equation (c) It is easily checked that T = 100x2 /3 satisfies the Laplace’s equation (any linear function would) as well as the temperature boundary conditions √ on ST It does not satisfy, however, the zero-flux conditions on Sq For example on AB qn = −(k/ 2)(100/3) = Therefore that guess temperature distribution is not the exact solution of the boundary value problem EXERCISE 2.4 There are several solutions Two of them are shown in Figure E2.3 and E2.4 The second one is more in the spirit of Tonti’s dual diagrams for the potentials C φ=C on S θ φ θ = M/(GJ) M J calculated from solution g =∇ φ σ= −1 ∇ 10 g g σ −1 σ + 2G θ = 0 not used empty Figure E2.3 Tonti diagram for St.-Venant torsion problem C φ=C on S φ θ θ = M/(GJ) M J calculated from solution σ =∇ φ σ ∇ σ + 2Gθ = ∇σ = grad σ ∇σ not used empty "flux variable" Figure E2.4 Alternative Tonti diagram for St.-Venant torsion problem 2–17 2–18 Chapter 2: DECOMPOSITION OF POISSON PROBLEMS EXERCISE 2.5 φ^ φ = φ^ σ φ −∇ φ = v ∇m = σ v ρv=m m ^n m.n= m m^ n Figure E2.5 Tonti diagram for potential flow problem EXERCISE 2.6 ^ Φ ^ Φ=Φ ρ Φ −∇ Φ = E ∇D = ρ εE=D D E ^ D.n = Dn Figure E2.3 Tonti diagram for electrostatics problem 2–18 ^ Dn ... however, the zero-flux conditions on Sq For example on AB qn = −(k/ 2)(100/3) = Therefore that guess temperature distribution is not the exact solution of the boundary value problem EXERCISE 2.4 There... (−x1 , x2 ) Hence the normal temperature gradient gn = ∂ T /∂n = ∂ T /∂ x1 vanishes there and so does the flux Consequently one can reduce the problem to one half by placing the boundary condition... EXERCISE 2.4 There are several solutions Two of them are shown in Figure E2.3 and E2.4 The second one is more in the spirit of Tonti’s dual diagrams for the potentials C φ=C on S θ φ θ = M/(GJ) M