32 the inner product and norm are de ned in l2 on the

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Unformatted text preview: the canonical element as procedure gram(N) N1 := N1=kN1k0 0 for k := 1 toPp do n ; t := Nk ; k=11(Nk Ni)0Ni i Nk := t=ktk0 0 bf end for return N Figure 10.3.2: Gram-Schmidt process to construct an orthogonal basis Nk k = 1 2 : : : np from a basis of monomials Nk , k = 1 2 : : : np . (u v)0 = Z 1 Z 1; 0 0 uvd d kuk 00 = (u u)1=2: 0 (10.3.6a) The result of the Gram-Schmidt process is a basis Nk , k = 1 2 : : : np that satis es the orthogonality condition (Ni Nk ) = ik i k = 1 2 : : : np : (10.3.6b) The actual process can be done using symbolic computation using a computer algebra system such as MAPLE or MATHEMATICA (cf. Remacle et al. 22] and Problem 2 at the end of this section). Example 10.3.1. We will illustrate some results using the discontinuous Galerkin method to solve two- and three-dimensional compressible ow problems involving the 36 Hyperbolic Problems Euler equations. This complex nonlinear system has the form of (10.3.1a) with 23 2 3 m n l 6m7 6 m2 = + p nm= lm= 7 67 6 7 7 u=6 n 7 f (u) = f (u) g(u) h(u)] = 6 mn= n2= + p ln= 67 6 7 2 4l5 4 ml= nl= l = +p 5 e (e + p)m= (e + p)n= (e + p)b= 23 0 607 67 b(x t u) = 6 0 7 : (10.3.7a) 67 405 0 Here, is the uid density m, n, and l are the Cartesian components of the momentum vector per unit volume e is the total energy per unit volume and p is the pressure, which must satisfy an equation of state of the form p=( ; 1) e ; (m 2 + n2 + l2 )=2 ]: (10.3.7b) This equation of state assumes an ideal uid with gas constant . Let us consider a classical Rayleigh-Taylor instability which has a heavy ( = 2) uid above a light ( = 1) uid (Figure 10.3.3). This hydrostatic con guration is unstable and any slight perturbation will cause the heavier uid to fall and the lighter one to rise. The uid motion is quite complex and Remacle et al. 22] simulated it using discontinuous Galerkin methods. They considered two-dimensional motion (l = 0, @=@z = 0 in (10.3.7)) with the initial perturbation 1 ;y 12 = 1 if 0=2 y < <=2 p = 3=2 ;...
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