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54      1 Linear algebra



                        × 1
                      1

                      12
                       1


                     s
                    v
                                dd  1    e   1

                      2

                              2                           1
                                                       × 1

                   Figure 1.13 Velocity profile for 1-D laminar flow with a moving upper plate.

                   Solving the 1-D fluid flow problem in MATLAB

                   simple flow 1D.m solves the 1-D flow example above where the fluid is water ( ρ =
                                  −3
                          3
                     3
                   10 kg/m ,µ = 10 Pa s), the upper plate is stationary, and the separation between plates
                   is 1 mm. The dynamic pressure gradient is selected to give a Reynolds’ number near 1. The
                   computed velocity profile is shown in Figure 1.13.

                   Fill-in (why Gaussian elimination is sometimes impractical)
                   It is important to note that many sparse systems cannot be placed in a banded form, and
                   elimination remains costly. For such systems, the iterative techniques discussed later in our
                   discussion of boundary value problems are preferred. Even if a matrix is banded; however,
                   elimination may be too costly due to fill-in.
                     The flow example (1.237) and (1.238) was of the form of a 1-D boundary value problem,
                                       2
                                      d ϕ
                                    −   2  = f (y)  ϕ(0) = ϕ 0  ϕ(B) = ϕ B           (1.256)
                                      dy
                   Each row of the linear system resulting from finite differences on a uniform grid of spacing
                    y has only three nonzero elements
                                  −1            2             −1
                         A k,k−1 =      A k,k =      A k,k+1 =      b k = f (y k )   (1.257)
                                ( y) 2        ( y) 2         ( y) 2
                   As is shown in Chapter 6, for the analogous problem on a 2-D domain,
                                       2
                                             2
                                      ∂ ϕ   ∂ ϕ
                               2
                            −∇ ϕ =−       −     = f (x, y)  0 ≤ x ≤ L    0 ≤ y ≤ H
                                      ∂x  2  ∂y  2
                                     BC1 ϕ(0, y) = 0  0 ≤ y ≤ H
                                     BC2 ϕ(L, y) = 0  0 ≤ y ≤ H
                                                                                     (1.258)
                                     BC3 ϕ(x, 0) = 0  0 ≤ x ≤ L
                                     BC4 ϕ(x, H) = 00 ≤ x ≤ L
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