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and in order to solve these equations, we need as well as boundary conditions, initial conditions of the form and u(x,0) = f (x) ∂u (x,0) = g (x) ∂t When B 2 −AC = 0 we have a parabolic problem, an example of which is the heat equation whose form is shown in equation (2.2). Parabolic equations require boundary conditions and an initial condition of the form u(x,0) = f (x) The difference between parabolic and hyperbolic equations is that hyperbolic problems have a finite propagation speed, whereas in parabolic problems the effects of propagation are felt immediately throughout the domain. 2.1.2 Well-posed problems In addition to the conditions described above, to be able to solve a partial differential equation, we should also have a well-posed problem, in the sense described by Hadamard (1923). We have a well-posed problem if 1. a solution to the problem exists. 2. the solution is unique. 3. the solution depends continuously on the problem data, that is, small changes in data yield small changes in the solution. For an elliptic problem, we need to have the partial differential equation defined in the interior of some region, with the solution subject to a single boundary condition at each point of the boundary. The type of boundary 8
condition may vary from one point on the boundary to another, but only one condition may be specified at any point. Hyperbolic equations require the same constraints on the boundary conditions as elliptic equations, but in addition they require two initial conditions, one to describe the initial state of the system and the other to describe the initial velocity. Parabolic equations also require the boundary conditions described for the previous two cases, but they need only a single initial condition specify- ing the state of the system at t = 0. 2.2 Methods of solution for the heat equation 2.2.1 Analytic solutions Assuming the thermal diffusivity α to be constant, we write the heat equation as ∂u ∂t = α∂2 u ∂x2 (2.3) The first method we consider is the method of separation of variables, men- tioned by Crank (1979) . In this case we look for solutions of the form u = X (x)T (t) (2.4) where X(x) and T(t) are functions of x and t respectively. Substituting equation (2.4) into equation (2.3) we get X dT dt = αT d2X dx2 and on separating the variables we have 1 dT T dt α d = X 2X dx2 so that the left hand side depends only on t and the right hand side depends only on x. Both sides therefore must be equal to the same constant and to 9
Page 1 and 2: Aspects of the Laplace transform is
Page 3 and 4: Acknowledgements It is my greatest
Page 5 and 6: Abstract There are many different m
Page 7 and 8: Contents 1 Introduction 1 2 The con
Page 9 and 10: 4.7.1 Contribution . . . . . . . .
Page 11 and 12: 9.2.3 To extend the use of Laplace
Page 13 and 14: 3.11 The positions of isotherm 8 fo
Page 15 and 16: 7.10 Diagram showing the position o
Page 17 and 18: List of Tables 3.1 The position of
Page 19 and 20: 6.1 The positions of the freezing f
Page 21 and 22: u temperature ũ dimensionless u co
Page 23 and 24: Chapter 1 Introduction We begin our
Page 25 and 26: system. In 1998 the Department was
Page 27 and 28: at a point per unit area. If we hav
Page 29: where G(x,y) is some function indep
Page 33 and 34: We can find the solution in differe
Page 35 and 36: this is the initial condition. The
Page 37 and 38: Figure 2.2: Typical grid for the fi
Page 39 and 40: fundamental solution which satisifi
Page 41 and 42: tending the technique to non-homoge
Page 43 and 44: 2.2.7 The method of separation of v
Page 45 and 46: to a set of linear equations of the
Page 47 and 48: labelled Qk, where k = 1,2,...,n
Page 49 and 50: in the case of melting ice (Crank a
Page 51 and 52: and the moving boundary differs. In
Page 53 and 54: 2.3.5 Fixed-domain methods In certa
Page 55 and 56: 2.4 Summary of Chapter 2 In this ch
Page 57 and 58: e easily evaluated, topological cha
Page 59 and 60: We wish to write the heat flow equa
Page 61 and 62: Equations (3.8), (3.9), (3.10) and
Page 63 and 64: In this example we consider the cas
Page 65 and 66: and ˜x 0.8 0.7 0.6 0.5 0.4 0.3 0.2
Page 67 and 68: ˜x 0.5 0.4 0.3 0.2 0.1 Isotherm 0
Page 69 and 70: w 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0
Page 71 and 72: ũ 10 9 8 7 6 5 0 0.2 0.4 0.6 0.8 1
Page 73 and 74: Table 3.5: Effect of errors in star
Page 75 and 76: Table 3.7: Effect of errors in star
Page 77 and 78: ˜x 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
Page 79 and 80: it is much more straightforward to
Page 81 and 82:
Chapter 4 The Laplace transform iso
Page 83 and 84:
available in tables. We do have an
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N� A = − 3. Gauss quadrature, w
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The value of M is chosen by the use
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are found where wj are the Stehfest
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where h is the difference in the va
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Table 4.3: Comparison of the result
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α = 1 + u. For the steady-state so
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Table 4.5: The positions of the iso
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To start this problem, we use an ac
Page 101 and 102:
4.7 Summary of Chapter 4 In this ch
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differences, and we shall show how
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˜x = 0, ũ = ũL, ˜t � 0 As in
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However, we find that doing this ca
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front and isotherm 8 respectively u
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straightforward and might require o
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freezing front is given by x0 = 2φ
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x 1 0.8 0.6 0.4 0.2 Freezing front
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ũ(˜x, ˜t) = 0, ˜x = ˜x0, ˜t
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so that when X = 0 Eliminating ũw
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˜x 1 0.8 0.6 0.4 0.2 Steady Convec
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˜x 1 0.8 0.6 0.4 0.2 Steady Convec
Page 125 and 126:
Page 127 and 128:
∂u ∂y = 0 y = 0 0 � x � 1 t
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6.3 The finite difference form We e
Page 131 and 132:
Results We see from table 6.1 that
Page 133 and 134:
where t0 is the initial time at whi
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x 1 0.8 0.6 0.4 0.2 Freezing front
Page 137 and 138:
Page 139 and 140:
grid line x = xM = 1 for all t > 0.
Page 141 and 142:
7.2 Solidification of a square pris
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which is our previous equation (6.8
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Figure 7.2: Diagram showing possibl
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Poots (1962) defines a function �
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y 1 0.8 0.6 0.4 0.2 t=0.05 t=0.1 t=
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so that we have ¯y (n) = y (t0) λ
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y 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0
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y 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0
Page 157 and 158:
Table 7.3: Values of the y co-ordin
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y 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6
Page 161 and 162:
y 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6
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y 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.
Page 165 and 166:
espaced using a given algorithm. We
Page 167 and 168:
Chapter 8 The use of multiple proce
Page 169 and 170:
In considering parallel computing m
Page 171 and 172:
The issue with shared memory system
Page 173 and 174:
Figure 8.3: Diagram showing a distr
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are several MPI implementations ava
Page 177 and 178:
communication in the program adds e
Page 179 and 180:
Sp 4 3.5 3 2.5 2 1.5 1 1 1.5 2 2.5
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work. Because the calculation invol
Page 183 and 184:
Table 8.3: Speed-up when allocating
Page 185 and 186:
not occur when only one process is
Page 187 and 188:
Therefore, as in examples 8.1 and 8
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upon which we focus. We mention the
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heating of a rod at zero degrees, n
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described by an equation involving
Page 195 and 196:
grid points. An example, described
Page 197 and 198:
simple. 9.2 Research objectives Our
Page 199 and 200:
Laplace transform, new complexities
Page 201 and 202:
orthogonal flow lines. To implement
Page 203 and 204:
Crank J (1957) Two methods for the
Page 205 and 206:
Davies A J and Mushtaq J (1998) Par
Page 207 and 208:
Kreider D L, Kuller R G, Ostberg D
Page 209 and 210:
Price R H and Slack M R (1954) The
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