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Unformatted text preview: F1.4ZH2 Numerical Solution of PDEs Page 7 1.4 Taylor series and difference operators We can use Taylor series expansions in one of the variables to see how well Finite Difference (FD) approximations work. Consider the approximation ( u ( x + Δ x, t ) u ( x, t )) / Δ x to ∂u/∂x . F x u ( x j , t n ) = u ( x j + Δ x, t n ) u ( x j , t n ) = u + Δ xu x + 1 2 Δ x 2 u xx + 1 3! Δ x 3 u xxx + . . . ( x j ,t n ) u ( x j , t n ) = Δ xu x + 1 2 Δ x 2 u xx + . . . ( x j ,t n ) ≈ Δ x u x ( x j , t n ) when Δ x is sufficiently small So F x Δ x u ( x j , t n ) = u x ( x j , t n ) + O (Δ x ) where O (Δ x ) means a quantity which → 0 as Δ x → 0. Similarly D x u = u ( x j + Δ x, t n ) u ( x j Δ x, t n ) = u + Δ x u x + 1 2 Δ x 2 u xx + 1 3! Δ x 3 u xxx + 1 4! Δ x 4 u xxxx + O (Δ x 5 ) ( x j ,t n ) u Δ x u x + 1 2 Δ x 2 u xx 1 3! Δ x 3 u xxx + 1 4! Δ x 4 u xxxx + O (Δ x 5 ) ( x j ,t n ) ⇒ D x u 2Δ x = u x + Δ x 2 6 u xxx + O (Δ x 4 ) ( x j ,t n ) So D x 2Δ x u ( x j , t n ) = u x ( x j , t n ) + O (Δ x 2 ) . This is better because Δ x 2 → 0 faster than Δ x as Δ x → 0. In a similar way we can show that δ 2 x u ( x j , t n ) = Δ x 2 u xx ( x j , t n ) + O (Δ x 4 ) so δ 2 x Δ x 2 u ( x j , t n ) = u xx ( x j , t n ) + O (Δ x 2 ) when Δ x is small . By expanding first in one variable and then in the other(s), we can analyse approximations involving terms like u ( x j + Δ x, t n + Δ t ). Exercise: Show that F x F t Δ x Δ t u ( x j , t n ) = u xt ( x j , t n ) + O (Δ x, Δ t ) . and D x D t 4Δ x Δ t u ( x j , t n ) = u xt ( x j , t n ) + O (Δ x 2 , Δ t 2 ) . so that again Central differences are better. We now have enough background to start looking at PDEs, starting with parabolic problems. F1.4ZH2 Numerical Solution of PDEs Page 8 2 Parabolic PDE’s 2.1 Introduction The most practical approach is to study a concrete example. Hence we start by looking at the heat or diffusion equation ∂U ∂T = K ∂ 2 U ∂X 2 , (2.1) where U = U ( X, T ) is measured in some units (say ◦ C) and the physical time and space variables T > 0, X ∈ (0 , L ) are also measured in appropriate units. K is a constant which depends on the material, called the Thermal Conductivity . We specify Dirichlet boundary conditions (BCs) at x = 0 , L , i.e. U (0 , T ) and U ( L, T ) given for all T > 0, and initial conditions (ICs) U ( X, 0) given for X ∈ (0 , L ). X L U(X) given T U T =U XX U(0,T) given U(L,T) given Need to find the solution in the open box T > , X ∈ (0 , L ) The heat equation models many things, for example • the temperature U in a long thin insulated bar with fixed temperatures at each end X L U(0,T) given U(L,T) given • The diffusion of a chemical in a solid or stationary fluid (in this case U is the concentration of the chemical)....
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 Spring '11
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 Approximation, Taylor Series

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