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Unformatted text preview: 12.6 1D Heat Equation: Solution by Fourier Integrals and Transforms In the previous section, we stated the onedimensional heat equation as a model for the heat conduction in a thin bar of length L > 0, which is laterally insulated. We may now consider the equation for an “infinitely long” bar. This leads to an initialvalue (or Cauchy) problem u t = c 2 u xx , in R × (0 , ∞ ) , (189) u = f, on R × { } . (190) The Cauchy problem ( 189 ), ( 190 ) is a fairly good model for the heat conduc tion in a very long, thin bar, where the temperature values on the boundary become unimportant. Separation of Variables We may still use separation of variables to solve the Cauchy problem ( 189 ), ( 190 ), so we write u ( x,t ) = F ( x ) G ( t ) , x ∈ R , t > . (191) The separate ODEs are the same as before ( 155 ), ( 156 ): F ′′ = kF, in R , (192) ˙ G = c 2 kG, in (0 , ∞ ) . (193) Solution of Separate ODEs Although we do not impose any boundary conditions on u , we still require a physically meaningful solution, such as that it remains bounded: ∃ M > 0 :  u ( x,t )  ≤ M, ∀ x ∈ R , t > . (194) This implies of course that both F and G must be bounded functions. With Proposition 1 , we conclude that the only bounded solution of ( 192 ) for k ≥ is the zero function F ≡ 0, so that we may again restrict our considerations to the case of a negative separation constant, k = ω 2 , ω > 0. In that case, solutions of ( 192 ), ( 192 ) are given by F ( x ; ω ) = A ( ω ) cos( ωx ) + B ( ω ) sin( ωx ) , x ∈ R , (195) G ( t ; ω ) = e − c 2 ω 2 t , t > . (196) 37 Note that any scaling factor in G may be moved over to A and B . Now the product ˜ u ( x,t ; ω ) := F ( x ; ω ) G ( t ; ω ) = ( A ( ω ) cos( ωx ) + B ( ω ) sin( ωx )) e − c 2 ω 2 t (197) satisfies ˜ u t = F ˙ G = c 2 ω 2 FG = c 2 ω 2 ˜ u, (198) c 2 ˜ u xx = c 2 F ′′ G = c 2 ω 2 FG = c 2 ω 2 ˜ u, (199) so that ˜ u ( · , · ; ω ) is a solution of the heat equation ( 189 ), for any ω > 0. Superposition Unlike in the case of a bounded domain, the eigenvalues c 2 ω 2 , ω > 0, do not form a countable set here. Therefore, we need a superposition of the functions ˜ u ( · , · ; ω ) in the form of an integral instead of a series: the general solution of ( 189 ) is given by u ( x,t ) = ∞ integraldisplay ˜ u ( x,t ; ω ) dω = ∞ integraldisplay ( A ( ω ) cos( ωx ) + B ( ω ) sin( ωx )) e −...
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 Spring '08
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 Math, Integrals, Partial differential equation, Cauchy Problem

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