Lecture06

Lecture06 - 12.6 1D Heat Equation: Solution by Fourier...

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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 one-dimensional 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 initial-value (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 =- 2 , > 0. In that case, solutions of ( 192 ), ( 192 ) are given by F ( x ; ) = A ( ) cos( x ) + B ( ) sin( x ) , x R , (195) G ( t ; ) = e c 2 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 ; ) := F ( x ; ) G ( t ; ) = ( A ( ) cos( x ) + B ( ) sin( x )) e c 2 2 t (197) satisfies u t = F G =- c 2 2 FG =- c 2 2 u, (198) c 2 u xx = c 2 F G =- c 2 2 FG =- c 2 2 u, (199) so that u ( , ; ) is a solution of the heat equation ( 189 ), for any > 0. Superposition Unlike in the case of a bounded domain, the eigenvalues- c 2 2 , > 0, do not form a countable set here. Therefore, we need a superposition of the functions u ( , ; ) 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 ; ) d = integraldisplay ( A ( ) cos( x ) + B ( ) sin( x )) e...
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This note was uploaded on 11/17/2011 for the course MATH 529 taught by Professor Staff during the Spring '08 term at UNC.

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Lecture06 - 12.6 1D Heat Equation: Solution by Fourier...

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