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Unformatted text preview: Green function for diﬀusion equation
• Consider the diﬀusion equation,
∂2u
1 ∂u
=2
2
∂x
α ∂t
• Let’s solve using the inverse Fourier transform,
∞ u (k , t )e ikx dx u (x , t ) =
−∞ • Substitution in the diﬀerential equation allows us to easily
integrate the time dependence...
−k 2 u (k , t ) = 1 ∂ u (k , t )
α2 ∂ t
22 • We obtain by integration u (k , t ) = u (k , t = t )e −α k (t −t )
• We could also do it by separation of variables, practice it that
way too.... Green function for diﬀusion equation, continued
• Assume we have a point source at t = t , so that
u (x , t = t ) = δ (x − x )
• We can then ﬁnd u (k , t = t ) for the Fourier transform of the
point source
u (k , t = t ) = 1
2π ∞ δ (x − x )e −ikx dx = −∞ e −ikx
2π • Finally we ﬁnd u (x , t ) for t > t from the inverse Fourier
transform
u (x , t ) = 1
2π ∞ e ik (x −x ) e −k 2 α2 (t −t ) dk −∞ • The integral can be done by “completing the squares” Green function for diﬀusion equation, continued
• The result of the integral is actually the Green function
G (x , x ; t , t )
G (x , x ; t , t ) = 1
[4πα2 (t 1/2 − t )] e −(x −x )2 /4α2 (t −t ) • Notice that the Green function only depends on x − x and t − t
∞
• We ﬁnd that at all times, −∞ G (x , x ; t , t )dx = 1
• Then if we have a t = 0 distribution u (x , t = 0), we can ﬁnd
u (x , t ) just by doing the integral,
∞ u (x , t ) = G (x , x ; t , 0)u (x , t = 0)dx
−∞ ...
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This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.
 Spring '03
 Staff

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