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We show that the Green's function for the diffusion equation,
![\begin{displaymath}G(z-z';t)=\frac{1}{\sqrt{4 \pi t'}}e^{-\frac{(z-z')^2}{4t'}},
\end{displaymath}](img74.gif) |
(35) |
satisfies the equation and behaves like a delta function at
t'=0.
Plugging the Green's function into the canonical diffusion equation,
Eq. 16, gives on both sides
![\begin{displaymath}\frac{\partial G(z-z';t')}{\partial t'}=
-\frac{1}{2 t'} G(z-...
...4t'^2} G(z-z';t')=
\frac{\partial^2 G(z-z';t')}{\partial z^2},
\end{displaymath}](img75.gif) |
(36) |
verifying that it is a solution to the equation.
As
, for
, the argument of the exponent goes
to
, and
. For z=z', it goes to infinity
as
. The integral over z' can be found by
substituting
and gives
![\begin{displaymath}\int G(z-z';t') dz'=
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-q^2/2} dq =1,
\end{displaymath}](img82.gif) |
(37) |
showing that it is correctly normalized to be the solution for
a delta function source or point source at
z=z' when
.
Next: Bibliography
Up: Solution of the Black
Previous: Post-Analysis
Dennis Silverman
1999-05-20