🚀 go-pugleaf

In article <C169GD.GzM@cs.mcgill.ca> filippou@cs.mcgill.ca (Dimitrios FILIPPOU) writes:
>I'm trying to solve a peculiar problem by a method called "population
>balance" (sth. like "mass balance") and I'm having difficulties
>solving an ODE, where I thought I wouldn't have any problem at all.
>
>   dn/dr + (-1/tu)n = (-1/tu) N delta(r-r*)         (2)
>
>When I try to solve this ODE by Laplace transformations, I get the
>following solution 
>
>        /  
>       |   0,                       for r < r*
>n(r) = |                                             (5)
>       |   (-N/ut) exp{(r-r*)/tu},  for r >= r*  
>        \
>
>
>This can not be true; I cann't have a NEGATIVE answer!
>Intuition says that the solution is exactly the opposite
>
>
>        /  
>       |   (N/ut) exp{(r-r*)/tu},  for r <= r*  
>n(r) = |                                            (6)
>       |   0,                       for r > r*
>        \
>

******************************************************************

The solution to (2) is found like this:

- for r < r* or r > r*, (2) becomes:

   dn/dr + (-1/tu)n = 0  

with solution:

	n(r) = c1 exp{(r-r*)/tu},  for r < r* 

	     = c2 exp{(r-r*)/tu},  for r > r* 

where c1 , c2 are constants of integration

- now integrate (2) with respect to r over an infinitesimal
interval of width d, containing r*, and then let d -> 0. 
Obtain:

n(r*+) - n(r*-)  =  (-1/tu) N

which fixes c2 as a function of c1 and the size of the jump
in the function at r*. c1 is fixed by one initial condition.

The general solution of (2):

	n(r) = c1 exp{(r-r*)/tu},  for r < r* 

	     = (c1 - N/tu) exp{(r-r*)/tu},  for r > r* 

As you can see, both (5) and (6) are particular solutions
of this ODE with different initial condition.



--
From:  simardr@iro.umontreal.ca