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Diffusion equation & separation of variables

  1. Mar 26, 2006 #1
    i have been trying to solve the diffusion equation using separation of variables. i know the answer should turn out something like the normal probability density function but its just turns into a mess when i try it.
    i am given the following information:
    [tex]\frac{\partial p}{\partial t}=D\frac{\partial^2 p}{\partial x^2}[/tex]
    where [tex]p=p(x,t)[/tex]
    where D is a constant (its not specified whether its negative or positive). it would be safe to say the sign for D which gives the easiest solution is the right one.
    now this is what i did
    [tex]X\frac{dT}{dt}=DT\frac{d^2 X}{dx^2}[/tex]
    which gives 2 equations, namely
    [tex]\frac{1}{DT}\frac{dT}{dt}=\lambda[/tex] (1) and
    [tex]\frac{1}{X}\frac{d^2X}{dx^2}=\lambda[/tex] (2)
    solving (1) and applying the initial conditions and boundary conditions i get
    [tex]T(t)=\exp(D\lambda t)[/tex]
    for (2) i assumed [tex]\lambda>0[/tex] to get
    [tex]X(x)=a \exp(\sqrt{\lambda}x)+b \exp(-\sqrt{\lambda}x)[/tex]
    then finally you plug all back in to [tex]p(x,t)=T(t)X(x)[/tex]
    now i'm not sure where to go here, or how to apply the initial and boundary conditions and nor can i see how this would end up being something like a normal pdf (with an exp(-x^2) in there somwhere)

    thanks in advance
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 26, 2006 #2


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    Staff Emeritus
    Science Advisor

    Okay, why did you assume [itex]\lambda[/itex] is positive? What happens if it is negative? Also "p(0,0)= 1" is not sufficient. Initial values for partial differential equations are typically functions: what is p(x,0)? Unless there is a [tex]e^{-x^2}[/tex] in the initial condition you will not get one in the solution to the differential equation!

    The general solution will be a sum of the separate solutions:
    [tex]\Sigma X(x)T(t)[/tex]
  4. Mar 26, 2006 #3
    i assumed lambda to be positive so that the solution to (2) was an exponential (if its negative then i think the solution involves sin and cos, further i would not know what the solution is if lambda can be positive and negative). i am doing stuff to do with the distribution of random walking particles over a period of time, its behaviour is said to display a normal distribution. i thought i needed a function like p(x,0) but it was not given. could you explain why p(0,0)=1 is not sufficient? is the equation not solvable given the information in the first post?

    Last edited: Mar 26, 2006
  5. Mar 27, 2006 #4


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    Firstly, on the positive assumption of $\lambda$...

    In this problem, it's usual to take your $\lambda$ (seperation constant) as $-\lambda$ and $D$ as, say, $\alpha^2$, then you reqns become:




    You should now see that the different signs of (this new) $\lambda$ give constant, normal or hyberbolic trig fn answers.

    For you initial condition - you need to specify $p(x,0)=f(x)$ - I guess you want $f(x)=0$ which is fine but you should probably state $0\le x\le L$
    Last edited: Mar 27, 2006
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