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Diffusion equation, semi-infinite solution

  1. Aug 12, 2009 #1
    help w/ diffusion equation on semi-infinite domain 0<x<infinity

    Woo! First post! And I'm trying out/learning the latex code which is really neato!

    Okay, so... please help!

    I'm trying to solve

    [tex]\frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}

    for [tex] 0 < x < \infty [/tex]
    with initial condition such as [tex] T(x,0) = g(x) [/tex]

    and [tex] T(\infty,t) = C_{1} [/tex]
    and [tex] T(0,t) = f(t)[/tex]

    Is this achievable with separation of variables? I get stuck with the spatial problem and the B.Cs.

    The two equations i got using separation of variables were:

    T(x,t) = U(x)V(t)

    [tex] U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV' [/tex]

    [tex] V(U''+\frac{1}{x}U') = \frac{1}{\alpha}UV' [/tex]

    [tex]\frac{V'}{V} = \frac{\alpha}{U}(U''+\frac{U'}{x}) = -\lambda [/tex]

    so the spatial problem I get is [tex]U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0[/tex]

    I am unsure of the boundary conditions for the spatial problem

    time problem I get is [tex]V' = -\lambda V [/tex]

    Can this be solved with these B.Cs? I dunno cuz its non homogeneous B.Cs and now im stuck. I've tried a forum search but haven't had any luck.

    Any help or guidance would be appreciated. Let me know if anything is unclear.
    Last edited: Aug 12, 2009
  2. jcsd
  3. Aug 12, 2009 #2
    i noticed the latex code doesnt show up well in internet explorer... anyone else having that problem? in firefox it looks great!?
  4. Aug 13, 2009 #3
    T(inf,t) = constant and T(0,t) = f(t) are your spatial boundary conditions.

    Looks like it's separable to me...
  5. Aug 13, 2009 #4
    Yes, then does that imply T(inf) = constant and T(0) = f(t)??

    If T(inf,t) was = 0 then since T(x,t) = U(x)V(t) then you could say T(inf)=0 else its trivial solution. I wasnt sure if i could do that with these boundary conditions... Are you sure?

    With the non homogeneous b.c., if T(inf,t) = constant = U(x)V(t) then i dont know if you can use the same approach..?
    Last edited: Aug 13, 2009
  6. Aug 13, 2009 #5
    It doesn't imply T(inf) = constant, it says T(inf,0) = constant. Same with T(0) =/= f(t), T(0,t) = f(t).
  7. Aug 13, 2009 #6
    Oh.. Okay i see.

    So T(inf,0) = constant and T(0,0) = f(t)

    So what does this mean for my spatial problem from separation?

    T(0,0)=f(t) = U(0)V(0) so since its a function of time only then this implies U(0) = 1 ?

    And then U(inf) = C1

    So my spatial problem BVP will be the same as above with
    U(0) = 1
    U(inf) = C1

    does that look right? i havent done pdes for a while so im rusty..! haha thanks for the help though feldoh
    Last edited: Aug 13, 2009
  8. Aug 13, 2009 #7
    As far as I can tell those boundary conditions for the spatial portion looks right!
  9. Aug 13, 2009 #8
    im stuck with separation by variables. I think i need to use method of characteristics
  10. Aug 13, 2009 #9
    Hmm? I believe that the solution to the spatial portion of the problem is a linear combination of Bessel functions.

    Surely the time-dependent portion is straight forward enough.
  11. Aug 13, 2009 #10
    Yes the time dependent portion is fine, but I can't get the spatial portion...!

    If I multiple through by x^2 then it ends up looking like bessel solution will work

    [tex] x^{2} U'' + xU' + \frac{\lambda}{\alpha}x^{2}U = 0 [/tex]

    except for the [tex] \frac{\lambda}{\alpha} [/tex] term
    Last edited: Aug 13, 2009
  12. Aug 15, 2009 #11
    Can i do it with fourier transforms?

    What is the fourier transform of
    \frac{1}{x}\frac{\partial T}{\partial x}

    F(\frac{1}{x}\frac{\partial T}{\partial x}) = \int^{\infty}_{-\infty} \frac{1}{x}\frac{\partial T}{\partial x} e^{i \theta x}dx = ??
    Last edited: Aug 16, 2009
  13. Aug 17, 2009 #12
    okay scrap fourier.

    So back to separation of variables.

    [tex]T(x,t) = U(x)V(t)[/tex]

    [tex]U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV'[/tex]

    [tex]\frac{U''}{U}+\frac{U'}{xU} = \frac{1}{\alpha} \frac{V'}{V} = -\lambda^2[/tex]

    time problem is [tex] V' + \alpha\lambda^{2}V = 0[/tex]
    which has the solution form [tex] V(t)=Ae^{-\alpha\lambda^{2}t}[/tex]

    spatial problem is
    [tex]U''+\frac{U'}{x}+\lambda^{2}U = 0[/tex]

    multiple through by [tex]x^2[/tex]
    [tex]x^{2}U''+xU'+\lambda^{2}x^{2}U = 0[/tex]

    solution form using bessel functions is
    [tex]U(x) = BJ_{0}(\lambda x) + CY_{0}(\lambda x)[/tex]

    now I think i can say C=0 since [tex]Y_{0}[/tex] is singular at x=0 and i'm looking for a physical solution.. (not sure about this)

    then solution has the form

    [tex]T(x,t) = [De^{-\alpha\lambda^{2}t}]J_{0}(\lambda x)[/tex] where [tex]D = AB[/tex]

    with initial conditions [tex]T(x,0) = g(x)[/tex]
    and [tex]T(\infty,t) = C_{1}[/tex]
    and [tex]T(0,t) = f(t)[/tex]

    how should I approach this now?
    Last edited: Aug 17, 2009
  14. Aug 17, 2009 #13
    So far so good...

    Just take a limit as x -> infinity which must be equal to C1

    J0(0) = 1, so you can probably go from there.

    Two equations two unknowns (lambda, and the product of coefficients from the spatial and time solutions, A*B)
  15. Aug 17, 2009 #14
    The only problem I can see is, well... isn't x-->infinity = 0 since J0(infinity) =>0. So for non-zero C1 that wont work...

    [tex]T(0,0) = g(0) = f(0) =
    [De^{-\alpha\lambda^{2}(0)}]J_{0}(\lambda (0)) = D(1)(1)

    so [tex] D = g(0) = f(0) [/tex]
    Last edited: Aug 17, 2009
  16. Aug 17, 2009 #15
    Does C1 physically make sense if it's zero? I mean 0 is constant so mathematically it works I believe.
  17. Aug 17, 2009 #16
    C1=0 would make physical sense, but I would like to be able to use for example C1 = 20

    its almost like it needs to take the form
    T(0,0) = g(0) = f(0) =
    T(x,t) = [De^{-\alpha\lambda^{2}(t)}]J_{0}(\lambda (x)) + C_{1}


    i'm not sure because all bessels -->0 as x-->infinity

    I thought of maybe leaving Y0 in, but then there would be no solution at x=0
    Last edited: Aug 17, 2009
  18. Aug 18, 2009 #17
    Feldoh, I think I have to try a different method! What do you think. Would doing a laplace transform on the PDE help me out here?
    Last edited: Aug 18, 2009
  19. Aug 19, 2009 #18
    it's the non homogeneous boundary conditions that are making this tough. it would be a lot easier if T=0 at each boundary
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