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Homework Help: Stationary solution to reaction-diffusion equation with certain boundary conditions

  1. Aug 4, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the stationary (steady state) solution to the following reaction diffusion equation:

    [tex]
    \frac{\partial C}{\partial t}= \nabla^2C - kC
    [/tex]

    Subject to the boundary conditions C(x, y=0) = 1, C(x = 0, y) = C(x = L, y) (IE, periodic boundary conditions along the x-axis, the value at x=0 is the same as at x=L). Also, at y = 0 and y = L, [tex]\frac{\partial C}{\partial x} = \frac{\partial C}{\partial y} = 0[/tex].

    2. Relevant equations

    With

    [tex]\frac{\partial C}{\partial t} = 0[/tex],

    rearrange to:

    [tex]
    \nabla^2C = kC
    [/tex]
    ...

    3. The attempt at a solution

    I believe I can solve this PDE without the boundary conditions, at least the one equation is satisified by a sum of hyberbolic sine or cosine functions. I have absolutely no idea how to incorporate the boundary conditions though. That they are periodic across x tells me that the solution should be symmetric about x = L / 2, but I have no mathematical reasons for this. I have never taking a PDE class before so I am a bit out of my element... any help would be very useful. I know that there IS an analytic solution with these constraints, but I haven't a clue what it is.
     
  2. jcsd
  3. Aug 4, 2007 #2
    Steady state implies that [tex]\frac{\partial C}{\partial t} = 0[/tex]
    And so,
    [tex]\frac{\partial ^2 C}{\partial x^2} + \frac{\partial ^2 C}{\partial y^2} = kC[/tex]

    I do not understand the Dirichlet problem. Is this on a rectangle? Can you be more implicit with the boundary conditions?
     
  4. Aug 4, 2007 #3

    HallsofIvy

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    Science Advisor

    As Kummer said, your "stationary solution" implies
    [tex]\frac{\partial ^2 C}{\partial x^2} + \frac{\partial ^2 C}{\partial y^2} = kC[/tex]
    Now "separate variables"- Let C= X(x)Y(y) so that the equation becomes
    [tex]Y\frac{d^2X}{dx^2}+ X\frac{d^2Y}{dy^2}= kXY[/tex]
    Divide by XY to get
    [tex]\frac{1}{X}\frac{d^2X}{dx^2}+ \frac{1}{Y}\frac{d^2Y}{dy^2}= k[/tex]
    In order that that be true for all x the two parts involving only X and only Y must be constants (other wise, by changing x but not y, we could change the first term but not the second- their sum could not remain the same constant, k). That is, we must have
    [tex]\frac{1}{X}\frac{d^2X}{dx^2}= \lambda[/tex]
    or
    [tex]\frac{d^2X}{dx^2}= \lambda X[/tex]
    and
    [tex]\frac{1}{Y}\frac{dY^2}{dy^2}= k- \lambda[/tex]
    or
    [tex]\frac{d^2Y}{dx^2}= (k- \lambda )Y[/tex]

    The general solution will be a sum of [itex]X(x,\lambda)Y(y,\lambda)[/itex] summed over all possible values of [itex]\lambda[/itex].

    Can you see that, in order to satisfy periodic boundary conditions on the x-axis, [itex]\lambda[/itex] must be [itex]-2n\pi[/itex] for some integer n?
     
    Last edited by a moderator: Aug 4, 2007
  5. Aug 4, 2007 #4
    1)What are the boundary conditions? I tried reading the post several times, I do not understand what they are?

    2)Is it on a rectangle?
     
  6. Aug 4, 2007 #5
    I am sorry to have not been more descriptive. I have not yet had time to digest HallsofIvy's response, which seems to be the most complete. To respond otherwise, though, the equation is being solved for the stationary state on a square of length/width L. The boundary conditions are: no flux at y = 0 or y = L (that is, top and bottom of square = no flux); periodic boundary conditions along x (that is, N(x = 0) = N(x = L)), and N = 1 along y = 0 (at the top of the square).

    Thank you all; HallsofIvy, I will now proceed to consider what you've posted.
     
  7. Aug 5, 2007 #6
    What is that supposed to mean?

    Anyway, it seems to me this is a partial differential equations with a non-homogenous boundary value problem.

    Which means you will have to solve for [tex]u_1(x,y)[/tex] so to satisfy:
    [tex]\frac{\partial^2 u_1}{\partial x^2}+\frac{\partial^2 u_1}{\partial y^2} = 0 \mbox{ with }\left\{ \begin{array}{c}u_1(x,0)=u_1(L,y)=u_1(x,L)=0\\ u_1(0,y)=f(y) \end{array} \right.[/tex]

    And then you need to solve for [tex]u_2(x,y)[/tex] so to satify:
    [tex]\frac{\partial^2 u_2}{\partial x^2}+\frac{\partial^2 u_2}{\partial ^2 u_2}{\partial y^2}=0 \mbox{ with }\left\{ \begin{array}{c}u_2(x,0)=u_2(L,y)=u_2(x,0)=0\\u_2(L,y)=f(y) \end{array} \right.[/tex]

    Then, [tex]u(x,y)=u_1(x,y)+u_2(x,y)[/tex] will be the solution to this equation.

    But to solve for those two individually use the methods of seperation of variables.
     
  8. Aug 5, 2007 #7
    Kummer: N = 1 along y = 0 means: N(x, 0) = 1 (all of the points along the line at the top of the square, at y = 0, have unit concentration).

    I will now consider your response... thank you.
     
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