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Homework Help: Diffusion in Cone

  1. Aug 23, 2013 #1


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    Hi everyone!

    1. The problem statement, all variables and given/known data
    We're given a three dimensional cone with perimeters d0 at the top and d1 at the bottom and a substance that diffuses through the cone with diffusion constant D from top to bottom. The concentration of the substance is held constant at the top plane of the cone. The z axis is chosen to go through the center of the cone from top to bottom.

    2. Relevant equations
    I have already done some math.
    We start with the Diffusion Equation

    [itex]\frac{\partial \varphi}{\partial t}=D ( \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}})\varphi(x,y,z,t)[/itex]

    Because of the symmetry of the cone

    [itex]\frac{\partial^{2}\varphi(x,y,z,t)}{\partial x^{2}}=\frac{\partial^{2}\varphi(x,y,z,t)}{\partial y^{2}}[/itex]

    So the PDE reduces to

    [itex]\frac{\partial \varphi}{\partial t}=D ( \frac{2\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial z^{2}})\varphi(x,y,z,t)[/itex]

    3. The attempt at a solution

    Now one can use separation of variables:


    So the PDE becomes:

    [itex]\frac{1}{\gamma(t)}\frac{\partial \gamma}{\partial t}=D(\frac{1}{\alpha(x)}\frac{2\partial^{2}\alpha(x)}{\partial x^{2}}+\frac{1}{\beta(z)}\frac{\partial^{2}\beta(z)}{\partial z^{2}})[/itex]

    Now all three terms must be constants. So:

    [itex]\frac{\partial \gamma(t)}{\partial t}=\lambda \gamma(t)[/itex]

    [itex]\frac{\partial^{2} \alpha(x)}{\partial x^2}=\lambda_{x} \alpha(x)[/itex]

    [itex]\frac{\partial^{2} \beta(z)}{\partial z^2}=\lambda_{z} \beta(z)[/itex]


    [itex] \lambda=2\lambda_{x}+\lambda_{z}[/itex]

    The time-dependant eigenvalue equation is easy to solve.
    But I am not sure how to solve the second and third equation.
    I tried to solve the second equation by using

    [itex]\alpha(x)=A exp(-kx)+B exp(kx) [/itex]

    and by applying boundary conditions:

    [itex] \alpha(x=-r)=\alpha(x=r)=0 [/itex]

    But then I get A=-B=0

    Somebody please help me out
  2. jcsd
  3. Aug 23, 2013 #2
    This problem is not completely specified. What is the initial condition, and what is the boundary condition at the bottom of the cone? What is the included angle of the cone (is it small, so that the cone has a very gradual taper?)? A small included angle would permit great simplification to the problem.
  4. Aug 24, 2013 #3


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    Hi Chestermiller,

    Thank you for your reply. Yeah you are right it is not completely specified.

    The initial conditions are:



    And at the bottom of the Cone


    The angle is not small enough for a simplification.
  5. Aug 24, 2013 #4
    I assume there is no diffusion through the side walls of the cone as well. Also, just out of curiosity, what is the included angle of the cone?
  6. Aug 24, 2013 #5


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    That doesn't look valid to me. At the point (x, y, z), x≠y, the two second derivatives will take different values.
    Try using cylindrical coordinates instead.
  7. Aug 25, 2013 #6


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    Hi Chestermiller and Haruspex!

    Thank you for your help so far.
    I have done some math. I figured that the solution to the eigenvalue equations could not be solved with exponentials so I thought of a different solution with a cosine and sine function like the solution to a quantum particle in an infinite well. You can see the solution in the pdf file. View attachment Cone_cell_2.pdf

    But now I get a different problem. The eigenvalues [itex]\lambda_{x} \lambda_{y} \lambda_{z}[/itex] are negative so λ is negative too. Which means that you get an exponential decay.
    But shouldn't it be an exponential rise, because the source of the substance at the top of the cone is constant?

    thank you
  8. Aug 25, 2013 #7
    You get exponential decay when the concentration is non-zero initially, and the values at the boundaries are zero. To achieve this situation, you can represent the solution to your problem as the linear superposition of the solution to two other problems:

    Problem 1: The steady state concentration distribution for your problem (i.e. the solution at long times) which satisfies Laplace's equation
    Problem 2: The transient diffusion problem in which the initial condition is the solution to problem 1, and the boundary conditions are zero concentrations on the top and bottom.

    The solution to your problem is the solution to problem 1 minus the solution to problem 2. The solution to problem 2 involves exponential decay terms, so that only the solution to problem 1 is left at long times.

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