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Solution of diffusion equation with three independent variable (r,z,t)

  1. Sep 8, 2014 #1
    Hi,

    I want to solve the following diffusion equation:

    (d/dt) C(r,z,t)=D*∇^2 C(r,z,t)

    where C is the concentartion and D is the coefficient of diffusivity (constant)

    with initial condition C(r,z,0)=C0 (constant)
    and boundary condition dc/dr=0 at r=0; (dc/dz) at z=-L equal to (dc/dz) at z=L

    where I have considered axisymmetric tube of length L.

    Can anybody help to solve the above mentioned problem for the concentration C?
     
  2. jcsd
  3. Sep 8, 2014 #2

    HallsofIvy

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    Since you are using "symmetric tube" you will probably want to use "cylindrical coordinates", suppressing the [itex]\theta[/itex] dependence. In cylindrical coordinates the Laplacian is
    [tex]\nabla^2 f= \frac{\partial^2 f}{\partial r^2}+ \frac{1}{r}\frac{\partial f}{\partial r}+ \frac{\partial^2 f}{\partial \theta^2}+ \frac{\partial^2 f}{\partial z^2}[/tex]

    Assuming "axially symmetric" so f does not depend on [itex]\theta[/itex], that is
    [tex]\nabla^2 f= \frac{\partial^2 f}{\partial r^2}+ \frac{1}{r}\frac{\partial f}{\partial r}+ \frac{\partial^2 f}{\partial z^2}[/tex]

    So your equation says
    [tex]\frac{\partial C}{\partial t}= D\left(\frac{\partial^2 C}{\partial r^2}+ \frac{1}{r}\frac{\partial C}{\partial r}+ \frac{\partial^2 C}{\partial z^2}\right)[/tex]

    A standard method of solving that is to look for "separable solutions". That is, look for solutions of the from [itex]C(r,z,t)= R(r)Z(z)T(t)[/itex]. That will separate the equation into ordinary differential equations for R, Z, and T separately. Depending on the boundary and initial conditions, the solution can be written as a sum of such "separated" solutions.
     
  4. Sep 8, 2014 #3
    Thanks Hallsoflvy for your nice explanation.
     
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