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Equation of Diffusion, trouble simplifying, PDE

  1. Aug 22, 2008 #1
    isotropic equation, so k, ρ, and c are constant, where k is thermal conductivity, c is specific heat, and ρ is the density of the body.

    the equation boils down to

    [tex]\left( \frac{c\rho}{k}\right) \left(\frac{\partial u}{\partial t}\right) - \left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}} \right) = 0 [/tex]

    The book proceeds to simplify it in such a way that changes the time scale: t' = (k/cρ)t, dropping the prime giving:

    [tex]\frac{\partial u}{\partial t} - \left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}} \right) = 0 [/tex]


    What happened??

    If you can just talk me through what happens when the time scale is changed, I will try to work through the computation.

    Thanks in advance.
    Last edited: Aug 22, 2008
  2. jcsd
  3. Aug 22, 2008 #2


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    It's a different system of units for the time. The "new" t (or t'):

    1. Is proportional to the actual time
    2. Has units of Length2

    It makes the DE less cumbersome to solve, since you no longer have the c, k and ρ in there.
  4. Aug 23, 2008 #3
    Hello Somefantastik,

    It is really very simple, under the assumption that the heat capacity, the density and the thermal conductivity are constant you can bring them inside the partial derivative of the time by considering:

    [tex]\left(\frac{c\cdot \rho}{k}\right) \cdot \left(\frac{\partial u}{\partial t}\right)= \frac{\partial u}{\partial \left(\displaystyle \frac{t\cdot k}{c \cdot \rho}\right)}= \frac{\partial u}{\partial t'}[/tex]

    By setting:

    [tex]t'=\frac{t\cdot k}{c \cdot \rho}[/tex]

    after which you can drop the prime.
  5. Aug 23, 2008 #4
    Thank you for the responses so far. That helps a little.

    Can someone try to explain to me what is physically happening when you change the time?

    So previously the time was in simple units, and now that we have changed it to include k,rho, and c, it's in units of lenght^2? Can someone expound on this please?
  6. Aug 23, 2008 #5
    There is not so much to explain on the transformation, it is a mathematical way of changing to a new variable so that you don't need to write the physical parameters all the time. After you solve the equation change back to t by inverse substituting and you have the solution to the original equation.

    The unit is indeed a length squared, as is the right hand side of the equation. There you have [tex]\partial x^2[/tex] which also is a length squared. This means you have the same dimension on the left and right hand side of the equation as it should be.

    There are very interesting ways of changing to new variables. If you want to know more on this, google on dimensionless parameters, pi-theorem of Buckingham and you will find information on this. This is a very powerful way of looking at equations and solutions. The idea behind it is that a physical problem described with a number of parameters, can be presented with fewer parameters by taking combinations of them. This way your problem and solution will be easier to investigate. It takes some time to get used to the dimensionless parameters, but once you have, it opens a new way of looking at physical problems.

    best regards,

  7. Aug 23, 2008 #6
    I see. Thank you for the input :)
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