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Very evil diffusion equation

  1. Sep 6, 2016 #1
    1. The problem statement, all variables and given/known data
    $$\frac{\partial U}{\partial t}=\nu \frac{\partial^{2} U}{\partial y^{2}}$$
    $$U(0,t)=U_0 \quad for \quad t>0$$
    $$U(y,0)=0 \quad for \quad y>0$$
    $$U(y,t) \rightarrow {0} \quad \forall t \quad and \quad y \rightarrow \infty$$
    2. Relevant equations
    This is a diffusion problem on fluid mechanics, but it's more of a math problem so i posted it here.

    3. The attempt at a solution
    I'm trying to solve this via separation of variables (the textbook uses a "similarity" method i've never seen before, and concludes the function U must be erf) is it even possible to reach an analytic result via SV?
    The first boundary condition is what gets me, I tried
    $$U_{0} e^{{k^{2}t}} e^{{-\frac{k}{\sqrt{\nu}}y}}$$
    But it clearly doesn't work for any boundary condition except the last.
    I don't think sinusoidal is the answer here either because it must eventually converge to zero, for every t.
    Is there really no analytic answer?
     
  2. jcsd
  3. Sep 7, 2016 #1

    Orodruin

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    Please show your separation of variables and the resulting differential equations.
     
  4. jcsd
  5. Sep 7, 2016 #2

    Ray Vickson

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    Use Laplace transforms with respect to ##t##. Let
    $$W(y,s) = \int_0^{\infty} e^{-st} U(y,t) \, dt $$
    be the Laplace transform. Then, using standard properties of Laplace transforms, we get the DE
    $$\nu W_{yy}(y,s) = s W(y,s) - U(y,0) = s W(y,s),$$
    where ##W_{yy} = \partial^2 W / \partial y^2##.
    Also: ##U(0,t) = U_0## implies that
    $$W(0,s) = \frac{U_0}{s} $$
    Finally, the initial value theorem requires that ##\lim_{s \to \infty} s W(y,s) = 0## for ##y > 0##.

    These are enough to determine ##W(y,s)##. Then it is just a matter of taking the inverse Laplace transform of ##W(y,s)## to get ##U(y,t)##.

    Separation of variables will never work in this example, simply because it leads to the wrong kind of function.
     
    Last edited: Sep 7, 2016
  6. Sep 7, 2016 #3

    THANK YOU :D
     
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