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How to prove solution to Fick's second law by substitution

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data
    I am to prove that a solution to the differential equation Fick's second law is valid by substitution.

    2. Relevant equations
    Fick's second law:
    [tex] \frac{\partial C}{\partial t} = \frac{\partial}{\partial x} \left( D \frac{C}{\partial x} \right) [/tex]
    Solution to Fick's second law:
    [tex] C(x,t) = \left( \frac{C_1 + C_2}{2} \right) - \left( \frac{C_1 - C_2}{2} \right) \text{erf} \left( \frac{x}{2 \sqrt{Dt}} \right) [/tex]
    The (Gauss) error function (erf) which I found online:
    [tex] \text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_o^z e^{-y^2} dy [/tex]

    3. The attempt at a solution
    I presume I am supposed to differentiate the proposed solution with respect to t once and compare it to the proposed solution differentiated twice with respect to x? But I am not sure how i shall handle the integral.

    Can someone help me/point me to literature or give me some pointers on how to proceed?
     
  2. jcsd
  3. Sep 6, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you have taken Calculus, then you should know "the fundamental theorem of Calculus"-
    [tex]\frac{d}{dx} \int_a^x f(t) dt= f(x)[/tex]
     
  4. Sep 6, 2014 #3
    @HallsofIvy - Thanks for a reply. The problem bugging me though, is that i do not have [itex] x [/itex], but [itex] \frac{x}{2\sqrt{Dt}}[/itex], such that:

    [tex] \frac{2}{\sqrt{\pi}} \frac{\partial}{\partial x} \int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy [/tex]
     
  5. Sep 6, 2014 #4
    Is this the correct change of limits and function of the integral?
    [tex]

    \frac{\partial}{\partial x} C(x,t) = -B \frac{\partial}{\partial x}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -B \frac{\partial}{\partial x} \int_0^x e^{-y^2\left(4Dt \right)^{-1}}dy = -B e^{-x^2\left(4Dt \right)^{-1}}\\
    \frac{\partial^2}{\partial^2 x}C(x,t) = -B \frac{\partial}{\partial x} e^{-x^2\left(4Dt \right)^{-1}} = -B e^{-x^2\left(4Dt \right)^{-1}} \frac{\partial}{\partial x}\left( -x^2\left(4Dt \right)^{-1} \right) = B \frac{2}{Dt}\cdot x e^{-x^2\left(4Dt \right)^{-1}}
    [/tex]
     
    Last edited: Sep 6, 2014
  6. Sep 6, 2014 #5
    Solved it!

    I finally understood how it worked :wink:

    [tex]

    D\frac{\partial}{\partial x} C(x,t) = -BD \frac{\partial}{\partial x}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -BD e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \frac{\partial}{\partial x} \left(\frac{x}{2\sqrt{Dt}}\right) = -BD e^{-\frac{x^2}{4Dt}} \left(\frac{1}{2\sqrt{Dt}}\right) \\= \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} \\
    \frac{\partial}{\partial x} \left( D \frac{C(x,t)}{\partial x} \right) = \frac{\partial}{\partial x} \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} = \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} \frac{\partial}{\partial x} \left( - \frac{x^2}{4Dt} \right) \\= \frac{BDx}{4}\left( Dt \right)^{-\frac{3}{2}}e^{-\frac{x^2}{4Dt}}\\

    \frac{\partial}{\partial t} = -B \frac{\partial}{\partial t}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -B e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \frac{\partial}{\partial t} \left(\frac{x}{2\sqrt{Dt}}\right) = -B e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \left( - \frac{Dx}{4(Dt)^{3/2}} \right) \\= \frac{BDx}{4}\left( Dt \right)^{-\frac{3}{2}}e^{-\frac{x^2}{4Dt}} \qquad \square

    [/tex]
     
    Last edited: Sep 6, 2014
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