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2nd order PDE (I think?)

  1. Aug 18, 2010 #1
    Hi, I've spent days trying to solve some equations in a paper (referenced below) that describes it as a "straightforward, albeit lengthy integration," but I can't work out the "straightforward" bit. The notation is also odd, which doesn't seem to help my problem. Perhaps someone could help?

    I am trying to solve for the function [tex]\beta^{(0)}[/tex], which is a function of T and P (though P is held constant). Ultimately, I need a value of [tex]\beta^{(0)}[/tex] at a specific temperature T.

    I am given in the paper:
    [tex]\beta^{(0)L}=(\frac{\partial\beta^{(0)}}{\partial T})_{P}[/tex]
    [tex]\beta^{(0)J}=(\frac{\partial\beta^{(0)L}}{\partial T})_{P}+(2/T)\beta^{(0)L}[/tex]

    Also given,
    where the U values are empirical constants.

    I (think) I managed to solve the following expression for [tex]\beta^{(0)}[/tex]:
    [tex]\beta^{(0)}=\int\beta^{(0)J}\frac{T}{1+2ln(T)}\partial T[/tex]
    but I cannot figure out how to integrate that expression.

    Thus, I also tried to express the functions as a PDE:
    [tex]\beta^{(0)J}=(\frac{\partial^{2}\beta^{(0)}}{\partial T^{2}})_{P}+\frac{2}{T}(\frac{\partial\beta^{(0)}}{\partial T})_{P}[/tex]
    and then substitute the U-series empirical expression of [tex]\beta^{(0)J}[/tex] and subtract it from each side. However, I am as at much of a loss to solve that expression for [tex]\beta^{(0)}[/tex] as the integral expression above.

    Can anyone help me out at all? These equations are important to work out some geochemical thermodynamics I need to set up an experiment.

    Rogers, P. S. Z. and Pitzer, K. S., High-Temperature Thermodynamic Properties of Aqueous Sodium-Sulfate Solutions. Journal of Physical Chemistry 85 (20), 2886 (1981).
  2. jcsd
  3. Aug 19, 2010 #2
    In fact, the method that comes into my mind is VERY lenghty:

    1) First, decompose the T(T-263)^3 term with Hermite theorem:


    To find the coefficients a_1 match the two sides, you should find (I made the calculations very quickly, so I don't guarantee that's correct)




    [tex]a_4=\frac{2\cdot 526U_8}{(263)^3}[/tex]

    2) Now try the ansatz:

    [tex]\beta^{(0)L}=b_1T+b_2+b_3\log T+b_4/T+b_5/T^2+b_6/(T-263)+b_7/(T-263)^2+b_8/(T-263)^3[/tex]

    Calculate the derivative of this and substitute into the differential equation for [tex]\beta^{(0)L}[/tex], finding in this way the coefficients b_i.

    3) Finally, to find [tex]\beta^{(0)J}[/tex], simply integrate, which should be simple because [tex]\beta^{(0)L}[/tex] is a sum of things easy to integrate.

    4) I'm afraid this procedure will cost you half a day...!
  4. Aug 19, 2010 #3


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    Homework Helper

    Multiply the second equation by an integrating factor [itex]\mu(T)[/itex], demanding that the right hand side is of the product rule form u v' + u' v. This gives an easy DE for mu, but you can also just check that [itex]\mu = T^2[/itex] will work. You thus have

    [tex]T^2 \beta^{(0)J} = \frac{\partial}{\partial T}(T^2\beta^{(0)L}).[/tex]


    [tex]\beta^{(0)L}(T,P) = \left(\frac{T_0}{T}\right)^2\beta^{(0)L}(T_0,P) + \frac{1}{T^2} \int_{T_0}^T dt~t^2 \beta^{(0)J}(t)[/tex]
    where the pressure dependence enters as the initial condition on beta^L.

    Since [itex]\beta^{(0)L}[/itex] is just a derivative of [itex]\beta^{(0)}[/itex], you can just integrate again,

    [tex]\beta^{(0)}(T,P) = \beta^{(0)}(T_0,P) + T_0^2\beta^{(0)L}(T_0,P)\left(\frac{1}{T_0}-\frac{1}{T}\right) + \int_{T_0}^T \frac{d\tau}{\tau^2}~\int_{T_0}^\tau dt~t^2 \beta^{(0)J}(t)[/tex]

    This might not be the most pleasant set of integrals to do, but it looks like they should be doable in closed form.
    Last edited: Aug 19, 2010
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