How can I solve the 2nd order PDE for \beta^{(0)} in geochemical thermodynamics?

[presentt]

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 $$\beta^{(0)}$$, which is a function of T and P (though P is held constant). Ultimately, I need a value of $$\beta^{(0)}$$ at a specific temperature T.

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

Also given,
$$\beta^{(0)J}=6U_{5}+\frac{2U_{6}}{T}+\frac{U_{7}}{T^{2}}+\frac{526U_{8}}{T(T-263)^{3}}$$
where the U values are empirical constants.

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

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

Reference:
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).

 

[Petr Mugver]

a "straightforward, albeit lengthy integration,"

In fact, the method that comes into my mind is VERY lenghty:

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

$$\frac{526U_8}{T(T-263)^3}=\frac{a_1}{T}+\frac{a_2}{(T-263)^3}+\frac{a_3}{(T-263)^2}+\frac{a_4}{(T-263)}$$

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)

$$a_1=-\frac{526U_8}{(263)^3}$$

$$a_2=\frac{526U_8}{263}$$

$$a_3=-\frac{526U_8}{(263)^2}$$

$$a_4=\frac{2\cdot 526U_8}{(263)^3}$$

2) Now try the ansatz:

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

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

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

4) I'm afraid this procedure will cost you half a day...!

 

[Mute]

$$\beta^{(0)L}=(\frac{\partial\beta^{(0)}}{\partial T})_{P}$$
$$\beta^{(0)J}=(\frac{\partial\beta^{(0)L}}{\partial T})_{P}+(2/T)\beta^{(0)L}$$

Multiply the second equation by an integrating factor $\mu(T)$, 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 $\mu = T^2$ will work. You thus have

$$T^2 \beta^{(0)J} = \frac{\partial}{\partial T}(T^2\beta^{(0)L}).$$

Integrating,

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

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

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

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