# An integral relation in soil physics

1. May 27, 2014

### re444

Hi every one,

Here is my question: In soil physics, knowing the relation between suction head, h, and the soil water content, S, one can derive the hydraulic conductivity, K, of that soil using a formula like:

(ignore the superscripts "cap")
where in my problem, τ=0.5, κ=1, β=2.

Now what if we need the reverse procedure? If I have the relation between h and K, and want to calculate S, how it is possible?

2. May 27, 2014

### SteamKing

Staff Emeritus
Iteration is one way.

You know K on the LHS of the equation, and you have h as a function of x, along with the constants tau, kappa, and beta. The integral in the denominator seems to be a fixed quantity, but the value of the integral in the numerator depends on the value of S for the upper limit.

The engineering approach, would be to assume different values of the parameter S and calculate the LHS and the RHS of the equation, or re-write the equation as LHS - RHS = 0. Then you can iterate until a particular value of S makes this relation arbitrarily close to zero.

You could probably put together a spreadsheet to calculate the value of the expression for various input values of S.

3. May 27, 2014

### re444

The issue here is in the reverse procedure we don't have any knowledge about the relation between h and x (or in fact h and S), so the integrals could not be calculated. How about this?

4. May 27, 2014

### pasmith

If you know $h : [0,1] \to \mathbb{R}$, $\beta$, $\kappa$, $\tau$ and $K$, then you can calculate $$I(S) = \left(\frac{\int_0^S h^{-\kappa}\,dx}{\int_0^1 h^{-\kappa}\,dx}\right)^{\beta}$$ for any $0 \leq S \leq 1$ and from that you can solve $$K = S^\beta I(S)$$ numerically for $S$. (I'm assuming here that only values of $S$ between 0 and 1 inclusive make physical sense.)

$h$ is a function of a real variable. If you can't determine $h(x)$ for every $x$ in the domain then there is no sense in which you know $h$. Thus you can't calculate $I(S)$ for any $S$ other than the trivial $I(0) = 0$ (if $\beta > 0$) and $I(1) = 1$. If you can't calculate $I(S)$ you have no hope of solving $K = S^{\tau}I(S)$ for $S$.

5. May 27, 2014

### re444

In $h(x)$, x is a dummy variable of integration (in calculating K , when the known relation $h(S)$ is available, S is replaced by x in the integrands. So $h(x)$ or $h(S)$ is exactly what we are looking for.

6. Jun 11, 2014

### re444

I've gone a bit further to get this: (assuming τ=0.5, κ=1, β=2)

$\frac{\mathrm{d} }{\mathrm{d} S} \int_{0}^{S}\frac{1}{h(x))}dx=\int_{0}^{1}\frac{1}{h(x))}dx \cdot \frac{\mathrm{d} }{\mathrm{d} S} \left ( \sqrt{\frac{K_{r}}{\sqrt{S}}} \right )$

$\Rightarrow \frac{1}{h(S))}=\int_{0}^{1}\frac{1}{h(x))}dx \cdot \frac{\sqrt{S}\frac{dK}{dh}\cdot \frac{dh}{dS}-\frac{K}{2\sqrt{S}}}{2\sqrt{\frac{K}{\sqrt{S}}}}$

now knowing the K and $\frac{dK}{dh}$, the last equation seems to be an ODE but in this stage I don't know what to do with the $\int_{0}^{1}\frac{1}{h(x))}dx$ ?

Can I in someway handle that?