tzx9633 said:
Then, what should the vertical scale be? It's not provided, or I miss out something?
I don't like anything about what your book is doing here. There is really no characteristic length scale in the x direction, because the system is infinite in that direction. In the y direction, a logical characteristic length scale would be the height of the permeable formation, which is not specified in the figure. So talking about length scales here makes no sense to me.
What they are really doing here is using a mapping of the independent variable to transform the differential equation into a mathematically more tractable form. This is not the same as specifying a length scale for the system. I have never seen anyone call it this (before now).
Also, I don't like the transformation they have used. I would have done it much differently. I would have written the following:
$$x=x' \left(\frac{k_x}{k_z}\right)^{1/4}$$
$$z=z' \left(\frac{k_z}{k_x}\right)^{1/4}$$
That would transform the differential equation (more symmetrically) into:
$$\sqrt{k_xk_z}\left(\frac{\partial ^2 H}{\partial z'^2}+\frac{\partial ^2 H}{\partial x'^2}\right)=0$$
The original flow equations, in terms of the stream function and the head are:
$$\frac{\partial \psi}{\partial z}=-k_x\frac{\partial H}{\partial x}$$
$$\frac{\partial \psi}{\partial x}=+k_z\frac{\partial H}{\partial z}$$
Applying our coordinate mapping to these equations yields:
$$\frac{\partial \psi}{\partial z'}=-\sqrt{k_xk_z}\frac{\partial H}{\partial x'}$$
$$\frac{\partial \psi}{\partial x'}=+\sqrt{k_xk_z}\frac{\partial H}{\partial z'}$$
So, by applying this transformation of the independent variables, the equations transform into those for an isotropic formation, with a hydraulic conductivity equal to the geometric mean of the vertical- and horizontal hydraulic conductivities of the anisotropic formation.
Anyway, that's what I would do (and also what I actually have done in solving real-world groundwater problems).
