Linear and non-linear fitting of the Theis equation


The Theis equation

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The drawdown or change in hydraulic head, $s$ (a measure of the water pressure above some geodetic datum) a distance $r$ from a well at time $t$ from which water is being pumped at a constant rate, $Q$, can be modelled using the Theis equation, $$ s(r, t) = H_0 - H(r,t) = \frac{Q}{4\pi T}W(u), \quad \mathrm{where}\quad u = \frac{r^2S}{4Tt}. $$ Here $H_0$ is the hydraulic head in the absence of the well, $S$ is the aquifer storage coefficient (volume of water released per unit decrease in $H$ per unit area) and $T$ is the transmissivity (a measure of how much water is transported horizontally per unit time). The Well Function, $W(u)$ is simply the exponential integral, $E_1(u).$

In practice, it is often more convenient to use an approximation to the exponential integral: $W(u) \approx -\gamma - \ln u$ where $\gamma=0.577215664\cdots$ is the Euler-Mascheroni constant. This results in an expression for $s(r,t)$ known as the Jacob equation: $$ s(r,t) = -\frac{Q}{4\pi T}\left(\gamma + \ln u \right). $$

The following code uses scipy.special.exp1 to calculate the full Theis equation, returning $s(r,t)$ for parameters $Q$, $S$ and $T$.

import numpy as np
import matplotlib.pyplot as plt
from scipy.special import exp1

def calc_u(r, S, T, t):
    """Calculate and return the dimensionless time parameter, u."""

    return r**2 * S / 4 / T / t

def theis_drawdown(t, S, T, Q, r):
    """Calculate and return the drawdown s(r,t) for parameters S, T.

    This version uses the Theis equation, s(r,t) = Q * W(u) / (4.pi.T),
    where W(u) is the Well function for u = Sr^2 / (4Tt).
    S is the aquifer storage coefficient,
    T is the transmissivity (m2/day),
    r is the distance from the well (m), and
    Q is the pumping rate (m3/day).


    u = calc_u(r, S, T, t)
    s_theis = Q/4/np.pi/T * exp1(u)
    return s_theis

Let's simulate a curve for $s(t)$ for a fixed distance from the well, $r=25\;\mathrm{m}$, and pumping rate $Q = 2000\;\mathrm{m^3/day}$ for a well with parameters $S = 0.0003$ and $T = 1000\;\mathrm{m^2/day}$. We'll add a bit of synthetic normally-distributed noise.

Q = 2000        # Pumping rate from well (m3/day)
r = 10          # Distance from well (m)

# Time grid, days.
t = np.array([1, 2, 4, 8, 12, 16, 20, 30, 40, 50, 60, 70, 80, 90, 100])

# Calculate some synthetic data to fit.
S, T = 0.0003, 1000
s = theis_drawdown(t, S, T, Q, r)
# Add some noise.
noise_sd = 4.e-3
s += s * noise_sd * np.random.randn(s.shape[0])

# Plot the data
plt.plot(t, s, 'x')

enter image description here

Fitting the data

Linear least squares fit

It is often the case that $s$ can be measured over time at a fixed distance from the well, $r$, for a known pumping rate $Q$, and it is required that the parameters $S$ and $T$ be found. The Theis equation is clearly non-linear in $t$, but the Jacob approximation to it can produce a straight-line plot, since: $$ s(r,t) = -\frac{Q}{4\pi T}\left[\gamma + \ln\left(\frac{r^2S}{4T}\right)\right] + \frac{Q}{4\pi T}\ln t, $$ So a plot of $s$ against $\ln t$ is a straight line with intercept and gradient $$ c = -\frac{Q}{4\pi T}\left[\gamma + \ln\left(\frac{r^2S}{4T}\right)\right] \quad \mathrm{and}\quad m = \frac{Q}{4\pi T} $$ whence, $$ T = \frac{Q}{4\pi m}, \quad S = \frac{4T}{r^2}\exp\left[-\left(\frac{c}{m}+\gamma\right)\right] $$

The most straightforward approach is to use NumPy's polyfit routine to fit a first-order polynomial (i.e. a straight line) through the data points $(s_i, \ln t_i)$:

# fit s, ln(t)
lnt = np.log(t)
coeffs = np.polyfit(np.log(t), s, 1)
# plot the data and the fitted straight line
plt.plot(lnt, s, ls='', marker='o')
fit_line = np.poly1d(coeffs)(lnt)
plt.plot(lnt, fit_line)
# Output the fitted coefficients and report the rms residual
print('m = {}, c = {}'.format(*coeffs))
rms_residual = np.sqrt(np.sum((fit_line - s)**2))
print('rms residual =', rms_residual)

enter image description here

m = 0.15873714463615268, c = 1.785086631832805
rms residual = 0.0402157944671

The parameters $S$ and $T$ which best fit the data are found to be as follows, and seem to agree well with the values used to create the simulated data set.

m, c = coeffs
def get_S_and_T(m, c):
    Tfit = Q / 4 / np.pi / m
    Sfit = 4 * Tfit / r**2 * np.exp(-(c/m + np.euler_gamma))
    return Sfit, Tfit

print('S = {}, T = {} m2/day'.format(*get_S_and_T(m,c)))


S = 0.00029419771863093883, T = 1002.6320144330448 m2/day

The same analysis can be performed using numpy.linalg.lstsq which returns the sum of the square residuals along with other information about the fit:

A = np.vstack((lnt, np.ones_like(t))).T
x, sq_resid, rank, sing_vals = np.linalg.lstsq(A, s)
m, c = x
print('S = {}, T = {} m2/day'.format(*get_S_and_T(m,c)))
print('rms residual =', np.sqrt(sq_resid[0]))


S = 0.00029419771863094463, T = 1002.6320144330432 m2/day
rms residual = 0.0402157944671

Nonlinear least squares fit

The full Theis equation can only be fit with a nonlinear routine such as scipy.optimize.curve_fit or scipy.optimize import leastsq. curve_fit can fit a function directly; it calls leastsq which minimizes the sum of squares of a set of equations: in this context, the residuals between the observed data and modelled $s(t)$.

A. curve_fit

curve_fit does not need an initial guess for the fit parameters (it sets them equal to 1 if none is provided, which may or may not be appropriate depending on the function). The function to be fit must take the independent variable as its first argument (here, $t$) and the parameters to be fit as the remaining arguments. Since we don't want to fit $Q$ and $r$, we'll define a wrapper function to pick them up in global scope and send them to theis_drawdown which requires them.

from scipy.optimize import curve_fit

def theis_func(t, S, T):
    return theis_drawdown(t, S, T, Q, r)

popt, pcov = curve_fit(theis_func, t, s)
Sfit, Tfit = popt
print('S = {}, T = {} m2/day'.format(Sfit, Tfit))
theis_fit = theis_drawdown(t, Sfit, Tfit, Q, r)
rms_resid = np.sqrt(np.sum((s - theis_fit)**2))
print('rms residual =', rms_resid)


S = 0.0002942031272530988, T = 1002.6307076502651 m2/day
rms residual = 0.040215620786

We can also plot the fit:

plt.plot(t, s, 'x', label='data')
plt.plot(t, theis_fit, label='fit')

enter image description here

B. leastsq

leastsq requires an initial guess for the fit parameters but allows additional arguments (beyond the fit parameters) to be passed to the objective function. However, it requires a function to calculate the residuals: the difference between the "observed" and "modelled" data points:

from scipy.optimize import leastsq

def theis_resid(p, t, s, Q, r):
    """Calculate the residual between observed and calculated drawdown s(t).

    This version uses the exact Theis equation.
    The parameters S, T are packed into the tuple p.


    S, T = p
    return s - theis_drawdown(t, S, T, Q, r)

Both leastsq and curve_fit can be provided with the Jacobian: a vector of the first derivatives of the model function with respect to each of the fit parameters. In the case of the Theis function, these can be calculated analytically: $$ \frac{\mathrm{d}s(r,t)}{\mathrm{d}S} = -\frac{Q}{4\pi T S}e^{-u} \quad \mathrm{and} \quad \frac{\mathrm{d}s(r,t)}{\mathrm{d}T} = \frac{Q}{4\pi T^2}\left[-W(u) + e^{-u}\right] $$

def theis_Dfun(p, t, s, Q, r):
    """Calculate and return ds/dS and ds/dT for the Theis equation.

    The parameters S, T are packed into the tuple p.


    S, T = p
    u = calc_u(r, S, T, t)
    dsdS = -Q / 4 / np.pi / T / S * np.exp(-u)
    dsdT = Q / 4 / np.pi / T**2 * (-exp1(u) + np.exp(-u))
    return np.array((-dsdS, -dsdT)).T

# Initial guesses for the fit parameters
p0 = 0.01, 1500
popt, flag = leastsq(theis_resid, p0, Dfun=theis_Dfun, args=(t, s, Q, r))
Sfit, Tfit = popt
print('S = {}, T = {} m2/day'.format(Sfit, Tfit))
theis_fit = theis_drawdown(t, Sfit, Tfit, Q, r)
rms_resid = np.sqrt(np.sum((s - theis_fit)**2))
print('rms residual =', rms_resid)


S = 0.00029420317360281554, T = 1002.6306958749764 m2/day
rms residual = 0.040215620786
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Ole 4 years, 4 months ago

Why not s = (Q / 4πT) * W(u) in Theis Eq ?
s_theis = Q/2/np.pi/T * exp1(u) does not produce the correct result

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Current rating: 5

christian 4 years, 4 months ago

Hmm... I think you're right: I'm missing a factor of 2. I think I've fixed it now.

Thanks for getting in touch,

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