9.6 Solution of simultaneous  9.5 Roots of equations  9.5.2 Interpolation

9.5.3 Other iterative procedures

Suppose that the equation to be solved, f(x)=0, can be separated into two parts, one of which can be solved for x:  

f(x)=f₁(x)+f₂(x)=0  (9.43)

 

where the equation f₁(x)=-f₂(x) could be solved for specified f₂(x). If the inverse of f₁(x), f₁^-1(x), is known, then the solution can be written as

$$x = g(x) = f_1^{-1}\Bigl(-f_2(x)\Bigr)$$ (9.44)

 

where the right side of this equation defines g(x). In the case where f₂(x) varies slowly with x, it may be possible to use (9.44) iteratively to find the root of (9.43).

 

Figure 9.3:  The solution of the equation f₁(x)+f₂(x)is the intersection of the two functions shown (here, for f₁=x² and f₂=-0.01+x⁵+x⁷).

Figure 9.3 illustrates that the root corresponds to the intersection of the two functions. As an example, consider the equation

$$x^2 = 0.01 - x^5 - x^7 \ .$$ (9.45)

 

If f₂(x)=-0.01+x⁵+x⁷ and f₁(x)=x², an iterative solution can be found from

$$x_{n+1}=g(x)=\sqrt{0.01-x_n^5-x_n^7} \ .$$ (9.46)

 

If the starting guess is 0, the iterative sequence is 0,0.1,0.09995,0.0999496, ..., so convergence is very rapid. This technique is often useful in cases where some part of the equation is complicated but introduces only a weak dependence.

The equation used by Green (1975), (1), provides a good example of the usefulness of this approach. It may be written as

$$a_0^3 = k\left[\left({{a}\over{a_0}}\right)^6-2\left({{a}\over{a_0}}\right)+1\right]$$ (9.47)

 

where $k\approx ca$ where c=0.0765 cm^-1. The problem is to find a₀ given a. To obtain a form that will converge, rewrite the equation to use the term with largest derivative with respect to a₀ as f₁(x):

$$\left({{a_0}\over{a}}\right)^6 =\left[2\left({{a}\over{a_0}}\right)-1 + {{a_0^3}\over{k}}\right]^{-1}$$ (9.48)

 

An iteration that leads rapidly to a solution is then

$${a_0}_{n+1} = a\left[{{{a^3_0}_n}\over{k}}+{{2a}\over{{a_0}_n}}-1\right]^{-1/6}$$ (9.49)

 

For example, if a=0.3, the sequence for a₀ that starts with a as the first estimate is 0.3, 0.2635, 0.26566, 0.265632, ..., which converges even faster than Newton's method for this problem.

The examples of iterative sequences shown in Fig. 9.4 illustrate the necessity of isolating the slowest-varying component as function f₁(x), so that the function g(x) will have as small a derivative as possible. For $\vert dg(x)/dx\vert\ge 1$, the sequence does not converge.

 

Figure 9.4:  Schematic representation of iterative sequences for a convergent case (a) and a divergent case (b). In each case, x_n+1 = g(x_n) = f₁^-1(-f₂(x_n))gives the sequence of estimates.

Example 9.1: The calculation of the wet-bulb temperature illustrates the use of these techniques. The defining equation for the wet-bulb temperature T_wb is

$$f(T_{wb}) = T_{wb} - T - {{L_v}\over{C_p}}\left(r-r_s(T_{wb})\right)=0$$ (9.50)

 

where T is the temperature, L_v the latent heat of vaporization, C_p the specific heat of air at constant pressure, r the water vapor mixing ratio, and r_s(T_wb) the saturation mixing ratio at the wet-bulb temperature. The saturation mixing ratio is a complicated function of temperature that is exponential even in the Clausius-Clapeyron approximation, and is better expressed by the more complicated Goff-Gratch formula (cf. List, 1958, p. 350), so this equation does not have a simple analytical solution.

Newton's method provides a good solution to (9.50) under most conditions, and is particularly fast at low humidity. At high humidity, the (approximately exponential) dependence of the saturation mixing ratio on wet-bulb temperature can introduce instabilities. To increase the range over which the solution converges, it is sometime helpful to average consecutive estimates in an iterative procedure; for example, the series

$$T_{wb,n+1} = {{T_{wb,n} + T +{{L_v}\over{C_p}}[r-r_s(T_{wb,n})]}\over{2}}$$ (9.51)

 

converges for a wider range of temperatures than does the corresponding series without averaging of consecutive terms.

 

Exercise 9.1: Calculate the wet-bulb temperature as a function of dewpoint for a pressure of 800 mb and a temperature of 10.

Exercise 9.2: Calculate the liquid water content that would be produced in adiabatic ascent from a cloud base of 800 mb and 10 to a final pressure of 500 mb.

Example 9.2: Example for further discussion: calculate the threshold temperature for contrail formation at 400 mb, if the ratio of moisture to heat produced during combustion of jet fuel is 0.0295 g_H2Okg^-1_air(^oC)^-1.

   
 9.6 Solution of simultaneous  9.5 Roots of equations  9.5.2 Interpolation