New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
Hydrology 510
Quantitative Methods in Hydrology
Motive
Preview: Example of a function and its limits
Consider the solute concentration, C [ML-3], in a confined aquifer downstream of a continuous
source of solute with fixed concentration, C0, starting at time t=0. Assume that the solute advects in
the aquifer while diffusing into the bounding aquitards, above and below (but which themselves
have no significant flow). (A homology to this problem would be solute movement in a fracture
aquitard
(diffusion controlled)
Flow
aquifer
Solute
(advection controlled)
aquitard
(diffusion controlled)
x
with diffusion into the porous-matrix walls bounding the fracture: the so-called “matrix diffusion”
problem. The difference with the original problem is one of spatial scale.)
An approximate solution for this conceptual model, describing the space-time variation of
concentration in the aquifer, is the function:
  x  x  D 1 / 2 
C ( x, t ) = C0 erfc   
  
  B  vt − x  vB  
or
x

C
D
= erfc 

C0
 B v ( vt − x ) 
where t is time [T] since the solute was first emitted
x is the longitudinal distance [L] downstream,
v is the longitudinal groundwater (seepage) velocity [L/T] in the aquifer,
D is the effective molecular diffusion coefficient in the aquitard [L2/T],
B is the aquifer thickness [L]
and erfc is the complementary error function.
Describe the behavior of this function. Pick some typical numbers for D/B2 (e.g., D ~ 10-9 m2 s-1,
typical for many solutes, and B = 2m) and v (e.g., 0.1 m d-1), and graph the function vs. time at a
one more locations, x, and vs. space at one or more times, t.
Later we’ll examine derivatives and integrals of this function and its parent, which includes details
of concentrations in the aquitard. And later still we’ll look at its origin, through solutions of PDEs.
-1v.1.1, F2008
New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
Review of calculus
(This is review material, taken and expanded from Carol Ash and Robert B. Ash, The Calculus Tutoring Book, IEEE
Press, New York, 1986)
Functions
Introduction to functions
“A function can be thought of as an input-output machine.” Given a particular input, x, the function
f(x) is the corresponding output. Functions are usually denoted by single letters. We’ll often used f
and g in this review to denote functions. “If the function g produces the output 3 when the input is
2, we write g(2)=3.” Mathematicians represent this process by a mapping table or diagram, as
shown here.
TABLE
Input Output
2
3
8
4
9
4
10
-1
MAPPING DIAGRAM
2
8
9
10
3
4
-1
In hydrology the machine can represent a model of a process. Then, for example, x could be
location in an aquifer and f(x) spatially variable hydraulic conductivity as a function of location. Or
for streamflow, x could be time and f(x) stream discharge as a function of time at a stream gauge
location. Or x could be a state, such as pressure or temperature, and f(x) a state-dependent property,
such as fluid density or viscosity which are functions of pressure or temperature (this is called an
equation of state, or EOS). We often consider forcings as an input x, such as precipitation or solar
radiation. For precipitation over a watershed as in input, streamflow at the outlet is a typical output,
while for radiation over a land-surface plot as an input, evapotranspiration from that plot back to the
atmosphere is a typical output. Inputs can be properties (or parameters), location, time, or forcings.
Outputs can be states, fluxes, other properties (or parameters), location, or (travel, residence, or
arrival) times.
Input, x
Machine or Model, f
Output, f(x)
The input, x, is called an independent variable while the output, f(x), is a called a dependent
variable.1 Mathematicians say that “f maps x to f(x), and call f(x) the value of the function at x. The
set of inputs x is called the domain of f and the set of outputs is called the range.”
1
In this review of calculus we assume one input and one output. Later we’ll extend the review to multivariate calculus
with more that one input and often more than one output. In hydrologic applications the multivariate case is the norm.
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Hyd 510
Quantitative Methods in Hydrology
Formally, a function f(x) is not allowed to send one input to more than one output.
Not a function
Consider the domain of x, between the limits a and b. The set of all x such that a ≤ x ≤ b is denoted
by mathematicians by [a,b] and is called a closed interval. The set of all x such that a < x < b is
denoted by mathematicians by (a,b) and is called an open interval. “Similarly we use [a,b) for the
set of x where a ≤ x < b, (a,b] for a < x ≤ b, [a,∞) for x ≥ a, and (-∞,a] for x ≤ a, and (-∞,a) for x
< a. In general, the square bracket, and the solid dot in … the figure below … , means that the
endpoint belongs to the set; a parentheses, and the small circle in … the figure …, means that the
end point does not belong to the set. The notation (-∞,∞) refers to the set of all real numbers.”
[a,b]
(a,b)
[a,∞)
(-∞,b)
Issues to review on your own:
Equations v. functions
One-to-one functions
Increasing and decreasing functions
Elementary functions (all of which are common functions in hydrology; also, see the elfun
directory in Matlab)
Type
Constant function
Examples
f(x) = 2 for all x, g(x) = -π for all x
Power function
x-1, x0.995, x, x2, x2.7, x12
Trigonometric functions
sine, cosine, tangent, secant
Inverse trigonometric
Functions
sin-1 x, cos-1 x, tan-1 x
Exponential functions
2x, (1/4)-x, 104, and especially ex,
where e = 2.71828 …
Logarithmic functions
log2 x, log10 x, and especially loge x = ln x
Trigonometric functions
Commonly encountered in, e.g.,
-3-
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Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
-problems with periodic forcing (e.g., diurnal, seasonal, decadal)
-cylindrical and spherical coordinates (e.g., radial well hydraulics,
intra-particle spherical diffusion and adsorption)
-geometric definition of object shapes
-certain (finite) spatial domain problems (e.g, Fourier series solutions)
Definitions of sine, cosine, and tangent
sin θ =
y sin θ
y
x
, cosθ = , tan θ = =
r
r
x cos θ
y
(x,y)
r
(1)
θ
x
where (x,y) = Cartesian coordinates
(r,θ) = radial coordinates
Radius r is always positive, but the signs of x and y
depend on the quadrant, thus the signs of the trig
functions also depend on quadrant:
y
x
sign of sin θ
sign of cos θ
sign of tan θ
Degrees v. radians
We measure angles in both degrees and radians. Recall that an angle θ of 180º = π radians. More
generally,
π
number of radians
=
(2)2
number of degrees 180
Examples of important angle and related trig functions:
Degrees Radians sin cos tan
30º
π/6
1/2 √3/4 √1/3
45º
π/4
√1/2 √1/2
1
60º
π/3
√3/4 1/2
√3
Degrees Radians sin cos tan
0º
0
0
1
0
90º
π/2
1
0 none
180º
0 -1
0
π
270º
3π/2
-1 0 none
360º
2π
0
1
0
2
The equation numbers refer to the numbers in Ash and Ash (1986).
-4-
New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
We prefer to use radians and, in some cases, we must use radians, such as
measuring the arc length along a circle. Let s equal the arc length, a
fraction of the circumference. The circumference is 2πr, therefore the arc
length is
s= θ r,
(5)
r
θ
s=rθ
where θ is in radians.
Reference angle
Trig tables list sin θ, cos θ, tan θ for 0º < θ <90º. To find trig functions for other angles use signs
given in the box above, plus reference angles. For example, if θ is in the second quadrant (upper left
quadrant) then the reference angle is 180º - θ. In particular, if θ is 150º then the reference angle is
30º. Etc.
Right angle trigonometry
opposite leg
adjacent leg
sin θ =
cos θ =
hypotenuse
hypotenuse
opposite leg
tan θ =
adjacent leg
(6a,b)
(6c)
hypotenuse
opposite
θ
adjacent
Graphs of sin x, cos x, and tan x
Exercise: Use Matlab to graph these three functions from x= -4π to +4π
Graphs of a sin(bx + c)
You should be familiar with:
-defns. of amplitude (a), period (2π /b), frequency (b), and phase lag (c)
(for example, the phase lag describes a shift (in radians) of the peak).
-applications to
> harmonic motion
> (earth and ocean) tides
> approximations to diurnal, seasonal, and other periodic temporal signals
(applications: temperature, evapotranspiration, spring discharge)
Application: A monitoring well on Cape Cod, Massachusetts, located 700m from the
Atlantic Ocean coast, observes that water table elevations fluctuate over time. The water
level data is fit to the model a sin(bx + c) where a is the amplitude (one-half the total
range) of the water level fluctuation, the period is 12 hours, x is time (in hours), and the
observed phase lag is 3 hours (compared to the local ocean tide; where for a lag of 3 hours
c is expressed in radians = π /2). The ratio of tidal amplitude in the well to that in the local
ocean, and the phase lag, are compared to a model of aquifer response to a tidal forcing in
order to estimate the aquifer parameters hydraulic conductivity and specific yield. (Similar
calculations are performed for other periodic forcings, like earth tides and fluctuating
stream stage.)
-5-
New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
Graphs of g(x)= f(x) sin x
If you need to sketch the function g(x) by hand you would first sketch the curve y=f (x) and the
curve y= -f (x), its reflection in the x-axis, to serve as an envelope. Then change the amplitude of the
sine curve with x, so that it just fits within the envelop. In addition, reflect the sine curve in the xaxis whenever f (x) is negative.
Exercise: Use Matlab to graph the function g for f(x)= exp(-|x|), from x= -4π to +4π
Application (con’t): The model of aquifer response to tidal forcing, mentioned in the
previous box, yields a different solution for each distance ℓ from the ocean. That is the
water level response is a(ℓ) sin[bx + c(ℓ)] where the amplitude a and the phase lag c depend
on location relative to the ocean. Wells that are closer to the ocean have larger amplitude
and smaller phase lag. For example, the amplitude decreases exponentially with distance,
a(ℓ) ∝ exp(- ℓ).
Secant, cosecant, and cotangent
1
1
1
cos θ
sec θ =
, csc θ =
, cot θ =
=
cos θ
sin θ
tan θ sin θ
(7)
or, for a right triangle:
sec θ =
hypotenuse
adjacent leg
hypotenuse
opposite leg
adjacent leg
cot θ =
opposite leg
csc θ =
(8a,b)
(8c)
Notation issues: It is common practice to write sin2x for (sin x)2, and sin x2 to mean sin(x2), etc.
Standard trigonometric identities. Below are a few examples of common identities. They
illustrate the various categories of identities. See standard math tables (e.g., as referenced on next
page) for complete list.
(9) Negative angle formulas
sin( − x ) = − sin x; cos( − x ) = + cos x
That is, sine is an odd function and cosine is an even function
(10) Addition formulas
sin( x + y ) = sin x cos y + cos x sin y
(11) Double angle formulas
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Hyd 510
Quantitative Methods in Hydrology
sin(2 x ) = 2 sin x cos x
tan( 2 x ) =
2 tan x
1 − tan 2 x
(12) Pythagorean identities
sin 2 x + cos2 x = 1
(13) Half-angle formulas
 x  1 − cos x
sin 2   =
2
2
 x  1 + cos x
cos 2   =
2
2
(14) Product formulas
sin( x + y ) + sin( x − y )
sin x cos y =
2
(15) Factoring formulas
sin x + sin y = 2 cos
x− y
x+ y
sin
2
2
(16) Reduction formulas
π

cos − θ  = sin θ

2
B
(17) Law of Sines
sin A sin B sin C
=
=
a
b
c
a
c
(18) Law of Cosines
c 2 = a 2 b 2 − 2ab cos C
A
b
C
(19) Area formula
Area of triangle ABC =
1
ab sin C
2
Reference re Math Tables:
Standard reference on algebra, calculus, and matrix methods, and ODEs, but no PDEs:
CRC Standard Mathematical Tables and Formulae, 31st Edition
Zwillinger, D., CRC Press, Boca Raton, FL, 2003 (or later edition).
see www.crcpress.com for latest edition.
Important note: this class, and all your subsequent hydrology classes, assume
that you have your own copy of this or another standard math tables and know
how to use it!
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Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
Inverse functions and inverse trig functions
The inverse function
Given a one-to-one function, f(a) = b, that maps a to b
Then a = f -1(b) is its inverse that maps b to a.
f
a
b
f -1
Example: A partial table for f (x) = 3x, and its inverse f -1(x) = x/3 :
f -1(x)
x
f(x)
x
6
2
2
6
15 5
5
15
21 7
7
21
Exercise: Use Matlab to graph f (x) and f -1(x), for this example, over the domain 1≤ x ≤25, on the
same plot.
The graph of f -1(x). One of the advantages of an inverse function is that its properties, such as its
graph, often follow easily from the properties of the original function. Comparing graphs of f (x)
and f -1(x) amounts to comparing points such as (2,6) and (6,2) in the example above. The points are
reflections of one another in the line y = x, so that the pair of graphs is symmetric with respect to
the line.
Exercise: If f (x) = x2, and x ≥ 0 so that f is one-to-one, then f -1(x) = √x. Use Matlab to graph f (x) and
f -1(x) for this new example, over the domain 0 ≤ x ≤ 1.2, on the same plot. Also on the plot include
the straight line y=x as a dashed line.
Common inverses of trig functions are
-The inverse sine function (sin-1 x or arcsin x)
-The inverse cosine function (cos-1 x or arccos x)
-The inverse tangent function (tan-1 x or arctan x)
Exponential and logarithmic functions
Exponential functions
Examples, 2x, (1/2)x, 7-x
-contrast these with power functions like x2, x1/2, x -7
Note that negative bases, e.g. (-4)x can be a problem. Be careful. Try to avoid.
Exercise: Use Matlab to plot example graphs of f (x) =3x, 2x and (1/2)x over the domain -3≤ x ≤+3.
Do this in two plots. First, use a linear plot and then a semilog plot (ln[f(x)] v. x).
-8-
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Hyd 510
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Most popular bases?
Computer science uses base 2. We’ll visit this in lab.
Algebra and much of science favors base 10
Calculus, ODEs, and PDEs favors base e
Powers and roots (examples)
ax ay = a (x + y)
a0 = 1 [if a≠ 0 ]
a -x = 1/ax
(ab)x = ax bx
ax / ay = a (x – y)
(ax )y = axy
a 1/x = x√a
a x/y = y√ax
x
√ab = x√a × x√b
The exponential function ex = exp x and especially e-x = exp -x
e is an irrational number between 2.71 and 2.72 (= 2.71828 …)
(with a particular defn, in terms of a derivative,
to be given later/elsewhere)
e or exp is known as “The Exponential Function”
Graph of exp x
¾ exp x is defined for all x
¾ exp x > 0; in fact, the range of exp x is (0,∞)
¾ exp x is an increasing function
Exercise: Use Matlab to plot exp x, and compare to graphs of 2x and 3x,
for the domain -3≤ x ≤+3
Exercise: Use Matlab to graph e-x = exp (–x) = 1/(exp x) for the domain 0≤ x ≤+2.
Compare on the same plot to Matlab graphs of (1/2)x, x-1 , x-2
Application: It is common to find hydrologic systems that respond exponentially in time,
t. The response function has the form exp(-kt), where k is a rate coefficient (per unit time)
and k-1, which has units of time, is sometimes called the “time constant”. Notice this
function decreases in time. Example applications include radioactive decay, first order
biotransformation of organic contaminants, and discharge from a lake or manmade surface
impoundment.
The natural log function ln x = loge x
ln x is the inverse of ex ; i.e., ln (ex) = x
ln a = b only if eb = a
since e0 = 1 and e1 = e, then
ln 1 = 0 and ln e = 1
-9-
New Mexico Tech
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Hyd 510
Quantitative Methods in Hydrology
Graphs of ln x
¾ ln x is defined for x > 0; you cannot take the logarithm of a negative number or
zero.
¾ The range of ln x is (-∞,∞)
¾ ln x is negative if 0 < x < 1, and positive if x > 1
¾ ln x is increasing.
Exercise: Use Matlab to graph ln x for the domain 0 ≤ x ≤+3
Exercise: Use Matlab to graph both ln x and its inverse, ex, on the same plot for the domain
-3≤ x ≤+3. Also include on the plot the straight line y=x as a dashed line. The pair of graphs
should be symmetric with respect to the dashed line. Note that the function ln x is undefined
for x ≤ 0.
Application: Pumping water from an aquifer results in drawdown of hydraulic head. After
a while the time behavior of drawdown in an observation well is approximated by a
logarithmic function of the form ln(βt), where β is a coefficient (per unit time) that
depends on aquifer properties, and distance away from the pumping.
Laws of exponents and logarithms
ex ey = ex + y
ex / ey = ex – y
e-x = 1/ex
(ex )y = exy
ln ab = ln a + ln b
a 
ln   = ln a − ln b
b
1
ln   = − ln b since ln 1 = 0
b
ln a b = ln ( a ) b = b ln a
Logarithms with other bases, especially bases 2 and 10
log2 x is the inverse of 2x
log10 x is the inverse of 10x
Solving equations and inequalities involving ex and ln x
To solve the equation ex = 7, take ln of both sides and use ln ex = x to get x = ln 7.
To solve the equation ln x = -6, take exp of both sides and use exp(ln x)= x to get x = e-6.
-10-
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Hyd 510
Quantitative Methods in Hydrology
Solving inequalities involving ex and ln x requires care, but takes advantage of both ex and ln x
being increasing functions.
Combinations (e.g., sums, products, compositions) of elementary functions are
elementary functions.
Examples: x2+x, x2 sin x¸
sin x
Solving equations and inequalities involving elementary functions
Review how to use algebra including factorization and zero finding, exercising special caution for
inequalities, such as f(x) > 0, when the functions are not increasing.
Example: Solve the equation, 4 ln (2x +5) = 8.
Solution:
ln (2x +5) = 8/4 = 2
2x +5 = exp(2)
2x = e2 -5
x = (½)(e2 -5)
{divide by 4}
{take exp}
{subtract 5}
{divide by 2}
Consider the example in the sketch below. The function f is zero at x=-2 and +2. These two points
both satisfy the equation f(x) = 0. That is, there are two solutions to this equation; we say that the
f
f is positive
f is positive
f jumps
f is zero
-3
-2
-1
f is zero
0
1
2
3
4
5
6
x
f is negative
f is negative
solution for the independent variable x is non-unique. (Exercise: Pick another value of f and find the
solution(s) for x.)
x=+2
f=0
x= -2
Notice that to the left of x=-2 the function is positive, as it is between x = +2 and up to +4. At x =
+4 and to its right the function is negative. If we seek a solution to the inequality f(x) ≥ 0 the
solution is x ≤ -2 and +2 ≤ x < 4. The solution to f(x) < 0 is -2 < x < +2 and x ≥ +4
There are some suggested guidelines for solving inequalities. Illustrated by the sketch, these will be
useful later.
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Hyd 510
Quantitative Methods in Hydrology
Step 1. Find the values of x where f is discontinuous, as at x=4 in the sketch. For an
elementary function these typically occur where f is not defined, in practice because of a
zero in the denominator (as at x=0 for the function1/x).
Step 2. Find the values of x where f is zero, as at x=±2 in the sketch; that is, solve the
equation f(x)=0.
Step 3. Look at the open intervals in between. On each of these intervals f maintains only
one sign. To find the sign you can test one number for each interval.
Exercise: Decide where the function f = (x + 1)/( x - 1) is positive and where it is negative. Then
plot in Matlab3.
Applications: There are many applications in hydrology seeking the solution to an
inequality. Where are contaminant concentrations greater than an MCL (regulated “maximum
contaminant level”)? At what times during a storm do water levels exceed flood stage?
Does the suction (negative pressure head) in a soil the wilting point of vegetation?
Some applications involve the design or operation of facilities. These often lead to
inequalities, referred to as constraints. Well pumping rates are often constrained by the
amount of drawdown they cause. Streamflow releases from dams are constrained by their
impact on downstream aquatic life and geomorphology.
Graphs of translations, reflections, expansions, and sums
In many applications it is useful to understand how the graphs of a function change under
transformations, mainly horizontal translations in x, horizontal and vertical expansions, and sum or
superposition of functions. Reflections are also encountered. You perform many of these operations
when you use a drawing program like Adobe Illustrator or the drawing features in MSWord.
In the following examples consider the function y=f(x), then transform it. In the first two examples
an operation is performed on the variable x. In the next two the operation is performed on the
function , i.e., on the entire right side. The fifth example is a composite operation.
Horizontal translation: y=f(x - 3) to translate (shift) the graph (otherwise unchanged) along the xaxis three units to the right.
Example: This is precisely what happens when a solute plume in a river advects
downstream without mixing, where x is location along the river, and the product of
stream velocity and time equals 3 (in the same distance units as x).
Horizontal expansion: y=f(x/2) doubles the x-coordinates of the graph so as to expand the graph
(since it doubles all parts of the graph it also shifts it to the right).
3
You may think “Why don’t I just plot in Matlab first and not try to figure this out in my head at all.” The trouble is
that Matlab (and other programs) can be fooled. It is a good idea to have a notion for what a function looks like before
trusting a program to plot it. Besides, it helps to improve your reasoning ability and understanding.
-12-
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Hyd 510
Quantitative Methods in Hydrology
Vertical contraction: y=(1/2) f(x) contracts the y value by a factor of 2.
Vertical reflection: y= - f(x) refects the graph in the x-axis.
Example: Suppose y=f(x) represents aquifer drawdown at location x due to a
pumping well. Then y= - f(x) represents the negative drawdown, or “drawup,” if the
well were used to inject water, instead of pump it, at the same rate.
Composite horizontal translation, horizontal expansion, and vertical contraction
1 x

y = f  − 3
2 2

reduces the graphs value to one half and expands it horizontally by a factor of
two, then corrects for the shift caused by the expansion, and finally translates the
resulting graph three units to the right.
This crudely mimics a solute plume in a river that advects downstream while mixing,
where the mixing deceases concentration by a factor of one half and spreads it out by
the factor of two, thus preserving solute mass.
Warning re f(x -1) v. f(x) – 1. The first of these involves a translation of 1 unit to the right. The
second translates the graph down.
Graph of f(x) + g(x): simply add the “heights” of the respective functions. Use Matlab to plot sin(x),
cos (x), and their sum, over the domain x= -4π to +4π.
Some more advanced functions
We encounter a large family of more advanced functions in hydrology. Three of the most common
advanced functions are mentioned here. Advanced functions can sometimes be found in standard
math tables, while the best reference for them is the more advanced and highly cited Abramowitz
and Stegun, Handbook Mathematical Functions, Dover Press, New York, 1964. This handbook was
originally published by the U.S. National Bureau of Standards, now call NIST (National Institute of
Standards and Technology). NIST is currently organizing a web-based successor to the NBS
Handbook, called the Digital Library of Mathematical Functions (DLMF). The web site for this
uncompleted project is http://dlmf.nist.gov/. Advanced functions can also be found in Matlab (see
the specfun directory), Maple, and Mathematica software, but Abramowitz and Stegun remains the
main reference. We’ll use the routines in Matlab (from the specfun directory) for the functions
below.
Error Function, erf x, and Complementary Error Function, erfc x = 1 – erf x,
are commonly encountered in groundwater flow, solute transport, and heat transport as a solution to
a parabolic partial differential “diffusion” equation4.
4
We’ll learn about these equations, and the meaning of terms like “parabolic”, later in the semester.
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New Mexico Tech
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Hyd 510
Quantitative Methods in Hydrology
2
erf x =
∫
π
erfc x =
x
0
2
π
∫
2
e −t dt
∞
x
the error function
2
e −t dt = 1 − erf x
complementary error function
erf (-x) = -erf x
Graph of erf x:
¾
¾
¾
¾
¾
¾
erf x is defined for all x
erf x > 0 and erfc x > 0
the range of erf x and erfc x is (0,1)
erf x is an increasing function; erfc x is a decreasing function
lim erf x = 1; lim erfc x = 0;
x →∞
x →∞
lim erf x = −1;
x→ −∞
lim erfc x = 2
x →−∞
Exercise: Using Matlab graph erf x and erfc x on the same plot over the domain -2≤ x ≤+2
Exponential Integral, Ei(x) and E1(x)
Ei (x) is most commonly encountered in aquifer well hydraulics. The commonly used model for
transient drawdown in response to pumping is called the Exponential Integral Model or the Theis
Equation.
There are two different, alternative definitions for the exponential integral. The first5,
−t
−t
x e
∞ e
Ei( x) = − ∫
dt = ∫
dt , x>0 , exponential integral function,
−∞ t
−x t
where t is a dummy variable of integration, is the version encountered in well hydraulics. The other
version is
−t
∞e
E1 ( x ) = ∫
dt , x>0
x
t
The two versions are related by
− Ei( x ) = E1 (- x )
Note that Matlab uses E1, the second definition, and calls that funcition expint. You need to
transform it using this last equation to get Ei. But you have to be careful. From the Matlab help page
for EXPINT(x):
By analytic continuation, EXPINT is a scalar-valued function in
the complex plane cut along the negative real axis.
Another common definition of the exponential integral function is
the Cauchy principal value integral from -Inf to X of (exp(t)/t)
5
We assume x is real here, but the exponential integral is more generally defined for complex arguments, z=x+iy, as is
clear in the relationship between the two definitions given in Matlab.
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dt, for positive X. This is denoted as Ei(x). The relationships
between EXPINT(x) and Ei(x) are as follows:
EXPINT(-x+i*0) = -Ei(x) - i*pi, for real x > 0
Ei(x) = REAL(-EXPINT(-x)), for real x > 0
Application: The space-time distribution of drawdown, s [m], near a fully penetrating
pumping well in a homogeneous, isotropic, confined aquifer of infinite extent, is given by the
Theis model,
Q
Q
r2S
s( r, t ) = −
Ei( −u ) =
W (u ) , u =
,
W (u ) = − Ei( −u ),
4πT
4πT
4Tt
where Q [m3s-1] is the pumping rate, r [m] is radius from the well, t [s] is time, T [m2s-1] is
aquifer transmissivity, S [-] is the aquifer storage coefficient, u [-] is a dimensionless
similarity variable, and W(u) is the “Theis Well Function”.
Note: this problem has two independent variables, s and t. It’s an example of multivariate calculus
and the solution of a (parabolic) partial differential equation.
Exercise: Using Matlab graph Ei(x) and –Ei(-x) on the same plot over the domain 0≤ x ≤+2. The
second of these graphs represents the Theis Well Function (see box).
Bessel Function of the first kind6, Jn(x)
Bessel functions are commonly encountered in problems involving a radial geometry, such as in
well hydraulics, or flow of water to a tree root in the vadose zone. Bessel functions constitute a
family of functions. For illustration purposes we mention only the Bessel Functions of first kind.
Bessel Function of the first kind of integer order n are defined by
π
( 1 x)n
J n ( x) = 1 2
cos( x cosθ ) sin 2 n θ dθ , x>0,
∫
0
1
2
π Γ( n + 2 )
where Γ is the gamma function,
∞
Γ( z ) = ∫ t z −1e −t dt , Re(z > 0),
1
1
with corresponding specific values for zero and first order Bessel Functions of Γ( 12 ) = π 2 and
1
Γ( 12 ) = 12 π 2 , respectively. In particular the zero order function is
1 π
J 0 ( x ) = ∫ cos( x cosθ ) dθ
π
0
( 12 x ) n
In the limit, as x→0, J n ( x ) ~
Γ( n + 1)
6
The Bessel Function argument can be complex, z=x+iy, although we won’t be considering that case.
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Exercise: Using Matlab graph J0(x) (besselj in Matlab) and K0(x) (besselk) on one plot, for the
domain 0 < x ≤ 10, where K0(x) is the Modified Bessel function of the second kind of order zero
(i.e., another Bessel Function).
Application: Suppose the confined aquifer, mentioned in the previous box, is leaky. That is,
bounding the top of the aquifer is an aquitard and above that is a phreatic aquifer. Pumping
the confined aquifer will induce leakage through the aquitard, recharging it from above.
Eventually, the pumping will be balanced by leakage and the drawdown in the confined
aquifer will reach a steady state. That equilibrium drawdown is described by a Modified
Bessel function of the second kind of order zero:
Q
B' T
s=
K0 (r β ) , β =
,
K'
2πT
where β [m] is a “leakage coefficient”, B’ [m] is aquitard thickness, K’ [m s-1] is the vertical
hydraulic conductivity of the aquitard, and s, Q, r & T were defined in the previous box.
The plot of K 0 ( r β ) shows how drawdown varies with location and leakage coefficient.
-16-
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Hyd 510
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Limits
Limits are used to describe:
1. discontinuities
2. the ends of graphs where x→∞ , x→ -∞
3. asymptotes
4. definitions for derivative and integral (later)
Limits are commonly encountered in hydrology, for example, when examining conditions near or
very far from boundaries, when considering what happens as time tends to infinity, and when
considering whether property values can be considered constant or must be allowed to vary.
Introduction
Limit definition for function f(x)
lim f ( x ) = L
x →a
if, for all x sufficiently close to, but not equal, to a, f(x) is forced to stay as close as we like,
and possibly equal to L.
Review on your own:
One-sided limits
Infinite limits
Limits as x→∞ , x→ -∞
Limits of continuous functions
Various types of discontinuities, especially jumps and “blow ups”
Exercise: Consider limits for the sketch:
3
f
A solid dot means that
the point belongs to the
set; and the small circle
means that the point
does not belong to the
set.
2
1
-3
-2
-1
0
0
1
2
3
4
5
6
x
-1
-2
Exercise: What happens near x = -∞, -3, -2, 0, +2, +4, +6, +∞ ?
Limits for combinations of functions
Review finding limits of combinations of functions. Find the limits of components of the expression
and put them together sensibly.
Example:
x 2 + 5 + ln x 0 + 5 + ( −∞) − ∞
=
=
= −∞
x →0 +
2e x
2
2
lim
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Recall that in taking limits of components you end up with sums, products and quotients which
must be resolved, as in the example above. Some more examples from Ash and Ash (1986), in
which each term is the result of a “sublimit”:
0×0=0
∞ - 4= ∞
0+0=0
∞/8 = ∞
0/3 = 0
-2 × ∞ = -∞
∞∞ = ∞
40 = 1
∞+∞=∞
5/0+ = +∞
5/0- = -∞
∞ × -∞ = -∞
∞
2/-∞ = 0
3 = +∞
Warnings: A limit problem of the form 2/0 does not necessarily have the answer ∞. Rather 2/0+ =
+∞ while 2/0- = -∞. In general, in a problem of the form (non 0)/0 it is important to examine the
denominator carefully.
Indeterminate limits
Limits of indeterminate form are commonly encountered:
0 ∞ ∞ ∞ ∞
, , , , , 0 × ∞, 0 × −∞, ∞ − ∞, ( −∞) − ( −∞), (0+ ) 0 , 1∞ , ∞ 0
0 ∞ ∞ ∞ ∞
Each of these can be resolved. None are truly “indeterminate”. We can use the rule below, for some
problems. Others require differential calculus (later).
Highest power rule
Uses the following principles:
1. As x → ∞ or x → -∞, a polynomial has the same value as its term of highest degree.
For example, lim ( x 4 + 2 x 2 + 3x − 2) = lim ( x 4 ) = ∞
x→ −∞
x→−∞
2. As x → ∞ or x → -∞, a quotient of polynomials has the same limits as the quotient …
term of highest degree in numerator
term of highest degree in denominator
which cancels to an expression whose limit is easy to evaluate.
x2
x5 + x3 + 1
x5
=
=
= +∞
lim
lim
x→−∞ 6 x 3 − 7 x 2 + x + 4
x→−∞ 6 x 3
x→−∞ 6
For example, lim
-18-
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The Derivative I
Preview
From freshman physics and calculus recall the concepts of Velocity and Slope:
Velocity of a particle = change in position / change in time ≅ [f(t+∆t) – f(t)] / ∆t
Slope of the line f(x) =y is given by the change in y-coordinate / change in x-coordinate, or
slope ≅ [f(x+∆x) – f(x)] / ∆x , or
slope =∆y / ∆x = tangent ≅ secant (uses B and A) (if ∆x is “small enough”)
tangent
secant
B=(x+∆x, f(x+∆x))
A= (x, f(x))
Negative slope
Positive slope
What is a
positive slope?
negative slope?
zero slope?
Zero slope
Applicaton: A number of “diffusion like” fluxes in hydrologic applications are driven by a
gradient (slope of a state variable, such as solute concentration, with respect to distance). A
flux density is an intensive measure of an amount of “something” passing through a surface,
per unit time per unit area.
Fluxes of solute mass due to diffusion are described by Fick’s First Law where
solute mass flux density [kg m-2 s-1] = -Dm dC/dx.
where C [kg m-3] is concentration,
x [m] is location,
d( )/dx [m-1] is the gradient “operator”, and
Dm [m2 s-1] is a diffusion coefficient.
That is, the diffusive flux of solute is driven by a concentration gradient.
Notes:
(a) Notice the negative sign in Fick’s Law. Why?
(b) The units of each variable have been written out in SI units. Observe that the units on
the left and right hand sides of Fick’s Law are in balance. Balancing units is a useful way to
check an equation and its solution for errors.
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Definition and some applications of the derivative
df
f ( x + ∆x ) − f ( x )
= f ' ( x ) = lim
Definition of the derivative of f(x) :
∆x → 0
dx
∆x
Recall the difference between speed and velocity where speed = |velocity|. The speed gives the
magnitude but not the direction (of slope), while the velocity gives both magnitude and direction.
Exercise:
B
Use the letters to
identify the following
for the function f(x):
Largest positive slope?
f
A
Graph of y=f (x)
C
-3
-2
-1
0
1
2
100
x
Largest negative slope?
D
E
Zero slope?
Zone of small negative
slope?
Zone(s) of
~ constant slope
~ Point(s) of inflection
101
A
f'
Graph of y=f ' (x)
-3
-2
-1
0
1
2
101
x
B
Concave up
Concave down
(f '' issue)
100
D
E
C
Notation: if y=f(x) then the derivative can be written as:
df d
dy
,
f ' , f ' ( x ),
f ( x ), Dx f , Df , y ' , or
dx dx
dx
In hydrology we typically use the notation
df d
dy
,
f ( x ), or
dx dx
dx
A more physical interpretation of the derivative f ' (x) is instantaneous rate of change of f wrt x,
where the “average rate of change” of f wrt x, in the interval between x and x+∆x, is
df change in f
f ( x + ∆x ) − f ( x )
=
=
dx change in x
∆x
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Applications: Other “diffusive” fluxes, or gradient laws, are heat conduction (Fourier’s Law,
which describes a diffusive flux of thermal energy due to a temperature gradient), viscous
momentum diffusion (Newton’ Law of viscosity, which describes the diffusive flux of fluid
momentum due to velocity shear gradients), and groundwater flow (Darcy’s Law, which
describes the hydraulic diffusive flux of porous media flow due to gradients of hydraulic
head). Insomuch as these processes, and Fick’s Law, have the same math but different physics
they are called homologies. Other processes invoking a similar homologous model include
turbulent diffusion (momentum, mass, and thermal fluxes due to turbulent velocity
fluctuations), dispersion in rivers (due to velocity variations across the channel cross-section),
and mechanical dispersion and macrodispersion in aquifers (due to upscaling and averaging of
velocity fluctuations caused, respectively, by pore structure/connectivity and aquifer
heterogeneity).
In all of these models the flux is proportional to the gradient of a state (e.g., concentration,
flux density [amount m-2 s-1] = - coefficient × gradient
temperature, or velocity), with a proportionality coefficient that must be determined
empirically. For example, with Darcy’s Law we observe fluid fluxes (specific discharge) and
head gradients and take their ratio to get an estimate of hydraulic conductivity, the
proportionality coefficient for that expression.
Higher derivatives:
The second derivative of f , denoted by f '', is the instantaneous rate of change of f ' wrt x.
Notation: f ' ' , f ' ' ( x ),
d2 f d2
d2y
2
2
,
f
(
x
),
D
f
,
D
f
,
y
'
'
,
or
x
dx 2 dx 2
dx 2
Typical application from freshman physics: Acceleration, the derivative of velocity
Warning: if f '' is positive then it is not necessarily true that the object is speeding up.
Units: example from acceleration, unit of accelerations are m2 s-1, while units of speed or velocity
are m s-1.
Concavity measure:
concave up (f '' positive), concave down (f '' negative), straight line (f '' zero)
Point of inflection: a point at which f '' changes sign.
Examples: see graphs on previous page
-21-
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Applicaton: Recall the box on Fick’s First Law, using the symbol N to represent flux
density. Then N = − Dm dC dx . How can we use this to find out how much flux changes
with location x? Take the derivative of N:
dN
d 
dC 
-3 -1
=
 − Dm
 [kg m s ]
dx 
dx dx 
If Dm is constant, move it outside the outer derivative, or
d  dC  dDm dC
d 2C
dN
D
−
=
−
= − Dm 

m
dx  dx  {
dx dx
dx 2
dx
=0
If the diffusion is steady (doesn’t change in time) and if there are no sources or sinks of
solute, then N should be constant and its derivative zero. Under this condition we expect
that d 2C dx 2 = 0 . For this special case, this is a model of conservation of solute mass.
Derivatives of basic, elementary functions (consider their graphs)
Derivative of a constant function
d ( const.)
=0
, or written in another notation, Dx(const) = 0
dx
Derivative of the functions x and xr
d ( x)
d (xr )
=1 ;
= r x r −1
dx
dx
Derivative of sin x, cos x, and other trigonometric functions
d sin( x )
d cos( x )
d tan( x )
= cos x ;
= − sin x ;
= sec 2 x
dx
dx
dx
Radian vs. degrees? Radians preferred (it’s easier).
Derivative of ex, and definition of e:
Defn. of e: “e is the base such that the graph of bx , where b is a fixed positive number, has a
slope of one at the point (0,1).” This also leads to the definition of the derivative of e.
d exp( x ) de x
=
= ex
dx
dx
or, Dx ex = ex
The derivative of an inverse function
dy
1
=
dx
dx
dy
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It’s used to find derivatives for ln x, sin-1x, cos-1x, now that we have derivatives of ex, sin x and cos
x. For example, …
Derivative of ln x
dx de y
=
= ey = x .
dy
dy
1
1 1
d (ln x ) dy
=
=
= y =
or Dx ln x =1/x
Thus
x
dx
dx dx e
dy
Derivatives of inverse trigonometric functions, use the same idea.
Let y = ln x, then x=ey and
Dx c = 0
Table of basic derivatives7 …
Dx sin x = cos x
Dx x = 1
Dx cos x = - sin x
Dx xr = r xr-1
Dx tan x = sec2 x
Dx ln x = 1/x
Dx cot x = -csc2 x
Dx ex = ex
Dx sec x = sec x tan x
Dx csc x = - csc x cot x
Dx sin-1 x =
Dx cos-1 x =
Dx tan-1 x =
1
1 − x2
−1
1 − x2
1
1 + x2
Non-differential functions
Discontinuous functions
If f is discontinuous at x=x0, then f is not differentiable at x=x0. (Equivalently, if f is
differentiable then f is continuous).
Cusps
A cusp arises when a graphs is continuous (in value) but suddently changes direction (so that
the curve is not “smooth”), and in this case f is not differentiable. Differentiability is a more
exclusive property that continuity. A differentiable function is continuous but a continuous
function is not necessarily differentiable. Leads to piecewise continuous functions.
7
Be cautious with the tables presented in these notes. They have not been proofed nearly as well as what you will find
in standard math tables and, in any event, they are not complete. You should probably refer to standard tables instead,
when solving problems. In any event, if you find an error in these notes bring it to my attention.
-23-
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A
f(x)
Exercise:
What is
happening at
the indicated
points?
B
A solid dot means
that the point
belongs to the set;
and the small
circle means that
the point does not
belong to the set.
E
C
D
x
Derivatives of constant multiples, sums, products and quotients
Constant multiple rule for c f(x), where c is a constant:
Dx (c f ) = c Dx f = c f '
Sum rule for the derivative of f(x)+g(x):
Dx (f + g) = Dx f + Dx g = f ' + g'
Product rule for derivative of f(x)g(x):
Dx (f g) = f Dx g + g Dx f = f g'+ g f '
Warning: don’t differentiate f and g separately and multiply
Product rule for more than two factors, e.g., three factors: Dx(f g h) = f g h' + f g' h + f 'g h
Dx (f /g) =
Quotient rule for the derivative of f(x)/g(x):
g Dx f − f Dx g g f ' − f g '
=
g2
g2
Derivative of a function “with two formulas”; do it by “intervals”
Example: Suppose f(x) = |ln x|. Then f(x) = ln x when ln x ≥ 0, but f(x) = -ln x when ln x < 0.
Thus
− ln x if 0 < x < 1
− 1 / x if 0 < x < 1
f ( x) = 
so that, f ' ( x ) = 
 ln x if x ≥ 1
 1 / x if x ≥ 1
Derivative of a composition
The chain rule for the derivative of a composition y=y(u) is
dy dy du
=
dx du dx
or, stated another way for function f,
Dxf(u(x)) = f '(u) u'(x)
Example:
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What is d(sin x2)/dx?
Let y = sin x2.
Then u=x2 and y = sin u.
The derivatives are
dy/du = cos u and
du/dx = 2x.
Thus d(sin x2)/dx = cos u ⋅ 2x = 2x cos x2
Warning: don’t omit a step in this approach.
Table of basic derivatives incorporating the chain rule:
Dx u = r ur-1 u'(x)
Dx sec u = sec u tan u ⋅ u' (x)
r
1
u' ( x )
u
Dx eu = eu u'(x)
Dx ln u =
Dx sin u = cos u ⋅ u' (x)
Dx cos u = - sin u ⋅ u' (x)
Dx tan u = sec2 u ⋅ u' (x)
Dx cot u = -csc2 u ⋅ u' (x)
Dx csc u = - csc u cot u ⋅ u' (x)
Dx sin-1 u =
Dx cos-1 u =
Dx tan-1 u =
1
1 − u2
−1
1 − u2
⋅ u' (x)
⋅ u' (x)
1
⋅ u' (x)
1 + u2
Implicit differentiation and logarithmic differentiation
Implicit differentiation
Example: What is the slope of the graph of y3 – 6x2 = 3 at the point (-2,3)? This equation
defines y(x) implicitly. Solve algebraically for y to obtain y = (6x2 + 3)1/3. This equation
expresses y explicitly as a function of x. We can take the derivative of this explicit function
and apply it at the subject point to get the desired slope. Try it. The answer is -8/9. But you
don’t have to approach the problem this way. In fact there are many cases where you won’t
be able to transform an equation to an explicit form. You can find the derivative y'
implicitly.
Recall the implicit equation, y3 – 6x2 = 3. Differentiate with respect to x on both sides of the
equation. In this procedure y is treated as a function of x, so that the derivative of y3 with
respect to x is 3y2y' by the chain rule. Thus …
-25-
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3y2y' – 12x = 0
y' = 12x /3y2 = 4x /y2
y'(-2,3) = 12⋅(-2) /3⋅32 = -24 /27 = -8/9
yielding the same slope as the explicit approach.
The process of finding y' without first solving for y is called implicit differentiation.
Be careful, don’t omit extra occurrences of y' demanded by the chain rule.
Logarithmic differentiation
Given a function y=f(x) whose base is not e (e.g., not ex) and with an exponent that contains the
variable x. Two examples are 2x and (sin x)x. Derivatives of this type of function are approached by
first taking the logarithm of both sides of y=f(x) and then finding f ' by implicit differentiation. The
approach is called logarithmic differentiation.
Example: Consider y=(sin x)x. Take the log of other sides to obtain
ln y = x ln (sin x)
This describes y implicitly but does so without any exponents. Differentiate implicitly and use the
product rule on x ln(sin x) to get
1
d (ln sin x )
d ( x)
1 d (sin x )
cos x
y' = x
+ ln sin x ⋅
=x
+ ln sin x = x
+ ln sin x = x cot x + ln sin x
y
dx
dx
sin x dx
sin x
Therefore,
d (sin x ) x
= y ' = y ( x cot x + ln sin x ) = (sin x ) x ( x cot x + ln sin x )
dx
Antidifferentiation
Above we concentrated on the differentiation process of finding f ', given f. Now let’s reverse that
and seek the function f, given the derivative f '. The process is called antidifferentiation and has
several types of application, including integration.
Commonly encountered antiderivatives of basic functions:
∫ kdx = kx + C where k stands for a constant
∫ sin x dx = − cos x + C
∫ cos x dx = + sin x + C
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∫ exp x dx = + exp x + C
r
∫ x dx =
x r +1
+ C,
r +1
1
∫ x dx = ln x + C ,
∫e
x
or
∫e
x
dx = + e x + C
(4)
r ≠ −1
(5)
x≠0
(6)
dx = e x + C
Less commonly encountered antiderivatives of basic functions:
∫ sec x dx = tan x + C
∫ csc x dx = − cot x + C
∫ sec x tan x dx = sec x + C
∫ csc x cot x dx = − csc x + C
∫
2
(7)
2
(8)
1
(10)
dx = sin −1 x + C
1− x
1
−1
∫ 1 + x 2 dx = tan x + C
2
(9)
(11)
(common in gw hydrology) (12)
Antiderivatives of elementary functions
∫c
f ( x ) dx = c ∫ f ( x ) dx
(13)
and
∫ [ f ( x ) + g ( x )] dx = ∫ f ( x ) dx + ∫ g ( x ) dx
(14)
So, for example,
∫ [c1 f ( x ) + c2 g ( x )] dx = c1 ∫ f ( x ) dx + c2 ∫ g ( x ) dx
Extending known antiderivative formulas
In general, if F(x) is an antiderivative of f(x), then
1
(15)
∫ f (ax + b) dx = a F (ax + b) + C
In other words, if x is replaced by ax+b in (1) – (12), anti-differentiate “as usual” but insert the extra
factor 1/a.
Example: Consider ∫cos(πx + 7) dx, which is of the form in (15), where a=π, b=7, and f = cos( ).
Apply (15) to find
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1
∫ cos(π x + 7) dx = π sin(π x + 7) + C
Check by taking the derivative of the RHS (right hand side).
Introduction to parametric equations
When a process is described by two dependent variables, say x and y, and their equations x(t) and
y(t) in terms of a third variable, in this case t, the equations are parametric equations and the third
variable is called the parameter.
The most common appearance on parametric equations involves kinematics, such as keeping
tracking of a fluid parcel or tracer packet as it moves through a multidimensional hydrologic
system. You’ve encountered this problem before, in freshman physics …
Example: Consider the ballistic model of a bullet fired from a gun. The acceleration, velocity, and
position of the bullet are, respectively, described by the following expressions (in English units of
feet and seconds, otherwise not shown). They assume that the gun muzzle is aimed at an angle of
30° with the horizontal and that the muzzle velocity is 60fps. From the initial muzzle velocity and
muzzle orientation, and given Earth’s gravity, the acceleration at all times and the initial velocity in
both the horizontal and vertical directions is known. The velocity at later times (t) and the position
(x,y) is determined by sequential antidifferentiation.
Vertical movement:
y''(t) = -32 for all t
y'(t) = -32 t + 30
y(t) = -16 t2 + 30 t + 40
(17)
Horizontal movement:
x'(t) = 30 √3 for all t > 0
x(t) = 30 √3 t
(18)
Equations (17) and (18) are parametric equations and t is the parameter. If (18) is solved for t and
substituted into (17) we have a non-parametric equation for y(x), height as a function of horizontal
distance, the path of a parabola.
2
x
4x2
 x 
 x 
y ( x ) = −16
+
+ 40
 + 30
 + 40 = −
675
3
 30 3 
 30 3 
-28-
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Hyd 510
Quantitative Methods in Hydrology
The Derivative II
Relative maxima and minima
It is useful for a wide variety of
reasons to be able to locate the
“peaks” and “valleys” of a function.
Definition of relative extrema
x1
x2
x3
x4
x5 x6
x7
A function has a relative maxima at
x0 if f(x0) ≥ f(x) for all x near x0.
A function has a relative minina at x0
if f(x0) ≤ f(x) for all x near x0.
Where are the relative minima and maxima in the graph?
If f is differentiable and f has a relative extreme value at x0 then f '(x0) = 0. Equivalently, if f '(x0) is a
nonzero number then f cannot have a relative extreme value at x0.
On the other hand, if f ' (x0) =0, then a relative extreme value may (see x2, x3, x4) but need not (see
x1) occur.
Critical numbers
If f '(x0) = 0 or f '(x0) does not exist then x0 is called a critical number.
Includes all relative minima, relative maxima, and possible nonextrema (see x1, x6, x7 ) as well.
First derivative tests
Let f be continuous. To identify a critical number x0 as a relative minima or maxima, examine the
sign of the first derivative to the left and right of x0.
Example: f(x) = 4x5 - 5x4 - 40x3
Set f '(x0) =0 and solve for critical numbers (the roots).
f '(x0) = 20x4 - 20x3 - 120x2 = 20x2 (x2 - x – 6) = 20x2 (x - 3) (x + 2)
x = 0, 3, -2 are the critical points
Now examine the sign of f '(x) between the critical points.
-29-
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Second derivative tests
Applicable to critical points x0 at which f '(x0) =0.
(1) If f '(x0) = 0 and f '' (x0) < 0
then f has a relative maximum at x0.
(2) If f '(x0) = 0 and f '' (x0) > 0
then f has a relative minimum at x0.
(3) If f '(x0) = 0 and f '' (x0) = 0
then no conclusion can be drawn.
Continuing previous example where critical points are x = 0, 3, -2 and f(x) = 4x5 - 5x4 - 40x3
f '' (x0) = 80x3 - 60x2 - 240x
f '' (-2) = -400 ; f '' (0) = 0 ; f '' (3) = 900
Suggesting that x = -2 is a local maxima, x = 3 is a local minimum, and we can draw no conclusion
about the other point.
Application: How do apply these maxima and minima concepts in hydrology? Here are three
types of application.
i. First, consider physical extrema. Suppose the function f represents topography and x
represents map coordinate. A local maxima defines a ridge top and drainage divide,
while a local minima defines the drainage, which is often the location of a stream. In
groundwater hydrology, the local maxima of a water-table aquifer is the water table
divide, separating subsurface flow systems, which the local maxima of solute
concentration in a contaminant plume identifies the spine of that plume.
ii. For our second type of application imagine designing an engineered facility, like a well
field, contaminant remediation scheme, or dam on a stream. These problems are often
set up to (globally) maximize or minimize some objective function such as cost,
benefits, or frequency of failure, where x is called a decision variable and represents the
design choices being made, such as size of the dam or location and pumping rate for a
well. These are called design optimization problems.
iii. The third application type refers to using data to build and parameterize (assign
numbers to parameters) a hydrologic model. The parameters, represented by x, are
varied until some measure of model performance, represented by f, approaches an
extreme value. The performance measure is usually a sum of squared differences
between observed and modeled states, like head or flow rate, and we seek an absolute
or global minima. This application is called an inverse problem since you are “solving”
for parameters given states, rather than the other way around.
-30-
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Quantitative Methods in Hydrology
Absolute maxima and minima
As mentioned in the box, we often we want to find the absolute or global maxima or minima, not
the local values.
Finding maxima and minima
Restrict x to the interval of interest, then the absolute maxima or minima will lie at …
(a) A critical value of f
- Solve for f '(x)= 0 and also locate places where derivative does not exist.
- Let f (x0) represent the resulting list of critical values.
- This candidate list contains the relative maxima and minima, as well as other values.
Expand the candidate list to also include:
(b) The end values of f (ends of the interval)8
(c) Infinite values of f.
MAX
9
1
-3
-2
-1
0
MIN
1
2
3
4
5
Example: Find maximum value of f(x) = x4 + 4x4 - 6x2 – 8 for 0 ≤ x ≤ 1
-
Note that f is continuous in the interval [0,1]
-
Find critical values of f in [0,1] ; get x = 0,
-
Given the constrained domain [0,1] for x, keep only the two positive roots,
− 3 + 21
f(0) = -8 and f(
) ≅ f(0.79) ≅ -9.4. The largest of these is f(0).
2
8
− 3 ± 21
2
When solving an optimization or inverse problem (see box) we often define constraints which restrict the domain of
the decision variable x (i.e., the dam can be only so big or the parameter value can’t exceed such-and-such a value).
These constraints can be somewhat arbitrary. When the solution lies on one of these constraints then we may have to
ask if the domain size has been overly restricted.
-31-
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-
Find end points; for x = 0, f(0) = -8 is both a critical and an end point; for the end
point x = 1, f(1) = -9.
-
Evaluate f at critical values and end points; f(0) = -8, f(0.79) ≅ -9.4, f(1) = -9..
Pick x for largest f, that f is the maximum value and the corresponding x is its
location: f(0) = -8, the largest value of f in the domain [0,1] is -8 and its located at
x=0.
Warning: sometimes we seek f and sometimes x at the maximum (or minimum)
L’Hopital’s Rule and orders of magnitude
A way to evaluate indeterminant limit forms.
L’Hopital’s Rule
f ( x)
0
Suppose lim
is one of the inderterminant forms
x →a g ( x )
0
that is, involving indeterminate quotients. Then …
Switch to lim
x →a
∞
∞
−∞
∞
∞
−∞
−∞
,
−∞
f ' ( x)
g' ( x)
If the new limit is L , ∞, or -∞, then the original limit is L, ∞, or -∞, respectively.
If the new limit does not exist because f '(x)/g'(x) oscillates badly then we have no information about
the original quotient. L’Hopital’s rule doesn’t help in this situation.
If the new limit is still an indeterminate quotient, L’Hopital’s rule may be used again.
The rule is also valid for limit problems in which x→a+, x→a-, x→ ∞, x→ -∞
Warning:
L’Hopital’s rule applies only to indeterminant quotients. Is should not be used (nor is it necessary)
for limits in the form of 2/∞ (the answer is immediately zero) or 3/0- (the answer is immediately -∞)
or 6/2 (the answer is immediately 3) and so on.
Examples:
3x 3 + 6 x 2 − 5
which is of indeterminate form ∞/∞ [Ans: 3/2]
x →∞ 2 x 3 + 5 x 2 − 3 x
Find lim
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sin x
which is of indeterminate form 0/0 [Ans: 1]
x →0
x
Find lim
Order of magnitude
f ( x)
is of form ∞/∞.
g( x)
If the limit is ∞ then f(x) is said to be a higher order of magnitude than g(x), that is, f grows faster
than g.
Suppose f(x) and g(x) both approach ∞ as x→∞ so that lim
x →∞
If the limit is 0 then f(x) is said to be a lower order of magnitude than g(x),
If the limit is a positive number L then f(x) and g(x) have the same order of magnitude.
“Pecking order” of functions that approach ∞ as x→∞, listing them in increasing order of
magnitude, from slower to faster:
ln x, (ln x ) 2 , (ln x ) 3 , ..., x , x, x 3 / 2 , x 2 , x 3 , ..., e x
(4)
Order of magnitude of a constant multiple:
In general, f(x) and cf(x) have the same order of magnitude for any positive constant c.
Highest order of magnitude rule. In general, as x→∞ (only), a quotient involving functions on the
list in (4) has the same limit as
term with the highest order of magnitude in the numerator
term with the highest order of magnitude in the denominator
For example,
3 − ex
ex
−∞
− ex
lim 3
=
= lim 3 = − lim 3 = −∞
x →∞ x + 2 x
x →∞ x
x →∞ x
∞
Since ex has a higher order of magnitude than x3.
Indeterminant products, differences, and exponential forms
For the forms 0 × ∞, or 0 × -∞ use algebra or a substitution to transform to a quotient form and
apply L’Hopital’s rule.
Warning: Don’t use L’Hopital’s rule indiscriminately. Also, verify your result whenever possible.
For the forms ∞ - ∞, or (-∞) - (-∞), use other methods.
For the forms (0+)0, ∞0, 1∞, use logarithms to change exponential problems into products.
-33-
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Newton’s Method
Newton’s method uses calculus to try to solve equations of the form f(x) = 0.
Solving f(x) = 0 is equivalent to finding where the graph of the function f(x) crosses the x-axis.
Procedure:
- Guess a root of f, calling the first guess x1.
- Draw a tangent line to the graph of f at the point (x1, f(x1)).
- Let x2 be the x-coordinate of the point where the tangent line crosses the x-axis.
- Now start again with x2.
- Draw a tangent line to the graph of f at the point (x2, f(x2)).
- Let x3 be the x-coordinate of the point where the new tangent line crosses the x-axis.
- Now start again with x3, and so on.
Convergent
Divergent
Root
x2
x3
x2
x1
Root
(x1, f(x1))
x1
x3
(x2, f(x2))
The numbers x1, x2, x3, … will approach the root if the method is convergent, or if not, x1, x2, x3, …
will diverge. See figures. More often than not the method is convergent (it converges “exactly” if f
is quadratic or linear in x). If it is not convergent, try another initial guess.
The equation (point-slope formula) for a tangent line is y − f ( x1 ) = f ' ( x1 )( x − x1 ) .
f ( x1 )
Set y=0 and solve for x to find that when the line crosses the x-axis, x = x1 −
.
f ' ( x1 )
This value of x is taken to be x2.
Generalize to the procedure:
new x = last x −
f ( xn )
f (last x )
, or for iteration counter n, xn +1 = x n −
f ' ( xn )
f ' (last x )
Exercise for later: We’ll apply this method using Matlab, using standard Matlab Newton routines
and also writing our own programs.
-34-
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Differentials
Recall the definition of a differential
dy = f '(x) dx = change in y when x changes by dx.
(1’)
Example: d(sin x) = cos x dx, that is, the differential of sin x is cos x dx. If x changes by dx then sin x
changes by approximately cos x dx.
Example:
The volume of a sphere is V= (4/3)πr3, where r is its radius. What about a spherical shell?
volume of a spherical shell = vol. outer sphere – vol. inner sphere
4
4
1
= π ( r + dr ) 3 − πr 3 = 4π [ r 2 dr + r ( dr ) 2 + ( dr ) 3 ]
(6)
3
3
3
where the inner sphere is of radius r, and the outer sphere has a larger radius, r + dr.
To find an approximate formula use the differential,
dV = 4π r 2 dr
(7)
Note that this equals the first term in (6) and is the desired approximation. It is often easier
to find such an approximation, using (1’) than it is to solve the problem exactly.
The difference between (7) and (6) is the error of the approximation, here equal to
1
ε = 4π [ r ( dr ) 2 + ( dr ) 3 ]
3
which is small if dr is small and approaches zero as dr → 0.
Application: Most fundamental models of hydrologic processes involve conservation of mass,
momentum, and/or energy. Sometimes we use an extensive model dealing with amount of these
quantities over a spatially lumped model of, for example, a hydrologic body like a plot, hillslope,
watershed, river reach, lake, aquifer, region, continent, or even the entire planet. More often we
use a spatially distributed, intensive model, where states like hydraulic head, fluid velocity,
thermal energy, and solute concentration vary over space, and we examine intensive quantities
(usually quantity per unit area or unit volume). To develop these distributed models we use
differentials. We use them to describe how fluxes of mass, momentum or energy change in space,
where in equation (1’) y is the flux density and x is location, or to describe how intensive
quantities vary over time, where y is the amount of a quantity per unit volume (or area) and x is
time.
-35-
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Separable differential equations
It is often possible to separate the variables in the (DE) differential equation dy = f '(x,y) dx, so that
the equation has the differential form
(expression in x) dx = (expression in y) dy
Then the equation is called separable, and is solved by antidifferentiating both sides. (This works on
first order equations only !!!!) We call this approach separation of variables (SOV).
Example:
x
. Rewrite as y 2 ( x ) y ' ( x ) = x .
y ( x)
1
1
Integrate (antidifferentiate) both sides to yield y 3 = x 2 + C .
3
2
dy
x
More conventionally and conveniently,
= 2 , leading to separation as y 2 dy = xdx .
dx y
1
1
Then antidifferentiate, ∫ y 2 dy = ∫ xdx to get y 3 = x 2 + C .
3
2
Consider the DE y ' ( x ) =
2
These are implicit solutions for y. You can solve algebraically to get explicit solutions.
Example: exponential decay or growth …
This type of problem occurs widely in hydrology. For example, decay can be due to
radioactive decay or 1st order biotransformation or hydrolysis of an organic compound, both
of which involve decay of concentration. Another application is discharge from an aquifer to
base flow in a stream, or discharge from a lake, where water levels “decay” over time
(approximately) exponentially. In all these cases the rate of change of something (the
dependent variable) depends on the quantity of that something remaining. As the quantity
decreases over time the rate of change slows down. This type of decay is called exponential
decay, as is made apparent below. If, on the other hand, something is added and the quantity
is increasing over time, according to this model, it is called exponential growth.
In the exponential decay model the rate of change of the dependent variable, y, with respect
to time, t, is described by9
dy
= − ky
dt
where k is a rate coefficient10. The model also needs an initial value (IV) of y. Call the IV
y(t=0) = y0. Solving by SOV we get
9
The minus sign on the RHS is significant; it leads to decay instead of growth.
For example, in radioactive decay the coefficient k is a linear function of radioactive “half-life”.
10
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1
dy = − k dt
y
ln y = −k t + C
Using the IV to determine constant C,
ln y0 = 0 + C
ln y = − k t + ln y 0
ln y − ln y 0 = ln
y
= −k t
y0
y = y 0e−k t
The dependent variable decays exponentially in time. The dependent variable, y, starts with
the initial value y0 and decays exponentially to zero.
Exercise: Plot ratio y/y0 using Matlab for three different values of k = 0.5, 1, 2 for the
domain 0 ≤ t ≤ 5.
Suppose instead of decaying, the process is one of exponential growth of say, population, y
(people and their demand for water, bacteria, etc). If λ represents the growth rate constant,
then the model is
dy
= + λy
dt
You should be able to show that the population is described by
y = y 0 e + λ ( t −t0 )
where the initial value, at time t0, is y(t= t0) = y0.
-37-
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Quantitative Methods in Hydrology
The Integral I
Preview
Integrals appear as models themselves or as part of solutions to models. Often we define new
functions (e.g., Erf and Ei) to replace integrals which frequently appear in these solutions. Integrals
as models or their solution appear in the calculation of time, length, area, and volume, and in
extensive properties like the amount of mass, energy and momentum, or their fluxes. Integrals
appear in calculating the amount of something (an extensive quantity) or, by dividing by the interval
over which it is calculated, the average amount of something (an intensive quantity, e.g., a flux
density) over that interval. Integrals appear in moment equations, and are used to calculate physical
moments (like the moment of inertia) or probabilistic moments (like the variance, or square of the
standard deviation, which is a second central moment).
Definition and some aspects of the integral
The integral is defined by
∫
b
a
f ( x )dx = lim ∑ f ( x )dx
(1)
dx →0
For a simplistic, but useful viewpoint, we can ignore the limit and consider
∫
b
a
f ( x )dx as merely Σ
f(x)dx, found using many subintervals of [a,b]. In other words, think of the integral as adding many
representative values of f, each value weighted by the length of the subinterval it represents
(i.e., the dx’s can vary, as sketched below).
∫
b
a
∑ f ( x )dx
f ( x )dx
f(x)
a
b
a
dx
b
The process of computing the integral is referred to as integration. The integral symbol is an
elongated S for “sum” (the same symbol is used in a different context for antidifferentiation). The
symbols a and b attached to the it indicate the interval of integration. The numbers a and b are
called the limits of integration, and f is called the integrand. The sums of the form Σ f(x)dx are
called Reimann sums.
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Consider the application of integrals and average values:
∫
b
a
Average value of f in [a,b] =
f ( x )dx
b−a
(3)
Think of the numerator as the weighted (dx) values and the denominator as the sum of the weights,
b-a = Σdx’s.
Some properties of the integral
∫
∫
∫
b
a
b
a
b
a
[ f ( x ) + g ( x )]dx =
∫
k f ( x )dx = k
b
a
∫
b
a
f ( x )dx +
∫
b
a
g ( x )dx
f ( x )dx , where k is constant
c
c
b
a
f ( x )dx + ∫ f ( x )dx = ∫ f ( x )dx , if a<b<c
(9)
(10)
(11)
Reminder about dummy variables
∫
∫
2
0
2
0
x 3 dx is a number, without the variable x appearing anywhere in the answer. Could also write it as
2
t 3 dt or ∫ a 3da . The letter x (or t or a) is called a dummy variable because it is entirely arbitrary.
0
In general,
∫
b
b
b
a
a
f ( x )dx = ∫ f (t )dt = ∫ f (u )du , and so on.
a
The Fundamental Theorem of Calculus
The Fundamental Theorem
If f is continuous on [a,b] and F is an antiderivative of f, then
∫
b
a
Example 1:
∫
3
0
f ( x )dx = F (b) − F ( a )
(1)
x dx = ?
∫
3
0
3
1
9
9
x dx = x 2 = − 0 =
2 0 2
2
Example 2: Suppose x2/2 + 7 is the antiderivative of x¸ rather than x2. (Why not? 7 is an arbitrary
constant.)
3
1 2
9
9
∫0 x dx = ( 2 x + 7) 0 = ( 2 + 7) − (0 + 7) = 2
The 7 cancels out.
3
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∫
Example 3:
2
1
Hyd 510
Quantitative Methods in Hydrology
1
dx = ?
x2
∫
2
1
2
1
1
1
1
= − − ( −1) =
dx = −
2
x
x1
2
2
Integral of a constant function:
∫
b
k dx = k (b − a )
a
(2)
Integral of a zero function:
∫
b
a
0 dx = 0
(3)
Integral of a piecewise function11 with several formulas.
x2
if x ≤ 3

Suppose f ( x ) = 2 x + 3 if 3 < x < 7
17 − x if x ≥ 7

To find
10
∫
∫
0
10
0
f ( x )dx use (11) of previous section.
3
7
10
3
7
f ( x )dx = ∫ x 2 dx + ∫ ( 2 x + 3)dx + ∫ (17 − x )dx
0
3 3
10
7
x
x2
+ ( x 2 + 3x ) + (17 x − ) = 9+52+25.5 = 86.5
=
3
3 0
2 7
Definite v. indefinite integrals
The symbol ∫ is used in two different and distinct ways.
•
First
∫
b
a
f ( x )dx is an integral, defined as the limit of the Reimann sum Σ f(x)dx. In this
context dx stands for the length of a typical subinterval of [a,b]. The symbol ∫ is used to
signify summation.
•
Second,
∫ f ( x )dx is the collection of all antiderivatives of f(x). In this context, the symbol
dx is an instruction to antidifferentiate with respect to the variable x. The symbol ∫ is used
because one of the methods of computing an integral (using the Fundamental Theorem of
calculus) begins with antidifferentiation.
11
One important application of integration of piecewise functions is in finite element numerical models.
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b
∫
called an definite integral, while ∫ f ( x )dx
Frequently, both
∫
f ( x )dx and
a
f ( x )dx are referred to as integrals; in particular,
∫
b
a
f ( x )dx is
is called an indefinite integral (rather than integral and
antiderivative). No matter which terminology you choose, it will always be true, for example, that
∫ 3x dx = x
2
3
3
+ C , where C is an arbitrary constant, while
∫ 3x dx = 19
2
2
Numerical integration
The evaluation of
∫
b
a
f ( x )dx = F (b) − F ( a ) seems like it should be simple, but it is often difficult
and sometimes impossible to find an antiderivative F. We then resort to numerical integration to
approximate the (definite) integral. A variety of numerical integration techniques exist, each
involving lots of arithmetic, almost always done on a computer.
All are approximate and have errors. For some methods it is possible to calculate a theoretical
estimation of the errors. For others you simply increase (e.g., double) your resolution, try again, and
see how much improvement you get.
In calculus you learned about three of these. The first is simply Riemann Summation for finite size
dx. The second is the trapezoidal rule. The third is Simpson’s rule. Later we’ll introduce Guassian
Quadrature and possibly other methods.
Riemann sums: approximate the integral by a series of steps (rectangles). In essence each dx is
assumed to have a constant value of f, a constant function.
Trapezoidal rule: approximate the integral by a series of trapezoids. In essence you are
approximating f over each increment dx by a straight sloped line, that is, a linear function, defined
by the value of f at each end of the increment.
a
b
Rieman Sum
a
b
Trapezoidal Rule
a
b
Simpson’s Rule
Simpson’s rule: approximate the integral over two neighboring increments, fitting a parabola, that
is a quadratic function, to the three values (circles in the sketch) of f defining each of the two
increments. For the two increments fit a quadratic to the left, middle and right values of the
function. Then move on to the next pair of intervals and repeat. If the dx’s are constant (equally
spaced), then it can be shown that
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h
( y 0 + 4 y1 + 2 y 2 + 4 y 3 + 2 y 4 + ... + 2 y n −2 + 4 y n −1 + y n )
a
3
where h = dx, the size of the x increment. There is no easy error estimator for Simpson’s rule.
∫
b
f ( x )dx ≈
Nonintegrable functions
Suppose f(x) = 1/√x, and you want to integrate it from 0 to 1, by computing the Riemann sum. What
value of f do you apply for the first increment, which has its left edge on zero? What happens when
you change dx?
Discontinuous functions can be nonintegrable. However, some discontinous functions can be
integrated using the following:
Improper integrals
For intervals of the form [a, ∞), (-∞,b], (-∞,∞)
∫
Example:
∞
1
1
dx = ? →
x
∫
∞
1
b 1
1
dx = lim ∫ dx
b→ ∞ 1 x
x
i.e., to integrate on [1,∞), integrate from x=1 to x= b and then let b approach ∞.
∫
b 1
1
b
dx = lim ∫ dx = lim(ln x 1 ) = lim(ln b − ln 1) = ∞ − 0 = ∞
1
b→ ∞
b→ ∞
b→ ∞
x
x
∞
1
In general,
∞
∫ f ( x)dx = lim ∫
b
b→ ∞ a
a
f ( x )dx and
∫
b
−∞
f ( x )dx = lim
∫
b
a → −∞ a
f ( x )dx
In abbreviated notation, if F is an antiderivative of f then
∫
∞
a
∞
f ( x )dx = F ( x ) a
and
∫
b
−∞
b
f ( x )dx = F ( x ) −∞
Convergence vs. divergence.
Evaluating an improper integral always involves computing an ordinary integral and taking a limit.
If the limit is finite, the integral is said to be convergent. If the limit is plus or minus infinity, or no
value at all, the integral diverges.
Example:
∫
−2
−∞
1
dx = ?
x2
ANS: ½, convergent
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Integrating on the interval (-∞,∞)
The usual definition is
∫
∞
−∞
Example:
∫
f ( x )dx = lim
b
f ( x )dx or
a → −∞ a
b→ ∞
∫
∞
−∞
∫
∞
−∞
∞
f ( x )dx = F ( x ) −∞
∞
π
π
1
dx = tan −1 x = − ( − ) = π
2
−
∞
1+ x
2
2
Integrating functions which blow up at the end of the interval of integration.
If f blows up at x = a then
∫
b
If f blows up at x = b then
∫
b
a
a
f ( x )dx = F (b) − F ( a + )
f ( x )dx = F (b − ) − F ( a )
Integrating functions which blow up within the interval of integration.
Suppose that a< c < b, and that f blows up at c, then
∫
b
a
c−
b
a
c+
c−
f ( x )dx = ∫ f ( x )dx + ∫ f ( x )dx = F ( x ) a + F ( x ) c+
b
The Integral II
Integrals with variable upper limit
Suppose we define a new integral
x
I ( x ) = ∫ f (t )dt
(2)
a
Some functions of this form are
Erf x =
2
π
∫
x
0
2
e −t dt error function
Erfc x = 1 − Erf x
−t
x e
Ei x = ∫
dt
−∞ t
x sin t
Si x = ∫
dt
0
t
(3a)
complementary error function
(3b)
exponential integral function
(3c)
the sine- integral function
(3d)
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Computing I(x).
If f(t) has a readily available antiderivative, then an explicit formula for I(x) may be found by using
the Fundamental Theorem of calculus.
x
x
Example: I ( x ) = ∫ 3t 2 dt = t 3 = x 3 − 1
1
1
If f has a simple graph it may be possible to find I(x) by integrating f in sections.
The derivative of I(x)
Often sought. Example, if W(u)= - Ei(-u) gives well drawdown, its derivative gives groundwater
velocities needed to compute fluxes, flowpaths, and travel times.
The derivatives of I(x) wrt x are given by the integrand used in the original formulation of I(x). This
x
is not a coincidence. In general, if I(x) = ∫ f (t )dt then I’(x)= f(x) at all points where f is continuous.
a
In other words, if a continuous function f is integrated with a variable upper limit x, and then the
integral is differentiated with respect to x, the original function is obtained. This result is called the
Second Fundamental Theorem of Calculus.
Example: d [erfc(x)] /dx = d1/dx - d [erf(x)] /dx = 0 -
Example: d [Ei(x)] /dx = −
− e−x
d x e −t
=
dt
dx ∫− ∞ t
x
-44-
2 d x −t 2
− 2 − x2
e dt =
e
∫
π dx 0
π
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Antidifferentiation
Introduction
Standard tables, such as the CRC, contain only a limited number of antiderivatives12. Differentiation
is easier. There are many rules to help us (sums, products, quotients, compositions), but in
antidifferentiation there are fewer rules. There are no product, quotient or chain rules for
antidifferentiation. The best we have are sum and constant multiple rules.
∫ c f ( x ) dx = c ∫ f ( x ) dx
∫ [ f ( x ) + g ( x)] dx = ∫ f ( x ) dx + ∫ g ( x ) dx
(1)
(2)
In the absence of sufficient rules we consult tables of antiderivatives, but even the large volumes of
tables13 that you find in the library have their limits. You need to know how to “extend” them and
the simpler tables you find in a text book or the CRC math tables.
Below we review some basic methods to achieve this “extension”.
Substitution
Example: What is
∫ 2 x cos x dx ? We can find it from the derivative.
2
By the chain rule, D x sin x 2 = 2 x cos x 2 so that we know that ∫ 2 x cos x 2 dx = sin x 2 + C .
But how can we find the antideriviative formula without seeing the derivative first? (For
example, because we don’t have the derivative already and we can’t find it in the tables.)
Reverse the chain rule to simplify to a form in the tables of antiderivatives. The chain rule
for the derivative of a composition y=y(u) is
dy dy du
=
dx du dx
In this example, use the device u=x2, du=2x dx. Substitute this into the integral to get
2
2
∫ 2 x cos x dx = ∫ cos u du = sin u + C = sin x + C
More examples:
12
The antiderivative tables presented earlier in these notes are but a faint shadow of even the most basic tables, such as
you will find in the CRC tables, or a good calculus textbook. You’ll need to consult those tables and not rely only on
these notes.
13
There are two readily available exhaustive references for antiderivatives and also for an army of definite derivatives:
Gradshteyn, I.S. and I.M. Ryzhik, 1980, Table of Integrals, Series, and Products, Academic Press, New York, NY.;
Rectorys, K., 1969, Survey of Applicable Mathematics, MIT Press, Cambridge, MA.
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1
x3
dx ; let u2 =2x4+7 and x3dx =(1/8)du. Ans:
1. ∫
+C
4
2
( 2 x + 7)
8( 2 x 4 + 7)
1
2. ∫ cos 2 x sin xdx ; let u =cos x and du= -sin x. Ans: − cos 3 x + C
3
Comment: There is no set rule on when or what to substitute. One useful tactic is to search the
integrand for an expression whose derivative is a factor in the integrand, and let u be that
expression. In example 1, the expression 2x4+7 has the derivative x3 (give or take an 8), which is a
factor. In example 2, the expression is cos x; its derivative is sin x (give or take a negative sign) and
is a factor. More than one substitution may work.
Some Algebra including proper/improper fractions & partial fraction decomposition
Sometimes you can use algebra to reduce a function to another function listed in the tables.
Example 1:
this formula is in std. tables; a 2 = 1/2
∫
1
6x 2 + 3
dx =
1
6
∫
1
x 2 + 1/ 2
dx =
1
6
⋅
644744
8
1
∫ a 2 + u 2 du
=

1
1 
1
ln(u + a 2 + u 2 ) 2
ln  x +
+C =
+ x2  + C
a =1 / 2
2
6
6 

u= x
Improper fractions
x5
, are those where the degree of the numerator is greater or
Improper fractions, such as 2
x +x+2
3x
, are those where the
equal to the degree of the denominator. Proper fractions, such as 2
x +1
degree of the numerator is less than the degree of the denominator. The improper kind are rarely
listed in the antiderivative tables. To find an antiderivative for an improper fraction that is not listed,
begin with long division (divide out the improper fraction).
x 5 dx
∫ x2 + x + 2 .
the improper fraction integrand:
x3 − x2 − x + 3
x2 + x + 2 x5
with remainder
Example 2: Consider
Divide out14
− x−6
x +x+2
That is, the integrand is expanded as the sum of a polynomial and a proper fraction:
14
Refer to your algebra basics for details.
-46-
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x5
− x−6
= x3 − x2 − x + 3 + 2
2
(3)
x +x+2
x +x+2
proper fraction = polynominal + proper fraction
Let’s integrate the polynominal,
x4 x3 x2
3
2
(
x
−
x
−
x
+
3
)
dx
=
−
−
+ 3x + C
∫
4
3
2
(4)
Now for the proper fraction, break it up into digestible pieces,
− x−6
x
6
=− 2
− 2
2
x +x+2
x +x+2 x +x+2
Then integrate
1
1
x dx
dx
−∫ 2
= − ln x 2 + x + 2 + ∫ 2
2
2 x +x+2
x +x+2
1
1
2x + 1
tan −1
= − ln x 2 + x + 2 +
+C
2
7
7
and
− 12
2x + 1
dx
− 6∫ 2
=
tan −1
+C
x +x+2
7
7
(6)
(7)
Finally combine (4), (6), and (7),
∫
− 11 −1 2 x + 1 1
x 5 dx
x4 x3 x2
tan
=
−
−
+ 3x −
− ln x 2 + x + 2 +K
2
x +x+2 4
3
2
2
7
7
Partial Fractions
Tables often omit proper fractions as well, when the denominator is greater than 2. Partial fraction
decomposition is an algebraic technique that helps in this and other several other applications we
will visit later regarding ordinary and partial differential equations. In each instance it is easier to
work separately with the partial fractions than with their sum.
We want to decompose a proper fraction which is not in the tables into a sum of partial fractions
which are either in the tables or which may be antidifferentiated by substitution or inspection. The
decomposition is accomplished in several steps.
Step 1. Factor the denominator as far as possible.
Step 2. The nature of the decomposition depends on the factors in the denominator.
-
If a factor is linear, such as 2x+3, then a fraction of the form A/(2x+3) appears as one of
the partial fractions.
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-
If a non-factorable quadratic, such as x2 + x + 10 appears in the denominator then
Ax + B
appears in the decomposition.
2
x + x + 10
-
If a repeated non-factorable quadratic, such as (x2 + x + 10)4 appears in the denominator
then
Ax + B
Cx+D
Ex+F
Gx+H
+ 2
+ 2
+ 2
2
2
3
( x + x + 10)
( x + x + 10) 4
x + x + 10 ( x + x + 10)
appears in the decomposition.
Step 3. Determine A, B, C, … in the decomposition (various methods; see example for one
method)
Example:
2 x 2 + 3x − 1
and then antidifferentiate. The denominator has already
( x + 3)( x + 2)( x − 1)
been factored, accomplishing step 1. Regarding step 2, the three factors in the denominator
are all linear, so each has a partial fraction consisting of a constant divided by the respective
linear function. Thus
2 x 2 + 3x − 1
A
B
C
+
+
=
( x + 3)( x + 2)( x − 1) x + 3 x + 2 x − 1
Decompose
Multiply this equation by the denominator, ( x + 3)( x + 2)( x − 1) , to get
2 x 2 + 3x − 1 = A( x + 2)( x − 1) + B( x + 3)( x − 1) + C ( x + 3)( x + 2)
(3)
This equation is suppose to hold for all x, so we can solve it by taking three arbitrary but
convenient numerical value for x. Values -3, -2, 1 are convenient because each of them set
two of the three terms in (3) to zero.
If x = -3,
If x = -2,
If x = 1,
then 8 = 4A, A = 2
then 1 = -3B, B = -1/3
then 4 = 12C, C= 1/3
We could use any three values and solve the resulting three equations simultaneously, but it
wouldn’t be as convenient. The result is
2 x 2 + 3x − 1
2
1/ 3
1/ 3
=
−
+
( x + 3)( x + 2)( x − 1) x + 3 x + 2 x − 1
We can down antidifferentiate this term by term using tabulated antiderivatives.
2 x 2 + 3x − 1
1
1
∫ ( x + 3)( x + 2)( x − 1) dx = 2 ln x + 3 − 3 ln x + 2 + 3 ln x − 1 + K
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Integration by Parts
The idea behind integration by parts is to reverse the derivative product rule. Since Dxuv=uv’ + vu’
we have the integration formula ∫(uv’ + vu’)dx=uv. But problems don’t usually originate in this
form, so we continue to a more useful version of the integration formula. Write it is ∫uv’dx=
uv - ∫ vu’dx, and to make it easier to apply, use the differential notation dv=v’dx, du=u’dx to get
∫ u dv = uv − ∫ v du
(1)
With this formula you trade one problem for another, which may or may not help depending on how
good a trader you are. To apply (1) a factor in the integrand must be called u. The rest of the
integrand, including the “factor” dx is labled dv. Success of the method, called integration by parts,
depends on being able to find v from dv (this is itself antidifferentiation) and on being able to find
∫ v du.
Example:
∫e
x
cos xdx
Let u= ex and dv = cos x dx. Then du= exdx and v= sin x, and
∫e
x
cos xdx = e x sin x − ∫ e x sin xdx
That’s doesn’t appear to be helpful, but let’s stick with it and apply integration by parts
again, to the second term on the RHS. Let u= ex and dv = sin x dx. Then du= exdx and
v= -cos x, to get
x
x
x
x
∫ e cos xdx = e sin x − ( −e cos x + ∫ e cos xdx )
The integral on the RHS is our original integral, which seems circular. But note that it now
appears on both sides of this equation. Collect terms involving this integral to get
1 x
x
x
∫ e cos xdx = 2 (e cos x + e sin x) +C
Recursion formulas
Some antiderivative formulas, said to be recursive, can be applied repeatedly without a problem in
order to get an answer (especially for forms sinm x and cosm x).
Example:
∫x
3
sin x dx
Many tables list the pertinent formula
n
n
n −1
n −2
∫ x sin x dx = − x cos x + nx sin x − n(n − 1)∫ x sin x dx
Applying that here with n=3
3
3
2
∫ x sin x dx = − x cos x + 3x sin x − 3(2) ∫ x sin x dx
This last integral is usually available in tables and we can finish the job.
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Trigonometric substitution
A collection of integrals in the tables can be found using
substitution of a special type called a trigonometric
substitution. These are based on angle substitutions, e.g.,
tan u =
x
, cos u =
a
a
cot u = , sec u =
x
a
a +x
2
2
, sin u =
a2 + x2
, csc u =
a
x
a +x
2
a
a2 + x2
u
2
x
a
a2 + x2
x
Example:
∫x
dx
a +x
2
2
=∫
1
a
a tan u ⋅
cos u
a sec 2 u du = −
1
csc u du
a∫
1
1
a2 + x2 a
− ln csc u + cot u + C = − ln
+ +C
a
a
x
x
from standard math tables
=
{
where x=a tan u yields dx = a sec2u du, and
a2 + x2 =
a
.
cos u
Choosing a method
If a function is not listed in the tables:
(a) Complete the square if the problem involves ax 2 + bx + c but the only similar formula in the
tables does not contain the term x
(b) Substitute if there is an expression in the integrand whose derivative is also a factor in the
integrand.
(c) Use long division on improper fractions
(d) Decompose proper fractions if they aren’t in the tables
(e) Use integration by parts to get recursion formulas. Integration by parts may also work when
other methods don’t seem to apply.
(f) If a problem involving a 2 ± x 2 or x 2 − a 2 is not in the tables, try trigonometric substitution.
Series
Series representations are useful for a variety of reasons. For example, we may have a complex
function that is difficult to work with, but if we represent it by a series we can work with each of the
simpler terms in that series (integration, differentiation). We may also find that only the first few
terms of the series are all that is needed to approximate the function, making it ever easier. For
example, drawdown in a confined aquifer can be represented by an Exponential Integral (Theis
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Well Function), but if enough time has elapsed (typical in application) the higher order terms of the
series representation of the Exponential Integral are unimportant. The lower order terms involve a
logarithm. Drawdown can then be approximated by a logarithm (the “Jacob Approximation” of well
hydraulics).
Note: The following review is superficial, but it highlights issues of concern. Consult your calculus
text for details.
Introduction
The symbol a1 + a2 + a3 + a4 + … is called a series with terms a1, a2, a3 , a4, …
∞
Also written as
∑a
n =1
n
or even Σ an.
Partial sums of this series are
S1 = a1
S2 = a1 + a2
S3 = a1 + a2 + a3
etc
If partial sums approach a number S, that is, if lim S n = S , then S is the sum of the series, and we
n →∞
write Σ an = S. In this case the series is called convergent; in particular, it converges to S. Else the
series is divergent.
∞
1 1 1 1
+ + + + K has
2 4 8 16
n =1
1
1 1 3
1 1 1 7
1 1 1 1 15
S1 = , S 2 = + = , S 2 = + + = , S 4 = + + +
=
, etc.
2
2 4 4
2 4 8 8
2 4 8 16 16
Example:
1
∑(2)
n
=
Or, lim S n = 1 , and the series has sum 1, that is, converges to 1. We write
n →∞
Two convergent series can be added, term by term.
1 1 1 1
+ + + + ... = 1
2 4 8 16
1 1
1
1
1
− +
−
+ ... =
4 16 64 256
5
→
3 3
9
15
6
+ +
+
+ ... =
4 16 64 256
5
Dropping initial terms
-51-
∞
1
∑(2)
n =1
n
= 1.
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If the first m terms of a series are dropped, then the new series and the original series will both
converge or both diverge. In other words, chopping off the beginning of a series doesn’t change
convergence or divergence. It does change the sum of a convergence series.
Geometric series
∞
∑ ar
Form:
n
= a + ar + ar 2 + ar 3 + ar 4 + ..., a ≠ 0
n= 0
is called a geometric series with ratio r.
Geometric series test.
Has simple criterion for convergence and, if the series converges, the sum is easily found.
If r ≥ 1 or r ≤ -1 then
∞
∑ ar
n
diverges
n= 0
If -1 < r < 1 then
∞
∑ ar
n= 0
∞
converges to
a
1− r
5
2
n= 0
(Application? For example, geometric series are encountered in probability theory.)
Example:
1
n
∑ 3(− 5 )
n
=
Convergence Tests
Positive series
All positive terms; Four rules to test convergence. Not reviewed here.
Standard series
Increasing order of magnitude:
ln n, (ln n ) 2 , (ln n ) 3 , ..., n , n, n 3 / 2 , n 2 , n 3 , ...,2 n ,100 n ,..., n!
Reciprocals:
1/ ln n, (ln n ) −2 , (ln n ) −3 , ..., n −1 / 2 , n −1 , n −3 / 2 , n −2 , n −3 , ...,2 − n ,100 − n ,...,1 / n!
The entries approach 0 as n →∞. The order of magnitude decrease reading from left to right.
That is, they approach zero more rapidly.
Subseries of a positive convergent series
If Σan is a positive convergent series then every subseries also converges.
Limit comparison test
Suppose that an and bn , both positive, have the same order of magnitude. Then Σan and Σbn act alike
in the sense that either both converge or both diverge
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Alternating series
Let an be positive. A series of the form
∞
∑a
n
( −1) n +1 = a1 - a2 + a3 - a4 + … is called a alternating
n=1
series.
The partial sums of a positive series are increasing, so a positive series either converges or else
diverges to ∞. But the partial sums of an alternating series rise and fall since terms are alternately
added and substracted; therefore an alternating series either converges, diverges to ∞, diverges to
-∞, or diverges but not to ∞ or -∞.
Alternating Series Tests.
nth term test for divergence:
• If an doesn’t approach 0 then the series diverges. (The partial sums oscillate but are not
damped, and hence do not approach a limit.)
• If the series converges then an →0.
• If an does approach zero, then the alternating series may converge or may diverge. More
testing is necessary.
• If the series diverges then an may or may not approach 0.
Alternating series test:
Consider the alternating series
∞
∑a
n
( −1) n +1 . Suppose an →0. Then the series converges to a sum S
n=1
between 0 and a1. Furthermore, if the last term of a subtotal involves addition, then the subtotal is
greater that S; if the last term of a subtotal involves subtraction then the subtotal is less than S. In
either case if only the first n terms are used, then the error, the difference between the subtotal Sn
and the series sum S is less than the first term not considered. In other words, |S - Sn| < an+1 .
Absolute convergence.
Another way to test an alternating series is to remove the alternating signs and test the positive
series. If it converges then the alternating series also converges.
Conditional convergence.
If Σ|an | diverges it is still possible for Σan to converge. In this case Σan is called conditionally
convergent. …
Power series
a0 + a1 x + a 2 x 2 + + a3 x 3 + a 4 x 4 + ...
Sometimes used to create a new function when elementary functions are inadequate, for example, to
solve an ode. That is, assume a power series form for the dependent variable. Solve the resulting
algebric equations to satisfy the ode and define the coefficients ai . The resulting series solution
essentially defines a new function. Many advanced functions originated this way. Caution. The new
function may converge for only a limited interval; you must tests for convergence.
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Power series representations of elementary functions15
It is also useful to have power series expansions for old functions. Polynomials are easy to work
with and representing an old function as an “infinite polynomial” is often convenient. Consider a
few examples of power series expansions of simple functions. The derivation (not given) of these
examples depends on finding a connection between the desired function, f(x), and a known
expansion.
Example: Consider first the following power series which is also a geometric series:
1
= 1 + x + x 2 + x 3 + x 4 + ... for -1 < x < 1
(1)
1− x
Binomial series
(1 + x ) q = 1 + qx +
q( q − 1) 2 q( q − 1)( q − 2) 3
x +
x ...
2!
3!
for -1 < x < 1 (4)
Another example, this one of an alternating series:
ln(1 + x ) = x −
x 2 x3 x 4 x5
+ − + − ...
2
3 4
5
for -1 < x < 1
(10)
Maclaurin Series
We seek a less arbitrary approach to finding a series representation of a function.
Given the power series
f ( x ) = a0 + a1 x + a2 x 2 + + a3 x 3 + a4 x 4 + ...
A Maclaurin series has coefficients
an =
f ( n ) (0)
n!
It can be shown that there are two possibilities. Either f has no power series of this form (i.e., with
these coefficients) or
f (0) f ′(0)
f ′′(0) 2 f ′′′(0) 3
+
x+
x +
x + ...
0!
1!
2!
3!
This expression can be used to find power series representations of lots of elementary functions.
Here are some typical results.
f ( x) =
15
See standard math tables for a listing of various power series representations and their interval of convergence.
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x3 x5 x7
sin( x ) = x − + − + ...
3! 5! 7!
for all x
(3)
x 2 x 4 x5
+ − + ... for all x
2! 4! 6!
x2 x3 x4
x
exp( x ) = e = 1 + x + + + + ... for all x
2! 3! 4!
cos( x ) = 1 −
(4)
(5)
Taylor series introduction
The Taylor Series approach is used more often by hydrologists than any other. It is fundamental to
developing conservation models for mass, momentum and energy, to various solution methods
including the finite difference method, and to making extrapolations and predictions. Its
fundamental feature is that it provides an error estimate for its approximations. Here we first use it
to derive the value of e. In the next section we introduce the Taylor Series as it is familiar to
hydrologists.
Suppose we set x=1 in the power series for ex. Then
1 1 1
+ + + ...
(1)
2! 3! 4!
We can approximate e by a partial sum of the series, but since the series does not alternate we don’t
have an error bound. How can we introduce an error bound for the Maclaurin series and use it for
this special case?
e = 1+1+
Suppose that x is fixed and f(x) is approximated by the beginning of its Maclaurin series, that is, by
a Maclaurin polynomial, say of degree 8:
f ( x) ≈
f (0) f ′(0)
f ′′(0) 2 f ′′′(0) 3
f ( 8 ) ( 0) 8
+
x+
x +
x + ... +
x
0!
1!
2!
3!
8!
If the series alternates then the first term omitted (the 9th term in the series, in this example) supplies
the error bound. But whether or not the series alternates, the error in the approximation can be
f ( 9 ) (m) 9
bounded as follows. Consider all possible values of
x for m between 0 and x, and find the
9!
maximum of the values. Taylor’s remainder formula states that the error, in absolute value, is less
than or equal to that maximum.
In general, the error (in absolute value) in approximating f(x) by its Maclaurin polynomial of degree
n is less than or equal to the maximum value of
f ( n+1) ( m ) n+1
(2)
x
( n + 1)!
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New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
for m between 0 and x.
Returning to the problem of approximating e, we can use a graphical method (not shown here) to
get a crude estimate which we then refine with Taylor approach. The graphical approach, based on
integrating 1/x, yields the estimate that 1 < e < 4. Now suppose we add the first five terms (1) to get
e = 1+1+
1 1 1
+ + = 2.708
2! 3! 4!
(3)
To estimate the error of this approximation, consider (2) with f(x) = ex, x = 1, n = 4 (since we added
terms through x4 in the series for ex), and 0 ≤ m ≤ 1. Then f (5)(x) = ex and
f ( 5) ( m ) 5 e m
1 =
(5)!
5!
Since 1 < e < 4, the maximum occurs when m=1, and that maximum is less than 4/5! or 1/30.
Therefore the approximation in (3) is less than 1/30. Furthermore, when the expansion in (1) stops
somewhere, all the terms omitted are positive, so the approximation in (3) is an underestimate. By
adding more terms it can be shown that 2.718281 < e < 2.718282.
Taylor Series in powers of x-b
Certain basic functions, like ln x, √x, and 1/x, can’t be expressed as a power series of the form
Σanxn, because they have derivatives that blow up at x=0. Other functions which have a power
series form converge too slowly. These limitations can be overcome by considering a power series
of the form
∞
∑ a ( x − b)
n
n
= a0 + a1 ( x − b) + a2 ( x − b) 2 + a3 ( x − b) 3 + ...
(1)
n=0
called a power series about b. The previous section consider the special case for b = 0. For this more
general case the Maclaurin series has coefficients
f ( n ) ( b)
an =
n!
If f does not blow up, or have derivatives which blow up, at b, it has the power series representation
f ( x) =
f (b) f ′(b)
f ′′(b)
f ′′′(b)
+
( x − b) 3 + ... (2)
( x − b) 2 +
( x − b) +
3!
2!
0!
1!
which is called a Taylor series for f about b. The partial sums of (2) are called Taylor polynomials.
Graphs of successive Taylor polynomials are a line, a parabola, a cubic, and so on, tangent to the
graph of f(x) at point (b, f(b)). They supply better and better approximations to the graph. The
Taylor series converges more rapidly if x is near b, and more slowly if x is far away. In application,
there is a trade off: keep (x-b) small and use a only a couple of terms in (2), or increase (x-b) and
use more terms.
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New Mexico Tech
Hydrology Program
Hyd 510
Quantitative Methods in Hydrology
Application: In the finite difference method of solving a differential equation for heat
transport, the temperature, T, at location x is related to the temperature at location b=x-∆x,
where ∆x is the spacing of adjacent finite difference node points at which temperature is to be
calculated. Then with x-b=∆x we can write (2) as
T ( x − ∆x ) ∆x dT
T ( x) =
+
+ O ( ∆x ) 2
1
1 dx x −∆x
where the higher order terms have been truncated, and the order of their approximation
indicated by the last term on the RHS. Solving for the temperature gradient we have
dT
T ( x ) − T ( x − ∆x )
=
+ O ( ∆x ) 2
dx x −∆x
∆x
which is then used to approximate the gradient in Fourier’s law of heat conduction, and to
provide an estimate of the error in the approximation.
Similar finite difference approximations are used for solute diffusion and dispersion,
groundwater flow, and other gradient driven processes.
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