Yang X.-S. Introductory Mathematics for Earth Scientists

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150 Chapter 11. Ordinary Differential Equations

From the integration, we know that the gradient dy/dx is x

− x,

and the function

−

has the gradient x

− x. We can say that

y(x) =

−

satisﬁes the diﬀerential equation (11.1), and thus y(x) is

a solution to (11.1). Then the question is: Are there any other solutions

to this equation? The answer is an inﬁnite number. Fro m Fig. 6.2, we

know that there is a family of curves whose gradient is x

− x. The

solution in general should contain an arbitrary constant C. That is to

say, the general solution of (11.1) can be written as

y(x) =

−

+ C. (11.2)

Any solution that corresponds to a single speciﬁc curve is a particular

solution. Fo r example, both x

/4 − x

/2 and x

/4 − x

/2 − 1 are

particular solutions.

Another important concept is the or der of a diﬀerential equation.

The order of a diﬀerential equation is the highest derivative of the un-

known. For example, the order of (11.1) is 1 as the highest derivative is

the gradient, so this equation is called a ﬁrst-order diﬀerential equation.

The following equation

y(x)

− 2x

dy(x)

+ y(x) = x

, (11.3)

is the second-or der diﬀerential equatio n as the highest derivative is the

second derivative d

y/dx

All the above equations only contain ﬁrst and/or second derivatives,

and there is only a single independent variable x. Such diﬀerential

equations are called ordinary diﬀere ntial equations (ODE).

On the other hand, if a quantity of interest such as u depends on

two or more independent variables such x and t, then partial derivatives

such as

∂u

∂x

and

∂u

∂t

will a ppear. In this case, we usually have to deal

with a partial diﬀerential equation (PDE). For example, the following

equation

∂u

∂t

∂

∂x

, (11.4)

is a diﬀusion equation and we will discuss this in more detail in the next

chapter. In this chapter, we will focus solely on the ordinary diﬀere ntial

equations.

11.2 First-Order Equations

We know that dy/dx = x

− x is a ﬁrst-order ordinary diﬀerential

equation, which can be generalised as

dy(x)

= f (x), (11.5)

11.2 First-Order Equations 151

where f (x) is a given function of x. Its solution can be obtained by

simple integration

y(x) =

f(x)dx + C. (11.6)

Example 11.1: T h e air can be co n sidered as an ideal gas obeying the

ideal gas law

pV = nRT,

where P is the air pressure, V is the volume of the air of interest, and n

is the number of modes of air. R = 8.31 J/mo le·K is the universal gas

constant, and T is the absolute temperature. The ideal gas law can also

be written as

p =

ρR

where ρ is the density of the air and M ≈ 28.9 g/mole is the molar mass

of the air.

We assume that the air pressure is hydrostatic. That is, the increase

of the pressure dp is balanced by the increment of the weight −ρgdz for

a thin layer with a unit area. Here z is th e altitude above the Earth’s

surface and the − sign indicates the fact that the z increases and pressure

decreases in a tm osphere. Therefore, we have dp = −ρgdz or

= −ρg.

From p = ρRT /M , we have ρ = pM/RT , and we now have

= −

pMg

= −

dz.

Integrating from z = 0 to z = h, we ha ve

dp = ln p − ln p

= −

dz = −

where p

is the pressure on the Earth’s surface at z = 0.

This me a n s that

= −

Taking the logarithms, we have

p = p

−γh

, γ =

152 Chapter 11. Ordinary Differential Equations

We can see that the air pressure decreases exponentia lly as the height h

increases. We ca n deﬁne a characteristic height

L =

so that

p = p

−h/L

For the typical values of M = 0.0289 kg/mole, g = 9.8 m/s

, and T = 293

K (or 20

◦

C), we have

L =

8.31 × 393

0.0289 × 9.8

≈ 8597 m = 8.597 km,

which corresponds to γ ≈ 0.00116 m

−1

. T h is means that the air pressure

at z = L will becomes 1/e ≈ 36.8% of the pressure on the Earth’s su rface.

The exponential decrease of the ai r pressure was conventi onally used

to determine the altitude in aviation and atmospheric sciences. From

p = p

−γh

, we get

ln(

) = −γh,

which leads to

h = −

= L ln

So if the pressure has dropped to half of the ground pressure (assuming a

constant temperature), the alti tu d e h i s given by

h = L ln

0.5p

= L ln 2 ≈ 5.959 km.

In our calcula tions, we have assumed that the temperature is constant. In

reality, the temperature does change signiﬁcantly.

What happens if f also depends on y, so that we have

dy(x)

= f (x, y). (11.7)

It is not possible to solve it by integration in general; however, in a

sp e c ial ca se when f(x, y) = g(x)h(y) is separable we can solve it by the

separation of variables. That is to rearrange the equation so that the

function of x is on one side, and the function of y is on the other side.

We have

h(y)

= g(x). (11.8)

Then, we can integrate both sides with respect to x, and we have

h(y)

dx =

h(y)

dy =

g(x)dx, (11.9)

11.2 First-Order Equations 153

where we have used dy = (dy/dx)dx or

()

dx =

()dy. If we provide

appropria te conditions for y or y

(0) at x = 0, we can also determine

the unknown constant. Let us look a t an example.

Example 11.2: To solve the ﬁrst-order diﬀerential equation

= (x

+ e

−y

under the condition y(0) = 0 at x = 0, we can rewrite it as

−y

= x

+ e

whose integration leads to

−y

dx =

+ e

)dx.

This becomes

dy =

dx +

dx,

+ e

+ C,

Using x = 0 and y(0) = 0, we have

+ e

+ C,

which gives 1 = 0 + 1 + C or C = 0. Now the solution becomes

+ e

Ta king the logarithm, we have

y = ln(

+ e

We will u se this solution in th e nu m e rical solution of ordinary diﬀerential

equations in later c h a p te rs.

When the function f(x, y) is a nonlinear function of y such as the

case of e

−y

, it is not so straightforward to solve such equations. In the

rest of the se c tion, we will only focus on the general linear ﬁr st-order

ordinary diﬀerential equation in the following form

+ p(x)y = q(x), (11.10)

154 Chapter 11. Ordinary Differential Equations

where p(x) and q(x) are known functions of x. We will attempt to solve

this generic diﬀerential equation. How do we start? We know from the

diﬀerentiation rule for a product that

d(Q(x)y)

= Q(x)

+ y

dQ(x)

= Q(x)y

+ Q

(x)y. (11.11)

If we could in some way write the left hand side of (11.10) as a derivative

of a product similar to the above, we should be able to rewrite the orig-

inal diﬀere ntial equation (11.10) as a simpler form (11.5). Therefore,

multiplying both sides of (11.10) by a function Q(x) to be determined

later, we have

Q(x)

+ Q(x)p(x)y = Q(x)q(x). (11.12)

The aim is to write it a s the for m

d(Q(x)y)

= Q(x)

dQ(x)

y = Q(x)q(x). (11.13)

By comparing with (11.12), we get

dQ(x)

= Q(x)p(x). (11.14)

This is a separable form as dis c ussed earlier. We have

Q(x)

= p(x), (11.15)

Q(x)

dx =

p(x)dx. (11.16)

This gives

ln Q(x) =

p(x)dx, (11.17)

Q(x) = e

p(x)dx

, (11.18)

which is often called the integrating factor. Using this integrating fac-

tor, equation (11.12) becomes

d[e

p(x)dx

= Q(x)q(x) = q(x)e

p(x)dx

. (11.19)

Integrating it with respect to x, we have

p(x)dx

y =

q(x)e

p(x)dx

dx + A, (11.20)

11.2 First-Order Equations 155

where A is the consta nt of integra tion. Dividing both sides of the above

equation by the integrating factor, we have the general solution

y = e

−

p(x)dx

(

q(x)e

p(x)dx

)

+ Ae

−

p(x)dx

. (11.21)

Let us solve the diﬀerential equation

+ 5y = e

−x

, (11.22)

given that y(0) = 1/2. The integrating factor Q(x) is

Q(x) = e

5dx

= e

. (11.23)

Multiplying the original equation (11.22) by Q(x), so we have

+ 5e

y =

d(e

= e

−x

= e

, (11.24)

whose integration leads to

y =

+ A. (11.25)

Therefore, the general solution is

y =

−x

+ Ae

−5x

. (11.2 6)

The constant A can be determined by y(0) = 1/2 at x = 0. That is

−0

+ Ae

−5×0

, (11.27)

or A = 1/4. The ﬁnal solution becomes

y =

−x

−5x

. (11.28)

Let us look at another example.

Example 11.3: In hydrol ogy, the runoﬀ model for linear reservoir pro-

vides a ﬂow equation

+ αq = αR,

where q is th e runoﬀ or di scharge, and R is the eﬀective rainfall and c a n be

considered as a constant for a short ti m e . The coeﬃcient α is a response

factor which can be taken as a constant.

156 Chapter 11. Ordinary Differential Equations

We know that the integrating factor is

Q = e

αdt

= e

αt

Therefore, using the formula (11.21), we have the general solution

q = e

−αt

(αR)e

αt

dt + Ae

−αt

= e

−αt

(αR)

αt

+ Ae

−αt

which bec omes

q = Ae

−αt

+ R.

If the discharge at t = 0 is q

, we have

= Ae

−α×0

+ R,

A = q

− R.

The ﬁnal solution now becomes

q = q

−αt

+ R(1 − e

−αt

As t → ∞, we have q → R.

11.3 Second-Order Equations

For second-order ordinary diﬀerential equations (ODEs), it is generally

more tricky to ﬁnd the general solution. However, a special case with

signiﬁcantly practical importance and mathematical simplicity is the

second-order linear diﬀerential equation with constant coeﬃcients in

the following form

+ b

+ cy(x) = f(x), (11.29)

where the coeﬃcients b and c are constants, and f (x) is a known smooth

function of x. Obviously, the more general form would be

+ b

+ cy(x) = f(x), (11.30)

however, if we divide bo th sides by a, we will re ach our sta ndard form.

Here we assume a 6= 0. In a spe c ial ca se of a = 0, it reduces to a

ﬁrst-order linear diﬀere ntial equation which ha s been discussed in the

previous section. So we will start our discussion from (11.29).

11.3 Second-Order Equations 157

A diﬀerential equation is said to be homogeneous if f(x) = 0. For a

given generic second-order diﬀerential equation (11.29), a function that

satisﬁes the homogeneous equation

+ b

+ cy(x) = 0, (11.31)

is called the complementary function, denoted by y

(x). Obviously, the

complementary function y

alone cannot satisfy the original equation

(11.29) because there is no way to produce the required f (x) on the

right-hand side. Therefore, we have to ﬁnd a speciﬁc function, y

∗

(x)

called the particular integral, so that it indeed satisﬁes the original

equation (11.29). The combined general solution

y(x) = y

(x) + y

∗

(x), (11.32)

will automatically satisfy the o riginal equation (11.29). The general

solution of (11.29) consists of two parts: the complementary function

(x) and the particular integral y

∗

(x). We can obtain these two parts

separately, and simply a dd them together bec ause the original equation

is linear, so their solutions are linear combinations.

First things ﬁrst, how to obtain the complementary function? The

general technique is to assume that it takes the form

(x) = Ae

λx

, (11.33)

where A is a constant, and λ is an exponent to b e determined. Substi-

tuting this assumed form into the homogeneous equation, we have

Aλ

λx

+ bAλe

λx

+ cAe

λx

= 0. (11.34)

Since Ae

λx

should not be zero (otherwise, we have a trivial solution

= 0 everywhere), we can divide all the terms by Ae

λx

, and we have

+ bλ + c = 0, (11.35)

which is the characteristic equation for the homogeneous equation. It

is also called the aux iliary equation of the ODE. The solution of λ in

this case is simply

λ =

−b ±

√

− 4c

. (11.36)

For simplicity, we can take A = 1 as it does not aﬀect the results.

In the discussion of quadratic equations in the ﬁrst chapter of the

book, we know tha t there are three possibilities for λ. They are: I) two

real distinct roots, II) two identical roots, and III) two c omplex roots.

In the case of two diﬀerent r oots: λ

6= λ

. Then, both e

and e

satisfy the homogeneous equation, so their linear combination forms the

complementary function

(x) = Ae

+ Be

, (11.37)

158 Chapter 11. Ordinary Differential Equations

where A and B are c onstants.

In the special case of identical roots λ

= λ

, or

c = λ

, b = −2λ

. (11.38)

we cannot simply write

(x) = Ae

+ Be

= (A + B)e

, (11.39)

because it still only one part of the complementary function y

= Ce

where C = A + B is just a nother constant. In this c ase, we should

try a diﬀerent combination, say, y

= xe

to see if it satisﬁes the

homogeneous equation or not. Since y

(x) = e

+ xλ

, and

(x) = λ

+ λ

+ xλ

, we have

(x)+ by

(x)+ cy

(x) = e

(2λ

+xλ

+be

(1+xλ

)+cxe

= e

[(2λ

+ b)] + xe

[λ

+ bλ

+ c] = 0, (11.40)

where we have used b + 2λ

= 0 (identical roots) and λ

+ bλ

+ c = 0

(the auxilia ry equa tion). This indeed implies that xe

also satisﬁes

the homogeneous equation. Therefore, the complementary function for

the identical roots is

(x) = Ae

+ Bxe

= (A + Bx)e

. (11.4 1)

Example 11.4: T h e second-order homogeneous equation

+ 5

− 6y = 0,

has a corresponding auxiliary equation

+ 5λ − 6 = (λ − 2 )(λ + 3) = 0.

It has two real roots

= 2, λ

= −3.

So the complementary function is

(x) = Ae

+ Be

−3x

But for the diﬀerential equation

+ 6

+ 9y(x) = 0,

its auxiliary equation becomes

+ 6λ + 9 = 0,

11.3 Second-Order Equations 159

which has two identical roots λ

= λ

= −3. The complementary function

in this case can be written as

(x) = (A + Bx)e

−3x

As complex roots always come in pairs, the case of complex roots

would give

1,2

= α ± iβ, (11.42)

where α and β are r e al numb e rs. The complementary function becomes

(x) = Ae

(α+iβ)x

+ Be

(α−iβ)x

= Ae

αx

iβx

+ Be

αx

−iβx

= e

αx

[Ae

iβx

+ Be

−iβx

]

= e

αx

{A[cos(βx) + i sin(βx)] + B[cos(−βx) + i sin(−βx)]}

= e

αx

[(A + B) cos(βx) + i(A − B) sin(βx)]

= e

αx

[C cos βx + D sin βx], (11.43)

where we have used the Euler formula e

θi

= cos θ + i sin θ and absorb

the constants A and B into C = A + B and D = (A − B)i.

A special case is when α = 0, s o the roots are purely imaginary. We

have b = 0, and c = β

. E quation (11.29) in this case becomes

+ β

y = 0, (11.44)

which is a diﬀerential equation for harmonic motions such as the oscil-

lations of a pendulum or a small-amplitude seismic detector. Here β is

the angular frequency of the system.

Example 11.5: For a simple pendulum o f mass m shown in Figure

11.1, we now try to derive its equation of oscillations and its period.

Since the motio n is circular, the tension or the centripetal force T is

thus given by

T = m

= m

where

θ = dθ/dt is the angular velocity.

Forces must be balanced both vertically and horizontally. The co m po -

nent of T in the vertical direction is T cos θ which must be equivalent to

mg, though in the opposite direction. He re g is the acceleratio n due to

gravity. That is

T cos θ = mg.

Since θ is small or θ  1, we have cos θ ≈ 1. This means that T ≈ mg.