MATHEMATICAL MODELLING WITH

DIFFERENTIAL EQUATIONS

Dierential equations are equations involving derivatives.

Basic Definitions

Ordinary or Partial

Order

Linear or Nonlinear

Solution:

Exact or Approximate

Analytical or Numerical

Definitions

With an equation such as

y = 3x

the variable y is a function of the variable x; the notation to

indicate this is y = y(x). y is termed the dependent variable

and x the independent variable. This means that any value can

be given to x and the corresponding value of y will be determined

from the equation. With a dierential equation, the variablebeing dierentiated is the dependent variable.

Examples of dierential equations

dy

dx= 5y

is a dierential equation with x as independent variable and yas its dependent variable, and so is the equation

d2y

dx2 3dy

dx+ 2y = 0

A dierential equation is called ordinary if it only contains deriv-atives involving the dierentiation of a function with respect toa single independent variable. Otherwise, the equation is called

a partial dierential equation. Abbreviations commonly usedare ODE and PDE.

Examples of ODEs and PDEs

Newtons second law can be written in the form of an ordinary

dierential equation in terms of velocity as:

dv

dt=F

m

or, in terms of displacement, as

d2x

dt2=F

m

The equation representing simple harmonic motion is

md2x

dt2+ cdx

dt+ kx = 0

which is an ODE for the displacement x with respect to time t.

The two-dimensional Laplace equation, describing heat transfer

in a solid, is a PDE given by

2Tx2

+2Ty2

= 0

The dependent variable is the temperature T which depends on

its spatial position, i.e. T = T (x, y); the variables x and y are

independent.

The order of a dierential equation is the order of the highestderivative appearing in the equation. Therefore, Newtons law is

given by a first-order ODE in terms of velocity, and by a second-

order ODE in terms of displacement. The Laplace equation is a

second-order PDE.

Example 1

Consider the dierential equation:

dy

dx= 2

describing a straight line with gradient of 2. Many such lines

can be drawn in a graph, as shown below. All the family of lines

have a general equation of the form

y = 2x + k

where the constant k (the intercept) has a dierent value for eachof the lines. The above solution is thus called a general solution.

To obtain a unique solution, adequate conditions have to be spec-

ified. For instance, if we impose that y = 0 when x = 0, then the

only solution satisfying this condition is y = 2x. This solution

is called a particular solution. If all conditions are specified

at the same value of the independent variable, then the prob-

lem is called an initial-value problem. In contrast, problems

where specification of conditions occurs at dierent values of theindependent variable are called boundary-value problems.

Example 2

The second-order dierential equation:

d2y

dt2 3dy

dt+ 2y = 0

has the general solution

y = Aet +Be2t

which can be verified by direct substitution. Find the particular

solution for which y = 3 when t = 0 and dy/dt = 5 when t = 0.

Solution:

If

y = Aet +Be2t

then

dy

dt= Aet + 2Be2t

and

d2y

dt2= Aet + 4Be2t

Substituting in the original ODE gives:

Aet + 4Be2t 3Aet 6Be2t + 2Aet + 2Be2t = 0

which verifies the general solution. Substituting the first condi-

tion into the expression for y gives

3 = Ae0 +Be0 = A+B

while application of the second condition to the expression for

dy/dt gives

5 = Ae0 + 2Be0 = A+ 2B

Solving the above equations for A and B gives A = 1, B = 2.

Thus, the particular solution is

y = et + 2e2t

It can be noticed that, for second-order equations, two conditions

have to be specified for a particular solution to be found.

Linearity

A dierential equation is said to be linear if

The dependent variable and all its derivatives occur only tothe first power.

There are no products of terms involving the dependent vari-able.

There are no functions of the dependent variable or its deriv-atives which are nonlinear.

For example, the equation

d2y

dx2+ y = x2

is linear, since it does not matter that the independent variable

x is raised to the power of 2. However, the equation

dy

dx+ y2 = 0

is nonlinear because of the term y2.

Linear equations can be further classified as homogeneous or

non-homogeneous. When all the terms in a linear equation

containing the dependent variable appear on the left-hand side

of the equals sign and the terms containing the independent vari-

ables and any constants appear on the right-hand side, then the

equation is said to be homogeneous when the right-hand side

is zero and non-homogeneous when it is not. For example, the

equation

dy

dx+ y2 = 0

is homogeneous, but the equations

dy

dx+ y2 = 5

or

dy

dx+ y2 = x

are non-homogeneous.

Solution of dierential equations

A solution is said to be analytical if a closed-form expression

can be derived which allows values of the dependent variable to

be found, for any given values of the independent variable(s).

A numerical solution, usually found using the computer, will

provide values of the dependent variable at a finite number of

pre-assigned points; values at other points can only be found by

interpolation.

The most common analytical methods of solution of dieren-tial equations are Laplace transforms, separation of variables,

series expansions, Greens functions, and others.

Separation of Variables

Consider the equation

dy

dx= 2x

Multiplying both sides by dx gives

dy = 2xdx

The above equation has been separated into terms which are only

function of y, and terms which are only function of x. Integrating

both sides gives:

Zdy =

Z2xdx

y = x2 + c

where c is the constant of integration. This is the general solution

of the dierential equation.

Note that because there are integrations on both sides of the

equation we could have two constants of integration. However,

it is usual to combine these into a single constant c.

Example 3

Solve the dierential equation:

dy

dx= 4y

using separation of variables.

Solution:

dy

4y= dx

Integrating both sides gives:

ln y = 4x+ c

Example 4

Solve the nonlinear dierential equation:

ydy

dx= 4x

Solution:

ydy = 4xdx

Integrating both sides gives:

y2

2= 2x2 + c

Complementary Function and Particular Integral

Suppose we have the non-homogeneous equation

dy

dx+ y = 5

The solution to this dierential equation is of the form

y = Aex + 5

The solution to the corresponding homogeneous equation, i.e.

dy

dx+ y = 0

is given by

y = Aex

Thus, the general solution of the non-homogeneous equation is

equal to the sum of the solution of the homogeneous equation,

called the complementary function, plus another term, called

the particular integral.

The particular integral is any solution of the non-homogeneous

equation. In the above case, it can be seen that y = 5 is indeed

a solution of the non-homogeneous equation, since dy/dx = 0 in

this case.

If we consider the general form of a linear, first-order, non-

homogeneous ODE

dy

dx+ P (x)y = Q(x)

then its solution can be written in the form

y = yc + yp

where yc is the complementary function and yp the particular

integral. The complementary function is the solution of the cor-

responding homogeneous equation

dy

dx+ P (x)y = 0

while the particular solution can be determined by educated

guesswork. The usual method is to try a function of the same

form as that on the right-hand side of the equation.

Example 5

Solve the non-homogeneous dierential equation:

dy

dx+ 2y = 4x

Solution:

The corresponding homogeneous equation is

dy

dx+ 2y = 0

whose solution can be found by separation of variables to be

ln y = 2x + c

or

y = Ce2x

with C = ec. This is the expression of the complementary func-

tion yc.

Because the term on the right is 4x, the particular integral we

can try is y = A+Bx. With this solution, the non-homogeneous

equation becomes

B + 2(A+Bx) = 4x

Hence, equating coecients of x gives B = 2 and equating con-stant terms gives B + 2A = 0, and so A = 1. Therefore

yp = 1 + 2x

The solution to the non-homogeneous equation is then of the

form

y = Ce2x 1 + 2x

Runge-Kutta Methods

The Euler and Heun methods are versions of one-step approaches

calledRunge-Kuttamethods, i.e. techniques of the general form

yi+1 = yi + h (xi, yi, h)

where (xi, yi, h) is called an increment function which can beinterpreted as a representative slope over the interval.

The increment function (xi, yi, h) can be written as:

= a1k1 + a2k2 + . . . + ankn

where the as are constants and the ks are functions of the form:

k1 = f(xi, yi)

k2 = f(xi + p1h, yi + q11k1h)

k3 = f(xi + p2h, yi + q21k1h+ q22k2h)

...

kn = f(xi+pn1h, yi+qn1,1k1h+qn2,2k2h+. . .+qn1,n1kn1h)

Notice that the ks are computed from recurrence relationships

in that k1 appears in the equation for k2, which appears in the

equation for k3, and so forth. This recurrence makes RKmethods

ecient for computer calculations.

Various types of Runge-Kutta methods can be devised by em-

ploying dierent numbers of terms in the increment function.The first-order RKmethod with n = 1 is, in fact, Eulers method.

Once n is chosen, values for the as, ps and qs are evaluated by

setting the general RK equation equal to terms in a Taylor series

expansion.

Second-Order Runge-Kutta Methods

The second-order version of the general RK equation is:

yi+1 = yi + h (a1k1 + a2k2) (1)

We recall that the second-order Taylor series for yi+1 in terms of

yi and f(xi, yi) is given by

yi+1 = yi + h f(xi, yi) +h2

2f 0(xi, yi)

where f 0(xi, yi) must be determined by chain-rule dierentiation:

f 0(xi, yi) =fx+fydy

dx

Substituting the above into the previous expression:

yi+1 = yi + h f(xi, yi) +h2

2

fx+fydy

dx

(2)

The basic strategy of all Runge-Kutta methods is to use algebraic

manipulations to solve for a1, a2, p1 and q11 (in this case) that

make equations (1) and (2) equivalent.

The Taylor series for a two-variable function is defined in the

form:

g(x+ r, y + s) = g(x, y) + rgx+ s

gy+ . . .

Applying this formula to expand the expression for k2 gives:

k2 = f(xi + p1h, yi + q11k1h) =

f(xi, yi) + p1hfx+ q11k1h

fy+O(h2)

Substituting the above result into equation (1):

yi+1 = yi + a1hf(xi, yi) + a2hf(xi, yi) + a2p1h2fx+

a2q11h2f(xi, yi)

fy+O(h3)

or, by collecting terms,

yi+1 = yi + (a1 + a2) h f(xi, yi)+

h2a2p1

fx+ a2q11f(xi, yi)

fy

+O(h3) (3)

Comparing now like terms in equations (2) and (3), it can be

determined that in order for the two equations to be equivalent,

the following must hold:

a1 + a2 = 1

a2p1 =1

2

a2q11 =1

2

These three simultaneous equations contain the four unknown

constants. Because there is one more unknown than the number

of equations, there is no unique set of constants that satisfy the

equations. However, by assuming a value for one of the constants,

it is possible to determine the other three. This gives rise to a

family of second-order RK methods rather than a single version.

Assuming that the value of a2 is specified, the following equations

are then obtained for the other variables:

a1 = 1 a2 p1 = q11 =1

2a2

The two most commonly used second-order Runge-Kutta meth-

ods are the Heun method and the Ralston method. The first is

obtained making a2 = 1/2, for which a1 = 1/2, p1 = q11 = 1,

producing the equation

yi+1 = yi + h

1

2k1 +

1

2k2

with

k1 = f(xi, yi)

k2 = f(xi + h, yi + hk1)

as previously studied.

Ralstons method is obtained when a2 = 2/3, in which case

a1 = 1/3, p1 = q11 = 3/4:

yi+1 = yi + h

1

3k1 +

2

3k2

with

k1 = f(xi, yi)

k2 = f

xi +

3

4h, yi +

3

4hk1

Example

Solve the equation

dy

dx= 2x3 + 12x2 20x + 8.5

with the initial condition y = 1 at x = 0, using Ralstons method

with a step length h = 0.5.

Solution

f(x, y) = 2x3 + 12x2 20x + 8.5

y(0.5) = y(0) + 0.5

1

3k1 +

2

3k2

k1 = f(0, 1) = 8.5

k2 = f

0 +

3

4 0.5, 1 + 3

4 0.5 8.5

k2 = f(0.375, 4.1875) = 2.582

y(0.5) = 1 + 0.5

1

3 8.5 + 2

3 2.582

y(0.5) = 3.277

Proceeding one more step:

k1 = f(0.5, 3.277) = 1.25

k2 = f

0.5 +

3

4 0.5, 3.277 + 3

4 0.5 1.25

k2 = f(0.875, 3.746) = 1.1523

y(1.0) = 3.277 + 0.5

1

3 1.25 + 2

3 (1.1523)

y(1.0) = 3.101

Fourth-Order Runge-Kutta Methods

The most popular Runge-Kutta methods are fourth order. As

with the second-order approaches, there are an infinite number of

versions. The following is sometimes called the classical fourth-

order RK method:

yi+1 = yi + h

1

6(k1 + 2k2 + 2k3 + k4)

where

k1 = f(xi, yi)

k2 = f

xi +

1

2h, yi +

1

2hk1

k3 = f

xi +

1

2h, yi +

1

2hk2

k4 = f (xi + h, yi + hk3)

Example

Solve the previous problem using the classical fourth-order RK

method with a step length h = 0.5.

Solution

f(x, y) = 2x3 + 12x2 20x + 8.5

y(0.5) = y(0) + 0.5

1

6(k1 + 2k2 + 2k3 + k4)

k1 = f(0, 1) = 8.5

k2 = f

0 +

1

2 0.5, 1 + 1

2 0.5 8.5

k2 = f(0.25, 3.125) = 4.21875

k3 = f

0 +

1

2 0.5, 1 + 1

2 0.5 4.21875

k3 = f(0.25, 2.0547) = 4.21875

k4 = f(0.5, 1 + 0.5 4.21875)

k4 = f(0.5, 3.1094) = 1.25

y(0.5) = 1 + 0.5

1

6(8.5 + 2 4.21875 + 2 4.21875 + 1.25)

y(0.5) = 3.21875

which is the exact value. This was expected because the true

solution is quartic in this case, thus the fourth-order method is

able to predict the exact solution.

Systems of Equations

All methods for single equations can be extended to the solution

of systems of equations. The procedure simply involves applying

the one-step technique for every equation at each step before

proceeding to the next step.

Example

Solve the following system of first-order dierential equationsusing Euler method, assuming that at x = 0, y1 = 4 and y2 = 6.

Use a step length of 0.5.

dy1dx

= 0.5y1

dy2dx

= 4 0.3y2 0.1y1

Solution

The equations for application of Eulers method are:

y1(xi + h) = y1(xi) + h [0.5 y1(xi)]

y2(xi + h) = y2(xi) + h [4 0.3 y2(xi) 0.1 y1(xi)]

Therefore:

y1(0.5) = 4 0.5 0.5 4 = 3

y2(0.5) = 6 + 0.5 (4 0.3 6 0.1 4) = 6.9

y1(1.0) = 3 0.5 0.5 3 = 2.25

y2(1.0) = 6.9 + 0.5 (4 0.3 6.9 0.1 3) = 7.715

Example

Solve the following second-order dierential equation for a mass-spring system using Eulers method

d2x

dt2+ x = 8 cos t

assuming that x(0) = x0(0) = 0. Use a step length of 0.1 andcompare results with the exact solution x = 4t sin t.

Solution

Defining a new variable y such that:

dx

dt= x0 = y

the previous equation can be rewritten in the form

dy

dt= x+ 8 cos t

The equations for application of Eulers method are then:

xi+1 = xi + h yi

yi+1 = yi + h (xi + 8 cos ti)

Therefore:

x(0.1) = 0 + 0.1 0 = 0

y(0.1) = 0 + 0.1 [0 + 8 cos(0)] = 0.800

x(0.2) = 0 + 0.1 0.800 = 0.080

y(0.2) = 0.800 + 0.1 [0 + 8 cos(0.1)] = 1.596

x(0.3) = 0.080 + 0.1 1.596 = 0.240

y(0.3) = 1.596 + 0.1 [0.080 + 8 cos(0.2)] = 2.372

x(0.4) = 0.240 + 0.1 2.372 = 0.477

y(0.4) = 2.372 + 0.1 [0.240 + 8 cos(0.3)] = 3.112

x(0.5) = 0.477 + 0.1 3.112 = 0.788

y(0.5) = 3.112 + 0.1 [0.477 + 8 cos(0.4)] = 3.801

Example

Solve again the equation for the mass-spring system using the

classical fourth-order Runge-Kutta method.

Solution

The equations for application of the classical fourth-order RK

method are of the form:

dx

dt= f1(t, x, y) = y

xi+1 = xi + h1

6(k1,1 + 2k2,1 + 2k3,1 + k4,1)

dy

dt= f2(t, x, y) = x+ 8 cos t

yi+1 = yi + h1

6(k1,2 + 2k2,2 + 2k3,2 + k4,2)

First, we must evaluate all the k1s:

k1,1 = f1(0, 0, 0) = 0

k1,2 = f2(0, 0, 0) = 8

Next, it is necessary to calculate the values of x and y that are

needed to determine the k2s:

xi +1

2h k1,1 = 0 +

1

2 0.1 0 = 0

yi +1

2h k1,2 = 0 +

1

2 0.1 8 = 0.4

k2,1 = f1(0.05, 0, 0.4) = 0.4

k2,2 = f2(0.05, 0, 0.4) = 7.990

The process continues as follows:

xi +1

2h k2,1 = 0 +

1

2 0.1 0.4 = 0.02

yi +1

2h k2,2 = 0 +

1

2 0.1 7.990 = 0.3995

k3,1 = f1(0.05, 0.02, 0.3995) = 0.3995

k3,2 = f2(0.05, 0.02, 0.3995) = 7.970

xi + h k3,1 = 0 + 0.1 0.3995 = 0.03995

yi + h k3,2 = 0 + 0.1 7.970 = 0.7970

k4,1 = f1(0.1, 0.03995, 0.7970) = 0.7970

k4,2 = f2(0.1, 0.03995, 0.7970) = 7.920

x(0.1) = 0 + 0.1

1

6(0 + 2 0.4 + 2 0.3995 + 0.7970)

x(0.1) = 0.040

y(0.1) = 0 + 0.1

1

6(8 + 2 7.990 + 2 7.970 + 7.920)

y(0.1) = 0.797

Results for the RK method up to time t = 0.5 are shown in the

table below.

Time x (Euler) x (RK) x (Exact) y (Euler) y (RK) y (Exact)

0.1 0 0.040 0.040 0.800 0.797 0.797

0.2 0.080 0.159 0.159 1.596 1.579 1.579

0.3 0.240 0.355 0.355 2.372 2.328 2.328

0.4 0.477 0.623 0.623 3.112 3.031 3.031

0.5 0.788 0.959 0.959 3.801 3.673 3.673