1. outlineof the analytical solutionspart of the unit · the motion of a violin string is a good...

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Department of Mechanical Engineering, University of Bath

Modelling Techniques 2 ME20021

Supplementary Notes on Analytical Methods for solving PDEs

1. Outline of the analytical solutions part of the unit

In this unit our aim is the analytical solution of various partial differential equations. In particular

we will be concentrating on (i) Fourier’s equation,

∂θ

∂t= α

∂2θ

∂x2, (1.1)

where α is the thermal diffusivity and θ is the temperature, (ii) Laplace’s equation,

∂2θ

∂x2+

∂2θ

∂y2= 0, (1.2)

and (iii) the wave equation,

∂2y

∂t2= c2

∂2y

∂x2, (1.3)

where y is the displacement of a taut string from its equilibrium profile, and c is known as the

wavespeed. There are higher dimensional analogues of these equations, but they will be presented

at the end of these notes.

Two different techniques will be used to solve these equations: (i) separation of variables followedby Fourier Series, and (ii) Fourier Transforms. Although there is a technique called separation of

variables which is used for solving Ordinary Differential Equations (ODEs), the one we use forsolving Partial Differential Equations (PDEs) is different. Fortunately it is impossible to confuse

these two methods! The Fourier Series which arises most often in these contexts is what is called

a half-range series, details of which will be covered later. Fourier Transforms are found to play asimilar role to that played by Laplace Transforms in Maths 2 ME10305 in that they are used to

remove derivatives with respect to one of the variables. Thus each of the above example PDEs are

reduced to ODEs when the Fourier Transform is taken, while an ODE is reduced to an algebraicequation.

In what follows, any technical term which is introduced and which forms a new concept will be

typeset in bold and underlined.

ME20021 Modelling Techniques 2 2

2. Use of Fourier Series

Fourier series usually appear in these problems when we are dealing with a finite domain in at least

one direction. The motion of a violin string is a good example of a finite domain since a violinstring has a finite length. Another good example is the heating or cooling of a solid bar.

2.1. Fundamental solutions for Fourier’s equation.

The aim here is to give a fairly general way of finding solutions to the above PDEs. To fix ideaswe will solve Fourier’s equation, as given in (1.1), and the solution will be subject to the boundary

conditions that θ = 0 on both x = 0 and x = 1, and the initial condition that θ = f(x), a known

function, at t = 0. In practical terms, this mathematical description translates into the following:“A solid bar of unit length has the temperature profile, θ = f(x), at t = 0, whereupon the ends of

the bars have their temperatures changed suddenly to θ = 0; what happens next?”

We will proceed by means of the following ansatz (i.e. an educated guess):

θ = T (t) sinnπx. (2.1)

where n is a positive integer. There are various reasons why this formula needs to be discussed.

But first note that the t-dependence and the x-dependence in θ(x, t) have been separated, hence

the name of the method, separation of variables.

The next thing to note is the use of a sine. Fourier’s equation has a second x-derivative, and

therefore if θ is proportional to a sine function of x, then so is its second derivative with respect to

x.

However, this last observation is also true for cosines. In this case the analogous cosine will be

equal to 1 when x = 0, and therefore it does not satisfy the boundary condition that θ = 0 whenx = 0. Therefore we are stuck with sines.

Ah, but which sines? Well, the ones we have chosen fit with the second boundary condition (namely,

that θ = 0 when x = 1) because sinnπ = 0 when n is an integer. Therefore the formula given in(2.1) is very reasonable. The curves for the first three modes are given in Fig. 2.1, below.

Figure 2.1. The first three modes of the form sinnπx

which satisfy Dirichlet conditions at x = 0 and x = 1.

sinπx (n = 1)

sin 2πx (n = 2)

sin 3πx (n = 3)

At this point in the analysis the shape of the function T (t) is unknown. Therefore all we can do is

to substitute Eq. (2.1) into Eq. (1.1), and we get

dT

dtsinnπx = −αn2π2T sinnπx. (2.2)


The sines cancel, as we expect, leaving us with,

dT

dt= −αn2π2T, (2.3)

which has the solution,

T = Be−αn2π2t, (2.4)

where B is an arbitrary constant. Therefore we can say that

θ = Be−αn2π2t sinnπx (2.5)

is our desired solution because it satisfies the governing PDE and the two boundary conditions. It

is this form which I call a fundamental solution.

However, Eq. (2.5) does not satisfy the given initial condition, namely that θ = f(x) at t = 0,and therefore this single solution is not yet sufficiently general. This difficulty may be resolved by

adding together all of the solutions of the form, (2.5), to give the following,

θ =

∞∑

n=1

Bne−αn2π2t sinnπx, (2.6)

where I have introduced the subscript, n, to ensure that all the B’s may take different values should

they need to. This simple addition of the fundamental solutions is called superposition and itworks because the PDE is linear and the boundary conditions are homogeneous (i.e. equal to zero).

Now we apply the initial condition; this gives

f(x) =

∞∑

n=1

Bn sinnπx. (2.7)

This looks like a Fourier Series, but technically it is a half-range series consisting of sines. The

reason that it is called half-range is because the n = 1 term corresponds to half a sine wave inthe physical domain we are considering. An alternative name, one which we will adopt here, is

Fourier Sine Series, and this must always be used to mean a half-range series consisting of sines.

At this point we would need to apply a formula for the Fourier coefficients, but the formula will begiven later after we have considered fundamental solutions for the other two main PDEs we shall

be considering.

2.2. Fundamental solutions for Laplace’s equation.

We may solve Laplace’s equation,∂2θ

∂x2+

∂2θ

∂y2= 0,

using the same ideas. This equation represents a steady two-dimensional distribution in a solid. If

we were to consider a semi-infinite strip of solid material, i.e. one which is contained in the domain,

0 ≤ x ≤ 1 and 0 ≤ y < ∞, then this is the type of domain where we can make analytical progress.So we’ll take the following boundary conditions:

y = 0 : θ = f(x), y → ∞ : θ → 0, x = 0, 1 : θ = 0. (2.8)

Physically, these conditions correspond to a strip where the temperature of the y = 0 end is precisely

f(x), and where the infinitely long sides at x = 0, 1 are being maintained at θ = 0 — it is usually


worth sketching a diagram of this configuration. In one sense the y → ∞ boundary condition isn’t

necessary, because we would expect the temperature to decay to zero that far away from the sourceof heating/cooling.

Given that the strip lies between x = 0 and x = 1, and that the boundary conditions are both zeroat these points, we may again use sines as part of the separation of variables ansatz. They will be

sinnπx, for integer values of n, for exactly the same reasons as the previous example. Therefore

we will substituteθ = Y (y) sinnπx (2.9)

into Eq. (1.2), and this yields,d2Y

dy2− n2π2Y = 0, (2.10)

where the sines have already been cancelled. Equation (2.10) has the following solutions,

Y = Aenπy +Be−nπy, (2.11)

where A and B are currently arbitrary constants. Therefore the reconstructed general solution is

θ =[

Aenπy +Be−nπy]

sinnπx. (2.12)

We can see immediately that A must be zero because this part of the solution grows exponentially,

whereas we need the solution to decay to zero. The solution now becomes,

θ = Be−nπy sinnπx. (2.13)

The final condition to apply is that θ = f(x) at y = 0. To do this, we must first superpose all thepossible solutions of the form given in (2.13):

θ =∞∑

n=1

Bn e−nπy sinnπx, (2.14)

and then apply the boundary condition. This gives,

f(x) =

∞∑

n=1

Bn sinnπx, (2.15)

which is again a Fourier Sine Series.

Note that if we were to have a conducting strip orientated in the x-direction with θ = 0 on y = 0, 1

and θ = f(y) on x = 0, then the solution would be given by

θ =

∞∑

n=1

Bne−nπx sinnπy, (2.16)

where the Fourier coefficients are to be determined from the application of the x = 0 boundarycondition. Therefore we have

f(y) =

∞∑

n=1

Bn sinnπy, (2.17)


from which the Fourier coefficients will be calculated.

Finally, it is important to point out that the way in which we have been applying the separation

of variables method, namely, the factoring out of appropriate sines or cosines, is not the techniquewhich is given in most textbooks. Therefore I have placed a description of that method in §3 for

interest. The reason I have adopted the process described above is that it is much quicker andmore intuitive for the types of equation we are solving here.

2.3. Fundamental solutions for the wave equation.

We will solve the wave equation,

∂2y

∂t2= c2

∂2y

∂x2(2.18)

subject to the boundary conditions, y = 0 at x = 0 and x = 1, and the initial conditions, y = f(x)

and ∂y/∂t = 0 at t = 0. The initial conditions are equivalent to plucking a violin string in the

sense that the string has an initial deformation profile and it is released from rest. The boundaryconditions state that there is zero displacement at the two ends, which is quite natural, or, rather,

quite essential for a fully functioning violin string.

As with the previous two cases we will factor out a sinnπx dependence by setting,

y(x, t) = T (t) sin nπx, (2.19)

where n takes positive integer values, and therefore the equation reduces to

d2T

dt2= −n2π2c2T (2.20)

after cancellation of the sines. The solution for T is

T = A cosnπct+B sinnπct, (2.21)

and therefore the fundamental solution which we are seeking is,

y =[

A cosnπct+B sinnπct]

sinnπx. (2.22)

Thus far we have written down a solution of the PDE which satisfies both the boundary conditionsat x = 0 and x = 1.

We now need to superpose all the fundamental solutions and to apply the initial conditions. After

superposition we have,

y =∞∑

n=1

[

An cosnπct+Bn sinnπct]

sinnπx. (2.23)

It is easier to apply the condition, ∂y/∂t = 0 at t = 0, first. The quickest way to do this is to

observe that cosnπt already has a zero derivative at t = 0, but sinnπt does not, and therefore we

must suppress the latter by setting Bn = 0 for all values of n. The solution now reads,

y =

∞∑

n=1

An cosnπct sinnπx. (2.24)


The initial condition that y = f(x) at t = 0 yields the Fourier Sine Series,

f(x) =

∞∑

n=1

An sinnπx. (2.25)

Once the An values have been obtained we may substitute them into Eq. (2.24) and that will

complete the solution for y. [Note: given that the Fourier coefficients for a sine series are, byconvention, denoted by a B, we may replace An by Bn at this point should we wish.]

2.4. Some comments.

All of the above solutions have corresponded to systems in which the physical domain of interesthas unit length. Thus both the strip and the taut string occupy the range 0 ≤ x ≤ 1. What

happens to the fundamental solutions when the strip/string have length, d, say? Well, in this case

we have 0 ≤ (x/d) ≤ 1, and therefore we may use the function, sin(nπx/d), in the separation ofvariables ansatz.

If we now wish to solve Fourier’s equation in the range 0 ≤ x ≤ d, then Eq. (2.1) should be replacedby

θ = T (t) sin(nπx/d). (2.26)

The function, T , now satisfies the equation,

dT

dt= −α

n2π2

d2T, (2.27)

and has solution,

T = e−αn2π2t/d2

. (2.28)

Therefore the full solution, without the initial condition having been applied, is

θ =

∞∑

n=1

Bne−αn2π2t/d2

sin(nπx/d). (2.29)

An alternative derivation of which sine to use is the following. Let the sine be sinσx where σ isto be found. When x = d we have sinσd which must be zero in order to satisfy the boundary

condition there. Therefore σd = nπ, since sinnπ = 0 when n is an integer. Therefore we find thatσ = nπ/d, and finally the sine is sin(nπx/d), as before.

If we have the situation where the strip/string occupies the region, −d ≤ x ≤ d, then the simplest

tactic is to redefine the x–coordinate. We would set x = x + d, which means that 0 ≤ x ≤ 2d.Therefore we would use sin(nπx/2d). However, I will not give you any examples where this needs

to be done.

2.5. Definitions of the various Fourier series.

Here we shall state the definitions of the Fourier series (plural) in some of its many guises.

2.5.1. Fourier Series.

If the function f(x) has period equal to d, then the Fourier Series is


f(x) = 12A0 +

∞∑

n=1

[

An cos(2nπx/d) +Bn sin(2nπx/d)]

(2.20)

where

A0 =2

d

∫ d

0

f(x) dx, An =2

d

∫ d

0

f(x) cos(2nπx/d) dx, Bn =2

d

∫ d

0

f(x) sin(2nπx/d) dx.

(2.31)

Note that the range of integration is one period, and therefore the actual limits used will dependon the range over which f(x) is defined explicitly. Note also that the n = 1 cosine is cos(2πx/d)

which executes one full cosine wave over the period, d.

We will need this when solving Laplace’s equation in polar coordinates in a circular or even annular

domain.

2.5.2. Fourier Sine Series.

This is the half-range series consisting of sines, and it is used when solving equations where the

function is zero on the boundaries. Such conditions are known as Dirichlet conditions.

If the function f(x) is defined in the range, 0 ≤ x ≤ d, then the Fourier Sine Series is

f(x) =∞∑

n=1

Bn sin(nπx/d) (2.32)

where

Bn =2

d

∫ d

0

f(x) sin(nπx/d) dx. (2.33)

Note that the number by which the integral is multiplied is the same as for standard Fourier Series.

However, the sine terms are slightly different, and the n = 1 sine is sin(πx/d) which executes half

a sine wave in the given interval, 0 ≤ x ≤ d.

2.5.3. Fourier Cosine Series.

This is the half-range series consisting of cosines, and it is used when solving equations where the

derivative of the function is zero on the boundaries. Such conditions are known as Neumann

conditions. We will tackle such problems later.

If the function f(x) is defined in the range, 0 ≤ x ≤ d, then the Fourier Cosine Series is

f(x) = 12A0 +

∞∑

n=1

An cos(nπx/d) (2.34)

where

A0 =2

d

∫ d

0

f(x) dx, An =2

d

∫ d

0

f(x) cos(nπx/d) dx. (2.35)

As with the Fourier Sine Series, the n = 1 cosine is cos(πx/d) which executes half a cosine wave in

the given interval, 0 ≤ x ≤ d. The value, 12A0, is the mean value of the function, f(x).


2.5.4. A quarter-range Fourier Sine Series.

Such a series will arise when solving a problem where the solution should be zero at x = 0, but its

derivative should be zero at x = d. The first quarter of a sine wave is an example of a function

which has the same properties. In this range, the sine wave will be sin(πx/2d), where the argumentto the function, namely πx/2d varies from 0 to π/2 as x varies from 0 to d. The other sine waves

with this property are sin(nπx/2d) where n takes odd integer values only.

If the function f(x) is defined in the range, 0 ≤ x ≤ d, then the quarter-range Fourier Sine Seriesis

f(x) =

∞∑

n=1

n odd

Bn sin(nπx/2d) (2.36)

where

Bn =2

d

∫ d

0

f(x) sin(nπx/2d) dx. (2.37)

2.6. An illustration of the convergence of the four types of Fourier Series.

On the next page we will see how the four different Fourier series converge to the function, f(x) = x

in the range 0 ≤ x ≤ 1. The full Fourier Series will be periodic with period, 1, but the others are

defined solely over that range.

In all four cases shown the retention of just one term in the respective Fourier Series provides a

very poor approximation to f(x), which is the diagonal line shown in every graph.

For the full Fourier Series, f(x) is periodic with period, 1, and the Fourier Series has to try to

mimic the discontinuity at x = 0 and x = 1. This it does by giving an increasingly steep series

as the number of terms increases. One may also see increasingly narrow ‘wiggles’ close to thatdiscontinuity.

For the other three types of Fourier Series, a close inspection of the partial sums of the Fourier

Series at x = 0 and x = 1 shows that they satisfy certain boundary conditions. For the FourierSine Series the partial sums are zero at those points, while the derivatives are zero for the Fourier

Cosine Series. For the quarter range series the partial sums are zero at x = 0 but their derivativeis zero at x = 1.


Fourier Series Fourier Sine Series Fourier Cosine Series Quarter range Series

1 term 1 term 1 term 1 term

5 terms 5 terms 2 terms 2 terms




12−

∞∑

n=1

sin 2nπx

nπ

∞∑

n=1

2(−1)n+1

nπsinnπx 1

2−∞∑

n=1

n odd

4

n2π2cosnπx

∞∑

n=1

n odd

8 sin(12nπ)

n2π2sin

(nπx

2

)


2.7. Some typical separation of variables problems.

We will cover the use of the half-range and quarter-range series given above by solving Fourier’s

equation,

∂θ

∂t= α

∂2θ

∂x2, (2.38)

2.7.1. A Fourier Sine Series solution.

We will use the initial condition that θ = x(1 − x) in the range, 0 ≤ x ≤ 1, when t = 0. The

boundary conditions are that θ = 0 on both x = 0 and x = 1, and so the boundaries are being helda uniformly colde temperature. This problem is almost identical to that given earlier, and therefore

I’ll go through the separation of variables part much faster.

We set θ = T (t) sin(nπx), where n is a positive integer, in order to satisfy the boundary conditions.

Substitution into the PDE gives

dT

dt= −αn2π2T, (2.39)

the solution for which is

T = Be−αn2π2t. (2.40)

On reconstructing θ, and superposing all the valid solutions, we have

θ =∞∑

n=1

Bne−αn2π2t sin(nπx). (2.41)

On applying the given initial condition, we have

x− x2 =∞∑

n=1

Bn sin(nπx). (2.42)

This is a Fourier Sine Series, and the coefficients, Bn, may be obtained by applying Eq. (2.33) withd = 1. We get,

Bn = 2

∫ 1

0

(x− x2) sin(nπx) dx

= 2[(

x− x2)(

−cosnπx

nπ

)

−(

1− 2x))(

− sinnπx

n2π2

)

+(

−2)(cosnπx

n3π3

)]1

0

= − 4

n3π3

[

cosnπx]1

0

= − 4

n3π3

[

cosnπ − cos 0]

= − 4

n3π3

[

(−1)n − 1]

=8

n3π3for n odd, or 0 for n even. (2.43)


Therefore the final solution is,

θ =

∞∑

n=1

n odd

8

n3π3e−αn2π2t sinnπx, (2.44)

and it is depicted in Fig. 2.2 below.

Figure 2.2. Depicting the solution given by Eq. (2.44) forthe following values of αt: 0 (uppermost curve), 0.05, 0.1,

0.15 and 0.2 (lowest curve).

t = 0

0 1x

2.7.2. The Fourier Cosine Series solution.

The initial condition will again be θ = x(1 − x) in the range, 0 ≤ x ≤ 1, where the boundaryconditions are ∂θ/∂x = 0 on both x = 0 and x = 1, i.e. the boundaries are insulated and heat

cannot escape. Intuitively this means that the temperature must evolve until it reaches a constant

value.

We set θ = T (t) cos(nπx), where n is a positive integer in order to satisfy the boundary conditions.

We need to use a cosine so that the zero derivative at x = 0 is guaranteed to be satisfied. Theconstant multiplying x is such that the zero derivative at x = 1 is also satisfied. However, it is

important to note that we must also allow n = 0. This value of n corresponds to a constant, since

cos 0 = 1, and a function which is a constant also satisfies the zero derivative boundary conditions.The first four of these shapes (i.e. for n = 0, 1, 2, 3) are shown in Fig. 2.3, below.

Figure 2.3. The first four modes of the form cosnπx whichsatisfy Neumann conditions at x = 0 and x = 1.

1 (n = 0)

cos πx (n = 1)

cos 2πx (n = 2)

cos 3πx (n = 3)



dT

dt= −αn2π2T, (2.45)

after cancelling the cosines, and the solution is

T = Ae−αn2π2t. (2.46)

This, of course, is the same as we had for the Fourier Sine Series example, although I have now

denoted the constant by A because we are heading towards a Fourier Cosine Series.

When n = 0 we have T ′ = 0, for which the solution is T =constant. I will name this constant 12A0,

again because of the Fourier Cosine Series that lies ahead.


θ = 12A0 +

∞∑

n=1

Ane−αn2π2t cos(nπx). (2.47)


x− x2 = 12A0 +

∞∑

n=1

An cos(nπx). (2.48)

This is a Fourier Cosine Series, and the coefficients, An, may be obtained by applying the integrals

in Eq. (2.35) with d = 1. We get,

A0 = 2

∫ 1

0

(x− x2) dx = 13 , (2.49)

and

An = 2

∫ 1

0

(x− x2) cos(nπx) dx

= 2[(

x− x2)(sinnπx

nπ

)

−(

1− 2x))(

−cosnπx

n2π2

)

+(

−2)(

− sinnπx

n3π3

)]1

0

=2

n2π2

[

(1− 2x) cosnπx]1

0

=2

n2π2

[

− cosnπ − cos 0]

=2

n2π2

[

−(−1)n − 1]

= − 4

n2π2for n even, or = 0 for n odd. (2.50)


θ = 16 −

∞∑

n=1

n even

4

n2π2e−αn2π2t cosnπx. (2.51)


From this solution we see that θ → 16 as t → ∞. This value is the mean temperature of the original

temperature profile. In this case the summation takes place over even values of n, which reflectsthe symmetry of the original profile (i.e. it is even about x = 1

2).

The evolution of the initial profile with time is given in Fig. 2.4 where the approach to the ultimateconstant-temperature state may be seen. In addition, a very rapid adjustment of the profiles near

x = 0 and x = 1 to ones where the profile has a zero derivative may also be seen.

Figure 2.4. Depicting the solution given by Eq. (2.51) forthe following values of αt: 0 (initial parabola), 0.0002,

0.002, 0.02 and 0.2 (‘horizontal’ line).

0 1x

2.7.3. The quarter-range Fourier Sine Series solution.

Yet again the initial condition will again be θ = x(1 − x) in the range, 0 ≤ x ≤ 1, where the

boundary conditions are now θ = 0 when x = 0 and ∂θ/∂x = 0 when x = 1. Therefore the domainis cooled on the left, but insulated on the right. Intuitively this means that heat leaks from the

system via the left hand boundary, and the temperature will eventually become zero everywhere.

We set θ = T (t) sin(nπx/2), where n is a positive odd integer in order to satisfy both the boundaryconditions. The left hand boundary condition is satisfied automatically by the use of a sine, but

the sines we have chosen are such that the zero derivative condition on the right is also satisfied.

The first three modes (i.e. n = 1, 3, 5) are shown in Fig. 2.5 below.

Figure 2.5. The first three modes of the form sinnπx/2

which satisfy Dirichlet conditions at x = 0 and Neumann

conditions at x = 1.

sinπx/2 (n = 1)

sin 3πx/2 (n = 3)

sin 5πx/2 (n = 5)


dT

dt= − 1

4αn2π2T, (2.52)

the solution for which is


T = Ae−αn2π2t/4. (2.53)


θ =∞∑

n=1

n odd

Bne−αn2π2t/4 sin(nπx/2). (2.54)


x− x2 =∞∑

n=1

n odd

Bn sin(nπx/2). (2.55)

This is a quarter-range Fourier Sine Series, and the coefficients, Bn, may be obtained by applyingthe integral in Eq. (2.37) with d = 1. We get,

Bn = 2

∫ 1

0

(x− x2) sin(nπx/2) dx

= 2[(

x− x2)(

−2 cosnπx/2

nπ

)

−(

1− 2x)(

−4 sinnπx/2

n2π2

)

+(

−2)(8 cos nπx/2

n3π3

)]1

0

=32

n3π3− 8 sin(nπ/2)

n2π2(2.56)


θ =

∞∑

n=1

n odd

( 32

n3π3− 8 sin(nπ/2)

n2π2

)

e−αn2π2t/4 sinnπx/2. (2.57)

It might look a little unsatisfactory leaving the term, sin(nπ/2) in this formula, but given that this

term oscillates between +1 and −1 as n increases through consecutive odd values, we could simplifythis aspect of the solution by setting n = 2m+ 1. Therefore, as n takes the values, 1, 3, 5, 7 · · ·, mwill take the values, 0, 1, 2, 3 · · ·. Eq. (2.3) may now be rewritten in the slightly more cumbersome

form,

θ =∞∑

m=0

( 32

(2m+ 1)3π3− 8(−1)m)

(2m+ 1)2π2

)

e−α(2m+1)2π2t/4 sin(2m+ 1)πx/2. (2.58)

The solution is depicted in Fig. 2.6, where various features may be seen.

Figure 2.6. Depicting the solution given by Eq. (2.57) for

the following values of αt: 0 (initial parabola), 0.0002,0.002, 0.02, 0.07, 0.2 and 0.5 (lowest curve).

0 1x


It is clear that the solution tends to zero as t increases, which may also be determined from the

form of Eq. (2.8). In addition we have the curious behaviour that the temperature at x = 1 risesfrom zero at first in order to ensure that there is no heat loss from that boundary. But it then

achieves a maximum before decaying towards zero once more as the profile increasingly resemblesa decaying quarter sine wave.

2.8. Further results for Laplace’s equation on finite domains.

In this subsection we shall consider how to modify the theory given above for solving Laplace’s

equation in a semi-infinite strip, to one which yields the solution in a rectangle. This does involve

more work than for a semi-infinite strip, but it is also possible to superpose different solutions inorder to solve a more complicated set of boundary conditions; all will become clear!

Figure 2.7. Depicting the unit square, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1within which Laplace’s equations will be solved subject to the

given boundary conditions.

0 x = 1

y = 1

T = 1

T = 0

T = 0 T = 0

To begin with we shall solve Laplace’s equation for the unit square and with the boundary conditionsthat are shown in Fig. 2.7. Given that we have T = 0 on both x = 0 and x = 1, this suggests

that we may use sinnπx as part of the separation-of-variables ansatz, where n = 1, 2, 3 and so on.

Therefore we set,

T (x, y) = Y (y) sinnπx. (2.59)

As before this will yield the following equation for Y :

Y ′′ − n2π2Y = 0,

for which the solution is

Y = Aenπy +Be−nπy.

As is usual, we find T ,

T =(

Aenπy +Be−nπy)

sinnπx,

and then superpose the resulting solutions for all admissible values of n:

T =

∞∑

n=1

[

Anenπy +Bne

−nπy]

sinnπx. (2.60)


When the strip was of infinite length in the y-direction (see §2.2) we had to suppress the exponen-

tially growing terms by setting An = 0. Here we cannot do that, but we do have two boundaryconditions to apply in the y-direction. The easier one is the one at y = 1 where T = 0. This means

that,

Anenπ +Bne

−nπ = 0,

and therefore that An = −Bne−2nπ. Substitution into (2.60) gives,

T =∞∑

n=1

Bn

[

e−nπy − enπ(y−2)]

sinnπx. (2.61)

When y = 0 we have T = 1, and therefore

1 =∞∑

n=1

Bn

[

1− e−2nπ]

sinnπx.

Upon noting that everything which multiplies the sinnπx term is the Fourier coefficient, we may

apply the formula for the Fourier Sine Series to obtain,

Bn

[

1− e−2nπ]

= 2

∫ 1

0

1× sin(nπx) dx = 4/nπ (n odd) 0 (n even).

Hence,

T =∞∑

n=1n odd

4

nπ

(

e−nπy − enπ(y−2)

1− e−2nπ

)

sinnπx.

This somewhat ugly expression may be improved by multiplying both the numerator and denomi-nator by enπ. This gives us,

T =∞∑

n=1n odd

4

nπ

(

enπ(1−y) − enπ(y−1)

enπ − e−nπ

)

sinnπx,

which may be further simplified into the form,

T =∞∑

n=1n odd

4

nπ

(

sinhnπ(1− y)

sinhnπ

)

sinnπx. (2.62)

This last form is not only compact, but it shows us immediately that T = 0 when y = 1

2.8.1. Variation 1.

Consider what happens for the unit square which is shown in Fig. 2.8. The T = 1 boundary

condition is now on theleft hand boundary, rather than on the lower boundary.


Figure 2.8. Depicting the unit square, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

within which Laplace’s equations will be solved subject to the


0 x = 1

y = 1

T = 0

T = 0

T = 1 T = 0

There is a symmetry between the boundary conditions displayed in Figs. 2.7 and 2.8. In fact, if

we were to swap the roles played by the x and y coordinates, then the problems may be seen to

be essentially identical. Therefore we may swap also the roles played by x and y in the solution(2.62). Thus the solution to Laplace’s equation for the problem shown in Fig. 2.8 is,

T =

∞∑

n=1n odd

4

nπ

(

sinhnπ(1− x)

sinhnπ

)

sinnπy. (2.63)

Here we see that T = 0 when x = 1 and also when y = 0 or y = 1.

2.8.2. Variation 2.

If we were to add the two solutions given by Eqs. (2.62) and (2.63),

T =∞∑

n=1n odd

4

nπ

(

sinhnπ(1− y)

sinhnπ

)

sinnπx+∞∑

n=1n odd

4

nπ

(

sinhnπ(1− x)

sinhnπ

)

sinnπy, (2.64)

then a check of its values at the four boundaries shows that this expression satisfies the boundary

conditions shown in Fig. (2.9) below.


within which Laplace’s equations will be solved subject to thegiven boundary conditions.

0 x = 1

y = 1

T = 1

T = 0

T = 1 T = 0


2.8.3. Variation 3.


within which Laplace’s equations will be solved subject to the


0 x = 1

y = 1

T = 0

T = 1

T = 0 T = 0

The final variation refers back to Fig. 2.7 but where T = 1 has been set on the upper boundaryinstead of on the lower boundary. This too is easily found from Eq. (2.62) by replacing all instances

of y by 1− y. The fundamental components of that solution (namely exponentials in y multipliedby the sine in x) are not changed and this seeming subterfuge still yields a solution of Laplace’s

equation, but more importantly it also satisfies these new boundary conditions. Hence we have,

T =

∞∑

n=1n odd

4

nπ

(

sinhnπy

sinhnπ

)

sinnπx. (2.65)

Clearly T = 0 when y = 0 for this solution.

By following these ideas it is possible to build up a solution for the case of a more general rectangular

domain with four different inhomogeneous boundary conditions.

2.9. Some comments..

The half- and quarter-range series may be used to solve Laplace’s equation. All three types of series

(i.e. the two half-range and the quarter range) produce physically realisable systems. However, the

nature of the physical system corresonding to the wave equation, i.e. a taut string which has zerodisplacement at its ends, is such that only Fourier Sine Series cases will arise. Should it be countered

that one could try to model the movement of a cantilever, such as a ruler which is clamped to a

desk and then twanged, then I would counter back that the clamped end satisfies both a zerodisplacement and a zero slope, and the free end satisfies a zero curvature (i.e. second derivative

of displacement) and a zero rate of change of curvature. Thus we have four boundary conditionsinstead of the two that we have been using quite happily so far. The resolution of this difficulty is

that a cantilever satisfies what is called the beam equation, which may be written in the simplified

form,

∂2y

∂t2= −c4

∂4y

∂x4, (2.66)


where c is related to the material properties of the cantilever including its Young’s modulus and its

moment of inertia. One may adopt a separation of variables approach to solve Eq. (2.66), which isknown as the beam equation, but the function to use in the x-direction will not be a sine wave if

we wish to study the movement of the cantilever, but will be a combination of trigonometric andhyperbolic functions. However, we may adopt sines if the beam is pin-jointed; in this case both y

and its second derivative are zero at the ends of the beam — this will appear in a problem sheet.

Another equation which may be solved using the ideas we have developed is the damped wave

equation,

∂2y

∂t2+ d

∂y

∂t= c2

∂2y

∂x2, (2.67)

where c remains the undamped wavespeed, and d is the damping coefficient. We see here anunavoidable clash of notation because c is often used as a damping coefficient in other contexts.

Equation (2.67) may be used to model the motion of a real-life violin string, especially when the

damping coefficient is small so that none of the modes of vibration is overdamped. Again, thisequation will appear in a problem sheet.

2.10. Heat Transfer in polar coordinates.

Laplace’s equation in Cartesian coordinates is given in Eq. (1.2) and its polar coordinate counterpart

is,

∂2T

∂r2+

1

r

∂T

∂r+

1

r2∂2T

∂θ2= 0. (2.68)

In this subsection we will adopt T to denote temperature in order free up θ to be the angularcoordinate. The variable, r, is the radial coordinate. The derivation of Eq. (2.68) from substituting

x = r cos θ and y = r sin θ into Eq. (1.2) is long and involved, and therefore it is omitted here.

This equation may be solved analytically in a variety of domains: circle, semicircle, quadrant and

any segment of a circle, such as 0 ≤ θ ≤ α. In addition it may also be solved inside an annulusor similarly defined annular segments. Finally, it may be solved in a region which is external to a

circle or semicircle, where, for a semicircle, the region is defined as being 0 ≤ θ ≤ π and r ≥ d,

where d is the radius. We will consider some of these.

2.11. Fundamental solutions for heat transfer in a semicircle.

Suppose that a semicircle lies in the region, 0 ≤ r ≤ d and 0 ≤ θ ≤ π, as shown in Fig. 2.11.

Suppose also that the temperature is zero on the straight edged boundary, i.e. on both θ = 0 and

θ = π where the black disc shows the location of the origin, and that T = θ(π − θ) on r = d.Then the zero boundary conditions at θ = 0 and θ = π suggest the use of sines in a separation of

variables ansatz. Such sines will be sinnθ where n is again an integer.


Figure 2.11. Depicting a semicircle of radius, d, withinwhich Laplace’s equations will be solved subject to the


r = d

θ = 0θ = π

T = θ(π − θ)

T = 0•

Therefore we proceed by setting,

T (r, θ) = R(r) sinnθ, (2.69)

where R(r) is a currently unknown function of r. Substitution of (2.69) into (2.68) and cancellationof the sines yields,

R′′ +1

rR′ − n2

r2R = 0. (2.70)

This equation may be improved cosmetically by multiplying each term by r; we get,

r2R′′ + rR′ − n2R = 0. (2.71)

Equation (2.71) is an example of a Cauchy-Euler equation, which is a category of linear ODE thatwas not covered in Mathematics 2 ME10305, last year. However, whatever form R takes, the fact

that the terms all balance means that rR′ must be of the same form as R. In other words, if we

differentiate R and then multiply the result by r we obtain something which is a multiple of whatwe started with. The most obvious function which fits that description is a power of r. Therefore

we shall set R = Ara, where we seek the value of a. Therefore Eq. (2.71) reduces to

Ara[

a(a− 1) + a− n2]

= 0, (2.72)

and we see that

a2 = n2 ⇒ a = ±n. (2.73)

We should have expected two possible values because the ODE is of second order. We shall use

both, in the same way as we do for λ after substitution of eλt into a second order time-dependent

constant-coefficient ODE.

R = Ar−n +Brn. (2.74)

We may reconstruct T from Eqs. (2.74) and (2.69), and then superpose all possible solutions. This

gives the following expression:

T =

∞∑

n=1

[

Anr−n +Bnr

n]

sinnθ. (2.75)


This solution satisfies Laplace’s equation and the boundary conditions at θ = 0 and θ = π, but

we haven’t yet applied the boundary condition at the curved boundary of the semicircle. However,this is only one boundary condition, whereas we have two sets of constants, An and Bn, which

might suggest that we need a further boundary condition. In problems of this type, namely oneswhere the origin is part of the solution domain, we may set An = 0 immediately, because the terms

they multiply are infinite at r = 0 — this might be terms a physical realisability condition: if

the boundary of a solid object has a fixed temperature profile, then we do not expect the internaltemperatures to become infinite. Hence Eq. (2.75) reduces to

T =∞∑

n=1

Bnrn sinnθ. (2.76)

Now we may apply the final boundary condition, that T = θ(π − θ) on r = d; we get

θ(π − θ) =

∞∑

n=1

Bndn sinnθ. (2.77)

Now this is a Fourier Sine Series but we have to regard Bndn as being the Fourier coefficient.

Therefore we have

Bndn =

2

π

∫ π

0

(πθ − θ2) sinnθ dθ

=2

π

[(

πθ − θ2)(− cosnθ

n

)

−(

π − 2θ)(− sinnθ

n2

)

+(

−2)(cosnθ

n3

)]π

0

=8

πn3for n odd, or 0 for n even. (2.78)

Therefore,

Bn =8

πn3dnfor n odd, or 0 for n even, (2.79)

and the solution is,

T =∞∑

n=1

n odd

8

πn3

( r

d

)n

sinnθ. (2.80)

Contours of the temperature are shown in Fig. 2.12, below.

Figure 2.12. Depicting the solution given by Eq. (2.80). Shown

are equally-spaced contours of the temperature where the lowerboundary corresponds to T = 0.

The rather boring isotherms show that the temperature almost increases linearly with height.


2.12. Heat transfer outside a semicircle.

Given the form of Eq. (2.80), what does the following expression represent?

T =

∞∑

n=1

n odd

8

πn3

(d

r

)n

sinnθ. (2.81)

The only difference between (2.80) and (2.81) is the r-dependent term. This solution satisfies allthree of the previously given boundary conditions, i.e. those at θ = 0, θ = π and r = d. It also

satisfies Laplace’s equation because the r-dependent terms that are present in (2.81) are those thatwe discarded above because they are infinite at the origin.

I suppose the hint is in the title of this subsection, that this solution corresponds to the temperature

profile outside a semicircle. More precisely, it corresponds to the solution in the domain, 0 ≤ θ ≤ πand r ≥ d, which is a half-plane with the original semicircle removed. We note that this solution

decays as r → ∞; the rn terms in (2.75) become infinitely large in that limit, and therefore theyhave to be removed for reasons of physical realisability.

Contours of the temperature are shown in Fig. 2.13, below.

Figure 2.13. Depicting the solution given by Eq. (2.81). Shownare equally-spaced contours of the temperature where the hor-

izontal boundary corresponds to T = 0. The inner radius is d

while the outer radius in the contour plot is 10d.

This slightly more interesting Figure shows that the temperature field decays as r becomes large,

in line with Eq. (2.81).

2.13. Heat transfer in a semicircular annulus.

The semicircular annulus lies in the range, 0 ≤ θ ≤ π and d ≤ r ≤ D. Let us suppose again thatT = 0 on θ = 0 and θ = π. However, we will set T = 0 on the inner radius, r = d, and T = θ(π− θ)

on the outer radius, r = D; see Fig. 2.14.


Figure 2.14. Depicting a semicircular annulus with outerradius, D, and inner radius, d, within which Laplace’s

equations will be solved subject to the given boundary

conditions.

r = D

r = d

θ = 0θ = π

T = θ(π − θ)

T = 0T = 0

T = 0

•

All of the arguments and analysis presented in §2.13 apply here up to the derivation of Eq. (2.75),which is repeated for convenience:

T =∞∑

n=1

[

Anr−n +Bnr

n]

sinnθ. (2.82)

Now we indeed have two boundary conditions in the r-direction and they must both be applied. The‘physical realisability’ argument is now not necessary because the origin isn’t part of the physical

domain.

• When r = d then T = 0. Equation (2.82) gives,

0 =∞∑

n=1

[

And−n +Bnd

n]

sinnθ. (2.83)

We could naıvely apply the formula for the Fourier Sine Series coefficients, but this merely tells us

that,

And−n +Bnd

n = 0 (2.84)

for all values of n.

• When r = D then T = θ(π − θ). Equation (2.82) gives,

θ(π − θ) =∞∑

n=1

[

AnD−n +BnD

n]

sinnθ. (2.85)

Application of the formula for the Fourier Sine Series coefficients will yield,

AnD−n +BnD

n =8


We now have a pair of simultaneous equations for An and Bn. Equation (2.84) gives,


An = −Bnd2n, (2.87)

and substitution of this into (2.86) gives,

Bn

(

Dn − d2n/Dn)

=8


A slight rearrangement of this yields,

Bn =8

πn3

1

dn1

(D/d)n − (d/D)nfor n odd, or 0 for n even, (2.89)

and so Eq. (2.87) gives,

An = − 8

πn3

dn

(D/d)n − (d/D)nfor n odd, or 0 for n even, (2.90)

The final solution is,

T =

∞∑

n=1

n odd

8

πn3

[ (r/d)n − (d/r)n

(D/d)n − (d/D)n

]

sinnθ. (2.91)

The formula has been presented in such a way that the boundary conditions at r = d and r = D

may easily be seen to be satisfied. Note that it is also possible to determine a solution analyticallyif the inner boundary of the annulus has a nonzero temperature profile; in this case we would need

to determine two Fourier series, the new one of which yield a nonzero right hand side to Eq. (2.84),

above.

Contours of this solutions are given on the next page for D = 1.5d and D = 5d.

Figure 2.15a. Depicting the solution given by Eq. (2.84). Shown

are equally-spaced contours of the temperature where the hor-izontal boundary corresponds to T = 0. The inner radius is d

while the outer radius in the contour plot is D = 1.5d.

Figure 2.15b. Depicting the solution given by Eq. (2.84). Shown

are equally-spaced contours of the temperature where the hor-izontal boundary corresponds to T = 0. The inner radius is d

while the outer radius in the contour plot is D = 5d.


2.14. Heat transfer in a quadrant and other domains.

The domain of interest now lies in the ranges, 0 ≤ θ ≤ π/2 and 0 ≤ r ≤ d. The separation of

variables ansatz is now,

T (r, θ) = R(r) sin 2nθ, (2.92)

where the n = 1 sine executes half a wavelength as θ increases from zero to π/2. Substitution intoLaplace’s equation yields,

r2R′′ + rR′ − 4n2R = 0. (2.93)

This has solution,

R = Ar−2n +Br2n. (2.94)

We may reconstruct T from Eqs. (2.94) and (2.92), superpose all possible solutions, and removethose that are physically unrealistic (i.e. the A-coefficients) and obtain,

T =

∞∑

n=1

Bnr2n sin 2nθ. (2.95)

Application of the boundary condition (not specified here) will yield the Bn-coefficients, and the

equation has been solved.

If the circular segment subtends the angle, α, instead of π or π/2, then the approriate sine to use

is sin(nπθ/α). On working through the same procedure we eventually obtain,

T =

∞∑

n=1

Bnrnπ/α sin(nπθ/α). (2.96)

2.15. Heat transfer in a circle.

Suppose the circumference of a circle of radius, d, is subject to the fixed temperature profile, T = θ2,

where −π ≤ θ ≤ π. Then we need to determine the temperature distribution inside the circle.

Now we are dealing with a full circle and not a segment, such as a semicircle or quadrant. The only

‘boundary condition’ we can apply in the θ-dirction is periodicity, and therefore we will expect to

use standard or full Fourier Series, rather than a half-range series.

Given the form of the boundary temperature, and the fact that it is an even function of θ, we

will also expect that this Fourier Series will consist solely of cosines. Therefore our separation of

variables ansatz will be to let,

T = R(r) cosnθ, (2.97)

where n = 0, 1, 2, · · ·, and where the n = 1 cosine varies over one whole period as θ varies between

−π and +π. Having introduced this ansatz, the analysis proceeds as usual to obtain T = Arn cosnθfor n = 1, 2, · · ·, where the r−n terms have been set to zero to maintain physical relevance. However,

since this is a cosine series, we also need to determine the n = 0 solution. If we substitute Eq. (2.97)

into Laplace’s equation, Eq. (2.68), when n = 0, then we obtain, r2R′′+ rR′ = 0. After division byr, the resulting equation is an exact differential, and we get (rR′)′ = 0. Integrating this, rearranging

and integrating once more yields,


R = A+B ln r. (2.98)

Clearly we must have B = 0 in order to avoid have infinitely negative values at r = 0. Therefore

all these solutions may be superposed to obtain the general form,

T = 12A0 +

∞∑

n=1

Anrn cosnθ, (2.99)

where the constant, A, in (2.98) has been replaced by 12A0 for convenience. Finally, the boundary

condition at r = d is applied and we get,

θ = 12A0 +

∞∑

n=1

Andn cosnθ. (2.100)

The An values are found using the Fourier Series formula,

An dn =2

2π

∫ π

−π

θ2 cosnθ dθ =4(−1)n

n2. (2.101)

Similarly, A0 = 2π2/3. Hence the final solution is,

T = 13π

2 +

∞∑

n=1

4(−1)n

n2

( r

d

)n

cosnθ. (2.102)

Contours of this solution are shown in Fig. 2.16, below.

Figure 2.16. Depicting the solution given by Eq. (2.102). Shown

are equally-spaced contours of the temperature. The tempera-ture is zero at the right hand extremity of the circle and takes

it maximum value at the left hand extremity.

Note: this example gave a cosine series, but it is not a half-range series; it is a standard full-range

series. The reason only cosines appear is that the even symmetry of the boundary conditions forces


the solution to be even too. If, on the other hand, we has set the temperature at r = d to be an odd

function, such as T = θ, then we would have obtained a Fourier Series consisting solely of sines.

2.16. Comments on multidimensional equations.

NOTE: not examinable, but is given here for background and interest. You may prefer to ignore

it!

The methods we have used may be applied to a wide range of more difficult partial differential

equations, and the following is a list of some of those where the above methods work with almost

no change in methodology.

Fourier’s equation in two dimensions is

∂θ

∂t= α

[∂2θ

∂x2+

∂2θ

∂y2

]

, (2.103)

and it governs unsteady heat transfer in a two-dimensional region or, if θ were to represent fluid

velocity, unsteady flow in a duct of uniform cross-section with a zero pressure gradient along the

duct. The three dimensional version is,

∂θ

∂t= α

[ ∂2θ

∂x2+

∂2θ

∂y2+

∂2θ

∂z2

]

. (2.104)

The three-dimensional Laplace’s equation,

∂2θ

∂x2+

∂2θ

∂y2+

∂2θ

∂z2= 0, (2.105)

governs steady three-dimensional heat transfer and is the steady version of Eq. (2.104).

The wave equation in two dimensions,

∂2u

∂t2= c2

[∂2u

∂x2+

∂2u

∂y2

]

, (2.106)

represents vibrations of a membrane such as a drumskin, where u is the displacement from equi-

librium. Clearly a three-dimensional version could be written down, and this will represent small-amplitude pressure waves in a compressible fluid such as air.

All of the above equations may be solved using separation of variables, but the resulting Fourierseries will take the form of double or triple series, and one could be faced with the possibly confus-

ingly named double half-range series. As an example for illustration purposes, we could solve the

three-dimensional Fourier’s equation given in Eq. (2.104). If the domain of interest is a unit cubewith all three coordinates lying in the range zero to 1, and if all six faces are maintained at θ = 0,

and if the initial temperature profile at t = 0 is θ = f(x, y, z), then the separation of variables

solution is

θ =∞∑

n=1

∞∑

m=1

∞∑

p=1

Bn,m,p e−α(n2+m2+p2)π2t sinnπx sinmπy sin pπz, (2.107)

where the Fourier coefficient, Bn,m,p, is given by the following triple half-range formula,


Bn,m,p = 8

∫ 1

0

∫ 1

0

∫ 1

0

f(x, y, z) sin nπx sinmπy sin pπz dx dy dz. (2.108)

The two-dimensional extension of the beam equation, which was quoted in Eq. (2.66), is the fol-lowing,

∂2u

∂t2= −c4

[∂4u

∂x4+

∂4u

∂x2∂y2+

∂4u

∂y4

]

. (2.109)

The presence of the mixed derivative might be a surprise, but the fourth derivative which occurs in

(2.66) is, for technical reasons, a second derivative of a second derivative, and the correct extensionwill be to replace each of these second derivatives by what is called the Laplacian. That is,

∂2

∂x2is replaced by

∂2

∂x2+

∂2

∂y2. (2.110)

Therefore

∂2

∂x2

(∂2u

∂x2

)

=∂4u

∂x4(2.111a)

is replaced by( ∂2

∂x2+

∂2

∂y2

)(∂2u

∂x2+

∂2u

∂y2

)

=∂4u

∂x4+ 2

∂4u

∂x2∂y2+

∂4u

∂y4. (2.111b)

The combination of derivatives which appear on the right hand side of (2.111b), namely,

∂4

∂x4+ 2

∂4

∂x2∂y2+

∂4

∂y4, (2.112)

is known as the biharmonic operator. It is also possible to write down more complicated versions

in three dimensions and in polar coordinates.

Finally, it is worth mentioning that similar higher dimensional extensions to the same equations

written in polar coordinates become very difficult to solve. Typically the straightforward polynomial

solutions we have obtained in the radial direction are replaced by Bessel functions which are quiteconsiderably more complicated to deal with. There are such entities as Fourier-Bessel series where

these Bessel functions replace the sines and cosines we have been dealing with. Likewise, forcantilevers, the solutions of the beam equation which are obtained when using separation of variables

are no longer sines and cosines. Again, there are Fourier–beam-function series, but all of these

matters are outside of the scope of the present unit.

3. The standard textbook method of Separation of Variables

I will illustrate what I call the long, slow and incredibly dull version of Separation of Variables by

repeating the analysis of §2.1, the solution of Fourier’s equation,

∂θ

∂t= α

∂2θ

∂x2, (3.1)

in the domain, 0 ≤ x ≤ 1.

First we assume a separation-of-variables solution, θ(x, t) = X(x)T (t), where neither X nor T are

known. This is substituted into Eq. (3.1) to get,


XT ′ = αX ′′T. (3.2)

Note that I have used primes to denote ordinary derivatives with respect to the appropriate variable,

so X ′ means dX/dx and T ′ means dT/dt. If we now divide both sides of (3.2) by XT we obtain,

T ′

T= α

X ′′

X. (3.3)

It turns out to be a little more convenient to divide both sides by α as well:

T ′

αT=

X ′′

X. (3.4)

This innocent-looking equation has important ramifications. The left hand side is a function of t

only, and we would normally expect that any change in the value of t would alter the value of theleft hand side. However, the right hand side is a function of x only, and therefore it is must be

unaffected by changes in the value of t. This means that the left hand side cannot vary as t varies.

A similar argument means that changes in x will not affect the value of either side of the equation.Therefore both sides must be equal to a constant, which we will take to be K, and which we will

call the separation constant. Equation (3.4) becomes,

T ′

αT=

X ′′

X= K, (3.5)

and therefore we must have both

X ′′ −KX = 0 and T ′ −KT = 0. (3.6a, b)

Further progress now requires us to consider the physics and/or the mathematics of the problem

in hand in order to choose not only the sign of K, but also suitable values.

Let us take K = p2 > 0. Then the solutions to (3.6a,b) are

X = Aepx +Be−px and T = Ceαp2t, (3.7a, b)

where A, B and C are arbitrary constants. Now apply the boundary conditions. If θ = 0 at x = 0

and x = 1, then X = 0 at these points too. Therefore the setting of x = 0 into (3.7a) gives

0 = A+B, ⇒ B = −A, (3.8a)

while the setting of x = 1 gives

0 = Aep +Be−p ⇒ B = −Ae2p. (3.8b)

The elimination of B between (3.8a) and (3.8b) shows that B = 0 and hence that A = 0. In turnthis implies that X = 0 and therefore θ = 0. While this answer is undoubtedly correct in the sense

that the equations and boundary conditions which have been applied so far are satisfied, it is not

particularly useful. However, we could have circumvented this mathematical analysis by notingthat we must set C = 0 in (3.7b) because we cannot have exponentially growing solutions in such

a heat transfer problem. This physical observation means that T = 0 and hence that θ = 0 again.

Let us now take K = 0. Equations (3.6) reduce to

X ′′ = 0 and T ′ = 0. (3.9a, b)


The solutions are that

X = Ax+B and T = C. (3.10a, b)

Although C = constant is physically reasonable, the application of the boundary conditions again

yields A = B = 0, implying that θ = 0. Thus this choice of the separation constant is no good

either.

Finally, let us take K = −p2. Equations (3.6) become,

X ′′ + p2X = 0 and T ′ + αp2T = 0. (3.11a, b)

The solutions are that

X = A cos px+B sin px and T = Ce−αp2t, (3.12a, b)

and from this we get

θ(x, t) = (A cos px+B sin px)Ce−αp2t. (3.13)

This equation looks as though it has three arbitrary constants, A, B and C, but in fact it has onlytwo: AC and BC, and therefore we’ll write (3.13) as

θ(x, t) = (A cos px+B sin px)e−αp2t. (3.14)

At x = 0 we have θ = 0 and substitution of this into (3.14) gives A = 0. Our solution is now,

θ(x, t) = B sin px e−αp2t. (3.15)

At x = 1 we also have θ = 0; equation (3.15) yields

0 = B sin p e−αp2t. (3.16)

Clearly we must not take B = 0 because this again leaves us with θ = 0. Therefore we must take

sin p = 0, (3.17)

from which we may deduce that p must be a multiple of π. So we may take

p = nπ for any integer value of n. (3.18)

Using this information in equation (3.15) means that our solution is

θ(x, t) = B sinnπx e−αn2π2t, (3.19)

for any integer value of n. We have now obtained Eq. (2.5) and therefore all that is necessary to

do now is to superpose all the possible solutions,

θ(x, t) =

∞∑

n=1

Bn sinnπx e−αn2π2t, (3.20)

and to apply the initial condition at t = 0.

This is the method which is generally found in textbooks and in my opinion it makes the determine

of the final solution quite cumbersome. Given the types of equations which we are confined to on

this unit I far prefer to use the sort of ansatz which is given in (2.1), the choice for depends on theboundary conditions to be satisfied.


4. Fourier Transforms

In this section we will discuss the use of Fourier Transforms in solving the same partial differentialequations as we met earlier. Attention will be confined to Cartesian coordinates.

4.1. Comparison with Laplace Transforms.

While Fourier Series may be used to solve certain PDEs in domains where at least one spatialdimension is of finite length, Fourier Transforms may be used in infinite domains. An example of

this would be steady heat conduction in the half-plane given by −∞ < x < ∞ and y ≥ 0, where

the temperature takes a given profile on the edge at y = 0. Fourier Series would be unable to dealwith this problem unless the temperature profile at the boundary is periodic in x.

Thus, Laplace Transforms and Fourier Transforms often play a similar role by allowing one to solve

differential equations. Their definitions have a great deal of similarity:

Laplace Transform: L[f(t)] =∫ ∞

0

f(t) e−st dt, (4.1)

Fourier Transform: F [f(t)] =

∫ ∞

−∞f(t) e−jωt dt. (4.2)

The lower limits of integration are different, and the exponentials also differ slightly in appearance.The Laplace Transform variable, s, is generally regarded as being real and positive, and therefore

the integral gives a finite value (i.e. a finite function of s) for a very wide class of functions, f(t);

this includes positive powers of t and functions which grow at most as fast as an exponential. Onthe other hand, the complex exponential in Eq. (4.2) does not decay as the magnitude of t becomes

large, and therefore it is generally the case that f(t) must be a decaying function of t as |t| becomesvery large. This will be discussed in a little more detail later.

4.2. From Fourier Series to Fourier Transforms.

Note that this section is for information only, and is designed solely to show the strong connection

there is between Fourier Series and Fourier Transforms.

It is possible to generate the Fourier Transform formally from Fourier Series by allowing the period

to become infinite. Strictly speaking, such a mathematical process (i) requires careful attentionto various limiting processes, details of which are well beyond the scope of this unit, (ii) may

nevertheless be seen as being quite feasible by appealing to the following example. Please note that

this is for background purposes only.

Consider the Fourier Series representation of the unit pulse, f(t), which is defined as follows,

f(t) =

1

0

|t| < 12,

12 < |t| ≤ 1

2P,(4.3)

where f(t) has period equal to P . An example, for which P = 3, is given on the next page.


Figure 4.1. Depicting a periodic sequence of unit pulses.In this example the period is P = 3.

t

f(t)

0 3 6−3−6

This function is clearly even, and therefore the Bn coefficients are zero. Using the formula for theFourier Series coefficients, we obtain,

A0 =2

P

∫ P/2

−P/2

f(t) dt =2

P

∫ 1/2

−1/2

1 dt =2

P, (4.4a)

An =2

P

∫ P/2

−P/2

f(t) cos(2πnt

P

)

dt =2

P

∫ 1/2

−1/2

cos(2πnt

P

)

dt =2

nπsin

(πn

P

)

. (4.4b)

The fundamental frequency which corresponds to the fundamental period, P , is,

ω =2π

P, (4.5)

and therefore we may now rewrite the expressions for A0 and An in the form,

A0 =2

P, An =

2

P

sin(ωn/2)

(ωn/2), (4.6)

where ωn = nω represents multiples of the fundamental frequency. Alternatively we have

12PA0 = 1, 1

2PAn =sin(ωn/2)

(ωn/2). (4.7)

If we now apply the Fourier Transform definition given in (4.2) to the isolated pulse which isobtained when the period is infinite, then we obtain,

∫ ∞

−∞f(t) e−jωt dt =

∫ 1/2

−1/2

1 e−jωt dt nonzero part of f(t)

=

∫ 1/2

−1/2

[cosωt− j sinωt] dt expand complex exponential

=

∫ 1/2

−1/2

cosωt dt odd integrand yields zero

=sin(ω/2)

(ω/2). (4.8)

The comparison between (4.7) and (4.8) is exact. In more detail, the scaled Fourier coefficients

given by Eq. (4.7) have been plotted against frequency in Fig. 4.2, and the Fourier Transform(Eq. (4.8)) is also shown for comparison; this Figure is on the next page.


ο

οο

οο

οο

οοοο

οοο

Figure 4.2. Depicting the Fourier coefficients for the periodic array of unit pulseswith period, P = 2, 4 and 8. The scaled coefficients are represented as circles

while the background curve is the Fourier Transform of the isolated unit pulse

centred at the origin. All quantities are plotted as a function of frequency, ω.

ω/π

P = 2

1 2 4 5 6 8

n=1

n=2

n=3

12P An

οοο

οο

οο

οο

οο οοοο

οοοο οο οο

οο οο οο ο ω/π

P = 4

1 2 4 5 6 8

n=1

n=2

n=3

n=4

n=6

12P An

ο οοοο

οοοο

οοοο

οοοο

οοοο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο οο ο ω/π

P = 8

1 2 4 5 6 8

1 2345678

12

12P An

Given that An represents the amplitude of modes of a set of discrete frequencies, the FourierTransform represents the amplitude of modes over a continuous set of frequencies.

The above Fourier Transform is real — this always happens when the chosen f(t) is even. When

f(t) is odd, the Fourier Transform is purely imaginary. For general cases the Fourier Transform iscomplex, and therefore it encodes both the frequency and the phase content of the signal, f(t).

4.3. The Fourier Transform pair.

It is possible to write down a simple formula for the inverse Fourier transform and the following iswhat is know as the Fourier Transform pair. If,

F (ω) = F [f(t)] =

∫ ∞

−∞f(t) e−jωt dt, (4.9)

then the inverse Fourier Transform is

f(t) = F−1[F (ω)] =1

2π

∫ ∞

−∞F (ω) ejωt dω. (4.10)

Note that the complex exponential in the inverse transform formula is the complex conjugate of

that in the transform formula. It is also important to note that these definitions are not unique;


some textbooks have the numerical factor, 1/√2π, multiplying both of the integrals in (4.9) and

(4.10). This is perfectly fine, except that one then carries the√2π through what could be a lot of

analysis, and therefore I feel that the above is cleaner to deal with. Fortunately most undergraduate

textbooks adopt the same policy.

Note also that we didn’t have a formula for the inverse Laplace Transform in year 1, but there is

one. Unfortuately this turns out to be a line integral in the complex plane. There are rules for

dealing with this and some of these are quite powerful. However we have avoided these mattersand will continue to do so in the present unit.

If we see F [1/(t2 + 1)] written down, then we say “Fourier Transform of 1/(t2 + 1).”

4.4. Some examples of Fourier Transforms.

Example 1: Find the Fourier Transform of e−|t|.

Note that this function decays as t → ∞ and as t → −∞, and that it is an even function. On

applying the definition of the Fourier Transform we have,

F [e−|t|] =

∫ ∞

−∞e−|t|e−jωt dt

=

∫ ∞

−∞e−|t| cosωt dt odd integrates to zero

= 2

∫ ∞

0

e−t cosωt dt symmetry and e−|t| = e−t when t > 0

=2

1 + ω2(4.11)

Note that since e−|t| is even, the Fourier Transform is real.

Example 2: Find the Fourier Transform of the unit pulse, f(t), given by

f(t) =

1

0

0 < t < 1,

otherwise.

F [f(t)] =

∫ ∞

−∞f(t)e−jωt dt

=

∫ 1

0

1×[

cosωt− j sinωt]

dt

=sinω

ω+ j

(cosω − 1

ω

)

. (4.12)

Note that, since f(t) is neither even nor odd, the Fourier Transform is complex.


Example 3: Find the Fourier Transform of the unit impulse based at t = a, namely, δ(t− a).

We’ll need to use the following property of integrals which involve the unit impulse:

∫ ∞

−∞δ(t− a) g(t) dt = g(a). (4.13)

Therefore,

F [δ(t − a)] =

∫ ∞

−∞δ(t − a) e−jωt dt = e−jωa. (4.14)

When a = 0, then we obtain,

F [δ(t)] = 1. (4.15)

Bearing in mind that the Fourier Transform represents the frequency content of the given function,

this result tells us that the unit impulse contains an equal amount of all frequencies. No doubt thisis why we hit TVs, cars, PCs and washing machines when they don’t work!

4.5. Existence of the Fourier Transform.

It has already been mentioned that the range of functions for which the Fourier Transform exists

is much more limited than for the Laplace Transform. For example, if we were to attempt to findthe Fourier Transform of f(t) = t, then we would obtain,

F [t] =

∫ ∞

−∞t e−jωt dt =

[

t][e−jωt

−jω

]∞

−∞−

[

1][e−jωt

j2ω2

]∞

−∞, (4.16)

but it is impossible to evaluate the limits. The presence of the complex exponentials means thattheir values keep changing from purely real to purely imaginary and back as t increases. Further,

the first term contains t, and therefore the magnitude of this oscillation continues to increase aswell. Therefore we say that the function, t, does not have a Fourier Transform.

Clearly it would be useful to have a set of rules whereby we can see almost at a glance if a functionhas a Fourier Transform or not — unfortunately such a set of rules will not help us to evaluate a

transform if it were to exist! Such a set of rules are as follows.

For the Fourier Transform of f(t) to exist, it is sufficient that f(t) satisfies the following:

(i) f(t) decays to zero as t → ±∞;

(ii) f(t) must be finite everywhere (i.e. it does not have poles/singularities).

The following is a Table of example functions.


f(t) FT? comment

e−t2 yes satisfies (i) and (ii)

e−|t| yes satisfies (i) and (ii)

(t2 + 1)−1 yes satisfies (i) and (ii)

e−t no violates (i) as t → −∞t−2 no violates (ii) at t = 0

t no violates (i) as t → −∞

1 yes∗ violates (i), but the FT exists

cos at yes∗ violates (i), but the FT exists

|t|−1/2 yes∗ violates (ii) but the FT exists

cos(t2) yes∗ violates (i) but the FT exists

In this Table I have quoted some of the more popular violations of our two rules (indicated by

asterisks in the Table) , but the rules are only meant to be sufficient; they are not necessary.

This means that, if they are satisfied, then the function very definitely has a Fourier Transform,but there are, nevertheless, cases of ‘mild’ violations, so to speak, where the functions also have

transforms. Generally these latter transforms tends to be strange, such as F [1] = 2πδ(ω), which is

a unit impulse in frequency.

4.6. Some useful results for Fourier Transforms.

When I say, useful results, what I mean is that some of them are theorems. However, they are

indeed very useful.

4.6.1. Fourier transform of a derivative.

Given that,

F (ω) = F [f(t)] =

∫ ∞

−∞f(t)e−ωt dt, (4.17)

then the transform of f ′(t) is

F [f ′(t)] =

∫ ∞

−∞f ′(t)e−jωt dt

=[

f(t)e−jωt]∞

−∞−

∫ ∞

−∞f(t)[−jωe−jωt] dt, (4.18)

on integrating by parts, and hence

F [f ′(t)] = jωF [f(t)] = jω

∫ ∞

−∞f(t)e−jωt dt = jωF (ω) (4.19)

provided that f −→ 0 as t −→ ±∞.


4.6.2. Fourier transforms of higher derivatives.

Similarly, one may use two integrations by parts to prove that

F [f ′′(t)] = (jω)2F (ω), (4.20)

provided that both f −→ 0 and f ′ −→ 0 as t −→ ±∞. Or, if one were being particularly lazy (orclever!), one could use Eq. (4.19) recursively:

F [f ′′(t)] = (jω)F [f ′(t)] = (jω)2F [f(t)] = −ω2F (ω). (4.21)

This may be extended easily to higher derivatives: for the third derivative we have,

F [f ′′′(t)] = (jω)3F (ω), (4.22)

provided that f, f ′, f ′′ −→ 0 as t −→ ±∞, and so on.

4.6.3. Shift theorem in t.

This result gives the effect of a time–shift in the function f(t) on its Fourier transform.

F [f(t− t0)] =

∫ ∞

−∞f(t− t0)e

−jωt dt,

by definition. Now we change variables from t to τ using τ = t− t0. Hence

F [f(t− t0)] =

∫ ∞

−∞f(τ)e−jω(τ+t0) dτ,

=e−jωt0

∫ ∞

−∞f(τ)e−jωτ dτ

=e−jωt0F (ω). (4.23)

So, for example, if we wish to find the function whose Fourier transform is

2e3ωj

1 + ω2

we first use the fact that

F [e−|t|] =2

1 + ω2

and then apply the theorem with t0 = −3:

F [e−|t+3|] =2e3ωj

1 + ω2.

4.6.4. Shift theorem in ω.

This result shows the effect of a frequency change in the transformed function.


F [f(t)ejat] =

∫ ∞

−∞f(t)ejate−jωt dt

=

∫ ∞

−∞f(t)e−j(ω−a)t dt

=F (ω − a). (4.24)

Given that F [e−|t|] is 2/(1 + ω2), this theorem gives the result that,

F [e−|t|ejat] =2

1 + (ω − a)2.

4.6.5. The Symmetry Theorem.

This theorem is motivated by the question, “If F [f(t)] = F (ω), then what is F [F (t)]? ”, and usesthe fact that the definitions of the Fourier transform and its inverse are very similar in form. An

example problem is the following: given that we know F [e−|t|] = 2/(1+ω2), what is F [2/(1+ t2)]?

By definition,

F [f(t)] =

∫ ∞

−∞f(t)e−jωt dt = F (ω) (4.25)

and we want to find

F [F (t)] =

∫ ∞

−∞F (t)e−jωt dt. (4.26)

NOTE that the following derivation is for interest only.

The inverse Fourier transform corresponding to (4.25) is

f(t) = F−1[F (ω)] =1

2π

∫ ∞

−∞F (ω)ejωt dω. (4.27)

It is important to notce that the argument to the exponential in (4.27) has the opposite sign to

what we need in (4.26) and that the variables, t and ω, are in the wrong places. These problemscan be overcome by a series of shifty manoeuvres. The first thing we must do is to interchange the

roles of t and ω; this is realised by first changing the (dummy) variable of integration from ω to s

to get

f(t) =1

2π

∫ ∞

−∞F (s)ejst ds.

Now rewrite this expression in terms of ω instead of t:

f(ω) =1

2π

∫ ∞

−∞F (s)ejsω ds,

and change the dummy variable from s to t:


f(ω) =1

2π

∫ ∞

−∞F (t)ejtω dt.

Finally we replace every occurrence of ω by −ω:

f(−ω) =1

2π

∫ ∞

−∞F (t)e−jtω dt,

and hence

F [F (t)] =

∫ ∞

−∞F (t)e−jωt dt = 2πf(−ω). (4.28)

Example 1: For the function given above we have, therefore,

F[ 2

1 + t2

]

= 2πe−|−ω| = 2πe−|ω|. (4.29)

Example 2: Given that F [δ(t)] = 1, then the Symmetry Theorem yields,

F [1] = 2πδ(−ω) = 2πδ(ω), (4.30)

which is one of the examples of a function which does not satisfy the sufficient conditions for the

existence of the Fourier Transform.

Example 3: We’ll try to use the Symmetry Theorem to find the Fourier Transform of cos at,

another mild violator of the rules of existence of the Fourier Transform.

Equation (4.14) shows that,

F [δ(t− a)] = e−jωa,

and therefore

F [δ(t+ a)] = ejωa

is also true. We may now add these two results together (which is a legitimate operation because

we at adding two integrals over the same range of integration) to give,

F [δ(t − a) + δ(t+ a)] = ejωa + e−jωa = 2cosωa. (4.31)

We now apply the Symmetry Theorem and divide the result by 2 to get,

F [cos at] = π[

δ(ω − a) + δ(ω + a)]

. (4.32)

The physical interpretation of this is that the frequency content of cos at is concentrated solely atω = ±a, which is why the unit impulses are present on the right hand side.


4.6.6. The Convolution Theorem.

The following derivation is also for interest only, but the result is applied very frequently when

solving ODEs and PDEs.

If f(t) and g(t) are given functions, both of which have Fourier transforms, then their convolutionis defined to be

f ∗ g =

∫ ∞

−∞f(τ)g(t− τ) dτ or

∫ ∞

−∞f(t− τ)g(τ) dτ, (4.33)

and we are interested in the Fourier transform of this function. Note that this is a different

definition from that which is used in Laplace Transforms, where the integral runs from 0 to t only.

By definition,

F [f ∗ g] =∫ ∞

−∞e−jωt

[

∫ ∞

−∞f(τ)g(t− τ) dτ

]

dt

=

∫ ∞

−∞

∫ ∞

−∞e−jωtf(τ)g(t− τ) dτ dt

=

∫ ∞

−∞

∫ ∞

−∞e−jω(t−τ)e−jωτf(τ)g(t− τ) dt dτ, (4.34)

on changing the order of integration and modifying the form of the integrand. Now change variables

from t to s according to s = t− τ (ds = dt) and hence

F [f ∗ g] =∫ ∞

−∞

∫ ∞

−∞g(s)e−jωsf(τ)e−jωτ ds dτ,

=[

∫ ∞

−∞g(s)e−jωs ds

][

∫ ∞

−∞f(τ)e−jωτ dτ

]

=G(ω)F (ω). (4.35)

The Fourier transform of the convolution of two functions is, therefore, the product of their respec-tive transforms. Examples of the use of this will be given in the lecture notes when we tackle the

solution of PDEs. But, for now, here is an example of the evaluation of a convolution.

Example: Find H(t) ∗H(t) where H(t) is the unit step-function.

Recall that H(t) = 1 when t > 0 and H(t) = 0 when t < 0. On applying the definition of the

convolution, we have

H(t) ∗H(t) =

∫ ∞

−∞H(τ) ∗H(t− τ) dτ. (4.36)

This integral may be evaluated by first determining the ranges over which the step functions in the

integrand are equal to 1. For H(τ) this range is τ > 0. For H(t − τ) the range is t − τ > 0, or,

equivalently, τ < t.


Figure 4.3a. Showing how to deal with the convolution of theunit step function with itself when t > 0.

τ

H(τ)H(t− τ)

t

1

t > 0

Figure 4.3b. Showing how to deal with the convolution of theunit step function with itself when t < 0.

τ

H(τ)H(t− τ) 1

t

t < 0

Therefore these two ranges have an overlap only when t > 0, and so the range of values of τ forwhich both step functions are equal to 1 is 0 < τ < t. Therefore

∫ ∞

−∞H(τ) ∗H(t− τ) dτ =

∫ t

0

1 dτ = t. (4.37)

But when t is negative, then there is no overlap, and therefore the integrand is zero for all valuesof t. Consequently, the integral is also zero. Therefore we have,

H(t) ∗H(t) =

t when t > 0,

0 when t < 0,

= tH(t). (4.38)

4.6.7. Linearity.

F [f(t) + g(t)] = F (ω) +G(ω), (4.39)

F [af(t)] = aF (ω). (4.40)

These results follow almost immediately from the definitions of the Fourier Transform.


4.7. A solution of an ordinary differential equation.

This purpose of this subsection is simply to show how Fourier Transforms may simplify the solution

of a differnential equation. For simplicity (and this will not be examinable for our chief purpose

is the solution of partial differential equations), we will solve an ordinary differential equation,although the true aim of the unit will be on the solutions of partial differential equations.

First, we will consider briefly the Fourier Transform of the function,

f(t) =

e−at when t > 0,

0 when t < 0,(4.41)

where a is taken to be positive. We have

F [f(t)] =

∫ ∞

−∞f(t) e−jωt dt by definition

=

∫ ∞

0

e−at e−jωt dt note lower limit

=

∫ ∞

0

e−(a+jω)t dt

=1

a+ jω. (4.42)

Now we are in a position to solve the ordinary differential equation,

dy

dt+ 2y = e−t. (4.43)

We do this by taking the Fourier Transform of every term in the equation. On using the results ofEqs. (4.19) and (4.43), we get,

(jω)Y + 2Y =1

1 + jω, (4.44)

where we have assumed that F [y(t)] may be written as Y (ω). The left hand side may be rewritten

in the following way, and the analysis proceeds as follows:

(2 + jω)Y =1

1 + jω

⇒ Y =1

(2 + jω)(1 + jω)

⇒ Y =1

(1 + jω)− 1

(2 + jω)partial fractions

⇒ y =

e−t − e−2t when t > 0,

0 when t < 0,(4.45)


This last line is obtained by taking the inverse Fourier Transform of the previous line, and then

using the result of Eq. (4.42) with a = 1 and a = 2. It may be a bit of surprise that one canemploy the method of partial fractions here! Those who have already met the solution of ordinary

differential equations using Laplace Transforms may notice how very similar this analysis was.

4.8. Fourier Transform solutions of PDEs.

4.8.1. Solutions of Fourier’s equation.

We will solve Fourier’s equation,

∂θ

∂t= α

∂2θ

∂x2, (4.46)

subject to the initial condition that

θ = f(x) at t = 0. (4.47)

We have to assume that the initial temperature profile, f(x), must decay to zero as x → ±∞ inorder to be able to use Fourier Transforms. The solution procedure follows three steps: (i) take the

Fourier Transform of the given equation, (ii) solve the transformed equation, (iii) take the inverseFourier Transform to find the desired solution. In many cases the final answer has to be written in

terms of an integral, but sometimes one may be able to find an explicit expression for the solution.

First we let Θ(ω, t) be the Fourier Transform of θ(x, t) with respect to x. Thus,

Θ =

∫ ∞

−∞θ e−jωx dx. (4.48)

Now we’ll take the Fourier Transform of the second derivative:

F[ ∂2θ

∂x2

]

=

∫ ∞

−∞

∂2θ

∂x2e−jωx dx

=[∂θ

∂x

][

e−jωx]∞

−∞−

[

θ][

−jωe−jωx]∞

−∞+

∫ ∞

−∞

[

θ][

−ω2e−jωx]

= −ω2Θ. (4.49)

Note that we have used the facts that both θ and its x-derivative tend to zero as x → ±∞.

The Fourier Transform of the time-derivative term is relatively straightforward:

F[∂θ

∂t

]

=

∫ ∞

−∞

∂θ

∂te−jωx dx

=∂

∂t

∫ ∞

−∞θ e−jωx dx

=∂Θ

∂t. (4.50)

Note that we have effectively swapped the order of differentiation with respect to t and integration

with respect to x when deriving this result.


Now Fourier’s equation, (4.46), is transformed to

∂Θ

∂t= −αω2Θ. (4.51)

This ordinary differential equation has the solution,

Θ = A(ω)e−αω2t, (4.52)

where the ‘constant of integration’ may be treated as a function of ω. While this treatment mightseem strange, the ordinary differential equation has t as its independent variable, and therefore ω

is simply a passive parameter. Therefore it makes sense to ensure that A is as general as possibleby allowing it to vary with ω.

The constant of integration may be found by applying an appropriate initial condition, but the only

one which is available at present is Eq. (4.47), for θ. However, we may take the Fourier Transformof this to obtain,

Θ = F [f(x)] = F (ω) at t = 0. (4.53)

Therefore the setting of t = 0 in Eq. (4.52) yields,

A = F, (4.54)

and so the Fourier Transform of the desired solution is,

Θ = F (ω)e−αω2t. (4.55)

We may now apply the formula for the inverse Fourier Transform (see Eq. (4.10)) to obtain the

final solution,

θ(x, t) =1

2π

∫ ∞

−∞F (ω)e−αω2tejωx dω. (4.56)

It is difficult to simplify this final integral in any meaningful way, and therefore we have to leave it

as it is. Clearly, for any chosen pair of values of x and t the integral may be evaluated numerically

with ease and to any desired accuracy. If a sufficient number of pairs of x and t are used it becomespossible to obtain either a contour plot of the evolution of θ in space and time, or else to show how

the temperature profile varies with time.

4.8.2. Solution of Laplace’s equation.

We now solve,

∂2u

∂x2+

∂2u

∂y2= 0, (4.57)

in the ranges −∞ < x < ∞ and 0 ≤ y < ∞. The boundary conditions are that,

u = f(x) on y = 0 and u → 0 as y → ∞. (4.58)

This system represents steady two-dimensional conduction in a half-plane where the temperature

profile on the edge is f(x).


Once more it is assumed that u → 0 as x → ±∞ so that it is possible to take Fourier Transforms

with respect to x. We now use U(ω, y) to denote the Fourier Transform of u(x, y) with respect tox. Given our experience for Fourier’s equation, it is clear that Laplace’s equation transforms to

∂2U

∂y2− ω2U = 0. (4.59)

The solution of this equation may be written as,

U = A(ω)eωy +B(ω)e−ωy. (4.60)

We may now apply the large-y boundary condition, namely that u → 0 or, equivalently, thatU → 0. Although this might seem straightforward to do (and most people would immediately say

that A = 0 in order to remove the exponentially growing solution) it is not quite as straightforward

as one might think. The reason is that ω may also take negative values. So when ω > 0 we needA = 0 so that U = B(ω)e−ωy, and when ω < 0 we need B = 0 so that U = A(ω)eωy .

We may combine these two cases into one formula in the following way. Let,

U = C(ω)e−|ω|y, (4.61)

where C is the arbitrary constant. We may now apply the boundary condition at y = 0, namely

that u = f(x) or that U = F (ω). Hence C = F , and we obtain the transformed solution,

U(ω, y) = F (ω) e−|ω|y. (4.62)

It is easy to write down the inverse Fourier Transform of this:

u(x, y) =1

2π

∫ ∞

−∞F (ω) e−|ω|y ejωx dω. (4.63)

Now Eq. (4.62) is the product of two functions of ω, and therefore it is possible, in principle, to

apply the Convolution Theorem to find its inverse transform. Earlier in the notes (see Eq. (4.29))we used the Symmetry Theorem to show that,

F[ 2

1 + x2

]

= 2πe−|ω|,

where it is now assumed that Fourier Transforms are taking place with respect to x, rather than t.

In a similar way it is also possible to show that,

F[ 2a

a2 + x2

]

= 2πe−|ω|a, (4.64)

where a is a constant. For a Fourier Transform with respect to x, we might use y as an alternative

constant to a to obtain,

F[ 2y

y2 + x2

]

= 2πe−|ω|y, (4.65)

and hence,

F[ 1

π× y

y2 + x2

]

= e−|ω|y. (4.66)

So the application of the Convolution Theorem to Eq. (4.62) yields,


u = f(x) ∗[ 1

π

y

y2 + x2

]

=

∫ ∞

−∞f(ξ − x)

[ 1

π× y

y2 + ξ2

]

dξ. (4.67)

Clearly this provides an alternative integral expression for the final solution to the one in Eq. (4.62).

Equation (4.67) may be used very easily if the initial condition is a unit impulse, f(x) = δ(x). The

integral in (4.67) now becomes,

u =1

π× y

y2 + x2. (4.68)

Of course, it is worth checking whether this simple-looking solution is reasonable. On y = 0 the

function is clearly zero, which is consistent with the delta function boundary condition, except

perhaps at the origin itself. As we move away from the origin along a straight line (where we mayset x = r cos θ and y = r sin θ and therefore we are moving along the line θ =constant and r → ∞),

then u is inversely proportional to the distance from the origin, i.e. it decays, which is also what we

expect. Finally, if we set x = 0, then u = 1/y, which becomes infinite as we approach the origin,which is where we have sited the unit impulse. Indeed, we also get an infinite limit for all lines of

approach to the origin, except on y = 0. This, unfotunately, is one of the many odd things thatcan happen with a unit impulse.

Contours of u = constant are circular arcs, as may be seen by the following analysis. If we

substitute u = 1/(2πα) into Eq. (4.68), where α is a constant, and where 2π is there for purenumerical convenience, then we obtain,

y

π(x2 + y2)=

1

2πα⇒ x2 + y2 = 2αy ⇒ x2 + (y − α)2 = α2. (4.69)

This is the equation of a circle of radius α which is centred at x = 0 and y = α. So, bizarrely, and

because of the presence of the unit impulse, all these circular contours pass through the origin, asmay be seen in the following Figure.

Figure 4.4. Depicting the contours (i.e. isotherms) of the solu-

tion given by Eq. (4.68). The x-axis is horizontal, and the unitimpulse is placed at the origin, which is the point through which

all the circular contours pass.

x0


4.9. Fourier Sine and Cosine Transforms..

When solving partial differential equations in an infinite domain it is often necessary to use the

Fourier Transform. However, not all problems are defined on an infinite domain (i.e. in −∞ < x <∞), but some are defined on a semi–infinite domain (0 ≤ x < ∞). This where we need to use either

the Fourier Sine Transform (FST) or the Fourier Cosine Transform (FCT).

The FCT and the FST are intimately related to the Fourier Transform and they and their inverses

may be derived from it and its inverse. The Fourier Cosine Transform pair is given by

Fc(ω) = Fc[f(x)] =

∫ ∞

0

f(x) cosωxdx (4.70)

and

f(x) = F−1c [Fc(ω)] =

2

π

∫ ∞

0

Fc(ω) cosωxdω. (4.71)

The Fourier Sine Transform pair is given by

Fs(ω) = Fs[f(x)] =

∫ ∞

0

f(x) sinωxdx (4.72)

and

f(x) = F−1s [Fs(ω)] =

2

π

∫ ∞

0

Fs(ω) sinωxdω. (4.73)

If we are solving an equation for u(x, y), say, where transforms are being taken in the x-direction,then the FST is used when u(x, y) is given on the boundary x = 0, and that the FCT is used when

the first x-derivative of u(x, y) is given on x = 0. The reasons for this are technical, and they arisenaturally during the integration by parts process for evaluating the transforms of derivatives.

The similarity between the formulae for the Fourier Sine transform and its inverse, and betweenthe Fourier Cosine Transform and its inverse is even stronger than between the Fourier Transform

and its inverse, as given in Eqs. (4.9) and (4.10). The reason I have mentioned this is that the

equivalent Symmetry Theorems for the sine and cosine transforms become almost trivial to writedown. Thus if Fc[f(x)] = Fc(ω) then Fc[Fc(t)] =

12πf(ω), with a similar-looking formula for the

sine transform. So, for example, it is straightforward to show that,

Fc[e−x] =

∫ ∞

0

e−x cosωxdx =1

1 + ω2. (4.74)

The Symmetry Theorem then tells us that,

Fc

[ 1

1 + x2

]

= 12πe

−ω. (4.75)

4.10. Some Fourier Sine Transform examples..

4.10.1. Example 1.

We will consider the following unsteady one-dimensional heat transfer problem. A semi-infinite

solid, which occupies the region, 0 ≤ x < ∞, has the temperature profile, θ = f(x), at t = 0.However, the x = 0 end of this region is maintained at the temperature, θ = 0. Determine the

evolution of the temperature profile.


Let Θs(ω, t) be the Fourier Sine Transform of θ(x, t) with respect to x, i.e. that

Θs = Fs[θ] =

∫ ∞

0

θ sinωxdx.

We will be solving Fourier’s equation, as given by Eq. (4.46), and therefore we will need to take

the Fourier Sine Transform of this equation. Beginning with the time-derivative term, we get

Fs

[∂θ

∂t

]

=

∫ ∞

0

∂θ

∂tsinωxdx =

∂

∂t

∫ ∞

0

θ sinωxdx =∂Θs

∂t. (4.76)

For the term with the second derivative in x, we have

Fs

[ ∂2θ

∂x2

]

=

∫ ∞

0

∂2θ

∂x2sinωxdx

=[∂θ

∂x

][

sinωx]∞

0−

[

θ][

ω cosωx]∞

0+

∫ ∞

0

[

θ][

−ω2 sinωx]

dx (4.77)

= −ω2Θs. (4.78)

Note that we had to use the “θ = 0 at x = 0” boundary condition in (4.77) in order to derive (4.78).

Note also that, if the boundary condition at x = 0 had been a Neumann condition (the gradientof θ is specified) then we would not be able to proceed further because the value of θ at x = 0 is

needed there. Thus the Fourier Sine Transform must be used with Dirichlet boundary conditions.

We also assumed that θ and its derivatives decay to zero as x → ∞. Therefore Fourier’s equation

transforms to,

∂Θs

∂t= −αω2Θs, (4.79)


Θs = A(ω)e−αω2t, (4.80)

where A(ω) is the arbitrary constant (from the point of view of y). The given initial condition is

that θ = f(x) when t = 0; on taking the Fourier Sine Transform of this, we obtain the transformedversion, namely,

Θs = Fs[f(x)] = Fs(ω) when t = 0. (4.81)

Substitution of (4.81) into (4.80) yields A = Fs, and hence

Θs = Fs(ω)e−αω2t. (4.82)

The final solution is obtained by taking the inverse Fourier Sine Transform of this; we get,

θ = F−1s [Θs] =

2

π

∫ ∞

0

Fs(ω)e−αω2t sinωxdω. (4.83)

4.10.2. Example 2.

We will again solve Fourier’s equation, but this time the initial temperature profile is θ = 0, and

the boundary at x = 0 is held at the temperature θ = 1. This is an example of a sudden heating


problem, the final solution representing the manner in which heat diffuses into a formerly cold

domain.

Proceeding as for Example 1, we again get to the stage represented by Eq. (4.77), but now we need

to use the fact that θ = 1 at x = 0. Therefore the present equivalent of Eq. (4.78) is,

Fs

[ ∂2θ

∂x2

]

= ω − ω2Θs. (4.84)

In this case, Fourier’s equation transforms to,

∂Θs

∂t= α

[

ω − ω2Θs

]

, (4.85)


Θs =1

ω+A(ω)e−αω2t, (4.86)

At t = 0 we have θ = 0, and hence Θs = 0 too. Subsitution of this fact into Eq. (4.86) gives

A = −1/ω, and therefore,

Θs =1− e−αω2t

ω. (4.88)

The inverse Fourier Sine Transform yields,

θ =2

π

∫ ∞

0

1− e−αω2t

ωsinωxdω. (4.89)

As a final note on this example, there is another way in which this problem might be solved andit will be met in one of the units concerned with Heat Transfer. For now I will merely quote the

solution:

θ =2√π

∫ ∞

x/2√αt

e−ξ2 dξ = erfc[ x

2√αt

]

. (4.90)

This function is known as the complementary error function; you’ll meet it formally one day.

4.10.3. Example 3.

This third and final example is a reprise of Example 1, but now we shall assume that the x = 0boundary is insulated, i.e. that ∂θ/∂x = 0 there.

Let Θc(ω, t) be the Fourier Cosine Transform of θ(x, t) with respect to x, i.e. that

Θc = Fc[θ] =

∫ ∞

0

θ cosωxdx.

Again, we will be solving Fourier’s equation as given by Eq. (4.46), and we will take the Fourier

Cosine Transform of this equation. Beginning with the time-derivative term once more, we get

Fc

[∂θ

∂t

]

=

∫ ∞

0

∂θ

∂tcosωxdx =

∂

∂t

∫ ∞

0

θ cosωxdx =∂Θs

∂t. (4.91)


For the term with the second derivative in x, we have

Fc

[∂2θ

∂x2

]

=

∫ ∞

0

∂2θ

∂x2cosωxdx

=[∂θ

∂x

][

cosωx]∞

0−

[

θ][

−ω sinωx]∞

0+

∫ ∞

0

[

θ][

−ω2 cosωx]

dx (4.92)

= −ω2Θc. (4.93)

Note that we had to use the “∂θ/∂x = 0 at x = 0” boundary condition in (4.92) in order toderive (4.93). Note also that, if the boundary condition at x = 0 had been a Dirichlet condition

(where θ is specified) then we would not be able to proceed further because the value of ∂θ/∂x at

x = 0 is needed there. Thus the Fourier Cosine Transform must be used with Neumann boundaryconditions.

Fourier’s equation transforms to,

∂Θc

∂t= −αω2Θc, (4.94)


Θc = A(ω)e−αω2t. (4.95)

The given initial condition is that θ = f(x) when t = 0; on taking the Fourier Cosine Transform of

this, we obtain the transformed version, namely,

Θc = Fc[f(x)] = Fc(ω) when t = 0. (4.96)

Substitution of (4.96) into (4.95) yields A = Fs, and hence

Θc = Fc(ω)e−αω2t. (4.97)

The final solution is obtained by taking the inverse Fourier Cosine Transform of this; we get,

θ = F−1c [Θc] =

2

π

∫ ∞

0

Fc(ω)e−αω2t cosωxdω. (4.98)

4.11. A final comment.

The last three examples have involved solutions of only Fourier’s equation. It is also possible to

use the Fourier Sine and Cosine Transforms to solve Laplace’s equation and the wave equation, butthese feature in Problem Sheet 5 and fairly detailed solutions are provided for these.

1. outlineof the analytical solutionspart of the unit · the motion of a violin string is a good...

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