Powers D.L. Boundary Value Problems and Partial Differential Equations

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198 Chapter 2 The Heat Equation

u(0, t) = 0, 0 < t,

u(x, 0) = f (x), 0 < x,

can be expressed as

u(x, t) =

√

4πkt



∞

f (x



)



exp



−(x



−x )

4kt



−exp



−(x



+x)

4kt





Hint: Start from the problem of this section with initial condition

u(x, 0) = f

(x), −∞ < x < ∞,

where f

is the odd extension of f .ThenuseEq.(7),andsplittheinterval

of integration at 0.

5. Verify by differentiating that the function

u(x, t) =

√

4kπt

exp



−

4kt



is a solution of the heat equation

∂

∂x

∂u

∂t

, 0 < t, −∞ < x < ∞.

What can be said about u at x = 0? at t = 0+?Whatislim

t→0+

u(0, t)?

Sketch u(x, t) forvariousﬁxedvaluesoft.

6. Suppose that f (x) is an odd periodic function with period 2a. Show that

u(x, t) deﬁned by Eq. (7) also has these properties.

7. If f (x) = 1 for all x, the solution of our heat conduction problem is

u(x, t) = 1. Use this fact together with Eq. (7) to show that

1 =

√

4πkt



∞

−∞

exp



−(x



−x)

4kt





8. Solve the problem that follows using Eq. (7).

∂

∂x

∂u

∂t

, −∞< x < ∞, 0 < t,

u(x, 0) =



1, x > 0,

−1, x < 0.

9. Can Exercise 8 be solved in the form of Eq. (4)? Note that



∞

sin(λx)

dλ =



1, 0 < x,

−1, x < 0.

2.12 The Error Function 199

Figure 10 Graph of the error function erf(z) for −3 < z < 3.

2.12 The Error Function

InSection11wemadetransformationsofaFourierintegraltoobtaintheso-

lution of the heat problem

∂

∂x

∂u

∂t

, −∞ < x < ∞, 0 < t, (1)

u(x, 0) = f (x), −∞ < x < ∞, (2)

in the form of a single integral,

u(x, t) =

√

4πkt



∞

−∞

f (x



−(x−x



)

/4kt



. (3)

Even for the simplest functions f , this integration cannot be carried out in

closed form, mainly because the indeﬁnite integral



−x

dx is not an elemen-

tary function. We can improve our understanding of the solution Eq. (3) if we

introduce the error function,deﬁnedas

erf(z) =

√



−y

dy. (4)

Agraphoferf(z) is shown in Fig. 10. Convenient tables, together with approx-

imations to the error function, will be found in Handbook of Mathematical

Functions, by Abramowitz and Stegun.

Several important properties of the error function follow immediately from

the deﬁnition. First, it is clear that erf(0) = 0, and it is easy to show that erf is

an odd function (Exercise 1). Second, by the fundamental theorem of calculus,

the derivative of the error function is

erf(z) =

√

−z

. (5)

200 Chapter 2 The Heat Equation

And ﬁnally, the error function supplies the integral



−y

dy =

√



erf(b) −erf(a)



. (6)

The reason for the choice of the constant in front of the integral in Eq. (4) is

to make

lim

z→∞

erf(z) =1. (7)

To see that this is true, deﬁne

A =



∞

−y

dy.

We are going to show that A =

√

π/2. First write A

as the product of two

integrals,



∞

−y



∞

−x

dx.

Remember that the name of the variable of integration in a deﬁnite integral

is immaterial. This expression for A

can be interpreted as an iterated double

integral over the ﬁrst quadrant of the x,y-plane, equivalent to



∞



∞

−(x

)

dx dy.

Now change to polar coordinates. The ﬁrst quadrant is described by the in-

equalities 0 < r < ∞,0<θ <π/2, and the element of area in polar coordi-

nates is rdrdθ.Thus,wehave



π/2



∞

−r

rdrdθ. (8)

This integral, which can be evaluated by elementary means (see Exercise 2),

has value π/4. Hence A =

√

π/2 and Eq. (7) is validated.

Many workers also use the complementary error function,erfc(z),deﬁnedas

erfc(z) =

√



∞

−y

dy. (9)

By using Eq. (7) we obtain the identity

erfc(z) =1 −erf(z). (10)

Some properties of the complementary error function are found in Exercise 3.

2.12 The Error Function 201

We are interested in the error function because of its role in solving the heat

equation. First we shall show that the solution of the problem

∂

∂x

∂u

∂t

, −∞< x < ∞, 0 < t, (11)

u(x, 0) = sgn(x), −∞ < x < ∞ (12)

is u(x, t) = erf(x/

√

4kt).(Recallthatsgn(x) has the value −1ifx is negative or

+1ifx is positive.) The easy way to prove this statement is to verify it directly.

(See Exercises 4 and 5.) Here, we shall arrive at the same conclusion, starting

from Eq. (3),

u(x, t) =

√

4πkt



∞

−∞

sgn(x



−(x−x



)

/4kt



. (13)

First, change the variable of integration to y = (x



− x)/

√

4kt.Thendy =



√

4kt,and

u(x, t) =

√



∞

−∞

sgn



x +y

√

4kt



−y

dy. (14)

Now the function e

−y

is even, and the sgn function changes from −1to+1at

y =−x/

√

4kt. Thus, the integrand of Eq. (14) is as shown in Fig. 11. The tail

to the left of −x/

√

4kt has the same area as the tail to the right of x/

√

4kt but

opposite sign. These two areas cancel, leaving

u(x, t) =

√



√

4kt

−x/

√

4kt

−y

dy.

Finally, use the symmetry of the integrand to halve the interval of integra-

tion and double the result:

u(x, t) =

√



√

4kt

−y

dy =erf



√

4kt



This is the result we wanted to arrive at. Figure 12 shows graphs of u(x, t) =

erf(x/

√

4kt) as a function of x for several values of kt.

Because erf(0) = 0, the function u(x, t) = erf(x/

√

4kt) must also be the so-

lution of the problem

∂

∂x

∂u

∂t

, 0 < x, 0 < t,

u(0, t) = 0, 0 < t,

u(x, 0) = 1, 0 < x.

202 Chapter 2 The Heat Equation

(a)

(b)

Figure 11 (a) Graph of exp(−y

) and (b) graph of sgn(x + y

√

4kt) exp(−y

The tails beyond ±x/

√

4kt have the same areas, with opposite signs.

Figure 12 Graphs of the solution of the problem in Eqs. (11) and (12),

u(x, t) = erf(x/

√

4kt),forx in the range −2to2andforkt =0.01, 0.1, 1, and 10.

As kt increases, the graph of u(x, t) collapses toward the x-axis.

A simple modiﬁcation leads to the conclusion that the complementary error

function, u(x, t) = erfc(x/

√

4kt) is the solution of this problem with zero ini-

tial condition and constant boundary condition,

∂

∂x

∂u

∂t

, 0 < x, 0 < t,

u(0, t) = 1, 0 < t,

u(x, 0) = 0, 0 < x.

2.12 The Error Function 203

EXERCISES

1. Show that erf(−z) =−erf(z),thatis,thaterfisanoddfunction.

2. Carry out the integration indicated in Eq. (8).

3. Verify these properties of the complementary error function:

erfc(z) =−e

−z

√

;

b. erfc(0) =1;

c. lim

z→∞

erfc(z) =0;

d. lim

z→−∞

erfc(z) =2;

e. erfc(z) is neither even nor odd.

4. Verify by differentiating that u(x, t) = erf(x/

√

4kt) satisﬁes the heat equa-

tion (1).

5. Ve r i f y t h at u(x, t) = erf(x/

√

4kt) satisﬁes the initial condition

u(x, t) =



1, 0 < x,

−1, x < 0.

6. In probability and statistics, the normal,orGaussian, probability density

function is deﬁned as

f (z) =

√

2π

−z

, −∞< z < ∞,

and the cumulative distribution function is

(x) =



−∞

f (z) dz.

Show that the cumulative distribution function and the error function are

related by (x) = [1 +erf(x/

√

2)]/2.

7. Express this integral in terms of the error function:

I(x) =



−x

√

dx.

8. Use error functions to solve the problem

∂

∂x

∂u

∂t

, 0 < x, 0 < t,

204 Chapter 2 The Heat Equation

u(0, t) = U

, 0 < t,

u(x, 0) = U

, 0 < x.

(Hint: What conditions does u(x, t) −U

satisfy?)

9. Assuming that U

< 0andU

> 0, the problem in Exercise 8 might be

interpreted as representing the temperature in a freezing lake. (Think of

x as measuring depth from the surface.) Deﬁne x(t) as the depth of the

ice–water interface; then u(x(t), t) = 0. Now ﬁnd x(t) explicitly.

10. Use error functions to solve the problem

∂

∂x

∂u

∂t

, −∞ < x < ∞, 0 < t,

u(x, 0) = f (x), −∞ < x < ∞,

where f (x) = U

for x < 0andf (x) = U

for x > 0.

2.13 Comments and References

In about 1810, Fourier made an intensive study of heat conduction problems,

in which he used the product method of solution and developed the idea of

Fourier series. Sturm and Liouville made their clear and simple generalization

of Fourier series in the 1830s. Among modern works, Conduction of Heat in

Solids, by Carslaw and Jaeger, is the standard reference. The Mathematics of

Diffusion,byCrank,andThe Heat Equation, by D.V. Widder, are also useful

references. (See the Bibliography.)

Although we have motivated our study in terms of heat conduction and,

to a lesser extent, by diffusion, many other physical phenomena of interest in

engineering are described by the heat/diffusion equation: for example, voltage

and current in an inductance-free cable and vorticity transport in ﬂuid ﬂow.

The heat/diffusion equation and allied equations are being employed in biol-

ogy to model cell physiology, chemical reactions, nerve impulses, the spread of

populations, and many other phenomena. Two good references are Differen-

tial Equations and Mathematical Biology, by D.S. Jones and B.D. Sleeman, and

Mathematical Biology,byJ.D.Murray.

The diffusion equation also turns up in some classical problems of probabil-

ity theory, especially the description of Brownian motion. Suppose a particle

moves exactly one step of length x in each time interval t.Thestepmay

be either to the left or to the right, each equally likely. Let u

(m) denote the

probability that, at time m t, the particle is at point i x (m = 0, 1, 2,...,

i = 0, ±1, ±2,...). In order to arrive at point i x at time (m + 1)t,the

particle must have been at one of the adjacent points (i ±1)x at the preced-

ing time m t and must have moved toward i x. From this, we see that the

2.13 Comments and References 205

m −3 −2 −10123

00 0 0 1 0 0

10 0 0.500.50 0

20 0.25 0 0.50 0.25 0

30.125 0 0.375 0 0.375 0 0.125

Tab le 3 Random-walk probabilities

probabilities are related by the equation

(m +1) =

i−1

(m) +

i+1

(m).

The u’s are completely determined once an initial probability distribution is

given. For instance, if the particle is initially at point zero (u

(0) = 1, u

(0) = 0,

for i = 0), the u

(m) are formed by successive applications of the difference

equation, as shown in Table 3.

The equation may be transformed into a close relative of the heat equation.

First, subtract u

(m) from both sides:

(m +1) −u

(m) =



i+1

(m) −2u

(m) +u

i−1

(m)



Next divide by x

/2ontherightandbyt ·(x

/2 t) on the left to obtain

(m +1) −u

(m)

t

2 t

(x)

i+1

(m) −2u

(m) +u

i−1

(m)

(x)

If both the time interval t and the step length x are small, we may think

of u

(m) as being the value of a continuous function u(x, t) at x = i x, t =

m t. In the limit, the difference quotient on the left approaches ∂u/∂t.The

right-hand side, being a difference of differences, approaches ∂

u/∂x

.The

heat equation thus results if, in the simultaneous limit as x and t tend

to zero, the quantity 2 t/(x)

approaches a ﬁnite, nonzero limit. In this

context, the heat equation is called the Fokker–Planck equat ion.Moredetails

and references may be found in Feller, Introduction to Probability Theory and

Its Applications.

We have us ed th e t erm linear partial differential equation several times. The

most general such equation, of second order in two independent variables, can

be put in the form

∂

∂x

∂

∂x∂t

∂

∂t

∂u

∂x

∂u

∂t

+Fu +G =0,

where A, B,...,G are known — perhaps functions of x and t but not of u or

its derivatives. If G is identically zero, the equation is homogeneous. Of course,

206 Chapter 2 The Heat Equation

the ordinary heat equation has this form if we take A = 1, E =−1/k and all

other coefﬁcients equal to zero.

Some astute students will have wondered why we should seek solutions in

product form. The simplest answer is that in many cases it works. A more

subtle rationale is that of seeking solutions that are geometrically similar func-

tions of x at different times. The idea of similarity — related to dimensional

analysis — has been most fruitful in the mechanics of ﬂuids.

Chapter Review

See the CD for Review Questions.

Miscellaneous Exercises

Also see Review Questions on the CD.

In Exercises 1–16, ﬁnd the steady-state solution, the associated eigenvalue

problem, and the complete solution for each problem.

∂

∂x

∂u

∂t

,0< x < a ,0< t,

u(0, t) = T

, u(a, t) = T

,0< t,

u(x, 0) = T

,0< x < a.

∂

∂x

−γ

u =

∂u

∂t

,0< x < a ,0< t,

u(0, t) = T

, u(a, t) = T

,0< t,

u(x, 0) = T

,0< x < a.

∂

∂x

+r =

∂u

∂t

,0< x < a,0< t,

u(0, t) = T

, u(a, t) = T

,0< t,

u(x, 0) = T

,0< x < a (r is constant).

∂

∂x

∂u

∂t

,0< x < a ,0< t,

u(0, t) = T

∂u

∂x

(a, t) = 0, 0 < t,

u(x, 0) =

,0< x < a.

∂

∂x

−γ

u =

∂u

∂t

,0< x < a ,0< t,

Miscellaneous Exercises 207

∂u

∂x

(0, t) = 0,

∂u

∂x

(a, t) = 0, 0 < t,

u(x, 0) =

,0< x < a.

∂

∂x

∂u

∂t

,0< x < a,0< t,

u(0, t) = 0, u(a, t) = T

,0< t,

u(x, 0) = 0, 0 < x < a.

∂

∂x

∂u

∂t

,0< x < a,0< t,

u(0, t) = T

, u(a, t) = T

u(x, 0) = T

,0< x < a .

∂

∂x

∂u

∂t

,0< x < a,0< t,

∂u

∂x

(0, t) =

T

∂u

∂x

(a, t) =

T

,0< t,

u(x, 0) = T

,0< x < a .

∂

∂x

∂u

∂t

,0< x < a,0< t,

u(0, t) = T

∂u

∂x

(a, t) = 0, 0 < t,

u(x, 0) = T

,0< x < a .

10.

∂

∂x

∂u

∂t

,0< x < a ,0< t,

∂u

∂x

(0, t) = 0,

∂u

∂x

(a, t) = 0, 0 < t,

u(x, 0) =







, 0 < x <

< x < a.

11.

∂

∂x

∂u

∂t

,0< x < ∞,0< t,

u(0, t) = T

,0< t,

u(x, 0) = T



1 −e

−αx



,0< x.

12.

∂

∂x

∂u

∂t

,0< x < ∞,0< t,