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

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138 Chapter 2 The Heat Equation

c ρκk =

ρc

Material



cal

◦



cal

scm

◦





Aluminum 0.21 2.70.48 0.83

Copper 0.094 8.90.92 1.1

Steel 0.11 7.80.11 0.13

Glass 0.15 2.60.0014 0.0036

Concrete 0.16 2.30.0041 0.011

Ice 0.48 0.92 0.004 0.009

Tab le 1 Typical values of constants

of u(x, t

∗

), with t

∗

a ﬁxed time. If a portion of the graph is shaped like U, J or

backwards J, the graph is concave there — that is, ∂

u/∂x

is positive. Then by

the heat equation, ∂u/∂t must be positive as well. Vice versa, when the graph

is convex, ∂

u/∂x

and hence ∂u/∂t must be negative. Thus, a solution of the

heat equation tends to straighten out.

This equation alone is not enough information to completely specify the

temperature, however. Each of the functions

u(x, t) = x

+2kt,

u(x, t) = e

−kt

sin(x)

satisﬁes the partial differential equation, and so do their sum and difference.

Clearly this is not a satisfactory situation either from the mathematical or

physical viewpoint; we would like the temperature to be uniquely determined.

More conditions must be placed on the function u. The appropriate additional

conditions are those that describe the initial temperature distribution in the

rod and what is happening at the ends of the rod.

The initial condition is described mathematically as

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

where f (x) is a given function of x alone. In this way, we specify the initial

temperature at every point of the rod.

The boundar y conditions may take a variety of forms. First, the temperature

at either end may be held constant, for instance, by exposing the end to an ice-

water bath or to condensing steam. We can describe such conditions by the

equations

u(0, t) = T

, u(a, t) = T

, t > 0,

where T

and T

may be the same or different. More generally, the temperature

at the boundary may be controlled in some way, without being held constant.

If x

symbolizes an endpoint, the condition is

u(x

, t) = α(t), (6)

Chapter 2 The Heat Equation 139

where α is a function of time. Of course, the case of a constant function is

included here. This type of boundary condition is called a Dirichlet condition

or condition of the ﬁrst kind.

Another possibility is that the heat ﬂow rate is controlled. Since Fourier’s

law associates the heat ﬂow rate and the gradient of the temperature, we can

write

∂u

∂x

, t) = β(t), (7)

where β is a function of time. This is called a Neumann condition or condition

of the second kind.Wemostfrequentlytakeβ(t) to be identically zero. Then

the condition

∂u

∂x

, t) = 0

corresponds to an insulated surface, for this equation says that the heat ﬂow is

zero.

Still another possible boundary condition is

u(x

, t) +c

∂u

∂x

, t) = γ(t), (8)

called third kind or a Robin condition. This kind of condition can also be real-

ized physically. If the surface at x = a is exposed to air or other ﬂuid, then the

heat conducted up to that surface from inside the rod is carried away by con-

vection. Newton’s law of cooling says that the rate at which heat is transferred

from the body to the ﬂuid is proportional to the difference in temperature

between the body and the ﬂuid. In symbols, we have

q(a, t) = h



u(a, t) −T(t)



, (9)

where T(t) is the air temperature. After application of Fourier’s law, this be-

comes

−κ

∂u

∂x

(a, t) = hu(a, t) −hT(t). (10)

This equation can be put into the form of Eq. (8). (Note: h is called the con-

vection coefﬁcient or heat transfer coefﬁcient; [h] = H/L

tT.)

AlloftheboundaryconditionsgiveninEqs.(6),(7),and(8)involvethe

function u and/or its derivative at one point. If more than one point is in-

volved, the boundary condition is called mixed. For example, if a uniform rod

is bent into a ring and the ends x =0andx =a are joined, appropriate bound-

ary conditions would be

140 Chapter 2 The Heat Equation

u(0, t) = u(a, t), t > 0, (11)

∂u

∂x

(0, t) =

∂u

∂x

(a, t), t > 0, (12)

both of mixed type.

Many other kinds of boundary conditions exist and are even realizable, but

the four kinds already mentioned here are the most commonly encountered.

Animportantfeaturecommontoallfourtypesisthattheyinvolvealinear

operation on the function u.

The heat equation, an initial condition, and a boundary condition for each

endformwhatiscalledaninitial value–boundary value problem. For instance,

onepossibleproblemwouldbe

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t, (13)

u(0, t) = T

, 0 < t, (14)

−κ

∂u

∂x

(a, t) = h



u(a, t) −T



, 0 < t, (15)

u(x, 0) = f (x), 0 < x < a. (16)

Notice that the boundary conditions may be of different kinds at different

ends.

Although we shall not prove it, it is true that there is one, and only one,

solution to a complete initial value–boundary value problem.

We have derived the heat equation (4) as a mathematical model for the tem-

perature in a “rod,” suggesting an object that is much longer than it is wide.

The equation applies equally well to a “slab,” an object that is much wider than

it is thick. The important feature is that we may assume in either case that the

temperature varies in only one space direction (along the length of the rod

or the thickness of the slab). In Chapter 5, we derive a multidimensional heat

equation.

It may come as a surprise that the partial differential equations of this sec-

tion have another completely different but equally important physical inter-

pretation. Suppose that a static medium occupies a region of space between

x = 0andx = a (a slab!) and that we wish to study the concentration u,mea-

sured in units of mass per unit volume, of another substance, whose molecules

or atoms can move, or diffuse, through the medium. We assume that the con-

centration is a function of x and t only and designate q(x , t) to be the mass

ﬂux ([q] =m/tL

). Then the principle of conservation of mass may be applied

to a layer of the medium between x and x +x to obtain the equation

q(x, t) + xg= q(x +x, t) +x

∂u

∂t

(x, t). (17)

Chapter 2 The Heat Equation 141

When we rearrange Eq. (17) and take the limit as x approaches 0, it becomes

−

∂q

∂x

+g =

∂u

∂t

. (18)

In these equations, g is a “generation rate” ([g] = m/tL

),afunctionthatac-

counts for any gain or loss of the substance from the layer by means other than

movement in the x-direction. For example, the substance may participate in a

chemical reaction with the medium at a rate proportional to its concentration

(a ﬁrst-order reaction) so that in this case the generation rate is

g =−ku(x, t). (19)

The concentration and the mass ﬂux are linked by a phenomenological re-

lation called Fick’s ﬁrst law, written in one dimension as

q =−D

∂u

∂x

. (20)

In words, the diffusing substance moves toward regions of lower concentra-

tion at a rate proportional to the gradient of the concentration. The coefﬁcient

of proportionality D, usually constant, is called the diffusivity. By combining

Fick’s law with Eq. (18) arising from the conservation of mass, we obtain the

diffusion equation

∂

∂x

∂u

∂t

. (21)

At a boundary of the medium, the concentration of the diffusing substance

may be controlled, leading to a condition like Eq. (6), or the ﬂux of the sub-

stance may be controlled, leading via Fick’s law to a condition like Eq. (7); an

impermeable surface corresponds to zero ﬂux. If a boundary is covered with a

permeable ﬁlm, then the ﬂux through the ﬁlm is usually taken to be propor-

tional to the difference in concentrations on the two sides of the ﬁlm. Suppose

that the surface in question is at x =a. Then these statements may be expressed

symbolically as

q(a, t) = h



u(a, t) −C(t)



, (22)

where the proportionality constant h is called the ﬁlm coefﬁcient and C is the

concentration outside the medium. Using Fick’s law here leads to the equation

−D

∂u

∂x

(a, t) = hu(a, t) −hC(t). (23)

This equation is analogous to Eq. (10) and can be put into the form of Eq. (8).

142 Chapter 2 The Heat Equation

EXERCISES

1. Give a physical interpretation for the problem in Eqs. (13)–(16).

2. Verify that the following functions are solutions of the heat equation (5):

u(x, t) = exp



−λ



cos(λx),

u(x, t) = exp



−λ



sin(λx).

3. Suppose that the rod exchanges heat through the cylindrical surface by con-

vection with a surrounding ﬂuid at temperature U (constant). Newton’s law

of cooling says that the rate of heat transfer is proportional to exposed area

and temperature difference. What is g in Eq. (1)? What form does Eq. (4)

take?

4. Suppose that the end of the rod at x = 0 is immersed in an insulated con-

tainer of water or other ﬂuid; that the temperature of the ﬂuid is the same

as the temperature of the end of the rod; that the heat capacity of the ﬂuid

is C units of heat per degree. Show that this situation is represented math-

ematically by the equation

∂u

∂t

(0, t) = κA

∂u

∂x

(0, t),

where A is the cross-sectional area of the rod.

5. Put Eq. (10) into Eq. (8) form. Notice that the signs still indicate that heat

ﬂows in the direction of lower temperature. That is, if u(a, t)>T(t),then

q(a, t) is positive and the gradient of u is negative. Show that, if the surface

at x = 0 (left end) is exposed to convection, the boundary condition would

read

∂u

∂x

(0, t) = hu(0, t) −hT(t).

Explain the signs.

6. Suppose the surface at x = a is exposed to radiation. The Stefan–Boltzmann

law of radiation says that the rate of radiation heat transfer is proportional

to the difference of the fourth powers of the absolute temperatures of the

bodies:

q(a, t) = σ



(a, t) −T



Use this equation and Fourier’s law to obtain a boundary condition for

radiation at x = a to a body at temperature T.

7. The difference cited in Exercise 6 may be written

−T

=(u −T)



T +uT



2.2 Steady-State Temperatures 143

Under what conditions might the second factor on the right be taken ap-

proximately constant? If the factor were constant, the boundary condition

would be linear.

8. Interpret this problem in terms of diffusion. Be sure to explain how the

boundary conditions could arise physically.

∂

∂x

+K =

∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = C,

∂u

∂x

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

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

2.2 Steady-State Temperatures

Before tackling a complete heat conduction problem, we shall solve a simpli-

ﬁed version called the steady-state or equilibrium problem. We begin with this

problem:

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t, (1)

u(0, t) = T

, 0 < t, (2)

u(a, t) = T

, 0 < t, (3)

u(x, 0) = f (x), 0 < x < a. (4)

We m ay th ink of u(x, t) as the temperature in a cylindrical rod, with insulated

lateral surface, whose ends are held at constant temperatures T

and T

Experience indicates that after a long time under the same conditions, the

variation of temperature with time dies away. In terms of the function u(x, t)

that represents temperature, we thus expect that the limit of u(x, t),ast tends

to inﬁnity, exists and depends only on x,

lim

t→∞

u(x, t) = v(x),

and also that

lim

t→∞

∂u

∂t

=0.

The function v(x),calledthesteady-state temperature distribution,muststill

satisfy the boundary conditions and the heat equation, which are valid for all

t > 0.

144 Chapter 2 The Heat Equation

Example.

For the preceding problem, v(x) should be the solution to the problem

=0, 0 < x < a, (5)

v(0) = T

,v(a) = T

. (6)

On integrating the differential equation twice, we ﬁnd

=A,v(x) = Ax + B.

The constants A and B aretobechosensothatv(x) satisﬁes the boundary

conditions:

v(0) = B = T

,v(a) = Aa +B = T

When the two equations are solved for A and B, the steady-state distribution

becomes

v(x) = T

+(T

−T

)

. (7)

Of course, Eqs. (5) and (6), which together form the steady-state problem

corresponding to Eqs. (1)–(4), could have been derived from scratch, as was

done in Chapter 0, Section 3. Here, however, we see it as part of a more com-

prehensive problem.



We can establish this rule for setting up the steady-state problem corre-

sponding to a given heat conduction problem: Take limits in all equations that

are valid for large t (the partial differential equation and the boundary condi-

tions), replacing u and its derivatives with respect to x by v and its derivatives,

and replacing ∂u/∂t by 0.

Example.

Find the steady-state problem and solution of Eqs. (13)–(16) of Section 2.1,

which were

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t, (8)

u(0, t) = T

, 0 < t, (9)

−κ

∂u

∂x

(a, t) = h



u(a, t) −T



, 0 < t, (10)

u(x, 0) = f (x), 0 < x < a. (11)

2.2 Steady-State Temperatures 145

When the rule given here is applied to this problem, we are led to the following

equations:

=0, 0 < x < a,

v(0) = T

, −κv



(a) =h



v(a) −T



The solution of the differential equation is v(x) = A +Bx. The boundary con-

ditions require that A and B satisfy

v(0) = T

: A = T

−κv



(a) =h



v(a) −T



: −κB = h(A +Ba −T

Solving simultaneously, we ﬁnd

A =T

, B =

h(T

−T

)

κ +ha

Thus the steady-state solution of Eqs. (8)–(11) is

v(x) = T

xh(T

−T

)

κ +ha

. (12)



In both of these examples, the steady-state temperature distribution has

been uniquely determined by the differential equation and boundary condi-

tions. This is usually the case, but not always.

Example.

For the problem

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t, (13)

∂u

∂x

(0, t) = 0, 0 < t, (14)

∂u

∂x

(a, t) = 0, 0 < t, (15)

u(x, 0) = f (x), 0 < x < a, (16)

which describes the temperature in an insulated rod that also has insulated

ends, the corresponding steady-state problem for v(x) = lim

t→∞

u(x, t) is

=0, 0 < x < a,

(0) =0,

(a) =0.

146 Chapter 2 The Heat Equation

It is easy to see that v(x) = T (any constant) is a solution to this problem.

However, there is no information to tell what value T should take. Thus, this

boundary value problem has inﬁnitely many solutions.



It should not be supposed that every steady-state temperature distribution

has a straight-line graph. This is certainly not the case in the problem of Exer-

cise 1.

While the steady-state solution gives us some valuable information about

the solution of an initial value–boundary value problem, it also is important

as the ﬁrst step in ﬁnding the complete solution. We now isolate the “rest” of

the unknown temperature u(x, t) by deﬁning the transient temperature distri-

bution,

w(x, t) = u(x, t) − v(x).

The name transient is appropriate because, according to our assumptions

about the behavior of u for large values of t,weexpectw(x, t) to tend to zero

as t tends to inﬁnity.

In general, the transient also satisﬁes an initial value–boundary value prob-

lem that is similar to the original one but is distinguished by having a homo-

geneous partial differential equation and boundary conditions. To illustrate

this point, we shall treat the problem stated in Eqs. (1)–(4) whose steady-state

solution is given by Eq. (7).

By using the equality u(x, t) = w(x, t) +v(x) and what we know about v —

that is, Eqs. (5) and (6) — we make the original problem for u(x, t) into a new

problem for w(x, t),asshowninwhatfollows.

∂

∂x

∂

∂x

by rules of calculus,

∂

∂x

because of Eq. (5),

∂u

∂t

∂w

∂t

by rules of calculus,

∂w

∂t

v(x) does not depend on t,

∂

∂x

∂w

∂t

by substituting into Eq. (1),

u(0, t) = w(0, t) +v(0) by deﬁnition of the transient,

= w(0, t) +T

from Eqs. (2) and (6),

u(a, t) = w(a, t) +v(a) by deﬁnition of the transient,

= w(a, t) +T

from Eqs. (3) and (6),

u(x, 0) = w(x, 0) +v(x) by deﬁnition of the transient,

f (x) = w(x, 0) +T

+(T

−T

)x/a from Eqs. (4) and (7).

2.2 Steady-State Temperatures 147

Now we collect and simplify these transformations of Eqs. (1)–(4) to get an

initial value–boundary value problem for w:

∂

∂x

∂w

∂t

, 0 < x < a, 0 < t, (17)

w(0, t) = 0, 0 < t, (18)

w(a, t) = 0, 0 < t, (19)

w(x, 0) = f (x) −



+(T

−T

)



(20)

≡g(x), 0 < x < a. (21)

In the last line, we have just renamed the combination of f (x) and v(x) in

Eq. (20).

Inthenextsection,weshallseehowtheproblemforthetransienttemper-

ature can be solved. The mathematical purpose of setting up the steady-state

problem and then the transient problem is that the transient problem is ho-

mogeneous.Youcantestthisbytryingw(x, t) ≡ 0: This function satisﬁes the

partial differential equation (17) and the boundary conditions (18) and (19). It

is crucially important for the method we will develop to have a homogeneous

partial differential equation and boundary conditions.

EXERCISES

SeeextraexercisesontheCD.

1. State and solve the steady-state problem corresponding to

∂

∂x

−γ

(u −U) =

∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = T

, u(a, t) = T

, 0 < t,

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

Also ﬁnd a physical interpretation of this problem. (See Exercise 3, Sec-

tion 1.)

2. State the problem satisﬁed by the transient temperature distribution corre-

sponding to the problem in Exercise 1.

3. Obtain the steady-state solution of the problem

∂

∂x

+γ

(u −T) =

∂u

∂t

, 0 < x < a, 0 < t,

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

u(x, 0) = T

, 0 < x < a.