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

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188 Chapter 2 The Heat Equation

EXERCISES

1. Find the explicit form for v(x) in terms of the function in Eq. (8) assuming

a. α

=β

=0, c

=0;

b. α

> 0orβ

> 0, and no coefﬁcient negative.

Why are these two cases separate?

2. Justify each of the conclusions.

3. Derive the general form of u(x, t) if the boundary conditions are ∂u/∂x = 0

at both ends. In this case, λ

=0isaneigenvalue.

2.10 Semi-Inﬁnite Rod

Up to this point we have seen only problems over ﬁnite intervals. Frequently,

however, it is justiﬁable and useful to assume that an object is inﬁnite in length.

(Sometimes this assumption is used to disguise ignorance of a boundary con-

dition or to suppress the inﬂuence of a complicated condition.) Thus, if the rod

we have been studying is very long, we may treat it as semi-inﬁnite —thatis,as

extending from 0 to ∞. If properties are uniform and there is no “generation,”

the partial differential equation governing the temperature u(x, t) remains

∂

∂x

∂u

∂t

, 0 < x, 0 < t.

Let us suppose that at x = 0 the temperature is held constant, say, u(0, t) = 0

in some temperature scale. In the absence of another boundary, there is no

other boundary condition. However, it is desirable that u(x, t) remain ﬁnite —

less than some ﬁxed bound — as x →∞.

Thus, our mathematical model is

∂

∂x

∂u

∂t

, 0 < x < ∞, 0 < t, (1)

u(0, t) = 0, 0 < t, (2)

u(x, t) bounded as x →∞, (3)

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

The heat equation (1) and the boundary condition (2) are homogeneous.

The boundedness condition (3) is also homogeneous in an important way:

A (ﬁnite) sum of bounded functions is bounded. Thus, we can attack Eqs. (1)–

(3) by separation of variables. Assume that u(x, t) = φ(x)T(t),sothepartial

2.10 Semi-Inﬁnite Rod 189

differential equation can be separated into two ordinary equations as usual:



(x)

φ(x)



(t)

kT(t)

=const. (5)

There is just one boundary condition on u, which requires that φ(0) = 0. The

boundedness condition also requires that φ(x) remain ﬁnite as x →∞.Itis

easy to check (see Exercises) that a positive separation constant produces func-

tions φ(x) that cannot fulﬁll both the boundary and boundedness conditions

without being identically 0. Thus, we must choose a negative separation con-

stant, −λ

. The differential equation, together with the boundary and bound-

edness conditions, forms a singular eigenvalue problem (singular because of

the semi-inﬁnite interval),



+λ

φ = 0, 0 < x, (6)

φ(0) = 0,φ(x) bounded as x →∞. (7)

The general solution of the differential equation is

φ(x) = c

cos(λx) + c

sin(λx),

which is bounded for any choice of the constants and for any value of λ.The

boundedness condition told us to use a negative constant in Eq. (5) and now

contributes nothing further.

Applying the boundary condition at x = 0showsthatc

=0, leaving φ(x) =

sin(λx). In this singular eigenvalue problem, there are no “special” values

of λ: Any value produces a nonzero solution of the differential equation that

also satisﬁes the boundary and boundedness conditions. (But negative values

of λ produce no new solutions.) Recalling that any constant multiple of a so-

lution of a homogeneous problem is still a solution, we choose c

= 1and

summarize the solution of the singular eigenvalue problem as

φ(x;λ) = sin(λx), λ > 0. (8)

The solution of Eq. (5) for T(t), with constant −λ

,is

T(t) = exp



−λ



For any value of λ

,thefunction

u(x, t;λ) = sin(λx) exp



−λ



satisﬁes Eqs. (1)–(3). Equation (1) and the boundary condition Eq. (2) are ho-

mogeneous, and Eq. (3) is homogeneous in effect; therefore any linear combi-

nation of solutions is a solution. Since the parameter λ may take on any value,

190 Chapter 2 The Heat Equation

we must use an integral — the continuous analogue of a sum or series — to

include all possibilities. Thus u should have the form

u(x, t) =



∞

B(λ) sin(λx) exp



−λ



dλ. (9)

(We need not include negative values of λ. They give no new solutions.) The

initial condition will be satisﬁed if B(λ) is chosen to make

u(x, 0) =



∞

B(λ) sin(λx) dλ = f (x), 0 < x.

We recognize this as a Fourier integral; B(λ) is to be chosen as

B(λ) =



∞

f (x) sin(λx) dx. (10)

If B(λ) exists, then Eq. (9) is the solution of the problem. Notice that when

t > 0, the exponential function makes the improper integral in Eq. (9) con-

verge very rapidly.

Some care must be taken in the interpretation of our solution. If the rod re-

ally is ﬁnite (say, length L) the expression in Eq. (9) is, of course, meaningless

for x greater than L. The presence of a boundary condition at x = L would

inﬂuence temperatures nearby, so Eq. (9) can be considered a valid approxi-

mation only for x  L.

Example.

Solve the problem in Eqs. (1)–(4) using the initial temperature distribution

f (x) =



, 0 < x < b,

0, b < x.

This means that a section of length b at the left end of the rod starts out at

temperature T

, different from the temperature of the long right end, which

is at the same temperature as the left boundary. (We assume T

> 0.) The

solution is given by Eq. (9), with B(λ) calculated from Eq. (10):

B(λ) =



∞

f (x ) sin(λx) dx



sin(λx) dx

λπ



1 −cos(λb)



2.10 Semi-Inﬁnite Rod 191

Figure 8 Graphs of the solution of the example, u(x, t) as a function of x over

the interval 0 < x < 3b,whereb = 1andT

= 100 for convenience. The times

have been chosen so that the dimensionless time kt/b

takes the values 0.001, 0.01,

0.1, and 1. When kt/b

= 0.01, the temperature near x = b/2 has not changed

noticeably from its initial value.

Therefore, the complete solution is

u(x, t) =



∞

1 −cos(λb)

sin(λx) exp



−λ



dλ.

In Fig. 8 are graphs of u(x, t) as a function of x forvariousvaluesoft;an

animation can be seen on the CD.



EXERCISES

1. Find the solution of Eqs. (1)–(3) if the initial temperature distribution is

given by

f (x ) =



0, 0 < x < a,

T, a < x < b,

0, b < x.

2. Ve r i f y t h a t u(x, t) as given by Eq. (9) is a solution of Eqs. (1)–(3). What is

thesteady-statetemperaturedistribution?

3. Find the solution of Eqs. (1)–(4) if f (x) =T

−αx

, x > 0.

4. Find a formula for the solution of the problem

192 Chapter 2 The Heat Equation

∂

∂x

∂u

∂t

, 0 < x, 0 < t,

∂u

∂x

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

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

5. Determine the solution of Exercise 4 if f (x) is the function given in Exer-

cise 1.

6. Penetration of heat into the earth. Assume that the earth is ﬂat, occupying

the region 0 < x (so that x measures distance down from the surface). At

the surface, the temperature ﬂuctuates according to season, time of day, etc.

We cover several cases by taking the boundary condition to be u(0, t) =

sin(ωt), where the frequency ω canbechosenaccordingtotheperiodof

interest.

a. Show that u(x, t) = e

−px

sin(ωt − px) satisﬁes the boundary condition

and is a solution of the heat equation if p =



ω/2k.

b. Sketch u(x, t) as a function of t for x = 0, 1, and 2 m, taking ω = 2 ×

−7

rad/s (approximately one cycle per year) and k = 0.5 ×10

−6

/s.

c. With ω as in part b, ﬁnd the depth (as a function of k)atwhichseasons

are reversed.

7. Consider the problem

∂

∂x

∂u

∂t

, 0 < x, 0 < t,

u(0, t) = T

, 0 < t,

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

Show that, for our method of solution to work, it is necessary to have T

lim

x→∞

f (x ). Find a formula for u(x, t) if this is the case.

8. If the separation constant in Eq. (5) were positive (say, p

), we would at-

tempt to solve φ



− p

φ = 0 subject to the conditions, Eq. (7). Solve the

differential equation, and show that any nonzero, bounded solution is not

0atx = 0 and that any solution that is 0 at x =0 is not bounded.

9. R.C. Bales, M.P. Valdez and G.A. Dawson [Gaseous deposition to snow,

2: Physical-chemical model for SO

deposition, Journal of Geophysical Re-

search, 92 (1987): 9789–9799] develop a mathematical model for the trans-

port of SO

gas into snow by molecular diffusion. The governing partial

2.11 Inﬁnite Rod 193

differential equation is

∂C

∂t



∂

∂x

−a



where C is the concentration of SO

as a function of x (depth into the snow)

and time and D is a diffusion constant. The term containing C appears be-

cause the SO

takes part in a chemical reaction with water in the snow,

forming sulphuric acid, H

.Thecoefﬁcienta

depends on pH, temper-

ature, and other circumstances; we treat it as a constant. The problem is to

be solved for a wide range of values for the parameters.

If the snow is deep, the authors believe that it is reasonable to use a semi-

inﬁnite interval for x and to add the condition C (x, t) → 0asx →∞.In

addition, a natural boundary condition at the snow surface is that con-

centration in the snow match that in the air: C(0, t) = C

.Furthermore,

if the snow is fresh, we can assume that the concentration throughout is

initially 0,

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

a. Find a steady-state solution v(x) that satisﬁes the partial differential

equation and the boundary conditions.

b. State the problem (partial differential equation, boundary condition at

x = 0, condition as x →∞, and initial condition) to be satisﬁed by the

transient w(x, t) = C(x, t) −v(x).

c. Solve the problem for the transient. Note that the condition as x →∞

must be relaxed to: w(x, t) bounded as x →∞.Individualproductso-

lutions do not approach 0 as x increases.

2.11 Inﬁnite Rod

If we wish to study heat conduction in the center of a very long rod, we may as-

sume that it extends from −∞ to ∞. Then there are no boundary conditions,

and the problem to be solved is

∂

∂x

∂u

∂t

, −∞< x < ∞, 0 < t, (1)

u(x, 0) = f (x), −∞< x < ∞, (2)



u(x, t)



bounded as x →±∞. (3)

194 Chapter 2 The Heat Equation

Using the same techniques as before, we look for solutions in the form

u(x, t) = φ(x)T(t) so that the heat equation (1) becomes



(x)

φ(x)



(t)

T(t)

=constant.

As in the previous section, the constant must be nonpositive (say, −λ

)in

order for the solutions to be bounded. Thus, we have the singular eigenvalue

problem



+λ

φ = 0, −∞ < x < ∞,

φ(x) bounded as x →±∞.

It is easy to see that every solution of φ



/φ =−λ

is bounded. Thus, our fac-

tors φ(x) and T(t) are

φ(x;λ) = A cos(λx) +B sin(λx),

T(t;λ) = exp



−λ



We c ombi ne the sol uti ons φ(x)T(t) intheformofanintegraltoobtain

u(x, t) =



∞



A(λ) cos(λx ) +B(λ) sin(λx)



exp



−λ



dλ. (4)

At time t = 0, the exponential factor becomes 1, and the initial condition is



∞



A(λ) cos(λx) +B(λ) sin(λx)



dλ = f (x), −∞< x < ∞.

As this is clearly a Fourier integral problem, we must choose A(λ) and B(λ) to

be the Fourier integral coefﬁcient functions,

A(λ) =



∞

−∞

f (x) cos(λx) dx, B(λ) =



∞

−∞

f (x ) sin(λx) dx. (5)

Then the function u(x, t) in Eq. (4) satisﬁes the partial differential equation (1)

and the initial condition (2), provided that f is sectionally smooth and |f (x)|

has a ﬁnite integral. It can be proved that the boundedness condition (3) is

also satisﬁed, provided that the initial value f (x) is bounded as x →±∞.

Example.

Solve the problem posed in Eqs. (1)–(3) with

f (x ) =



0, x < −a,

, −a < x < a,

0, a < x.

2.11 Inﬁnite Rod 195

Figure 9 Solution of example problem. At t = 0, the temperature is T

> 0for

−a < x < a and is 0 in the rest of the rod; u(x, t) is shown as a function of x

on the interval −3a < x < 3a for three times. The times are chosen so that the

dimensionless time kt/a

takes the values 0.01, 1, and 10 (to get a clear picture of

the changes in u). Note that u(x, t) is positive everywhere for any t > 0. The values

=100 and a =1 have been used for convenience. Also see the CD.

In words, the rod has a center section of length 2a whose temperature is differ-

ent from that of the long sections to the left and right. We must compute the

coefﬁcient functions A(λ) and B(λ). The latter is identically 0 because f (x) is

an even function; and

A(λ) =



∞

−∞

f (x) cos(λx) dx



−a

cos(λx) dx

λπ

sin(λa).

Thus, the solution of the problem is

u(x, t) =



∞

sin(λa)

cos(λx) exp



−λ



dλ. (6)

This function is graphed as a function of x for several values of t in Fig. 9

and animated on the CD. The ﬁgure suggests that u(x, t) is positive for all x

when t > 0. This is indeed true and illustrates an interesting property of the

solutions of the heat equation: the instantaneous transmission of information.

The“hot”sectionintheinterval−a < x < a instantly raises the temperature

everywhere else from the initial value of 0 to a positive value.



Starting from the general form of a solution in Eq. (4), we can derive some

very interesting results. Change the variable of integration in Eq. (5) to x



and

196 Chapter 2 The Heat Equation

substitute the formulas for A(λ) and B(λ) into Eq. (4):

u(x, t) =



∞





∞

−∞

f (x



) cos(λx



) dx



cos(λx)



∞

−∞

f (x



) sin(λx



) dx



sin(λx)



exp



−λ



dλ.

Combining terms we ﬁnd

u(x, t) =



∞



∞

−∞

f (x



)



cos(λx



) cos(λx) +sin(λx



) sin(λx)





× exp



−λ



dλ



∞



∞

−∞

f (x



) cos



λ(x



−x )





exp



−λ



dλ.

If the order of integration may be reversed, we may write

u(x, t) =



∞

−∞

f (x



)



∞

cos



λ(x



−x )



exp



−λ



dλ dx



The inner integral can be computed by complex methods of integration. It

is known to be (Miscellaneous Exercises 32, Chapter 1)



∞

cos



λ(x



−x )



exp



−λ



dλ =



4kt

exp



−(x



−x )

4kt



, t > 0.

This gives us, ﬁnally, a new form for the temperature distribution:

u(x, t) =

√

4kπt



∞

−∞

f (x



) exp



−(x



−x)

4kt





. (7)

Using this form, we ﬁnd the solution of the example problem solved earlier

(seeEq.(6))tobe

u(x, t) =

√

4πkt



−a

exp



−(x



−x)

4kt





. (8)

Of the two formulas, Eqs. (4) and (7), for the solution u(x, t), each has its

advantages. For simple problems we may be able to evaluate the coefﬁcients

A(λ) and B(λ) in Eq. (5). However, it is a rare case indeed when the integral

in Eq. (4) can be evaluated analytically. The same is true for the integral in

Eq.(7).Thus,ifthevalueofu at a speciﬁc x and t is needed, either integral

would be calculated numerically. For large values of kt, the exponential factor

2.11 Inﬁnite Rod 197

in the integrand of Eq. (4) will be nearly zero, except for small λ.Thus,Eq.(4)

is approximately

u(x, t)

∼







A(λ) cos(λx ) +B(λ) sin(λx)



exp



−λ



dλ

for  not large, and the right-hand side may be found to a high degree of

accuracy with little effort.

On the other hand, if kt is small, the exponential in the integrand of Eq. (7)

will be nearly zero, except for x



near x. The approximation

u(x, t)

∼

√

4kπt



x+h

x−h

f (x



) exp



−(x



−x)

4kt





is satisfactory for h not large, and again numerical techniques are easily applied

to the right-hand side.

The expression in Eq. (7) also has a number of other advantages. It requires

no intermediate integrations (compare Eq. (5)). It shows directly the inﬂuence

of initial conditions on the solution. Moreover, the function f (x) need not

satisfy the restriction



∞

−∞



f (x )



dx < ∞

in order for Eq. (7) to satisfy the original problem.

EXERCISES

See the exercise Common Singular Eigenvalue Problems on the CD.

1. Find the solution of Eqs. (1)–(3) using the form given in Eq. (7) if the initial

temperature distribution is

f (x) =



, x < 0,

, 0 < x.

2. Find the solution of Eqs. (1)–(3) using the form given in Eq. (4) if

f (x) =





a −|x|



, −a < x < a,

0, otherwise.

3. Same task as in Exercise 2, with f (x) = T

−|x/a|

for all x.

4. Show that the solution of the problem studied in Section 10,

∂

∂x

∂u

∂t

, 0 < x, 0 < t,