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

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148 Chapter 2 The Heat Equation

Can you think of a physical interpretation of this problem? Note the differ-

ence between the partial differential equation in this exercise and in Exer-

cise 1. What happens if γ = π/a?

4. State the initial value–boundary value problem satisﬁed by the transient

temperature distribution corresponding to Eqs. (8)–(11).

5. Find the steady-state solution of the problem

∂

∂x



∂u

∂x



=cρ

∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = T

, u(a, t) = T

, 0 < t

if the conductivity varies in a linear fashion with x: κ(x) = κ

+βx,where

and β are constants.

6. Find and sketch the steady-state solution of

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t

together with boundary conditions

∂u

∂x

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

;

b. u(0, t) −

∂u

∂x

(0, t) = T

∂u

∂x

(a, t) = 0;

c. u(0, t) −

∂u

∂x

(0, t) = T

, u(a, t) +

∂u

∂x

(a, t) = T

7. Find the steady-state solution of this problem, where r is a constant that

represents heat generation.

∂

∂x

+r =

∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = T

∂u

∂x

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

8. Find the steady-state solution of

∂

∂x

+γ



U(x) −u



∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = U

∂u

∂x

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

where U(x) = U

+Sx (U

, S are constants).

2.3 Example: Fixed End Temperatures 149

9. This problem describes the diffusion of a substance in a medium that is

moving with speed S to the right. The unknown function u(x, t) is the con-

centration of the diffusing substance. Write out the steady-state problem

and solve it. (D, U,andS are constants.)

∂

∂x

∂u

∂t

∂u

∂x

, 0 < x < a, 0 < t,

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

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

2.3 Example: Fixed End Temperatures

In Section 1 we saw that the temperature u(x, t) in a uniform rod with insu-

lated material surface would be determined by the 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)

if the ends of the rod are held at ﬁxed temperatures and if the initial tempera-

ture distribution is f (x). In Section 2 we found that the steady-state tempera-

ture distribution,

v(x) = lim

t→∞

u(x, t),

satisﬁed the boundary value problem

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

v(0) = T

,v(a) = T

. (6)

Infact,wewereabletoﬁndv(x) explicitly:

v(x) = T

+(T

−T

)

. (7)

We also deﬁned the transient temperature distribution as

w(x, t) = u(x, t) − v(x)

150 Chapter 2 The Heat Equation

anddeterminedthatw satisﬁes the boundary value–initial value problem

∂

∂x

∂w

∂t

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

w(0, t) = 0, 0 < t, (9)

w(a, t) = 0, 0 < t, (10)

w(x, 0) = f (x) −v(x) ≡ g(x), 0 < x < a. (11)

Our objective is to determine the transient temperature distribution, w(x, t),

and — since v(x) is already known — the unknown temperature will be

u(x, t) = v(x) +w(x, t). (12)

The problem in w can be attacked by a method called product method, sepa-

ration of variables,orFourier’s method. For this method to work, it is essential

to have homogeneous partial differential equation and boundary conditions.

Thus, the method may be applied to the transient distribution w but not to

the original function u. Of course, because both the partial differential equa-

tion and the boundary conditions satisﬁed by w(x, t) are homogeneous, the

function w ≡ 0 satisﬁes them. Because this solution itself is obvious and is of

no help in satisfying the initial condition, it is called the trivial solution.We

are seeking the unobvious, nontrivial solutions, so we shall avoid the trivial

solution at every turn.

The general idea of the method is to assume that the solution of the par-

tial differential equation has the form of a product: w(x, t) = φ(x)T(t).We

require that neither of the factors φ(x) and T(t) be identically 0, since that

would lead back to the trivial solution. Now, each of the factors depends on

only one variable, so we have

∂

∂x

=φ



(x)T(t),

∂w

∂t

=φ(x)T



(t).

The partial differential equation becomes



(x)T(t) =

φ(x)T



(t),

and on dividing through by φT we ﬁnd



(x)

φ(x)



(t)

kT(t)

, 0 < x < a, 0 < t.

Here is the key argument: The ratio on the left contains functions of x alone

and cannot vary with t. On the other hand, the ratio on the right contains

functions of t alone and cannot vary with x. Since this equality must hold for

2.3 Example: Fixed End Temperatures 151

all x in the interval 0 < x < a and for all t > 0, the common value of the two

sides must be a constant, varying neither with x nor t:



(x)

φ(x)

=p,



(t)

kT(t)

=p.

Now we have two ordinary differential equations for the two factor functions:



−pφ =0, T



−pkT =0. (13)

The two boundary conditions on w may also be stated in the product form:

w(0, t) = φ(0)T(t) = 0,w(a, t) = φ(a)T(t) = 0.

There are two ways these equations can be satisﬁed for all t > 0. Either the

function T(t) ≡ 0 for all t, which is forbidden, or the other factors must be

zero. Therefore, we have

φ(0) = 0,φ(a) = 0. (14)

Our job now is to solve Eqs. (13) and satisfy the boundary conditions (14)

while avoiding the trivial solution.

Case 1:Ifp > 0, the solutions of Eqs. (13) are

φ(x) = c

cosh



√



sinh



√



, T(t) = ce

pkt

Now we apply the boundary conditions:

φ(0) = 0: c

=0,

φ(a) = 0: c

sinh



√



=0.

Because the sinh function is 0 only when its argument is 0 — clearly not true

√

pa —wehavec

=0andφ(x) ≡ 0, which is not acceptable.

Case 2:Ifwetakep = 0, the solutions of the differential equations (13) are

φ(x) = c

x , T(t) = c. The boundary conditions require

φ(0) = 0: c

=0,

φ(a) = 0: c

a = 0.

Again we have φ(x) ≡ 0.

Case 3: We now try a negative constant. Replacing p by −λ

in Eqs. (13)

gives us the two equations



+λ

φ = 0, T



+λ

kT =0,

whose solutions are

φ(x) = c

cos(λx) +c

sin(λx), T(t) = c exp



−λ



152 Chapter 2 The Heat Equation

If φ has the form given in the preceding, the boundary conditions require that

φ(0) = c

=0, leaving φ(x) = c

sin(λx).Thenφ(a) = c

sin(λa) = 0.

We now have two choices: either c

=0, making φ(x) ≡ 0 for all values of x,

or sin(λa) =0. We reject the ﬁrst possibility, for it leads to the trivial solution

w(x, t) ≡ 0. In order for the second possibility to hold, we must have λ =

nπ/a,wheren =±1, ±2, ±3,....Thenegativevaluesofn do not give any

new functions, because sin(−θ) =−sin(θ ).Henceweallown = 1, 2, 3,...

only. We shall set λ

=nπ/a.

Incidentally, because the differential equations (13) and the boundary con-

ditions (14) for φ(x) are homogeneous, any constant multiple of a solution is

still a solution. We shall therefore remember this fact and drop the constant c

in φ(x). Likewise, we delete the c in T(t).

To review our position, we have, for each n = 1, 2, 3,...,afunctionφ

(x) =

sin(λ

x) and an associated function T

(t) = exp(−λ

kt).Theproductw

(x, t)

=sin(λ

x) exp(−λ

kt) has these properties:

∂

∂x

=−λ

;

∂w

∂t

=−λ

; and therefore w

satisﬁes the heat equa-

tion.

2. w

(0, t) = sin(0)e

−λ

= 0 for any n and t; and therefore w

satisﬁes the

boundary condition at x = 0.

3. w

(a, t) = sin(λ

a)e

−λ

= 0 for any n and t because λ

a = nπ and

sin(nπ)= 0. Therefore w

satisﬁes the boundary condition at x = a.

Now we call on the Principle of Superposition in order to continue.

Principle of Superposition.

If u

, u

,...are solutions of the same linear, homogeneous equations, then so

u =c

+···.



In fact, we have inﬁnitely many solutions, so we need an inﬁnite series to

combine them all:

w(x, t) =

∞



n=1

sin(λ

x) exp



−λ



. (15)

Using an inﬁnite series brings up questions about convergence that we are go-

ing to ignore. However, it is easy to verify that the function deﬁned by the

series does satisfy the boundary conditions: At x = 0andatx = a ,eachterm

is 0, so the sum is 0 as well. To check the partial differential equation, we have

to differentiate w(x, t) by differentiating each term of the series. This done, it

is easy to see that terms match and the heat equation is satisﬁed.

2.3 Example: Fixed End Temperatures 153

Notice that the choice of the coefﬁcients b

does not enter into the check-

ing of the partial differential equation and the boundary conditions. Thus,

Eq. (15) plays the role of a general solution of Eqs. (8)–(10).

Of the four parts of the original problem, only the initial condition has not

yet been satisﬁed. At t = 0, the exponentials in Eq. (15) are all unity. Thus the

initial condition takes the form

w(x, 0) =

∞



n=1

sin



nπx



=g(x), 0 < x < a. (16)

We immediately recognize a problem in Fourier series, which is solved by

choosing the constants b

according to the formula



g(x) sin



nπx



dx. (17)

If the function g is continuous and sectionally smooth, we know that the

Fourier series actually converges to g(x) in the interval 0 < x < a,sothesolu-

tion that we have found for w(x, t) actually satisﬁes all requirements set on w.

Even if g does not satisfy these conditions, it can be shown that the solution

we have arrived at is the best we can do.

Once the transient temperature has been determined, we ﬁnd the original

unknown u(x , t) as the sum of the transient and the steady-state solutions,

u(x, t) = v(x) +w(x, t).

Example.

Suppose the original problem to be

∂

∂x

∂u

∂t

, 0 < x < a, 0 < t,

u(0, t) = T

, 0 < t,

u(a, t) = T

, 0 < t,

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

The steady-state solution is

v(x) = T

+(T

−T

)

The transient temperature, w(x, t) = u(x, t) −v(x),satisﬁes

∂

∂x

∂w

∂t

, 0 < x < a, 0 < t,

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

154 Chapter 2 The Heat Equation

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

w(x, 0) =−T

−(T

−T

)

≡g(x), 0 < x < a.

According to the preceding calculations, w has the form

w(x, t) =

∞



n=1

sin(λ

x) exp



−λ



(18)

and the initial condition is

w(x, 0) =

∞



n=1

sin



nπx



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

The coefﬁcients b

are given by





−T

−(T

−T

)



sin



nπx



cos(nπx/a)

(nπ/a)



−

−T

)

sin(nπx/a) −(nπ x/a) cos(nπx/a)

(nπ/a)



=−

nπ



1 −(−1)



2(T

−T

)

nπ

(−1)

−2

nπ



−T

(−1)



Now the complete solution (see Fig. 3) is

u(x, t) = w(x, t) +T

+(T

−T

)

where

w(x, t) =−

∞



n=1

−T

(−1)

sin(λ

x) exp



−λ



. (19)

ThesolutionofthisproblemisshownasananimationontheCD.



We can discover certain features of u(x, t) by examining the solution. First,

u(x, 0) really is zero (0 < x < a) because the Fourier series converges to −v(x)

at t = 0. Second, when t is positive but very small, the series for w(x, t) will

almost equal −T

− (T

− T

)x/a .Butatx = 0andx = a, the series adds

up to zero (and w(x, t) is a continuous function of x); thus u(x, t) satisﬁes

the boundary conditions. Third, when t is large, exp(−λ

kt) is small, and the

2.3 Example: Fixed End Temperatures 155

Figure 3 The solution of the example with T

= 100 and T

= 20. The function

u(x, t) is graphed as a function of x forfourvaluesoft, chosen so that the dimen-

sionless time kt/a

has the values 0.001, 0.01, 0.1, and 1. For kt/a

=1, the steady

state is practically achieved. See the CD.

other exponentials are still smaller. Then w(x, t) may be well approximated

by the ﬁrst term (or ﬁrst few terms) of the series. Finally, as t →∞, w(x, t)

disappears completely.

EXERCISES

Also see Separation of Variables Step by Step on the CD.

1. Write out the ﬁrst few terms of the series for w(x, t) in Eq. (19).

2. If k = 1cm

/s, a = 1 cm, show that after t = 0.5 s the other terms of the

series for w are negligible compared with the ﬁrst term. Sketch u(x, t) for

t = 0, t =0.5, t =1.0, and t =∞.TakeT

=100, T

=300.

3. We can see from Eq. (19) that the dimensionless combinations x/a and

kt/a

appear in the sine and exponential functions. Reformulate the partial

differential equation (8) in terms of the dimensionless variables. ξ = x/a,

τ =kt/a

.Setu(x, t) = U(ξ, τ).

4. Sketch the functions φ

, φ

,andφ

, and verify that they satisfy the bound-

ary conditions φ(0) = 0, φ(a) = 0.

In Exercises 5–8, solve the problem

∂

∂x

∂w

∂t

, 0 < x < a, 0 < t,

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

w(x, 0) = g(x), 0 < x < a

for the given function g(x).

156 Chapter 2 The Heat Equation

g(x) = T

(constant).

6. g(x) = βx (β is constant).

7. g(x) = β(a − x) (β is constant).

8. g(x) =











, 0 < x <

(a −x)

< x < a.

9. A.N. Virkar, T.B. Jackson, and R.A. Cutler [Thermodynamic and kinetic ef-

fects of oxygen removal on the thermal conductivity of aluminum nitride,

Journal of the American Ceramic Society, 72 (1989): 2031–2042] use the fol-

lowing boundary value problem to study the kinetics of oxygen removal

from a grain of aluminum nitride by diffusion:

∂C

∂t

∂

∂x

, 0 < x < a, 0 < t,

C(0, t) = C

, C(a, t) = C

, 0 < t,

C(x, 0) = C

, 0 < x < a.

In these equations, C is the oxygen concentration, D is the diffusion con-

stant, a is the thickness of a grain, C

and C

are known concentrations.

a. Find the steady-state solution, v(x).

b. State the problem (partial differential equation, boundary conditions

and initial condition) for the transient, w(x, t) = C(x, t) −v(x).

c. Solve the problem for w(x, t), and write out the complete solution

C(x, t).

d. The concentration in the center of the grain, C(a/2, t),variesfromC

at time t = 0towardC

as t increases. Suppose we want to ﬁnd out how

long it takes for this concentration to complete 90% of the change it will

make from C

to C

; that is, we want to solve this equation for t:



, t



−C

=0.9(C

−C

Show that this equation is equivalent to the equation



, t



=−0.1(C

−C

Find an approximate formula for the solution by using just the ﬁrst term

of the series for w(x, t).

2.4 Example: Insulated Bar 157

e. Use the formula in d to ﬁnd t explicitly for a = 5 × 10

−6

m, D =

−11

/s. Be careful to check dimensions.

2.4 Example: Insulated Bar

We shall consider again the uniform bar that was discussed in Section 1. Let

us suppose now that the ends of the bar at x = 0andx = a are insulated in-

stead of being held at constant temperatures. The boundary value–initial value

problem that describes the temperature in this rod is:

∂

∂x

∂u

∂t

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

∂u

∂x

(0, t) = 0,

∂u

∂x

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

u(x, 0) = f (x), 0 < x < a, (3)

where f (x) is supposed to be a given function.

We saw in Section 2 that the solution of the steady-state problem is not

unique. However, the mathematical purpose behind ﬁnding the steady-state

solution is to pave the way for a homogeneous problem (partial differential

equation and boundary conditions) for the transient. In this example the par-

tial differential equation and boundary conditions are already homogeneous.

Thus, we do not need the steady-state solution or the transient problem. We

maylookforu(x, t) directly.

Assume that u has the product form u(x, t) = φ(x)T(t), with neither factor

identically 0. The heat equation becomes



(x)T(t) =

φ(x)T



(t),

and the variables are separated by dividing through by φT,leaving



(x)

φ(x)



(x)

kT(t)

, 0 < x < a, 0 < t.

In order that a function of x equal a function of t,theirmutualvaluemust

be a constant. If that constant were positive, T would be an increasing expo-

nential function of time, which would be unacceptable. It is also easy to show

that if the constant were positive, φ could not satisfy the boundary conditions

without being identically zero.

Assuming then a negative constant, we can write



(x)

φ(x)

=−λ



(t)

kT(t)