Ch 10.6: Other Heat Conduction Problems

In Section 10.5, we considered the heat conduction problem

with solution

At this stage, this is a formal solution, since we have not provided a rigorous justification of the limiting processes.

While such a justification is beyond our scope, we discuss certain features of the argument on the next slide.

Lxxfxu

ttLutu

tLxuu txx

0),()0,(

0,0),(,0),0(

0,0,2

L

nn

tLnn dxLxnxf

LcLxnectxu

01

)/( /sin)(2

,/sin),(2

Formal Solution

Once we obtain the series

it can be shown that in 0 < x < L, t > 0, the series converges to a continuous function u(x,t), that uxx and ut can be computed by differentiating the series term by term, and that the heat conduction equation is indeed satisfied.

The argument relies on each term having a negative exponential factor, resulting in rapid convergence of the series.

A further argument shows that u(x,t) satisfies the boundary and initial conditions, and hence the formal solution is justified.

,/sin),(1

)/( 2

n

tLnn Lxnectxu

Heat Conduction as a Smoothing Process

Although the initial temperature distribution f satisfies the conditions of Theorem 10.3.1 (Fourier convergence theorem), it is piecewise continuous and hence may have discontinuities.

Nevertheless the solution u(x,t) is continuous for arbitrarily small values of t > 0.

This illustrates the fact that heat conduction is a diffusive process that instantly smooths out any discontinuities that may be present in the initial temperature distribution u(x,0) = f (x).

Diffusion of Heat

Finally, since the initial temperature distribution f is bounded, it follows from the equation

that the coefficients cn are also bounded.

Thus the presence of a negative exponential in each term of

guarantees that

regardless of initial condition.

L

n dxLxnxfL

c0

/sin)(2

1

)/( /sin),(2

n

tLnn Lxnectxu

0),(lim

txut

Nonhomogeneous Boundary Conditions (1 of 5)

We now consider the heat conduction problem with nonhomogeneous boundary conditions:

We solve this problem by reducing it to a problem having homogeneous boundary conditions.

The homogeneous problem is then solved as in Section 10.5.

The technique for reducing this problem to a homogeneous one is suggested by a physical argument, as presented in the following slides.

Lxxfxu

tTtLuTtu

tLxuu txx

0),()0,(

0,),(,),0(

0,0,

21

2

Steady-State Temperature Distribution (2 of 5)

After a long time (i.e., as t ), we anticipate that a steady-state temperature distribution v(x) will be reached, which is independent of time t and the initial conditions.

Since v(x) must satisfy the equation of heat conduction,

we have

Further, v(x) must satisfy the boundary conditions

Solving for v(x), we obtain

,0,2 Lxuu txx

Lxxv 0,0)(

21 )(,)0( TLvTv

112)( TL

xTTxv

Transient Temperature Distribution (3 of 5)

Returning to the original problem, we will try to express u(x,t) as the sum of the steady-state temperature distribution v(x) and another (transient) temperature distribution w(x,t). Thus

Since we have an expression for v(x), it remains to find w(x,t).

We first find the boundary value problem for w(x,t) as follows.

Substituting u(x,t) = v(x) + w(x,t) into 2uxx = ut, we obtain 2wxx = wt, since vxx= vt = 0.

Next, w(x,t) satisfies the boundary and initial conditions

),()(),( txwxvtxu

)()()()0,()0,(

0)(),(),(

0)0(),0(),0(

22

11

xvxfxvxuxw

TTLvtLutLw

TTvtutw

Transient Solution (4 of 5)

Therefore the boundary value problem for w(x,t) is

where

From Section 10.5, the solution to this problem is

where

Lxxvxfxw

ttLwtw

tLxww txx

0),()()0,(

0,0),(,0),0(

0,0,2

1

)/( /sin),(2

n

tLnn Lxnectxw

L

n dxLxnTL

xTTxf

Lc

0 112 /sin)(2

112)( TL

xTTxv

Nonhomogeneous Solution (5 of 5)

Recall our original nonhomogeneous boundary value problem

Thus the solution u(x,t) = v(x) + w(x,t) is given by

where

1

)/(112 /sin),(

2

n

tLnn LxnecT

L

xTTtxu

L

n dxLxnTL

xTTxf

Lc

0 112 /sin)(2

Lxxfxu

tTtLuTtu

tLxuu txx

0),()0,(

0,),(,),0(

0,0,

21

2

Example 1: Nonhomogeneous Heat Conduction Problem (1 of 3)

Consider the nonhomogeneous heat conduction problem

The steady-state temperature satisfies v''(x) = 0 and boundary conditions v(0) = 20 and v(30) = 50. Thus v(x) = x + 20.

The transient temperature distribution w(x,t) satisfies the homogeneous heat conduction problem

300,260)0,(

0,50),30(,20),0(

0,300,

xxxu

ttutu

txuu txx

300,34020260)0,(

0,0),30(,0),0(

0,300,2

xxxxxw

ttwtw

txww txx

Example 1: Solution (2 of 3)

The nonhomogeneous solution u(x,t) is given by the sum of the steady-state temperature distribution v(x) and the transient temperature distribution w(x,t).

Thus

where

30

030/sin340

15

1dxxnxcn

1

)30/( 30/sin20),(2

n

tnn xnecxtxu

Example 1: Graphs of Solution (3 of 3)

The figure below shows a plot of the initial temperature distribution u(x,0) = 60 – 2x, the final temperature distribution v(x) = x + 20, and the temperature distribution u(x,t) at two intermediate times.

Note that the intermediate temperature satisfies the boundary conditions for any time t > 0.

As t increases, the effect of the

boundary conditions gradually

moves from the ends of the bar

toward its center.

Bar with Insulated Ends (1 of 11)

Now suppose that the ends of the bar are insulated so that there is no passage of heat through them.

It can be shown (see Appendix A of this chapter) that the rate of flow of heat across a cross section is proportional to the rate of change of temperature in the x direction.

Thus, in the case of no heat flow, our problem takes the form

This problem can be solved using the method of separation of variables, as we examine next.

Lxxfxu

ttLutu

tLxuu

xx

txx

0),()0,(

0,0),(,0),0(

0,0,2

Separation of Variables Method (2 of 11)

As in Section 10.5, we begin by assuming

Substituting this into our differential equation

we obtain

or

where is a constant, as in Section 10.5.

We next consider the boundary conditions.

)()(),( tTxXtxu

txx uu 2

TXTX 2

,0

0122

TT

XX

T

T

X

X

Boundary Conditions (3 of 11)

Recall our original problem

Substituting u(x,t) = X(x)T(t) into boundary condition at x = 0,

Since we are interested in nontrivial solutions, we require X'(0) = 0 instead of T(t) = 0 for t > 0. Similarly, X'(L) = 0.

We therefore have the following boundary value problem

0)()0(),0( tTXtux

0)()0(,0 LXXXX

Lxxfxu

ttLutu

tLxuu

xx

txx

0),()0,(

0,0),(,0),0(

0,0,2

Boundary Value Problem for < 0 (4 of 11)

Thus we must solve the boundary value problem

It can be shown that nontrivial solutions exist only if is real.

Suppose < 0, and let = 2, where is real and positive.

Then our equation becomes

whose general solution is

In this case the boundary conditions require k1 = k2 = 0, and hence the only solution is the trivial one.

Therefore cannot be negative.

0)()0(,02 LXXXX

0)()0(,0 LXXXX

xkxkxX coshsinh)( 21

Boundary Value Problem for = 0 (5 of 11)

Our boundary value problem is

Suppose = 0. Then our equation becomes

whose general solution is

From the boundary conditions, k1 = 0 and k2 is not determined.

Hence = 0 is an eigenvalue, with eigenfunction X(x) = 1.

Also, from the equation below, T(t) = k3, with k3 a constant.

It follows that u(x,t) = C, where C = k2k3 is a constant.

0)()0(,0 LXXX

0)()0(,0 LXXXX

21)( kxkxX

02 TT

Boundary Value Problem for > 0 (6 of 11)

Our boundary value problem is

Suppose > 0, and let = 2, where is real and positive.

Then our equation becomes

whose general solution is

In this case the first boundary conditions requires k1 = 0, while

k2 is arbitrary as long as = n /L, n = 1, 2, ….

Thus our eigenvalues and eigenfunctions are

n = n22/L2, Xn(x) = cos(n x/L),

0)()0(,02 LXXXX

0)()0(,0 LXXXX

xkxkxX cossin)( 21

Fundamental Solutions (7 of 11)

For the eigenvalues n = n22/L2, the equation

has solution

Combining all these results, we have the following fundamental solutions to our original problem:

where arbitrary constants of proportionality have been omitted.

02 TT

constant. ,2)/(

ntLn

nn kekT

,,2,1,/cos),(

,1),(2)/(

0

nLxnetxu

txutLn

n

Initial Condition (8 of 11)

Because the original differential equation and its boundary conditions are linear and homogeneous, any finite linear combination of the fundamental solutions satisfies them.

We assume that this is true for convergent infinite linear combinations of fundamental solutions as well.

Thus, to satisfy the initial condition

we assume

Lxxfxu 0),()0,(

1

)/(0

10

0

/cos2

),(),(2

),(

2

n

tLnn

nnn

Lxnecc

txuctxuc

txu

Initial Condition (9 of 11)

Thus to satisfy the initial condition

we assume

where the cn are chosen so that the initial condition is satisfied:

Thus we choose the coefficients cn for a Fourier cosine series:

Lxxfxu 0),()0,(

1

)/(0 /cos2

),(2

n

tLnn Lxnec

ctxu

1

0 /cos2

)()0,(n

n Lxncc

xfxu

L

n ndxLxnxfL

c0

,2,1,0,/cos)(2

Solution (10 of 11)

Therefore the solution to the heat conduction problem for a bar with insulated ends (and insulated sides)

is given by

where

Lxxfxu

ttLutu

tLxuu

xx

txx

0),()0,(

0,0),(,0),0(

0,0,2

1

)/(0 /cos2

),(2

n

tLnn Lxnec

ctxu

L

n ndxLxnxfL

c0

,2,1,0,/cos)(2

Physical Interpretation (11 of 11)

The solution to our heat conduction problem

can be thought of as the sum of the steady-state temperature distribution (given by c0/2) which is independent of time, and

a transient solution (given by series) that tends to 0 as t .

Physically, we would expect the process of heat conduction to gradually smooth out the temperature distribution in the bar, as no heat escapes bar.

Note the average value of original temperature distribution is

1

)/(0 /cos2

),(2

n

tLnn Lxnec

ctxu

L

dxxfL

c0

0 )(1

2

Example 2: Heat Conduction Problem (1 of 4)

Find the temperature u(x,t) in a metal rod 25 cm long, insulated on the sides and ends and whose initial temperature distribution is u(x,0) = x for 0 < x < 25.

This heat conduction problem has the form

250,)0,(

0,0),25(,0),0(

0,250,2

xxxu

ttutu

txuu

xx

txx

Example 2: Solution (2 of 4)

The solution to our heat conduction problem is

where

Thus

1

)25/(0 25/cos2

),(2

n

tnn xnec

ctxu

1,even,0

odd,)/(10025/cos

25

2

,2525

2

225

0

25

00

nn

nndxxnxc

dxxc

n

,5,3,1

)50/(2

25/cos1100

2

25),(

2

n

tn xnen

txu

Example 2: Rapid Convergence (3 of 4)

Thus the temperature along the rod is given by

The negative exponential factor in each term causes the series to converge rapidly, except for small values of t or 2.

Therefore accurate results can usually be obtained by using only a few terms of the series.

In order to display quantitative results, let t be measured in seconds; then 2 has the units cm2/sec.

If we choose 2 = 1 for convenience, then the rod is of a material whose properties are somewhere between copper and aluminum (see Table 10.5.1).

,5,3,1

)50/(2

25/cos1100

2

25),(

2

n

tn xnen

txu

Example 1: Temperature Graph (4 of 4)

The graph of the temperature distribution u(x,t) in the rod at several times is given below.

Observe that as time t increases, the temperature distribution u(x,t) along the rod smooths out to the average value (12.5) of the initial temperature distribution u(x,0) = x, 0 < x < 25.

More General Problems (1 of 2)

The method of separation of variables can also be used to solve heat conduction problems with other boundary conditions than those discussed in this section.

For example, the left end of the bar might be held at a fixed temperature T while the other end is insulated.

In this case the boundary conditions are

The first step is to reduce the given boundary conditions to homogeneous ones by subtracting the steady-state solution.

The resulting problem is solved by essentially the same procedure as in the problems previously considered.

0,0),(,),0( ttLuTtu x

More General Problems (2 of 2)

Another type of boundary condition occurs when the rate of heat flow through the end of bar is proportional to temperature.

It is shown in Appendix A that the boundary conditions in this case are of the form

where h1 and h2 are nonnegative constants.

If we apply the method of separation of variables to the heat conduction problem with these boundary conditions, we have

As before, only certain nonnegative values of give rise to fundamental solutions, which can then be superposed to form an overall solution satisfying the initial condition. See text.

,0,0),(),(,0),0(),0( 21 ttLuhtLutuhtu xx

0)()(,0)0()0(,0 21 LXhLXXhXXX