1. 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.

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.

3. Heat Conduction as a Smoothing Process
Although the initial temperature distribution f satisfies the conditions of Theorem (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).

4. 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.

5. 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.

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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.

13. 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.

14. 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.

15. 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

16. 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.

17. 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.

18. 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 = n22/L2, Xn(x) = cos(n x/L),

19. Fundamental Solutions (7 of 11)
For the eigenvalues n = n22/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.

20. 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

21. 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:

22. 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

23. 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

24. 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

25. Example 2: Solution (2 of 4)
The solution to our heat conduction problem is
where
Thus

26. 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 ).

27. 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.

28. 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.

29. 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.