3. Second-Order Differential Equations
3.5 Boundary Value Problems and Heat Flow
Let us consider the following problem in steady-state heat conduction. A cylin- drical, uniform, metallic bar of lengthLand cross-sectional areaAis insulated on its lateral side. We assume the left face atx= 0 is maintained atT0degrees and that the right face at x=Lis held atTL degrees. What is the tempera- ture distributionu=u(x) in the bar after it comes to equilibrium? Hereu(x) represents the temperature of the entire cross section of the bar at positionx, where 0< x < L. We are assuming that heat flows only in the axial direction along the bar, and we are assuming that any transients caused by initial tem- peratures in the bar have decayed away. In other words, we have waited long
enough for the temperature to reach a steady state. One might conjecture that the temperature distribution is a linear function of x along the bar; that is, u(x) =T0+ TL−T0
L x. This is indeed the case, which we show below. But also we want to consider a more complicated problems where the bar has both a variable conductivity and an internal heat source along its length. An inter- nal heat source, for example, could be resistive heating produced by a current running through the medium.
The physical law that provides the basic model is conservation of energy.
If [x, x+dx] is any small section of the bar, then the rate that heat flows in at x,minus the rate that heat flows out at x+dx, plus the rate that heat is generated by sources, must equal zero, because the system is in a steady state.
See figure 3.5.
laterally insulated
Aφ(x) Aφ(x + dx)
A
x L x + dx
x 0
Figure 3.5 Cylindrical bar, laterally insulated, through which heat is flowing in thex-direction. The temperature is uniform in a fixed cross-section.
If we denote byφ(x) the rate that heat flows to the right at any section x (measured in calories/(areaãtime), and we letf(x) denote the rate that heat is internally produced atx, measured in calories/(volumeãtime), then
Aφ(x)−Aφ(x+dx) +f(x)Adx= 0.
CancelingA, dividing bydx,and rearranging gives φ(x+dx)−φ(x)
dx =f(x).
Taking the limit asdx→0 yields
φ(x) =f(x). (3.27)
This is an expression of energy conservation in terms of flux. But what about temperature? Empirically, the flux φ(x) at a section x is found to be pro- portional to the negative temperature gradient −u(x) (which measures the
steepness of the temperature distribution, or profile, at that point), or
φ(x) =−K(x)u(x). (3.28)
This is Fourier’s heat conduction law. The given proportionality factor K(x) is called thethermal conductivity, in units of energy/(lengthãdegrees
ãtime), which is a measure of how well the bar conducts heat at locationx. For a uniform bar K is constant. The minus sign in (3.28) means that heat flows from higher temperatures to lower temperatures. Fourier’s law seems intuitively correct and it conforms with the second law of thermodynamics; the larger the temperature gradient, the faster heat flows from high to low temperatures.
Combining (3.27) and (3.28) leads to the equation
−(K(x)u(x))=f(x), 0< x < L, (3.29) which is thesteady-state heat conduction equation. When theboundary conditions
u(0) =T0, u(L) =T1, (3.30) are appended to (3.29), we obtain aboundary value problem for the tem- perature u(x). Boundary conditions are conditions imposed on the unknown state u given at different values of the independent variable x, unlike initial conditions that are imposed at a single value. For boundary value problems we usually usexas the independent variable because boundary conditions usually refer to the boundary of a spatial domain.
Note that we could expand the heat conduction equation to
−K(x)u(x)−K(x)u(x) =f(x), (3.31) but there is little advantage in doing so.
Example 3.21
If there are no sources (f(x) = 0) and if the thermal conductivity K(x) =K is constant, then the boundary value problem reduces to
u = 0, 0< x < L, u(0) = T0, u(L) =T1.
Thus the bar is homogeneous and can be characterized by a constant conduc- tivity. The general solution ofu= 0 isu(x) =c1x+c2; applying the boundary conditions determines the constantsc1andc2and gives the linear temperature distributionu(x) =T0+TL−T0
L x, as we previously conjectured.
In nonuniform systems the thermal conductivityKdepends upon locationx in the system. And,Kmay depend upon the temperatureuas well. Moreover, the heat source term f could depend on location and temperature. In these cases the steady-state heat conduction equation (3.29) takes the more general form
−(K(x, u)u) =f(x, u),
which is a nonlinear second-order equation for the steady temperature distri- butionu=u(x).
Boundary conditions at the ends of the bar may also specify the flux rather than the temperature. For example, in a homogeneous system, if heat is injected at x = 0 at a rate of N calories per area per time, then the left boundary condition takes the formφ(0) =N,or
−Ku(0) =N.
Thus, a flux condition at an endpoint imposes a condition on the derivative of the temperature at that endpoint. In the case that the end at x=L, say, is insulated, so that no heat passes through that end, then the boundary condition is
u(L) = 0,
which is called an insulated boundary condition. As the reader can see, there are a myriad of interesting boundary value problems associated with heat flow. Similar equations arise in diffusion processes in biology and chemistry, for example, in the diffusion of toxic substances where the unknown is the chemical concentration.
Boundary value problems are much different from initial value problems in that they may have no solution, or they may have infinitely many solutions.
Consider the following.
Example 3.22
WhenK= 1 and the heat source term isf(u) = 9uand both ends of a bar of lengthL= 2 are held atu= 0 degrees, the boundary value problem becomes
−u = 9u, 0< x <2.
u(0) = 0, u(2) = 0.
The general solution to the DE is u(x) =c1sin 3x+c2cos 3x, where c1 and c2 are arbitrary constants. Applying the boundary condition at x = 0 gives u(0) = c1sin(3ã0) +c2cos(3ã0) = c2 = 0. So the solution must have the form u(x) = c1sin 3x. Next apply the boundary condition at x = 2. Then u(2) =c1sin(6) = 0,to obtain c1 = 0. We have shown that the only solution
isu(x) = 0. There is no nontrivial steady state. But if we make the bar length π, then we obtain the boundary value problem
−u = 9u, 0< x < π.
u(0) = 0, u(π) = 0.
The reader should check that this boundary value problem has infinitely many solutions u(x) = c1sin 3x, where c1 is any number. If we change the right boundary condition, one can check that the boundary value problem
−u = 9u, 0< x < π.
u(0) = 0, u(π) = 1, has no solution at all.
Example 3.23
Find all real values ofλfor which the boundary value problem
−u = λu, 0< x < π. (3.32)
u(0) = 0, u(π) = 0, (3.33)
has a nontrivial solution. These values are called the eigenvalues, and the corresponding nontrivial solutions are called theeigenfunctions. Interpreted in the heat flow context, the left boundary is held at zero degrees and the right end is insulated. The heat source is f(u) = λu. We are trying to find which linear heat sources lead to nontrivial steady states. To solve this problem we consider different cases because the form of the solution will be different for λ= 0, λ <0, λ >0.Ifλ= 0 then the general solution ofu= 0 isu(x) =ax+b.
Then u(x) = a. The boundary condition u(0) = 0 implies b = 0 and the boundary condition u(π) = 0 implies a = 0. Therefore, when λ= 0, we get only a trivial solution. Next consider the caseλ <0 so that the general solution will have the form
u(t) =asinh√
−λx+bcosh√
−λx.
The conditionu(0) = 0 forcesb= 0. Thenu(t) =a√
−λcosh√
−λx.The right boundary condition becomesu(π) =a√
−λcosh(√
−λã0) = 0, giving a= 0.
Recall that cosh 0 = 1.Again there is only the trivial solution. Finally assume λ >0.Then the general solution takes the form
u(t) =asin√
λx+bcos√ λx.
The boundary condition u(0) = 0 forces b = 0. Then u(t) = asin√ λx and u(x) =a√
λcos√
λx. Applying the right boundary condition gives u(π) =a√
λcos√ λπ= 0.
Now we do not have to choose a = 0 (which would again give the trivial solution) because we can satisfy this last condition with
cos√ λπ= 0.
The cosine function is zero at the valuesπ/2±nπ, n= 0,1,2,3, ...Therefore
√λπ=π/2 +nπ, n= 0,1,2,3, ...
Solving forλyields λ=
2n+ 1 2
2
, n= 0,1,2,3, ....
Consequently, the values of λ for which the original boundary value problem has a nontrivial solution are 14,94,254, .... These are the eigenvalues. The corre- sponding solutions are
u(x) =asin
2n+ 1 2
x, n= 0,1,2,3, ...
These are the eigenfunctions. Notice that the eigenfunctions are unique only up to a constant multiple. In terms of heat flow, the eigenfunctions represent possible steady-state temperature profiles in the bar. The eigenvalues are those valuesλfor which the boundary value problem will have steady-state profiles.
Boundary value problems are of great interest in applied mathematics, sci- ence, and engineering. They arise in many contexts other than heat flow, includ- ing wave motion, quantum mechanics, and the solution of partial differential equations.
EXERCISES
1. A homogeneous bar of length 40 cm has its left and right ends held at 30◦C and 10◦C, respectively. If the temperature in the bar is in steady- state, what is the temperature in the cross section 12 cm from the left end? If the thermal conductivity isK, what is the rate that heat is leaving the bar at its right face?
2. The thermal conductivity of a bar of length L = 20 and cross-sectional area A = 2 isK(x) = 1, and an internal heat source is given by f(x) = 0.5x(L−x). If both ends of the bar are maintained at zero degrees, what is the steady state temperature distribution in the bar? Sketch a graph of u(x). What is the rate that heat is leaving the bar at x= 20?
3. For a metal bar of lengthLwith no heat source and thermal conductivity K(x), show that the steady temperature in the bar has the form
u(x) =c1 x
0
dy K(y)+c2,
wherec1andc2are constants. What is the temperature distribution if both ends of the bar are held at zero degrees? Find an analytic formula and plot the temperature distribution in the case that K(x) = 1 +x. If the left end is held at zero degrees and the right end is insulated, find the temperature distribution and plot it.
4. Determine the values of λfor which the boundary value problem
−u = λu, 0< x <1, u(0) = 0, u(1) = 0, has a nontrivial solution.
5. Consider the nonlinear heat flow problem
(uu) = 0, 0< x < π, u(0) = 0, u(π) = 1,
where the thermal conductivity depends on temperature and is given by K(u) =u. Find the steady-state temperature distribution.
6. Show that if there is a solution u=u(x) to the boundary value problem (3.29)–(3.30), then the following condition must hold:
−K(L)u(L) +K(0)u(0) = L
0
f(x)dx.
Interpret this condition physically.
7. Consider the boundary value problem
u+ω2u= 0, u(0) =a, u(L) =b.
When does a unique solution exist?
8. Find all values ofλfor which the boundary value problem
−u−2u = λu, 0< x <1, u(0) = 0, u(1) = 0, has a nontrivial solution.
9. Show that the eigenvalues of the boundary value problem
−u = λu, 0< x <1, u(0) = 0, u(1) +u(1) = 0,
are given by the numbers λn =p2n, n= 1,2,3, ..., where thepn are roots of the equation tanp = 1/p. Plot graphs of tanp and 1/p and indicate graphically the locations of the values pn. Numerically calculate the first four eigenvalues.
10. Find the values ofλ(eigenvalues) for which the boundary value problem
−x2u−xu = λu, 1< x < eπ, u(1) = 0, u(eπ) = 0, has a nontrivial solution.