SOLUTION This is a steady one-dimensional heat conduction problem with constant thermal conductivity and no heat generation in the medium, and the heat conduction equation in this case can be expressed as (Eq)
\frac{d^{2} T}{d x^{2}} = 0
whose general solution was determined in the previous example by direct integration to be
T(x) = C_{1} x + C_{2}
where C_{1} and C_{2} are two arbitrary integration constants. The specific solutions corresponding to each specified pair of boundary conditions are determined as follows.
(a) In this case, both boundary conditions are specified at the same boundary at x = 0, and no boundary condition is specified at the other boundary at x = L. Noting that
\frac{d T}{d x} = C_{1}
the application of the boundary conditions gives
– k \frac{d T(0)}{d x} = \dot{q}_{0} \rightarrow – k C_{1} = \dot{q}_{0} \quad \rightarrow \quad C_{1} = – \frac{q_{0}}{k}
and
T(0) = T_{0} \quad \rightarrow \quad T_{0} = C_{1} \times 0 + C_{2} \rightarrow C_{2} = T_{0}
Substituting, the specific solution in this case is determined to be
T(x) = – \frac{q_{0}}{k} + T_{0}
Therefore, the two boundary conditions can be specified at the same boundary, and it is not necessary to specify them at different locations. In fact, the fundamental theorem of linear ordinary differential equations guarantees that a unique solution exists when both conditions are specified at the same location. But no such guarantee exists when the two conditions are specified at different boundaries, as you will see below.
(b) In this case different heat fluxes are specified at the two boundaries. The application of the boundary conditions gives
– k \frac{d T(0)}{d x} = \dot{q}_{0} \rightarrow – k C_{1} = \dot{q}_{0} \quad \rightarrow \quad C_{1} = – \frac{q_{0}}{k}
and
– k \frac{d T(L)}{d x} = \dot{q}_{L} \quad \rightarrow \quad – k C_{1} = \dot{q}_{L} \quad \rightarrow \quad C_{1} = – \frac{q_{L}}{k}
Since \dot{q}_{0} \neq \dot{q}_{L} and the constant C_{1} cannot be equal to two different things at the same time, there is no solution in this case. This is not surprising since this case corresponds to supplying heat to the plane wall from both sides and expecting the temperature of the wall to remain steady (not to change with time). This is impossible.
(c) In this case, the same values for heat flux are specified at the two boundaries. The application of the boundary conditions gives
– k \frac{d T(0)}{d x} = \dot{q}_{0} \rightarrow – k C_{1}=\dot{q}_{0} \quad \rightarrow \quad C_{1} = – \frac{\dot{q}_{0}}{k}
and
– k \frac{d T(L)}{d x} = \dot{q}_{0} \quad \rightarrow \quad – k C_{1} = \dot{q}_{0} \quad \rightarrow \quad C_{1} = – \frac{\dot{q}_{0}}{k}
Thus, both conditions result in the same value for the constant C_{1} , but no value for C_{2} . Substituting, the specific solution in this case is determined to be
T(x) = – \frac{\dot{q}_{0}}{k} x + C_{2}
which is not a unique solution since C_{2} is arbitrary. This solution represents a family of straight lines whose slope is – \dot{q}_{0} / k . Physically, this problem corresponds to requiring the rate of heat supplied to the wall at x = 0 be equal to the rate of heat removal from the other side of the wall at x = L. But this is a consequence of the heat conduction through the wall being steady, and thus the second boundary condition does not provide any new information. So it is not surprising that the solution of this problem is not unique. The three cases discussed above are summarized in Figure.
