Applied Partial Differential Equations with fourier series and boundary value problems 5th edition 1.4.7 part B

OK so first of all, to correct the textbook. In "Applied Partial Differential Equations: with Fourier Series and Boundary Value Problems, Fourth Edition"  (I was unable to obtain the 5th edition, but I assume they're similar), the authors state "We define an equilibrium or steady-state solution to be a temperature distribution that does not depend on time." Technically, equilibrium is distinct from steady-state. In equilibrium, entropy has been maximized, and so any temperature gradients are not allowed. So with $\frac{\partial u}{\partial x} (0,t)=1$ we by definition cannot have equilibrium and so just with that boundary condition alone I would say equilibrium is impossible. However, let's not dwell on equilibrium versus steady-state and let's proceed as if the authors did not make a terrible error.

So let's assume the authors mean "Determine a steady-state temperature distribution (if one exists). For what values of B are there solutions? Explain physically.

To start with, let's assume that sources and sinks of thermal energy within $u(x,t)$ are not allowed, which seems reasonable given that the system does not have any in $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$.

Therefore, in steady state, where $u$ does not depend on time, or $\frac{\partial u}{\partial t}= 0$, we get that $\frac{\partial^2 u}{\partial x^2} = 0$. So $\frac{\partial u}{\partial x}$ must be a constant to satisfy this relation. In other words, $\frac{\partial u}{\partial x}(0,t)=1 = \frac{\partial u}{\partial x}(L,t)=1 = B$

Therefore, $B = 1$ is the only valid solution to this equation with no sinks or sources.

So the steady-state temperature distribution with no sinks or sources of heat within the system is: $u(x) = x + C$ where $C$ is some constant.

In steady state (the state of the system after a sufficiently long time), all the oscillations in $f(x)$ will be smoothed out. The absolute magnitude of the temperature $C$ will depend on the initial conditions of the system, determined by $f(x)$. We may wonder, what is $C$? Well, since the flux of heat entering the system is equal to the flux leaving, since $\frac{\partial u}{\partial x}(0,t)= \frac{\partial u}{\partial x}(L,t)$, that means that the total energy within the system remains constant. Note, we'll have to restrict $f(x)$ to functions that have a slope of $1$ at both $x=0$ and $x=L$. But that's a detail for whoever is going to choose $f(x)$ down the road. The energy is conserved, and so $ \frac{1}{L}\int_{0}^{L} f(x) dx = \frac{1}{L} \int_{0}^{L} x+ C dx$. Solving this, we get:
$\int_{0}^{L} f(x) dx = \left (x^2/2 + Cx \right )|^{L}_{0} = L^2/2 + C L$. Solving for $C$:

$C = \left (\int_0^L f(x) dx - L^2/2 \right )/L$ or more intuivevly:
$C = \frac{1}{L} \int_0^L f(x) dx - L/2$

So the steady state temperature distribution is:
$u(x) = x + \frac{1}{L} \int_0^L f(x) dx - L/2$