this problem is giving me some trouble. It is a review problem given to us to study for our Final Exam this week. I would love some help in understanding how to solve it so that I can study it and solve a similar problem on my final exam:
Q: Find the equilibrium temperature in a narrow insulated rod described by the boundary value problem:
$\frac{\partial u}{\partial t}=k \frac{\partial^2 u}{\partial x^2}$, $\frac{\partial u}{\partial x}(0,t)=0$, $\frac{\partial u}{\partial x}(L,t)=0$, $u(x,0)=a \chi(L-x) $
[a is a given constant]
A: We know the equilibrium temperature has been reached when $\frac{\partial u}{\partial t}=0$, thus we can say that $\frac{\partial^2 u}{\partial x^2}=0$. Using integration, we can then solve for $u(x)$ as follows:
$\frac{\partial u}{\partial x}=a$
$u(x) = ax+b$
Now we can use the initial conditions to solve for a and b. We know that $\frac{\partial u}{\partial x}(0,t)=a=0$, thus $a=0$. This leaves us with $u(x)=b$
Now I do not know how to proceed from here. I cannot figure out how to solve for b.
I have two equations that may apply to this problem, but I don't know for sure.
$Q = \int_0^L c\rho A udx$
$P=\int_0^L udx$
Where Q is the total energy in the rod, and P is something that I don't know what it is unfortunately.
Any help would be appreciated on how to solve from here!
You are correct that from the two boundary conditions, we get $v(x)=b$. But how to determine the value of that constant $b$?
If you solve the initial-boundary value problem for $u(x,t)$ (by separation of variables) and take the limit as $t\to\infty$, you will see that every term in the solution vanishes except the leading (constant) term. So you have $$\lim_{t\to\infty}u(x,t)={1\over 2}a_0 \quad\text{ where }\quad a_0={2\over L}\int_0^L a_\chi(L-x)\,dx=a_\chi L. $$
Then $$\lim_{t\to\infty}u(x,t)={1\over 2}a_0={a_\chi L\over 2},$$ and thus the steady state temperature distribution across the rod is $$v(x)\equiv b={a_\chi L\over 2}.$$
Think about what this says: for a rod insulated at both ends, the steady-state temperature distribution across the rod is the average (in the calculus sense) of the initial temperature across the rod because $$ v(x)=b={1\over 2}a_0=\underbrace{{1\over L}\int_0^L \underbrace{a_\chi(L-x)}_\text{initial temp of rod}\,dx}_\text{average value of a function on an interval}. $$