Consider the one-dimensional heat equation with initial boundary conditions given by
$$u_{t}=ku_{xx}, \ 0<x<L, \ t >0,$$
$$u_{x}(0,t)=u_{x}(L,t)=0,$$
$$u(x,0)=f(x).$$ Using separation of variables, eigenvalues, and eigenfunctions, I obtained the solution $$u=\sum_{n=1}^{\infty}A_{n}\cos\left ( \frac{n\pi x}{L} \right )\exp\left [ -kt \left ( \frac{n \pi}{L} \right )^2\right ].$$ The question asks what happens so the solution as $t \rightarrow \infty$. More formally,
Determine the temperature distribution as $t \rightarrow \infty.$
Clearly, $$\lim_{t\rightarrow \infty}u=0.$$ This becomes $0$ because of the negative exponential. However, my friends and I were talking and we don't think this makes that much sense, physically, at least. Mathematically I see no problem.
Another reason I am slightly dubious about this answer is the fact that the next question asks about the temperature distribution as $t \rightarrow \infty$ with specific values of $L,k$ and $f(x).$ Again, the limit I obtained was $0.$
Am I not interpreting this correctly? Can anyone give me some feedback?
Thanks so much!
You've omitted $n=0$ from the summation, incorrectly, and that is what remains as $t\to\infty$. What happens is that the temperature distribution tends to a constant, the average of the initial condition, because the boundary conditions are insulating (no heat flow in or out).