Please confirm my work regarding heat through a one-dimensional rod. Is my process correct? If not, where did I go wrong?
As I said above, the problem involves the heat flow through a one-dimensional rod. The question then asks the following.
For the following problem, determine the equilibrium temperature distribution (if one exists). For what values of $\beta$ are these solutions?
$$ u_t(x,t)=u_{xx}(x,t) ; u(x,0)=f(x), u_x(0,t)=1, u_x(L,t)=\beta $$
Where $0<x<L$ and $u(x,t)$ is the temperature of the rod.
I began with the heat equation for a 1D rod.
$$ cpu_t(x,t)=K_0u_{xx}(x,t)+Q $$
Where $K_0$ is the thermal conductivity, which is constant, and $Q$ is the heat energy generated in the rod. In comparing the heat equation with the PDE given in the problem, we see that $Q=0$, $cp=1$, $K_0=1$. Since the equilibrium temperature distribution implies that the temperature in the rod is constant, or in other words, independent of time, we can then assume the following.
$$ u_t=0 \\\\\\\\ u(x,t)=u(x)\\u_x(0,t)=u_x(0)=1 \\u_x(L,t)=u_x(L)=\beta $$
So, our equation then becomes
$$ u_{xx}=0 \Rightarrow u_x=c_1 $$
To find the values for beta,
$$ u_x(0)=1=u_x(L)=\beta=c_1 \\\\ \therefore c_1=\beta=1 $$
By integrating $u_x$, we get that
$$ u(x)=c_1x+c_2=x+c_2 $$
Now, by using our initial conditions, we see that
$$ u(x)=f(x)=x+c_2 \Rightarrow c_2=f(x)-x $$
So then the equilibrium temperature distribution is
$$ u(x,t)=f(x) $$
Thank you!
Best Answer
Yes, this problem is correct. After reviewing old homework's for this course, and reviewing old unanswered questions, I decided that this post is correct.