0
$\begingroup$

So, this question probably has a very straight forward answer that I'm just not seeing. The question is:

Consider a 1D rod $0 \le x \le L $ of known length and known constant thermal properties without sources. Suppose the temperature is an unknown constant $T$ at $x = L$ . Determine $T$ if we know (in the steady state) both the temperature and the heat flow at $x=0$

I know that the steady state implies there is no time dependence, so the PDE becomes a simple ODE. $$ \frac{d u(x)}{d x^2} = 0 $$ The problem states that at $x=L, u(L)=T$. Being in the steady state, this would imply that $u(0) = T$. Now, I know the solution to the ODE is $$ u(x) = c_1x + c_2 $$ where $c_1$ and $c_2$ are constants. Using the boundary conditions laid out by the problem, I get the solution $$ u(x)=\frac{2T}{L}x + T $$ My first problem is that the units don't work out, the second problem is that I did not find $T$ explicitly. Where have I gone wrong?

Any hints would be appreciated, thank you.

EDIT: $$u(0)=u(L)=T \\ u(0)= T = c_2 \\ u(L) = c_1L+T = T \Rightarrow c_1=\frac{2T}{L} \\ u(x)= \frac{2T}{L}x+T$$

  • 1
    If you assume the same temperature at the left end you will actually only wind up with the constant solution of $u(x)=T$, it would be better to assume that $u(0)=u_0$ some value that you don't know explicitly. – 2017-01-26
  • 0
    @Triatticus, using $u(0)=u_0$ gives me the solution $u(x)= \frac{T-u_0}{L}x +u_0$. I still do not see how to get an equation for $T$ from this. Also, with the assumptions I made above, I made an edit to show my steps. I do not understand how you got $u(x) = T$ – 2017-01-26
  • 1
    In the line $u(L)=c_1L+T=T$ implies that $c_1=0$ you are adding instead subtracting. The solution would then be $u(x)=T$. – 2017-01-26
  • 0
    @Rumplestillskin, I see my sign error. I noticed it right as you posted this comment. So, is the final answer simply $u(x)=T$? Thank you. – 2017-01-26
  • 1
    @Kosta Yep. You are all set :) – 2017-01-26

0 Answers 0