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Determine the equilibrium temperature distribution for a 1D rod with constant thermal properties with following source and boundary conditions $$Q=0,u(0)=T,\frac{\partial u(L)}{\partial x}+u(L)=0$$

My work thus far: First in equilibrium $\frac{\partial u}{\partial t}=0$ so we find the equilibrium temperature by integrating $u''(x)=0$ twice, obtaining $u_{eq}(x)=c_1x+c_2$.

My approach is to apply the first boundary condition $u(0)=T$, we get $c_2=T$. My confusion then comes in applying the second boundary condition. \begin{align} \frac{\partial u(L)}{\partial x}+u(L)&=0\\ c_1+c_1(L)+T&=0\\ c_1&=\frac{-T}{1+L}\\ \end{align} Therefore $u_{eq}=\frac{-T}{1+L}x+T$.

Is this the correct approach?

Note I know that there are a lot of similar problems already posted but I have seen none with a Robin boundary condition

  • 1
    applying the first boundary condition $u(0) = T$ should give $c_2 = T,$ not $c_2 = 0,$ right? – 2017-01-17
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    @spaceisdarkgreen yes let me correct that. – 2017-01-17
  • 0
    Looks right. (Although you should edit out the part about both coefficients being zero.) – 2017-01-17
  • 0
    It looks correct. – 2017-01-21

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