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Say you have a slab of material occupying the region $0\leq x\leq a$. Heat is supplied at a constant unit rate so the temperature $T(x,t)$ satisfies $$\frac{\partial T}{\partial t}= k \frac{\partial^2 T}{\partial x^2} +1$$ The initial temperature is $0$, $x=0$ is maintained at zero temperature and face $x=a$ is thermally insulated.

  1. What are the initial and boundary conditions?
  2. How would you figure out the steady state solution $T_s(x,t)$?
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    $\begingroup$ This looks like homework... a) has nothing to do with the "+1", it's given in the last sentence of your description. b) The steady state is reached when $\partial T/\partial t = 0$, I think you know how to integrate the right-hand side. $\endgroup$ – Sheljohn Apr 18 '14 at 23:36
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a) As specified by your question, the initial condition is $T(x,0) = 0$ and the boundary conditions are $T(0,t) = 0$ ($x=0$ maintained at 0), and $T_x(a,t) = 0$.

b) The steady-state solution is reached when $T_t(x,t) = 0$, so: $$ \frac{\partial^2 T_s}{\partial x^2} = -\frac{1}{k} \implies \frac{\partial T_s}{\partial x} = - \frac{x}{k} + \mathrm{Const}_1(t) $$

Evaluating the rhs at the boundary $x=a$ for all $t$ yields $\mathrm{Const_1(t)} = \frac{a}{k}$. This leads to: $$ T_s(x,t) = \frac{ax}{k} - \frac{x^2}{2k} + \mathrm{Const}_2(t) $$ and evaluating the rhs this time at the boundary $x=0$ for all $t$ yields $\mathrm{Const}_2(t) = 0$, which finally result in: $$ T_s(x,t) = \frac{x}{k}\left( a - \frac{x}{2} \right) $$

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