1-D heat diffusion in a bar with fixed temperatures at the edges

Question

I'm trying to calculate the temperature gradient in a bar of metal with a heater at either end. Initially, the bar is at room-temperature $T_0$, then at $t=0$ the heaters are turned on: $u(0,t)=T_1$ and $u(L,t)=T_2$. As I understand it, eventually the temperature gradient will just be linear between $0$ and $L$. But how does this system evolve over time?

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The solution is given by: $$ {\rm T}\pars{x,t} = T_{1} + \pars{T_{2} - T_{1}}\,{x \over L} + \sum_{n = 1}^{\infty}A_{n}\pars{t}\sin\pars{k_{n}\,x}\,, \quad k_{n} = {n\pi \over L}\tag{1} $$ ${\rm T}\pars{x,t}$ obeys the Difussion Equation $\ds{\partiald{{\rm T}\pars{x,t}}{t} = D\,\partiald[2]{{\rm T}\pars{x,t}}{x}}$ where $D$ is the Difussion Constant.

From $\pars{1}$: $$ \sum_{n = 1}^{\infty}\totald{A_{n}\pars{t}}{t}\,\sin\pars{k_{n}\,x} = D\sum_{n = 1}^{\infty}A_{n}\pars{t}\pars{-\,k_{n}^{2}}\sin\pars{{n\pi \over L}\,x} \quad\imp\quad A_{n}\pars{t} = A_{n}\pars{0}\exp\pars{-Dk_{n}^{2}t} $$

$\pars{1}$ becomes: $$ {\rm T}\pars{x,t} = T_{1} + \pars{T_{2} - T_{1}}\,{x \over L} + \sum_{n = 1}^{\infty}A_{n}\pars{0}\sin\pars{k_{n}\,x}\exp\pars{-Dk_{n}^{2}t} $$

$$ T_{0} = {\rm T}\pars{x,0} = T_{1} + \pars{T_{2} - T_{1}}\,{x \over L} + \sum_{n = 1}^{\infty}A_{n}\pars{0}\sin\pars{k_{n}\,x} $$

$$ \int_{0}^{L}\bracks{T_{0} - T_{1} + \pars{T_{1} - T_{2}}\,{x \over L}}\, \sin\pars{k_{n}x}\,\dd x =\sum_{m = 1}^{\infty}A_{m}\pars{0} \overbrace{\int_{0}^{L}\sin\pars{k_{n}x}\sin\pars{k_{m}\,x}\,\dd x} ^{\ds{=\ L\,\delta_{n,m}/2}} =\half\,L\,A_{n}\pars{0} $$

$$ A_{n}\pars{0} = {2 \over L} \int_{0}^{L}\bracks{T_{0} - T_{1} + \pars{T_{1} - T_{2}}\,{x \over L}}\, \sin\pars{k_{n}x}\,\dd x =2\,{T_{0} - T_{1} + \pars{T_{2} - T_{0}}\pars{-1}^{n} \over n\pi} $$

\begin{align} \color{#00f}{\large{\rm T}\pars{x,t}} &= \color{#00f}{T_{1} + \pars{T_{2} - T_{1}}\,{x \over L}} \\[3mm]&\phantom{=}\color{#00f}{+ {2 \over \pi}\sum_{n = 1}^{\infty}{T_{0} - T_{1} + \pars{T_{2} - T_{0}}\pars{-1}^{n} \over n}\,\sin\pars{{n\pi \over L}\,x} \exp\pars{-\bracks{Dn^{2}\pi^{2}/L^{2}}t}} \end{align}