Applications of the heat equation PDE 
Question:


*

*A rod of length $1$ m, initially heated to
$100^{\circ}$ C before one end is inserted into the ice that maintains that end’s temperature at $0^{\circ}$ C.

*If we denote $\theta(x,t)$ to be the temperature at distance $x$ from the end with temperature at $0^o$C at time $t$, write down this initial condition.

*Solve the Fourier series for $\theta(x, t)$. What does the model predict for the long term temperature distribution in the rod $?$.

Recall that initial condition refers to
$$\theta(x,0) = f(x)$$
where $f$ gives the initial temperature distribution along the rod.
In this case, we know that $f$ satisfies
$$f(x) = 
\begin{cases}
0 & \text{ if } x = 0,\\
100 & \text{ if } 0<x\leq 1
\end{cases}
$$
I know that this is related to the homogeneous heat diffusion equation
$$\frac{\partial\theta}{\partial t} = \frac{1}{\alpha} \frac{\partial^2 \theta}{\partial x^2}$$
where $L = 1$.
I also know that boundary conditions are $\theta(0,t) = \frac{\partial \theta}{\partial x}(1,t) =0$.
However, I am not sure whether $\frac{\partial\theta}{\partial x}(x,1) = 0$.
Also, its Fourier series is
$$\theta(x,t) = \sum_{r=1}^\infty B_r \sin \left( r\pi x \right) \exp\left( -\alpha \left( \frac{r\pi}{2} \right)^2 t \right)$$
where
$$B_r = 2 \int_0^1 f(x)\sin(r\pi x)dx.$$
I am wondering whether my attempt above answer the question or not.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets $\ds{\theta\pars{x,t} = 
\sum_{n = 0}^{\infty}\on{a}_{n}\pars{t}\sin\pars{k_{n}x}}$ where
$\ds{k_{n} = \pars{2n + 1}\,{\pi \over 2}}$ which already satisfy
$\ds{\theta\pars{0,t} = \left.\partiald{\theta\pars{x,t}}{x}\right\vert_{\ x\ =\ 1} = 0,\ \forall\ t}$.

The above expression for $\ds{\theta\pars{x,y}}$ must satisfy the above differential equation. Namely,
\begin{align}
&\sum_{n = 0}^{\infty}\dot{\on{a}}_{n}\pars{t}\sin\pars{k_{n}x} =
-\,{1 \over \alpha}
\sum_{n = 0}^{\infty}\dot{\on{a}}_{n}\pars{t}k_{n}^{2}\sin\pars{k_{n}x}
\end{align}
With $\ds{\int_{0}^{1}\sin\pars{k_{m}x}\sin\pars{k_{n}x}
\,\dd x = {1 \over 2}\,\delta_{mn}}$, I got
$$
\dot{\on{a}}\pars{t} + {k_{n}^{2} \over \alpha}\on{a}\pars{t} = 0 \implies
\on{a}_{n}\pars{t} =
\on{a}_{n}\pars{0}\exp\pars{-\,{k_{n}^{2} \over \alpha}\,t}
$$
The general solution is reduced to:
\begin{align}
&\theta\pars{x,t} = \sum_{n = 0}^{\infty}
\on{a}_{n}\pars{0}\exp\pars{-\,{k_{n}^{2} \over \alpha}\,t}\sin\pars{k_{n}x}
\\[5mm] &\
\mbox{and}\quad
100 = \theta\pars{x,0} = \sum_{n = 0}^{\infty}
\on{a}_{n}\pars{0}\sin\pars{k_{n}x}
\\[5mm] & \ \implies
100\
\underbrace{\int_{0}^{1}2\sin\pars{k_{n}x}\,\dd x}
_{\ds{4/\pi \over 2n + 1}} =
\on{a}_{n}\pars{0}
\\[5mm] &\ \implies
\on{a}_{n}\pars{0} = {400/\pi \over 2n + 1} =
{200 \over k_{n}}
\\[5mm] &\ \implies
\bbx{\theta\pars{x,t} = 200
\sum_{n = 0}^{\infty}
{\sin\pars{k_{n}x} \over k_{n}}
\exp\pars{-\,{k_{n}^{2} \over \alpha}\,t}} \\ &
\end{align}
