Finding $\beta$ where $u_x(0)=\beta$ is a boundary value of a heat equation 
Suppose an ice core are perfectly insulated and heated at one end at a rate $\alpha$ and cooled from the other end at a rate $\beta$. Consider the following boundary value problem $$u_t-u_{xx}=0, \ u_x(0)=\beta, \ u_x(l)=\alpha,$$ where $u$ is temperature, $t$ is time and $l$ is the length of the core.
What should $\beta$ be such that the ice cores temperature remains stable ($u_t=0$)?

My attempt:
If $u_t=0$, then this implies $u_{xx}=0$. if we integrate both sides with respect to $x$, we get $$u_x=A$$
for some $A\in\mathbb{R}$. So,
$$u_x(0)=\beta\implies A=\beta,\\u_x(l)=\alpha\implies A=\alpha.$$
Hence, we conclude that $\beta=\alpha\in\mathbb{R}$.
For part (b), assuming $\alpha=1, l=10$ and the average temperature of the core is $-15$, what is the solution for $u$ in the stable case.
Finally, If the cooling mechanism were to fail ($\beta=0$) how long would it take before the ice core started to melt (i.e. when would $u$ rise above $0$ at any point)?
 A: $\def\d{\mathrm{d}}\def\peq{\mathrm{\phantom{=}}{}}$For the first question, your solution so far is correct, but it would be better to complete it through further verification.
If $β = α$, then $u(x, t) = αx + c$, where $c$ is a constant, are solutions and the uniqueness of solutions to PDE of the second type boundary condition (up to a constant) implies that these are all the solutions. It is easy to verify that $u_t = 0$.
For the second question, since $u(x, t) = x + c$ and the average temperature is$$
-15 = \frac{1}{l} \int_0^l u(x, t) \,\d x = \frac{1}{10} \int_0^{10} (x + c) \,\d x = c + 5,
$$
then $c = -20$ and $u(x, t) = x - 20$.
For the third question, suppose $u(x, t) = v(x, t) + \dfrac{αx^2}{2l}$, thus$$
\begin{cases}
v_t - v_{xx} = \dfrac{α}{l}\\
v(x, 0) = -\dfrac{αx^2}{2l} + αx + c\\
v_x(0) = v_x(l) = 0
\end{cases}. \tag{1}
$$
Using separation of variables, plugging in $v(x, t) = X(x) T(t)$ yields$$
\begin{cases}
\dfrac{X''}{X} = \dfrac{T'}{T} = -λ\\
X'(0) = X'(l) = 0
\end{cases}.
$$
For $λ = 0$, $X_0(x) = 1$. For $λ > 0$,$$
\begin{cases}
X'' + λX = 0\\
X'(0) = X'(l) = 0
\end{cases} \Longrightarrow \begin{cases}
λ_k = \dfrac{k^2 π^2}{l^2}\\
X_k(x) = \cos(\sqrt{λ_k} x)
\end{cases}.
$$
Thus the solution to (1) can be written as$$
v(x, t) = \sum_{k = 0}^∞ X_k(x) T_k(t).
$$
Because\begin{gather*}
\frac{α}{l} = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l \frac{α}{l} · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{α}{l} X_0(x), \quad c = cX_0(x),\\
x = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l x · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{l}{2} X_0(x) + \sum_{k = 1}^∞ \frac{2l}{k^2 π^2} ((-1)^k - 1) X_k(x),\\
x^2 = \sum_{k = 0}^∞ \frac{\displaystyle \int_0^l x · X_k(x) \,\d x}{\displaystyle \int_0^l X_k^2(x) \,\d x} X_k(x) = \frac{l^2}{3} X_0(x) + \sum_{k = 1}^∞ \frac{4l^2}{k^2 π^2} (-1)^k X_k(x),
\end{gather*}\begin{align*}
v_t - v_{xx} &= \sum_{k = 0}^∞ c_k X_k(x) T_k'(t) - \sum_{k = 0}^∞ X_k''(x) T_k(t)\\
&= \sum_{k = 0}^∞ X_k(x) T_k'(t) - \sum_{k = 0}^∞ (-λ_k X_k(x)) T_k(t)\\
&= \sum_{k = 0}^∞ X_k(x) (T_k'(t) + λ_k T_k(t)),
\end{align*}
then comparing coefficients of $X_k$'s yields$$
\begin{cases}
T_0'(t) = \dfrac{α}{l}\\
T_0(0) = \dfrac{1}{3} αl + c
\end{cases}, \quad \begin{cases}
T_k'(t) + λ_k T_k(t) = 0\\
T_k(0) = -\dfrac{2αl}{k^2 π^2}
\end{cases}\ (k \geqslant 1),
$$
which implies$$
T_0(t) = \frac{αt}{l} + \frac{1}{3} αl + c, \quad T_k(t) = -\frac{2αl}{k^2 π^2} \exp(-λ_k t)\ (k \geqslant 1).
$$
Therefore,\begin{align*}
&\peq u(x, t) = v(x, t) + \frac{αx^2}{2l}\\
&= \frac{αx^2}{2l} + \frac{αt}{l} + \frac{1}{3} αl + c - \sum_{k = 1}^∞ \frac{2αl}{k^2 π^2} \cos\left( \frac{kπx}{l} \right) \exp\left( -\frac{k^2 π^2 t}{l^2} \right).
\end{align*}
$u(x, t)$ is plotted below.

