Given the PDE $$u_t=u_{xx}$$ defined in $x \in [0,L], \, t\in [0,+\infty)$, can a solution $u(x,t)$ be found subject to an initial condition $u(x,0)=f(x)$ and two conditions:
\begin{cases} u(a_1,t)=b_1 \\ u(a_2,t)=b_2\end{cases}
for $a_1, \, a_2 \in (0,L)$ and $b_1, \, b_2 \in \mathbb{R}$? If not, could a solution be found with one boundary condition, one initial conidition and one of the above?
Assume without loss of generality that $a_2 > a_1$. Let $$ y(x) \;=\; b_1 \;+\; (x - a_1) \frac{b_2 - b_1}{a_2 - a_1} $$ be the straight line that passes through your two "condition points."
Define a new function $v(x,t)$ via $u(x,t) = v(x,t) + y(x)$. The new problem in terms of $v(x,t)$ becomes: \begin{align} v_t &= v_{xx}\\ v(x,0) &= g(x) = f(x) - y(x)\\ v(a_1,t) &= 0\\ v(a_2,t) &= 0 \end{align} The set of basis functions which obey these conditions at $x = a_1$ and $x = a_2$ is $$ \sin\left[\frac{n \pi (x - a_1)}{a_2 - a_1}\right]\, , $$ where $n$ is a positive integer. Thus the solution may be written as the sum $$ v(x,t) \;=\; \sum_{n = 1}^{\infty} c_n(t) \sin\left[\frac{n \pi (x - a_1)}{a_2 - a_1}\right]\, . $$
Do you know how to proceed from here to obtain the coefficients $c_n(t)$, or do you need more of a hint?