To begin, let $\Omega\subset\Bbb R^n$ be whatever kind of domain we like, and let $$\begin{align}f&:\Omega\times(0,+\infty)\to\Bbb R \\ d &:\partial\Omega\to\Bbb R \\ g&:\Omega\to\Bbb R\end{align}$$ be functions, also however regular we desire.
Say I want to solve the problem $$(\star)\begin{cases}u_t-\Delta u &= f &\text{ in }\Omega \\ u\rvert_{\partial\Omega} &= d & \forall t\in(0,+\infty) \\ u(\cdot,0) & = g\end{cases}$$
That is, to find a function $u : \overline\Omega\times(0,+\infty)\to\Bbb R, (x,t)\mapsto u(x,t)$ that fulfills the above. After checking my notes and fiddling around a bit, I found that if I had a solution $v:\overline\Omega\to\Bbb R$ to
$$\text{(S)}\begin{cases}\Delta v &= 0 &\text{ in }\Omega \\ v\rvert_{\partial\Omega} &= d &\end{cases}$$
and a solution $w:\overline\Omega\times(0,+\infty)\to\Bbb R$ to:
$$(\text H)\begin{cases}u_t-\Delta u &= f &\text{ in }\Omega \\ u\rvert_{\partial\Omega} &= 0 & \forall t\in(0,+\infty) \\ u(\cdot,0) & = g-v\end{cases}$$
$(\text S$ for stationary and $\text H$ for homogeneous boundary conditions$)$, then I could solve $(\star)$, by letting $\tilde v(x,t) := v(x)$, and the solution $u$ would be $$u(x,t) = (w+\tilde v)(x,t)$$
Unfortunately, I'm completely stuck on the problem of letting $d$ depend on $t$ as well (in the above you can observe it only depends on $x$). The method was applicable because $(\text S)$ is a Laplace equation, which I can theoretically solve using Green functions (but really with separation of variables for the questions I'm aksed in school), and $(\text H)$ is a heat equation with null Dirichlet conditions, so I theoretically have the heat kernel (but again, really just seperation of variables). In other words I want to solve $(\star)$, but now the domain of $d$ is $\partial\Omega\times(0,+\infty)$.
I can't see how to modify this method (playing around with known kernels and the likes) to include the case of $d$ depending on $t$. I have to say that I have no knowledge of the Dirac Delta.
Can someone help?
Question ends here
Response to comment (too long for a comment)
What you're suggesting in the example of the cube only works for the Laplace equation. I believe the reason for this is that, in using separation of variables, you end up solving $$\frac{\psi''(x)}{\psi(x)}=\frac{-(\Delta_{y,z})\phi(y,z)}{\phi(y,z)} = \lambda$$
(if we've chosen the face that lies on $\{x=0\}$ to be nonzero) And because you've arranged the zeros accordingly, the equation in $\phi$ is the eigenvalue problem (in one dimension lower), with null Dirichlet conditions. The original nonzero face is now actually an initial condition, considering $x$ as the time variable.
That was me trying to figure out some deeper reasons, but here is a very down to earth reason: $$\begin{cases}u_t - u_{xx} = 0 & (x,t)\in (0,L)\times(0,+\infty) \\ u(0,t)=e^{-t}\,, u(L,t) = 0 & t\in(0,+\infty) \\ u(x,0) = 0 & x\in(0,L)\end{cases}$$
This is the $(0,d,0)$ problem on the $1$-cube, where we've already selected the "face" $\{L\}$ to be $0$. Separation $u(x,t) = X(x)T(t)$ gives $$\begin{align}&\frac{T'(t)}{T(t)}=\frac{-X''(x)}{X(x)} = \lambda \\[3ex] & u(0,t) = X(0)T(t) = e^{-t} \\[3ex] &u(L,t) = X(L)T(t) = 0\Rightarrow X(L) = 0 \\[3ex] & u(x,0) = 0 = X(x)T(0)\Rightarrow T(0)= 0\Rightarrow T\equiv 0 \Rightarrow u\equiv 0\,!! \end{align}$$
unless $\lambda = 0$, but then $u(x,t) = (at+b)(\alpha x^2 +\beta x +\gamma)$, which can't meet the boundary conditions. Maybe I'm missing something, but everything indicates that plain old seperation of variables doesn't work in the case that the Dirichlet condition is $t$-dependent (because it comes down to solving the eigenvalue problem where $\phi$ isn't null on the boundary ergo not even over a vector space). The way I do know how to solve this is to let $$\tilde u = u(x,t) + \left(\frac xL - 1\right)e^{-t}$$ which solves the $(f,0,g)$ problem, with $$\begin{align}f &= \left(1-\frac xL\right)e^{-t} \\ g &= \frac xL\end{align}$$
I was able to extend $d$ into $\Omega = (0,L)$ quite easily, because in $1$ dimension I can always calculate a time-varying line between two time-varying points. In $2$ dimensions I have to "lay a sheet" on a time-varying "wire" ($\partial \Omega$). I don't know how to do this, let alone in $n$ dimensions.