Assume that $u(x,t)$ solves the heat equation:
\begin{equation} u_t(x,t) = u_{xx}(x,t),\hspace{1cm} (0<x<L) \end{equation}
subject to the time-dependent boundary conditions:
\begin{equation} u(0,t)=0, \\ u(L,t)=g(t). \end{equation}
In this post user JohnD has solved it by assuming the eigenfunction expansion form:
\begin{equation} u(x,t) = \sum_{n=0}^\infty b_n \sin{(\sqrt{\lambda_n}x)}e^{-\lambda_n t}, \end{equation}
and then proceeds to find the coefficients $b_n$.
Also There is a slightly different approach introduced in this page. At first the solution is written as the sum of two functions:
$$u(x,t)=s(x,t) + v(x,t),$$ $s(x,t)$ is the part that satisfies the nonhomogeneous boundary conditions, which for the problem at hand becomes: $$ s(x,t) = \frac{x}{L} g(t),$$ and then $v(x,t)$ satisfies the following equation: $$v_t (x,t) - v_{xx} (x,t) = - s_t(x,t), \hspace{1cm} v(0,t)=v(L,t)=0$$ Again the solution proceeds by expanding both sides of the equation in terms of the eigenfunctions of LHS, i.e. : $$\sin(n\pi \frac{x}{L}),$$
(this method is also explained here and in this post).
Questions:
1. In the first method mentioned above, the obtained solution doesn't meet the required boundary conditions:
$$ u(L,t) =g(t),$$
because $\sin(n \pi)$ is always zero, and so regardless of the value of $b_n$s, we always get $u(L,t)=0.$
Is that right or am I missing sth obvious? (for eg. maybe the limit of the obtained solution when $x$ approaches $L$ is $g(t)$, but I couldn't check this limit)
Is the obtained result by this method a solution of the boundary value PDE?
2. Regarding the second method, is expanding a linear function ( i.e. $\frac{x}{L} g(t)$) in terms of $\sin(n \pi \frac{x}{L})$s legitimate? because again regardless of their coefficients, they give $0$ for $x=L$.
3. As a consequence of #2, the obtained solution by using the second method doesn't satisfy the original PDE exactly at $x=L$, because on the one hand: $$u_t(L,t) = s_t(L,t) + v_t(L,t) = g_t(t),$$ and on the other hand: $$u_{xx}(L,t) = s_{xx}(L,t) + v_{xx}(L,t) = 0$$
the question is just like #1: Is the obtained result by this method a solution of the boundary value PDE?
4. Can we solve this PDE on the domain $[0,L]$, instead of $(0,L)$?
