Consider $$\text{PDE:}~u_t = \alpha^2 u_{xx} $$ $$BC: \left\{ \begin{array} uu(0,t) = g_1(t) \\ u_x(L,t) + hu(L,t) = g_2(t) \end{array} \right.$$ $$IC: ~ u(x,0)=\phi(x)$$
from my understanding for this type of problem we look for a solution of the type
$$u(x,t) = A(t)(1-\frac{x}{l}) + B(t)(\frac{x}{l}) + v(x,t)$$
in order to construct a (hopefully) homogeneous PDE and choosing A and B in order to set the Steady state solution (ie $u(x,\infty)$) to the BC's.
noting that we let $$S(x,t)=A(t)(1-\frac{x}{l}) + B(t)(\frac{x}{l})$$ and using BC we solve for A and B accordingly. this gives $$A(t) = g_1(t) ~\&~ B(t) = \frac{g_1(t)+Lg_2(t)}{1+Lh}$$
this gives solution
$$u(x,t) = g_1(t)(1-\frac{x}{l}) + \frac{g_1(t)+Lg_2(t)}{1+Lh}(\frac{x}{l}) + v(x,t)$$
and substituting this into our original problem should give us an PDE in
so doing that gives us
$$u_t = \alpha^2 u_{xx}$$ with $$u_t = g'_1(t)(1-\frac{x}{l}) + \frac{g'_1(t)+Lg'_2(t)}{1+Lh}(\frac{x}{l}) + v_t$$
and $$u_{xx} = v_{xx}$$
at this point im a little lost. from this point how i do dervive the PDE for v? ie i assume it to be something along the likes of
$$\text{PDE:}~v_t = \alpha^2 v_{xx} + f(x,t)$$ $$BC: \left\{ \begin{array} vv(0,t) = 0 \\ v_x(L,t) + hv(L,t) = 0 \end{array} \right.$$ $$IC: ~ v(x,0)=\phi(x)$$
but uncertain. any help would be great. thank you very much!
ps i understand i may be overlooking something very simple none the less if you could point that out id be very appreciative.
Look for solutions of the form
$$ v(x,t) = \sum_n X_n(x)T_n(t) $$
where $X(x)$ are eigenfunctions of the homogeneous problem and satisfies the given B.C.s
\begin{cases} X'' + \lambda^2 X = 0 \\ X(0) = 0 \\ X'(L) + hX(L) = 0 \end{cases}
Solving this gives
$$ X_n(x) = \sin(\lambda_n x) $$
where the eigenvalues $\lambda_n$ satisfy
$$ \lambda_n \cos(\lambda_n L) + h\sin(\lambda_n L) = 0 \implies \tan(\lambda_nL) = -\frac{\lambda_n}{h} $$
These eigenfunctions are mutually orthogonal, so you'll be able to decompose the equation into its eigenmodes
\begin{align} f(x,t) &= \sum_n f_n(t)\sin(\lambda_n x) \\ \phi(x) &= \sum_n \phi_n\sin(\lambda_n x) \end{align}
Then the PDE reduces to a family of ODEs
\begin{cases} {T_n}' + \alpha^2\lambda_n^2 T_n = f_n(t) \\ T_n(0) = \phi_n \end{cases}
which has solutions
$$ T_n(t) = \phi_ne^{-\alpha\lambda_nt} + e^{-\alpha\lambda_nt}\int_0^t f_n(\tau) e^{\alpha\lambda_n\tau} d\tau $$