Solving heat equation using Fourier-series with non-homogeneous assymetric bondary conditions Problem
I am having trouble with finding a solution which satisfies the boundary and initial conditions to this PDE:
$$\frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial^2x}$$
where $u=u(x,t)$, $0 \leq x \leq L$
with boundary & initial conditions:
BC1: $u(x=0,t>0)=T_f$;
BC2: $u(x=L,t>0)=T_i$;
IC: $u(x,t=0)=f(x)$


Related but different posts

*

*Motivation on Using Fourier Series to Solve Heat Equation: the answer to this uses BCs: $u(x=0,t)=u(x=L,t)=0 \forall t$ which is not the same as my BCs


*Solve Heat Equation using Fourier Transform (non homogeneous): solving a modified version of the heat equation, Dirichlet BC


*Solving the heat equation using Fourier series: relies on the same source as I do (wikipedia), but it does not advance the simpler version of the problem outlined there, and I am attempting to do it here.


*non homogeneous heat equation?: different IC, not much elaborated


What I do get
Following the strategy outlined here, I do separation of variables:
$$u(x,t) = X(x)T(t)$$
The PDE becomes:
$$\frac{T'}{\alpha T}=\frac{X''}{X}$$
LHS only time dependence, RHS only x dependence, so they must be equal to a  constant. Let this constant be $-\lambda$ (with $\lambda > 0$ so we get an exponential decay rather than growth for the temporal equation).
Temporal eq:
$$T'=-\lambda \alpha T$$
which implies:
$$T(t)=A e^{-\lambda \alpha t}$$
Spatial eq:
$$X''+\lambda X = 0$$
having a solution:
$$X(x) = B e^{\sqrt{-\lambda}x}+Ce^{-\sqrt{-\lambda}x}$$
which, since $\lambda>0$, can be rewritten to:
$$X(x) = B \sin(\sqrt{\lambda}x) + C \cos(\sqrt{\lambda}x)$$

Where issues begin
If we had $u(x=0,L;t) = 0$, I would be confident to use these to determine $\lambda$ & C to be:
$$\lambda = \frac{n^2 \pi^2}{L^2}$$
$$C=0$$
Then, proceed by setting $A=1$, so we have: $$u(x,t) = \sum_{n=0}^{\infty}B_{n} \sin\left(\frac{n\pi}{L}x\right) e^{-\frac{n^2 \pi^2}{L^2}\alpha t}$$
Then, determine $B_n$s using orthogonality of the different frequency sine functions:
$$B_n = \frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi}{L}x\right)dx$$

How do I attempt to tackle them
Despite my IC is not $u(x=0,L;t) = 0$, I have some hope for this path. Now my solution is:
$$u(x,t) = e^{-\frac{n^2 \pi^2}{L^2}\alpha t} \sum_{n=0}^{\infty}\frac{2}{L}\left(\int_0^Lf(x)\sin\left(\frac{n\pi}{L}x\right)dx\right)\sin\left(\frac{n\pi}{L}x\right)$$
Which I believe satisfies my IC but not my BCs. Don't worry, lets add to $X(x)$ a line which make it satisfy the BCs as well (which are: $u(x=0,t>0)=T_f$ & $u(x=L,t>0)=T_i$). Let's call this $X_p$:
$$X_p(x) = \frac{T_i-T_f}{L}x$$
Now if I just add this to the previously found $X$, I will obviously screw up the $B_n$s, which were calculated such a way that the weighted sum of sines will give me $f(x)$. If I just add $X_p$, the weighted sum of sines and $X_p$ will give me $f(x)+X_p$. Lets subtract $X_p$ from $f(x)$ when calculating the coefficeients of the sines, that way the weighted sum of the sines and $X_p$ will give me $f(x)-X_p+X_p = f(x)$ in $t=0$, which is good, and in $t>0$ I still satisfy my BCs because I have added $X_p$ to the general solution.

Where I arrive
So my final result is:
$$u(x,t) = e^{-\frac{n^2 \pi^2}{L^2}\alpha t} \left(\sum_{n=0}^{\infty}\frac{2}{L}\left(\int_0^L\left(f(x) - \frac{T_i-T_f}{L}x\right)\sin\left(\frac{n\pi}{L}x\right)dx\right)\sin\left(\frac{n\pi}{L}x\right)+\frac{T_i-T_f}{L}x\right)$$

Question
Is this a right way of obtaining the solution to the equation, or is it completely off track? Is there a name for the different steps I am using? If there is a standard way of solving these kind of equations which is not this way, I'd like to know.
(My guess would be that the $X_p$ is some kind of "particular solution", but I have used that term in different context, and my memory regarding terminology is pretty short.)
 A: Reduce the problem by looking at the equation solved by
$$
               v(x,t)=u(x,t)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i
$$
This function $v$ satisifes
$$
          v_t = v_{xx} \\
          v(x=0,t > 0) = u(x=0,t)-T_f=0 \\
          v(x=L,t > 0) = u(x=L,t)-T_i=0 \\
          v(x,0)= f(x)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i
$$
With the homogeneous conditions at $x=0,L$, the solution $v$ can be writte as a $\sin$ series in $x$:
$$
             v(x,t)= \sum_{n=1}^{\infty}A_n(t)\sin(n\pi x/L)
$$
The functions $A_n(t)$ are determined by $v_t = v_{xx}$ and the initial conditions.
$$
               \sum_{n=1}^{\infty}A_n'(t)\sin(n\pi x/L)=-\sum_{n=1}^{\infty}A_n(t)\frac{n^2\pi^2}{L^2}\sin(n\pi x/L) \\
         \implies A_n(t) = A_n(0)e^{-n^2\pi^2 t/L^2}
$$
Then $v(x,0)=f(x)-(1-x/L)T_f-(x/L)T_i$ gives
$$
       f(x)-(1-x/L)T_f-(x/L)T_i = v(x,0)=\sum_{n=1}^{\infty}A_n(0)\sin(n\pi x/L)
$$
This determines the coefficients $A_n(0)$ as Fourier $\sin$ coefficients, which completes the solution. (I'll leave it to you to find the $A_n(0)$ using the orthogonality of the functions $\sin(n\pi x/L)$.)
