Let the function $v(x,t)$ solve the PDE $$ v_t = v_{xx} + av_{x}, \qquad t \geq 0, \quad x \in [0,1], $$ where $a \in \mathbb{R}$, with an initial condition (i.c.) for $t = 0$ and zero Dirichlet boundary conditions.
In the sense of the standard $L_2[0,1]$ norm, is this problem well-posed?
My attempt:
Separate variables by writing $v(x,t) = E(t)g(x)$. Provided that the i.c. $g(x)$ possesses a Fourier series expansion for $x \in [0,1]$: $$g(x) = \sum_{m=1}^{\infty}\alpha_m \sin m\pi x$$
and, by substituting it into our PDE, the full solution can then be written as $$v(x,t) = \sum_{m=1}^{\infty}\alpha_m \exp\left(-m^2\pi^2+am\pi\frac{\sum_{m=1}^{\infty}\alpha_m\cos m\pi x}{\sum_{m=1}^{\infty}\alpha_m\sin m\pi x}\right)\sin m\pi x.$$ Now $$\vert v(x,t)\vert^2 = \int_0^1 \vert v(t,x)\vert^2 dx = \frac{1}{2}\sum_{m=1}^{\infty}\alpha_m \exp\left(-2m^2\pi^2+2am\pi\frac{\sum_{m=1}^{\infty}\alpha_m\cos m\pi x}{\sum_{m=1}^{\infty}\alpha_m\sin m\pi x}\right), $$ where we have used the fact that $$\int_0^1 \sin m\pi x \sin j\pi x dx$$ is only non-zero (and $=1/2$) when $j = m$. To demonstrate well-posedness, I want to try and show that $\vert E(t) \vert^2 \leq C$ uniformly for all $t$. This means we have to put a bound on the exponentials in our sum. The first term is promising, the second term with the messy looking fraction is less so.
Questions:
Is my approach so far correct (i.e. no mistakes)?
How do I finish my argument to show well-posedness?
Any help much appreciated.
The question being to prove the well-posedness in $L^2$ of the equation, we could avoid explicit computation of solution and use a priori estimates. Indeed if $v$ solves the equation, multiplying it by $v$ and integrating on $[0,1]$ gives: $$\int_0^1 v_t v dx -\int_0^1 v_{xx} v dx - a\int_0^1 v_x v dx =0.$$ which upon integration by parts in the second integral is: $$\int_0^1 \frac12 (v^2)_t dx +\int_0^1 v_x^2 dx+\left[v_x v\right]_0^1 - a\int_0^1 \frac12 (v^2)_x dx =0.$$ The bracket and the last integral are zero from homogeneous Dirichlet boundary conditions. Thus $$\frac{d\;}{dt}\int_0^1 v^2 dx+\int_0^1 v_x^2 dx=0$$ which gives upon time integration: $$\|v(t)\|_{L^2}\le\|v(0)\|_{L^2}\qquad\forall t\in \mathbb{R}^+.$$ From the linearity of the equation, this shows that the solution depends continuously on the initial condition, ie the problem is well-posed.