Studying the following problem
$ \left\{\begin{matrix} u_{t}=au _{xx}-u_x & 0<x<1;\ \ 0<t\\ u(x,0)=f(x) & x\in[0,1] & \\ u(0,t)= 0 & t\in[0,\infty) & \\ u(1,t)=1 & t\in[0,\infty) \\ \end{matrix}\right. $
I am advised to first study the steady-state solution, that is, the solution to the equation $av_{xx} -v_x=0$ (with same conditions). Having solved it, $v(x,t)=\frac{1}{e^{1/a}-1}(e^{x/a} -1)$; I would like to show $u$ converge in $L^2$ to the steady state solution: $\|u-v\|(t)\xrightarrow{t\rightarrow \infty} 0$, but i seem to miss something. My work so far:
$$ \begin{align} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\|u-v\|^2 =&\int_0 ^1 (u-v)(u_t -v_t) =\int_0^1(u-v)(au_{xx}-u_x) \\ =&a\int_0^1u u_{xx} -\int_0^1 uu_x -a\int_0^1 vu_{xx} +\int_0^1vu_x \\ =&\left(\underbrace{auu_x\vert_0^1}_{=0} - a\|u_x\|^2\right)-\left(\underbrace{\frac{u^2}{2}\vert^1_0}_{=1/2}\right) +\left(- \underbrace{avu_x\vert_0^1}_{=0}+a\int_0^1v_xu_x \right)+\int_0^1 vu_x \\ =&-a\|u_x\|^2 - 1/2+\left(\underbrace{avu_x\vert_0^1}_{=0} - a\int_0^1v_{xx}u \right)+\int_0^1vu_x \\ =&-a\|u_x\|^2-1/2 +\int_0^1 v_xu +\int_0^1 vu_x \\ =& -a\|u_x\|^2-1/2 +\int_0^1 \frac{\mathrm{d}}{\mathrm{d}x}(uv) \\ =& -a\|u_x\|^2 -1/2 +1 \\ =& - a\|u_x\|^2 + 1/2 \end{aligned} \end{align}$$
But I cannot figure out how to proceed from here, the bounds are not good enough, I would like to connect the last result with $v$ or $v_x$ and then show the decay of the solution to the steady-state solution.
You're welcome to help!