I consider the following problem on the half real line
$$ (\star) \begin{cases} u_t = u_{xx} + u_x, & \quad t > 0, \, x > 0, \\[2mm] -u_x|_{x=0} = u|_{x=0}, & \quad t > 0, \, x = 0, \\[2mm] u|_{t=0} = u_0, & \quad t = 0, \, x > 0, \end{cases} $$$$ \begin{cases} u_t = u_{xx} + u_x, & \quad t > 0, \, x > 0, \\[2mm] -u_x|_{x=0} = u|_{x=0}, & \quad t > 0, \, x = 0, \\[2mm] u|_{t=0} = u_0, & \quad t = 0, \, x > 0, \end{cases}\label{1}\tag{$\star$} $$
where $u_0$ is nonnegative, bounded and compactly supported. This problem preserves the $L^{1}$-norm of $u_0$, that is $\Vert u(t,\cdot)\Vert_{L^{1}(\mathbb{R}^*_+)}=\Vert u_0 \Vert_{L^{1}(\mathbb{R}^*_+)}$ for any $t>0$, and owns nonnegative steady states of the form
$$ U_\lambda(x) = \lambda e^{-x}, $$
where $\lambda\geq 0$ is the mass of $U_\lambda$.
I want to show that, for $\lambda=\Vert u_0 \Vert_{L^{1}(\mathbb{R}^*_+)}$, we have
$$ \boxed{\lim\limits_{t\to \infty} \Vert u(t,\cdot) - U_\lambda \Vert_{L^{1}(\mathbb{R}^*_+)} = 0} $$
by using entropy methods.
I tried the relative entropy
$$ H(t) : = \int_0^\infty u(t,x) \log \Big( \frac{u(t,x)}{U_\lambda(x)} \Big) dx. $$
For this entropy, I get for the dissipation
\begin{align} \frac{d}{dt}H (t) & = - \int_0^\infty \Bigg( \frac{u_x}{\sqrt{u}} +\sqrt{u} \Bigg)^{2} dx = - I(u|U_\lambda) \leq 0, \end{align}
where
$$ \begin{align} \frac{d}{dt}H (t) & = - \int_0^\infty \Bigg( \frac{u_x}{\sqrt{u}} +\sqrt{u} \Bigg)^{2} dx = - I(u|U_\lambda) \leq 0, \end{align} $$ where $$ I(u|U_\lambda) = \int_0^\infty u(t,x) \Bigg[ \partial_x \log \Big( \frac{u(t,x)}{U_\lambda(x)} \Big) \Bigg]^{2} dx $$
stands stands for the Fisher information.
This basically tells me that this entropy $H(t)$ dissipates but I have no idea on how to prove the convergence of $H$ to $0$ --- this fact would solve my problem thanks to Csiszár-Kullback inequality that basically says that $\Vert u(t,\cdot) - U_\lambda \Vert_{L^{1}}^{2}\leq k \times H(t)$.
For the following problem
$$ (\star\star) \begin{cases} v_t = v_{xx} + (x v)_x, & \quad t > 0, \, x > 0, \\[2mm] -v_x|_{x=0} = 0, & \quad t > 0, \, x = 0, \\[2mm] v|_{t=0} = v_0, & \quad t = 0, \, x > 0, \end{cases} $$$$ \begin{cases} v_t = v_{xx} + (x v)_x, & \quad t > 0, \, x > 0, \\[2mm] -v_x|_{x=0} = 0, & \quad t > 0, \, x = 0, \\[2mm] v|_{t=0} = v_0, & \quad t = 0, \, x > 0, \end{cases}\label{2}\tag{$\star\star$} $$
we can do similar computations to prove the convergence towards Gaussian steady states
$$ V_\lambda(x) = \lambda e^{-x^{2}/2} $$
by using the log-Sobolev inequality that gives
$$ \frac{d}{dt}H (t) \leq - I(v|V_\lambda) \leq - k \times H (t), $$
and therefore, $H(t)\leq H(0) e^{-k t} \to 0$.
However, this log-Sobolev inequality does not apply in case $(\star)$\eqref{1} since (if I have correctly understood) the potential
$$-\log (U_\lambda) (x) = x - \log (\lambda) $$
is not convex while in case $(\star \star)$\eqref{2},
$$-\log (V_\lambda) (x) = \frac{x^2}{2} - \log (\lambda) $$
is convex.
Any ideas or suggestions are welcome!