$L^{1}$-convergence to steady states for an advection-diffusion equation on the half real line

Question

I consider the following problem on the half real line

$$ \begin{cases} u_t = u_{xx} + u_x, & \quad t > 0, \, x > 0, \\[2mm] -u_x|_{x=0} = u|_{x=0}, & \quad t > 0, \, x = 0, \\[2mm] u|_{t=0} = u_0, & \quad t = 0, \, x > 0, \end{cases}\label{1}\tag{$\star$} $$

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 $U_\lambda$.

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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.

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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 $$ I(u|U_\lambda) = \int_0^\infty u(t,x) \Bigg[ \partial_x \log \Big( \frac{u(t,x)}{U_\lambda(x)} \Big) \Bigg]^{2} dx $$ stands 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

$$ \begin{cases} v_t = v_{xx} + (x v)_x, & \quad t > 0, \, x > 0, \\[2mm] -v_x|_{x=0} = 0, & \quad t > 0, \, x = 0, \\[2mm] v|_{t=0} = v_0, & \quad t = 0, \, x > 0, \end{cases}\label{2}\tag{$\star\star$} $$

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 \eqref{1} since (if I have correctly understood) the potential

$$-\log (U_\lambda) (x) = x - \log (\lambda) $$

is not convex while in case \eqref{2},

$$-\log (V_\lambda) (x) = \frac{x^2}{2} - \log (\lambda) $$

is convex.

Any ideas or suggestions are welcome!

Answer 1

Here's a solution by semigroup methods (rather than entropy methods) - but I don't know if the argument can be adapted to the non-autonomous situaion mentioned by the OP in the comments.

The solution of the equation is given by $u(t) = T(t)u_0$ for all $t \ge 0$, where $T = (T(t))_{t \ge 0}$ is a stochastic $C_0$-semigroup on $L^1(0,\infty)$ ("stochastic" means that for every initial function $0 \le u_0 \in L^1(0,\infty)$ one has $T(t)u_0 \ge 0$ and $\|T(t)u_0\| = \|u_0\|$ for all $t \ge 0$).

As pointed out in the question, the semigroup $T$ has a fixed point in $L^1$ which is $>0$ almost everywhere.

Moreover, one can show that each of the operators is, for $t > 0$, given by integration against an integral kernel (for general elliptic operators on $d$-dimensional domains an argument for this can be found in Section 5.2 of this article by Wolfgang Arendt; in the one-dimensional situation discussed here, one can probably find an easier argument).

So to sum up, the semigroup $T$ is positive and contractive (since it is stochastic), it has a fixed point that is $>0$ almost everywhere and it consists of operators given by integral kernels. This implies that $T(t)$ converges strongly as $t \to \infty$. This is a result from the 1980s that goes back to Günther Greiner. See for instance Theorem 1 in this article by Moritz Gerlach and myself for a quite straightforward proof (the article also contains references to more general results and to Greiner's original proof).