## [27.4.2016] Generalized entropy method for the renewal equation with measure data

In this paper of P. Gwiazda and E. Wiedemann, they apply the generalized entropy method, which is initated in a series of work of B. Perthame and collaborators, cf. this paper, to show the exponential convergence to equilibrium for the renewal equation with measure initial data.

$\partial_t n(t,x) + \partial_{x}n(t,x) = 0$ on $\mathbb R_+^2$

$n(t,x=0) = \int_0^{\infty}B(y)n(t,y)dy$

$n(t=0,x) = n_0(x)$

This equation has been extensively studied recently by many authors due to its application to biology.

The convergence to equilibrium by using relative entropy method for this problem was known for $L^1$-initial data. By denoting $N(x)$ and $\varphi(x)$ are the solutions to an eigenvalue problem and its dual, and defining

$h(t,x) = n(t,x) - N(x)\int_0^{\infty}n_0(y)\phi(y)dy$

with some suitable function $\phi$, then we have the large time behaviour of $n(t,x)$ as follows

$\int_{0}^{\infty}|h(t,x)|\varphi(x)dx \leq e^{-\mu(t-y_0)}\int_0^{\infty}|h(y_0,x)|\varphi(x)dx$.

This result was based on the so-called entropy method (see this paper for more details).

For measure initial data, the arguments applied to $L^1$-initial data seems not to be directly applicable. However, looking at the convergence result, we would expect similar results for measure data (with some suitable changes).

This is what was done in the paper of Gwiazda and Wiedemann. The main idea is to use recession function $f^{\infty}$ for a function $f$ defined as

$f^{\infty}(z) = \lim\limits_{s\rightarrow \infty}\dfrac{f(sz)}{s},\quad z\in \mathbb R^n-\{0\}$

provided that $f$ grows mostly linearly. By expoloiting these functions, the authors succeeded in choosing a convex function making the entropy method works in the case of measure initial data. Denote by

$g(t,x) = n(t,x) - N(x)\int_0^{\infty}\varphi(x)dn_0(x)dx$,

$\int_0^{\infty}\eta(x)d|g(t,x)| \leq e^{-\sigma(t-y_0)}\int_0^{\infty}\eta(x)d|g(0,x)|$
for some bounded function $\eta$ positive on $supp(\varphi)$.