Part IV: The nonlinear Cauchy problem

\begin{align} \partial_t f & = q(f,f) \text{ in } \mathbb{R}^n\times [0,T_s) \\ \
f_{\mid t=0} & = f_{\text{in}} \text{ in } \mathbb{R}^n \end{align}

From now on we normalize the initial mass, assuming without loss of generality that \begin{align} \int_{\mathbb{R}^n}f_{\text{in}}(v)\;dv = 1. \end{align}

Theorem I.
Take $n=3$ and $\alpha(r) = cr^{\gamma}$ for some $\gamma \in [-3,0]$. Consider $f(t,v)$ as above, and suppose that in addition, \begin{align} \int_{\mathbb{R}^3}\frac{|\nabla f_{\text{in}}|^2}{f_{\text{in}}}\;dv < \infty. \end{align} Then, the integral \begin{align} \int_{\mathbb{R}^3}\frac{|\nabla f(t,v)|^2}{f(t,v)}\;dv \end{align} is non-increasing for $t\in [0,T_s)$.

Theorem II.
Consider the same assumptions as in Theorem I, and assume further \begin{align} \int_{\mathbb{R}^n}f_{\text{in}}|v|^k\;dv <\infty \end{align} for $k>k(\gamma)$. Then there is no finite time blown up, i.e. $T_s=+\infty$ and thus there is a smooth solution for all $t>0$.

The method provides similar conclusions for other dimensions, for instance, if $n=2$ then the theorem holds for any $\gamma \in [-2,0]$. The full range of $\gamma$ covered by the theorem will be discussed later.

The integral quantity appearing in Theorem I is known as the Fisher information of $f$, and this will be a considerable focus of this chapter. Given a probability density $f:\mathbb{R}^n\to\mathbb{R}$ we define its Fisher information as the quantity \begin{align} i(f) := \int_{\mathbb{R}^n}\frac{|\nabla f|^2}{f}\;dv \end{align}

Proof sketch of Theorem II. By Theorem I, we have for every $t\in [0,T_s)$ \begin{align} i(f(t)) \leq i(f_{\text{in}}). \end{align} Now, \begin{align} \|\nabla \sqrt{f}(t)\|_{L^2}^2 = \int_{\mathbb{R}^3}|\nabla \sqrt{f}|^2(t,v)\;dv = \frac{1}{4}i(f(t))\leq \frac{1}{4}i(f_{\text{in}}) \end{align} Therefore, the Sobolev embedding for $H^1(\mathbb{R}^3)$ says that \begin{align} \|f(t)\|_{L^3} = \|\sqrt{f}(t)\|_{L^6}^2 \leq C_{\text{Sob}}^2 \|\nabla \sqrt{f}(t) \|_{L^2}^2 \leq \frac{1}{4}C_{\text{Sob}}^2i(f_{\text{in}}). \end{align} We conclude that $\|f(t)\|_{L^3}$ remains bounded as $t\to T_s$, i.e. $f \in L^\infty_t L^3_v((0,T_s)\times \mathbb{R}^3)$. As it turns out, this is enough to prevent finite time blow up. We will only sketch this, and refer to the literature, particularly Silvestre (2017). The proof there is based on non-divergence techniques, here we give a heuristic using energy methods (see also Gualdani-Guillen 2016).
Fix $p>1$, we have \begin{align} \frac{1}{p}\frac{d}{dt}\int f^p\;dv & = -(p-1)\int (f^{p-2}\nabla f,A\nabla f-\text{div}(A_f)f)\;dv \\ & = -\frac{4(p-1)}{p^2}\int (A\nabla f^{p/2},\nabla f^{p/2})\;dv + \frac{p-1}{p}\int h_f f^p\;dv \end{align} Here, we are going to make use of weighted Sobolev inequalities and Holder's inequality, namely \begin{align} \int h_f f^p\;dv \leq \left ( \int h_f^{3/2}\langle v\rangle^{3|\gamma|}\;dv\right )^{2/3}\left (\int \langle v\rangle^{3\gamma}f^{3p}\;dv \right )^{1/3} \end{align} Ultimately, using the weighted Sobolev inequality in $\mathbb{R}^3$, we have \begin{align} \frac{1}{p}\frac{d}{dt}\int f^p\;dv & \leq -\frac{2(p-1)}{p^2}\int (A\nabla f^{p/2},\nabla f^{p/2})\;dv + \frac{p-1}{p}M\int_{\{h_f<M \} } f^p\;dv \end{align} From here we see one can propagate all $L^p$ norms, in particular, if $f_{\text{in}}$ is initially in $L^\infty$, then all the $L^p$ norms of $f_{\text{in}}$ remain bounded as $t\to T_s^-$.

Lifting

Motivation: Boltzmann’s $H$-theorem.

\begin{align} \frac{d}{dt}\int_{\mathbb{R}^n}f\log f\;dv = \int_{\mathbb{R}^n}(\partial_tf)\log f(v)\;dv \end{align}

\begin{align} \int_{\mathbb{R}^n}(\partial_tf)\log f(v)\;dv = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\text{div}_{v-w}(\alpha a(v-w)\nabla_{v-w}[f(v)f(w)]) \log f(v)\;dwdv \end{align} The double integral on the right is unchanged if we exchange $(v,w)$, therefore, it is equal also to \begin{align} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\text{div}_{v-w}(\alpha a(v-w)\nabla_{v-w}[f(v)f(w)]) \log f(w)\;dv \end{align} We conclude that \begin{align} \int_{\mathbb{R}^n}(\partial_tf)\log f(v)\;dv = \frac{1}{2}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\text{div}_{v-w}(\alpha a(v-w)\nabla_{v-w}[f(v)f(w)]) \log [f(v)f(w)]\;dwdv \end{align} Integrating by parts we then have \begin{align} \int_{\mathbb{R}^n}(\partial_tf)\log f(v)\;dv = -\frac{1}{2}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}(\alpha a(v-w)\nabla_{v-w}[f(v)f(w)],\nabla_{v-w}\log [f(v)f(w)]) \;dwdv \end{align} This suggests it is helpful to study the linear differential operator \begin{align} Q(F)(v,w) := \text{div}_{v-w}( \alpha(|v-w|)a(v-w)\nabla_{v-w}F)(v,w) \end{align}

Consider the initial value problem \begin{align} \partial_t F & = Q(F) \text{ in } \mathbb{R}^{2n}\times (0,\infty), \\ \
F_{\mid t=0} & = f(t_0,v)f(t_0,w) \text{ in } \mathbb{R}^{2n}. \end{align} This linear evolution equation in $\mathbb{R}^{2n}$ is called the lifted equation. Observe that this is a linear and (weakly) parabolic equation.

\begin{align} \frac{d}{dt}_{\mid t=0} \int_{\mathbb{R}^{2n}}F\log F\;dwdv = 2\frac{d}{dt}_{\mid t=t_0}\int_{\mathbb{R}^n}f\log f\;dv \end{align}

The entropy is decreasing for the Landau equation because the entropy functional in $\mathbb{R}^{2n}$ is decreasing for the lifted equation. Turns out the same holds for any functional with what we call a lifting.

Definition. Consider a functional $j(f)$ defined over probability densities $f$ in $\mathbb{R}^n$, a lifting of $j$ is a functional $J(F)$ defined over probability densities $F$ in $\mathbb{R}^{n}\times \mathbb{R}^n$ satisfying the following two conditions:

1. For any two probability densities $f_1,f_2$ in $\mathbb{R}^n$ we have \begin{align} J(f_1\otimes f_2) = j(f_1)+j(f_2) \end{align} 2. If $F$, a probability density in $\mathbb{R}^{2n} = \mathbb{R}^n\times \mathbb{R}^n$ has marginals $f_1$ and $f_2$, then \begin{align} J(F) \geq j(f_1)+j(f_2) \end{align}

Lemma. If $J$ is a lifting of $j$, then with $f(t)$ and $F(t)$ as before we have \begin{align} 2\frac{d}{dt}_{\mid t=t_0}i(f(t)) = \frac{d}{dt}_{\mid t=0}I(F(t)) \end{align}

The Fisher information in $\mathbb{R}^n$ can be ‘‘lifted’’ to the respective Fisher information in $\mathbb{R}^{2n}$.

Lemma. The Fisher information in $\mathbb{R}^{2n}$ is a lifting of the Fisher information in $\mathbb{R}^n$.

Proof.(sketch)
First, we prove that \begin{align} I(f_1\otimes f_2) = i(f_1)+i(f2) \end{align} For this, note that if $F = f_1\otimes f_2$ then \begin{align} \log F(v,w) = \log f_1(v) + \log f_2(w) \end{align} and therefore \begin{align} |\nabla \log F(v,w)|^2 = |\nabla \log f_1(v)|^2+|\nabla \log f_2(w)|^2 \end{align} Multiplying both sides by $F$, and integrating \begin{align} & \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|\nabla \log F(v,w)|^2F(v,w)dvdw \\ & = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|\nabla \log f_1(v)|^2f_1(v)f_2(w)+|\nabla \log f_2(w)|^2f_1(v)f_2(w)\;dvdw. \end{align} The integral on the right is then the sum of two integrals that separate, and we have \begin{align} I(F) & = \left ( \int f_2(v)dv\right) i(f_1) + \left ( \int f_1(v)dv\right) i(f_2) \end{align} Since we are dealing with probability densities, this shows $I(F) = i(f_1)+i(f_2)$. Now let us prove that if $F$ has marginals $f_1,f_2$ then \begin{align} I(F) \geq i(f_1)+i(f_2). \end{align} By definition, $f_1(v) = \int_{\mathbb{R}^n}F(v,w)dw$, so \begin{align} \nabla \log f_1(v) = \frac{\int_{\mathbb{R}^n}\nabla_v F(v,w)dw}{\int_{\mathbb{R}^n}F(v,w)dw}. \end{align} Now, by Jensen's inequality \begin{align} \left |\int_{\mathbb{R}^n}\nabla_v F(v,w)\;dw\right |^2 & \leq \int_{\mathbb{R}^n}(\nabla_v \log F(v,w)) F(v,w)\;dw \\ & = \left ( \int_{\mathbb{R}^n}|\nabla_v \log F(v,w)|^2 F(v,w)\;dw\right )\int_{\mathbb{R}^n}F(v,w)\;dw \end{align} Therefore, \begin{align} |\nabla \log f_1(v)|^2 \leq \frac{\int_{\mathbb{R}^n}|\nabla_v \log F(v,w)|^2F(v,w)\;dw}{\int_{\mathbb{R}^n}F(v,w)dw} = \frac{1}{f_1(v)}\int_{\mathbb{R}^n}|\nabla_v \log F(v,w)|^2F(v,w)\;dw \end{align} Multipying both sides by $f_1(v)$ and integrating in $v$ we have \begin{align} i(f_1) = \int |\nabla \log f_1(v)|^2f_1(v)\;dv \leq \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|\nabla_v \log F(v,w)|^2F(v,w)dw \end{align} The same argument repeated now for the second marginal yields the respective bound for $i(f_2)$ but now involving the derivative with respect to $w$, \begin{align} i(f_2) \leq \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|\nabla_w \log F(v,w)|^2F(v,w)dw \end{align} Adding these two inequalities we have $i(f_1)+i(f_2) \leq I(F)$ as we wanted.

Corollary. \begin{align} 2\frac{d}{dt}_{\mid t=t_0}i(f(t)) = \frac{d}{dt}_{\mid t=0} I(F(t)) \end{align}

This Corollary is a significant reduction of the problem of computing the rate of change for $i(f)$: the expression on the left involves a nonlinear and nonlocal evolution equation, while the expression on the right involves a linear and local (weakly) parabolic equation. The trade off (if it can be even called that) is that the equation on the right is set in double the number of state variables.

Now then, we focus on showing the Fisher information is decreasing for the lifted equation for all the $\alpha(r)$’s of interest.

Theorem III. Suppose that \begin{align} \Lambda(\alpha) := \left ( \sup \frac{|r\alpha'(r)|}{2\alpha(r)}\right )^2 \leq \frac{11}{2} \end{align} Then for any solution of the lifted equation we have \begin{align} \frac{d}{dt}I(F(t)) \leq 0 \end{align} In particular, this will be the case if $\gamma(r) = cr^\gamma$ for $\gamma \in [-3,1]$.