Boltzmann and PDE: Lecture 5

 

(03 Feb 2026)

(You may see the rest of the lecture notes here).

Part I: Elliptic operators, local or not.


Last time

In the previous lecture we started our discussion of the existence theory. After some preliminaries, we proceeded to describe the method for the existence of weak solutions in a basic and simple setting: the heat equation.

We generated a sequence of functions $f_k(x)$ generated via a semi-discrete implicit Euler method with time step $h>0$, \begin{align} f_{k}-f_{k-1} = h\Delta f_k \text{ when } k \geq 1 \end{align} with $f_0$ chosen to be equal to $f_{\text{in}}$. From this sequence we built a function $f^{(h)}(t,x)$ by \begin{align} f^{(h)}(t,x) = f_k(x) \text{ if } t\in (kh,(k+1)h). \end{align}

Our aim now is to show that the following limit exists (for some notion of convergence) \begin{align} \lim \limits_{h\to 0} f^{(h)}(t,x), \end{align}

and that this limit solves the heat equation with initial data $f_{\text{in}}$.

Properties of $f^{(h)}$

We define the following discrete approximations to the operator $\partial_t \phi$,
\begin{align} \partial^{(h)}_t\phi(t,x) & = \frac{1}{h}(\phi(t+h,x)-\phi(t,x))\\ \
\partial^{(-h)}_t\phi(t,x) & = \frac{1}{h}(\phi(t,x)-\phi(t-h,x))
\end{align}

Then, the function $f^{(h)}$ solves the following Discrete Heat Equation

\begin{align} \partial_t^{(-h)}f^{(h)} = \Delta f^{(h)} \text{ in } (h,\infty)\times \mathbb{R}^n, \end{align}

this fact is the content of the next proposition.

Proposition. The function $f^{(h)}:(0,\infty)\times \mathbb{R}^n \to \mathbb{R}$ defined above belongs to the Hilbert space \begin{align} L^2((0,\infty),H^1(\mathbb{R}^n)), \end{align} and it solves the equation \begin{align} \partial^{(-h)}_t f^{(h)} = \Delta f^{(h)} \text{ in } (0,\infty)\times \mathbb{R}^n \end{align} in the following weak sense: for any test function $\phi \in C^\infty_c([0,\infty)\times \mathbb{R}^n)$ we have the identity \begin{align} \int_h^\infty\int_{\mathbb{R}^n} (\partial^{(-h)}_t f^{(h)})\phi + \nabla f^{(h)}\cdot \nabla \phi \;dxdt = 0. \end{align}

Proof. For any fixed $t\in [kh,(k+1)h)$, we have \begin{align} (f_k-f_{k-1})\phi(x,t) - h(\Delta f_k)\phi(x,t) = 0 \;\forall\;x\in\mathbb{R}^n \end{align} Integrating in $x$ and integrating by parts we have that \begin{align} \int_{\mathbb{R}^n}(f_k(x)-f_{k-1}(x))\phi(t,x)\;dx + h\int_{\mathbb{R}^n}\nabla f_k(x)\cdot \nabla \phi(t,x)dx = 0 \end{align} This identity holds for $k=1,2,\ldots$. We will add these identities, starting from $k=1$ and ending at $k=m$, for some $m\in\mathbb{N}$. For this we note that \begin{align} \sum \limits_{k=1}^m\int_{\mathbb{R}^n}(f_k-f_{k-1})\phi(t,x)\;dx & = \int_h^{mh}\int_{\mathbb{R}^m} \partial^{(-h)}_t f^{(h)} \phi(t,x)\;dx, \text{ and }\\\ \\\ \sum \limits_{k=1}^m h \int_{\mathbb{R}^n}\nabla f_k \cdot \nabla \phi(t,x)dx & = \int_h^{mh}\int_{\mathbb{R}^n}\nabla f^{(h)}(t,x) \cdot \nabla \phi(t,x)dxdt. \end{align} Adding these identities we have \begin{align} \int_h^{mh}\int_{\mathbb{R}^n}\partial_t^{(-h)}f^{(h)}(t,x)\phi(t,x) + \nabla f_k \cdot \nabla \phi(t,x)dxdt = 0 \end{align} Now, $m$ is arbitrary, and $\phi(t,x) \equiv 0$ for all $x$ if $t$ is large enough, so we actually have \begin{align} \int_h^{\infty}\int_{\mathbb{R}^n}\partial_t^{(-h)}f^{(h)}(t,x)\phi(t,x) + \nabla f_k \cdot \nabla \phi(t,x)dxdt = 0 \end{align} and the proposition is proved.

One can actually write the equation in alternative weak form, by approximately “integrating by parts” in $t$ to move $\partial^{(-t)}_h$ from $f$ to $\phi$. We leave the main step as an exercise, and state the alternative formulation as a corollary.

Exercise. Show that if $\phi\in C^\infty_c([0,\infty)\times \mathbb{R}^n)$ then for any $T>0$ we have \begin{align} \int_h^T\int_{\mathbb{R}^n} \partial_t^{(-h)}f^{(h)} \phi\;dxdt & = \int_h^{T-h}\int_{\mathbb{R}^n}f^{(h)}(-\partial_t^{(h)}\phi)\;dxdt \\\ \\\ & + \frac{1}{h}\int_{T-h}^T\int_{\mathbb{R}^n}f^{(h)}\phi\;dxdt - \frac{1}{h}\int_0^h\int_{\mathbb{R}^n}f^{(h)}\phi_h dxdt \end{align} where $\phi_h(t,x) := \phi(t+h,x)$. In particular, \begin{align} \int_h^\infty\int_{\mathbb{R}^n} \partial_t^{(-h)}f^{(h)} \phi\;dxdt & = \int_h^{\infty}\int_{\mathbb{R}^n}f^{(h)}(-\partial_t^{(h)}\phi)\;dxdt \\\ \\\ & - \frac{1}{h}\int_0^h\int_{\mathbb{R}^n}f^{(h)}\phi_h dxdt. \end{align}

Corollary. The function $f^{(h)}$ from the previous property has the following additional property: for any test function $\phi \in C^{\infty}_c([0,\infty)\times\mathbb{R}^N)$ we have \begin{align} \int_h^\infty\int_{\mathbb{R}^n} -f^{(h)}\partial^{(h)}_t\phi + \nabla f^{(h)}\cdot \nabla \phi \;dxdt = \frac{1}{h}\int_0^h\int_{\mathbb{R}^n}f_{\text{in}}(x)\phi(t+h,x) dxdt. \end{align}

Remark. Observe that the right hand side in the corollary only depends on $h$ through $\phi$, and that once the test function $\phi$ is fixed then as $h\to 0$ the right hand side converges to \begin{align} \int_{\mathbb{R}^n} f_{\text{in}}(x)\phi(0,x)\;dx. \end{align}

Remark. From the corollary, we should expect that if the functions $f^{(h)}$ have some weak limit as $h\to 0$ (even along a subsequence) then that limit $f$ will be a weak solution of the equation in the sense that if

\begin{align} \int_0^\infty\int_{\mathbb{R}^n} -f\partial_t\phi + \nabla f^{(h)}\cdot \nabla \phi \;dxdt = \int_{\mathbb{R}^n}f_{\text{in}}(x)\phi(0,x) dx. \end{align}

Accordingly, what we need to do now is obtain some bound on $f^{(h)}$ that will guarantee that some limit of $f^{(h)}$ exists (at least along a subsequence) as $h\to 0$.

The way we will guarantee the existence of a limit will be by obtaining some integral bounds on $f^{(h)}$ that will be independent of $h$. For this, we will use a slightly more general version of the last corollary, which follows by the same argument: for any $h>0$ and any $T>h$ we have \begin{align} \int_h^{T-h}\int_{\mathbb{R}^n} f^{(h)}(-\partial_t^{-(h)}\phi)+\nabla f^{(h)}\cdot \nabla \phi\;dxdt = \frac{1}{h}\int_0^h\int_{\mathbb{R}^n}f^{(h)}\phi\;dxdt - \int_{T-h}^T\int_{\mathbb{R}^n}f^{(h)}\phi\;dxdt. \end{align} (Recall from the previous lecture the identity \begin{align} \frac{1}{2}||f(t_2)||_2^2 + \int_{t_1}^{t_2}\int_{\mathbb{R}^n}|\nabla f|^2dxdt = \frac{1}{2}||f(t_1)||_2^2, \end{align} we are looking for a discrete version for our semi-discrete solution $f^{(h)}$).

Observe that for the sequence $\{f_k\}_k$ we have

\begin{align} \frac{1}{h} \left ( ||f_{k+1}||^2-||f_k||^2 \right ) = \frac{1}{h}(f_{k+1}-f_k,f_{k+1}+f_k) \end{align}

From this identity, and the last Proposition follows that for any $T = mh$ with $m \in \mathbb{N}$ we have

\begin{align} \frac{1}{2}||f^{(h)}(T)||_2^2 + \frac{1}{2}\int_h^T\int_{\mathbb{R}^n}|\nabla f^{(h)}|^2+(\nabla f^{(h)}\cdot \nabla f^{(h)}(\cdot-h,\cdot))dxdt = \frac{1}{2}||f_{\text{in}}||_2^2. \end{align}

Now, it is not difficult to see directly from the definition of $f_k$ in terms of $f_{k-1}$ that

\begin{align} \int_{\mathbb{R}^n} \nabla f_k\cdot \nabla f_{k-1}\;dx \geq \int_{\mathbb{R}^n}|\nabla f_k|^2\;dx \end{align}

and thus \begin{align} \frac{1}{2}||f^{(h)}(T)||_2^2 + \int_h^T\int_{\mathbb{R}^n}|\nabla f^{(h)}|^2dxdt = \frac{1}{2}||f_{\text{in}}||_2^2 \end{align}

This yields the following estimate for $f^{(h)}$,

\begin{align} \sup \limits_{t>0} \frac{1}{2}||f^{(h)}(t)||_2^2 + \int_h^\infty \int_{\mathbb{R}^n}|\nabla f^{(h)}|^2\;dxdt \leq \frac{1}{2}||f_{\text{in}}||_2^2. \end{align}

This shows in particular that $\{f^{(h)} \}_{h>0}$ is uniformly bounded (after restricting to $h<h_0$) in the semi norm \begin{align} f\mapsto \int_{h_0}^\infty \int_{\mathbb{R}^n}|\nabla f|^2\;dxdt. \end{align}

This will give us all the compactness we need to pass to the limit.

The Laplacian and the spaces $H^1$ and $(H^1)^*$

Given $f\in L^2(\mathbb{R}^n)$ we may define an element of the dual $(H^1(\mathbb{R}^n))^*$ by setting

\begin{align} \phi_f(g) = \int_{\mathbb{R}^n}fg\;dx. \end{align}

One can use a version of this to define $\Delta f$ as an element of $(H^1(\mathbb{R}^n))^*$ whenever $f \in H^1(\mathbb{R}^n)$, by noting that if $f \in H^1(\mathbb{R}^n)$ and $\phi \in C^\infty_c(\mathbb{R}^n)$,

\begin{align} \langle \Delta f, \phi \rangle = \int_{\mathbb{R}^n} f \Delta \phi \;dx = - \int_{\mathbb{R}^n} \nabla f\cdot \nabla \phi\;dx. \end{align}

Therefore, one could define a functional $\tilde \phi_f \in (H^1(\mathbb{R}^n))^*$ by

\begin{align} \tilde \phi_f(g) = -\int_{\mathbb{R}^n}\nabla f\cdot \nabla g\;dx \end{align}

and this functional extends the functional $\langle \Delta f,\cdot\rangle$ from test functions to all of $H^1(\mathbb{R}^n)$. That this functional is bounded follows from the Cauchy-Schwartz inequality and that $f\in H^1(\mathbb{R}^n)$,

\begin{align} |\tilde \phi_f(g)| \leq ||\nabla f||_{L^2}||\nabla g||_{L^2} \leq ||f||_{H^1}||g||_{H^1} \end{align}

It is in this way we will understand $\Delta f$ when $f\in H^1(\mathbb{R}^n)$, as a bounded linear functional in $H^1(\mathbb{R}^n)$ itself. The map $f\mapsto \Delta f$ thus understood defines a continuous operator $\Delta: H^1(\mathbb{R}^n) \to (H^1(\mathbb{R}^n))^*$, with

\begin{align} ||\Delta f||_{(H^1)^*} \leq ||\nabla f||_{L^2}\leq ||f||_{H^1} \end{align}

What have we learned so far?

If we fix $\tau_0>0$ (small) and $T_0>0$ (large) the families

\begin{align} \{ f^{(h)}\}_h \text{ and } \{\partial^{(-h)}_tf^{(h)}\}_h \end{align}

are uniformly bounded in the respective Hilbert spaces

\begin{align} L^2((\tau_0,T_0),H^1(\mathbb{R}^n)) \text{ and } L^2((\tau_0,T_0),(H^1(\mathbb{R}^n))^*) \end{align}

By weak compactness of bounded sets in a Hilbert space we know there is a subsequence $h_k \to 0$ and $f_\infty \in L^2((\tau_0,T_0), H^1(\mathbb{R}^n))$ such that

\begin{align} f^{(h_k)} \rightharpoonup f_\infty \end{align}

Lemma. Consider the function $f_\infty(t,x)$ defined from the above limit, then, given any $\phi \in C^\infty_c([0,\infty)\times \mathbb{R}^n)$ we have \begin{align} \int_0^\infty \int_{\mathbb{R}^n} f_\infty (-\partial_t \phi) +\nabla f\cdot \nabla \phi \; dxdt = \int_{\mathbb{R}^n}\phi(x,0)f_{\text{in}}(x,0)dx. \end{align}

We will prove this in the next lecture. For now, some consequences

(Why? Since they are both weak solutions with the same initial data, it follows by linearity that $f_1-f_2$ is a weak solution with zero initial data, applying the energy identity from the previous point we conclude that $f_1-f_2$ must be zero for a.e. $(t,x)$. )