Lecture 3. Diffusion operators

Posted on January 27, 2019 by Fabrice Baudoin

In this lecture, we illustrate some of the concepts introduced earlier in the context of diffusion operators in \mathbb{R}^n.

Throughout the lecture, we consider a second order differential operator that can be written

L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},
where b_i and \sigma_{ij} are continuous functions on \mathbb{R}^n and for every x \in \mathbb{R}^n, the matrix (\sigma_{ij}(x))_{1\le i,j\le n} is a symmetric and non negative matrix. Such operator is called a diffusion operator.

We will assume that there is Borel measure \mu which is equivalent to the Lebesgue measure and that symmetrizes L in the sense that for every smooth and compactly supported functions f,g : \mathbb{R}^n \rightarrow \mathbb{R},

\int_{\mathbb{R}^n} g Lf d\mu= \int_{\mathbb{R}^n} fLg d\mu.
In what follows, as usual, we denote by C_0^\infty(\mathbb{R}^n) the set of smooth and compactly supported functions f : \mathbb{R}^n \rightarrow \mathbb{R}.

Exercise: On C_0^\infty(\mathbb{R}^n), let us consider the operator
L=\Delta +\langle \nabla U, \nabla \cdot \rangle,

where U: \mathbb{R}^n \rightarrow \mathbb{R} is a C^1 function. Show that L is symmetric with respect to the measure
\mu (dx)=e^{U(x)} dx.

Exercise: (Divergence form operator) On C_0^\infty(\mathbb{R}^n), let us consider the operator
Lf=\mathbf{div} (\sigma \nabla f),
where \mathbf{div} is the divergence operator defined on a C^1 function \phi: \mathbb{R}^n \rightarrow \mathbb{R}^n by

\mathbf{div} \text{ } \phi=\sum_{i=1}^n \frac{\partial \phi_i}{\partial x_i}

and where \sigma is a C^1 field of non negative and symmetric matrices. Show that L is a diffusion operator which is symmetric with respect to the Lebesgue measure of \mathbb{R}^n.

For every smooth functions f,g: \mathbb{R}^n \rightarrow \mathbb{R}, let us define the so-called carre du champ operator which is the symmetric first-order differential form defined by:
\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).
A straightforward computation shows that
\Gamma (f,g)=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j},
so that for every smooth function f,
\Gamma(f,f) \ge 0.

 
In the sequel we shall consider the bilinear form given for f,g \in C_0^\infty(\mathbb{R}^n) by
\mathcal{E} (f,g)=\int_{\mathbb{R}^n} \Gamma (f,g) d\mu.
This is the quadratic associated to L. It is readily checked that \mathcal{E} is symmetric:
\mathcal{E} (f,g)=\mathcal{E} (g,f),
and non negative
\mathcal{E} (f,f) \ge 0.

We may observe that thanks to symmetry of L,
\mathcal{E}(f,g)=-\int_{\mathbb{R}^n} fLg d\mu=-\int_{\mathbb{R}^n} gLf d\mu.

The operator L on its domain \mathcal{D}(L)= C_0^\infty(\mathbb{R}^n) is a densely defined non positive symmetric operator on the Hilbert space L^2( \mathbb{R}^n, \mu). However, it is not self-adjoint (why?).
The following proposition provides a useful sufficient condition for essential self-adjointness that is easy to check for several diffusion operators. We recall that a diffusion operator is said to be elliptic if the matrix \sigma is invertible.

Proposition: If the diffusion operator L is elliptic with smooth coefficients and if there exists an increasing sequence h_n\in C_0^\infty(\mathbb{R}^n), 0 \le h_n \le 1, such that h_n\nearrow 1 on \mathbb{R}^n, and ||\Gamma (h_n,h_n)||_{\infty} \to 0, as n\to \infty, then the operator L is essentially self-adjoint.

Proof: Let \lambda >0. According to a previous exercise, it is enough to check that if L^* f=\lambda f with \lambda >0, then
f=0. As it was observed above, L^* f=\lambda f is equivalent to the fact that, in sense of distributions, Lf =\lambda f.
From the hypoellipticity of L, we deduce therefore that f is a smooth function. Now, for h \in C_0^\infty(\mathbb{R}^n),

\int_{\mathbb{R}^n} \Gamma( f, h^2f) d\mu  =-\langle f, L(h^2f)\rangle_{L^2( \mathbb{R}^n, \mu)}
=-\langle L^*f ,h^2f \rangle_{L^2( \mathbb{R}^n, \mu)}
=-\lambda \langle f,h^2f\rangle_{L^2( \mathbb{R}^n, \mu)}
=-\lambda \langle f^2,h^2 \rangle_{L^2( \mathbb{R}^n, \mu)}
 \le 0.

Since
\Gamma( f, h^2f)=h^2 \Gamma (f,f)+2 fh \Gamma(f,h),
we deduce that

\langle h^2, \Gamma (f,f) \rangle_{L^2( \mathbb{R}^n, \mu)}+2 \langle fh, \Gamma(f,h)\rangle_{L^2( \mathbb{R}^n, \mu)} \le 0.
Therefore, by Cauchy-Schwarz inequality
\langle h^2, \Gamma (f,f) \rangle_{L^2( \mathbb{R}^n, \mu)} \le 4 \| f |_2^2 \| \Gamma (h,h) \|_\infty.
If we now use the sequence h_n and let n \to \infty, we obtain \Gamma(f,f)=0 and therefore f=0, as desired \square

Exercise: Let
L=\Delta +\langle \nabla U, \nabla \cdot\rangle,
where U is a smooth function on \mathbb{R}^n. Show that with respect to the measure \mu(dx)=e^{U(x)} dx, the operator L is essentially self-adjoint on C_0^\infty(\mathbb{R}^n).

Exercise:  On \mathbb{R}^n, we consider the divergence form operator
Lf=\mathbf{div} (\sigma \nabla f),
where \sigma is a smooth field of positive and symmetric matrices that satisfies
a \|x \|^2 \le \langle x , \sigma x \rangle \le b \|x \|^2, \quad x \in \mathbb{R}^n,
for some constant 0 < a \le b. Show that with respect to the Lebesgue measure, the operator L is essentially self-adjoint on C_0^\infty(\mathbb{R}^n).

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