Lecture 5. The diffusion semigroup as a solution to the heat equation

In this lecture, we show that the diffusion semigroup that was constructed in the previous lectures appears as the solution of a parabolic Cauchy problem. Under an ellipticity and completeness assumption, it is moreover the unique square integrable solution.

Proposition: Let L be an essentially self-adjoint diffusion operator and let (\mathbf{P}_t)_{t \ge 0} be the corresponding diffusion semigroup. Let f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), and let u (t,x)= \mathbf{P}_t f (x), \quad t \ge 0, x\in \mathbb{R}^n. Then u is a weak solution of the Cauchy problem
\frac{\partial u}{\partial t}= L u,\quad u (0,x)=f(x).

Proof: For \phi \in \mathcal{C}_c ((0,+\infty) \times \mathbb{R}^n,\mathbb{R}), we have
\int_{\mathbb{R}^{n+1}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right) u(t,x) d\mu(x) dt
=\int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right)  \mathbf{P}_t f (x) d\mu(x) dt
= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}}   \mathbf{P}_t \left( \left( -\frac{\partial}{\partial t} -L \right) \phi (t,x) \right)  f (x) d\mu(x) dt
= \int_{\mathbb{R}} \int_{\mathbb{R}^{n}}   -\frac{\partial}{\partial t} \left(  \mathbf{P}_t \phi (t,x) f(x) \right) d\mu(x) dt =0.

If the operator L is furthermore assumed to be elliptic, then as we have seen in the previous lecture, the map (t,x) \to \mathbf{P}_t f(x) is smooth and therefore, the above solution is also strong.

We now address uniqueness questions. We need further assumptions that already have been met before. We consider an elliptic diffusion operator L with smooth coefficients on \mathbb{R}^n such that:

  • There is a Borel measure \mu, symmetric and invariant for L on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R});
  • There exists an increasing sequence h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), 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.

Under these assumptions we already know that L is essentially self-adjoint. The next proposition implies that (t,x) \to \mathbf{P}_t f(x) is the unique solution of the parabolic Cauchy problem.

Proposition Let L be a diffusion operator that satisfies the above assumptions. Let u(t,x) be a smooth solution of the Cauchy problem \frac{\partial u}{\partial t}= L u,\quad u (0,x)=0,. Assume that \| u (t , \cdot) \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} < +\infty. Then u(t,x)=0.

Proof: Let h_n be as above. On one hand, we have \int_0^\tau \int_{\mathbb{R}^n} h_n^2 u Lu d\mu dt =\frac{1}{2} \int_0^\tau \frac{\partial}{\partial t} \int_{\mathbb{R}^n} h_n^2 u^2 (t,x) d\mu dt = \int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu.
On the other hand, we have \int_{\mathbb{R}^n} h_n^2 u Lu d\mu =-\int_{\mathbb{R}^n} \Gamma(h_n^2 u, u) d\mu =-\int_{\mathbb{R}^n} h_n^2 \Gamma(u) d\mu -2 \int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu.
From Cauchy-Schwarz inequality, we now have \int_{\mathbb{R}^n} uh_n \Gamma(u,h_n)d\mu \ge - \left( \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu\right)^{1/2}\left( \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu\right)^{1/2}
\ge -\frac{1}{2} \int_{\mathbb{R}^n} u^2\Gamma(h_n) d\mu-\frac{1}{2} \int_{\mathbb{R}^n} h_n^2\Gamma(u) d\mu.

We deduce that \int_{\mathbb{R}^n} h_n^2 u Lu d\mu \le \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu. As a conclusion we obtain that \int_{\mathbb{R}^n} h_n^2 u^2 (\tau,x) d\mu \le \int_0^\tau \int_{\mathbb{R}^n} u^2 \Gamma(h_n) d\mu dt. Letting n \to \infty, yields \int_{\mathbb{R}^n} u^2  d\mu \le 0 and thus u=0 \square

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2 Responses to Lecture 5. The diffusion semigroup as a solution to the heat equation

  1. Muay Zane says:

    Dear Prof. Baudoin,

    I believe in Proposition 2, the phrase “where f is in L^2” is not supposed to be there.

    Best,

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