How does one show directly that the solution following parabolic partial differential equation (PDE) of $p(t,v)$ approaches its stationary solution which is a solution of an elliptic partial differential equation?

$$\frac{\partial}{\partial t}p = \frac{k^2}2\frac{\partial^2}{\partial v^2}(vp)+\frac{\partial}{\partial v}(\gamma(v-\theta)p) \tag1$$ with initial and boundary conditions $$P(t=0,v) = \delta(v-v_0),$$ $$P(t,v=0)=P(t,v=\infty)=0,$$ and all variables being real, $k, \gamma, \theta>0$. We can start with $k,\gamma,\theta$ being constant then graduating to they being functions of only $v$ then finally to being functions of $t$ and $v$.

We can solve this equation via Fourier transform and express the solution with special functions. Then we can see the solution approaches a Gamma distribution at time infinity. My question is without solving this equation explicitly, can we prove the solution approaches the stationary solution which is the elliptic PDE obtained from setting the time partial derivative of the original PDE to zero?

I am thinking about obtaining the properties of spectrum (eigenvalues) of the elliptic operator on the right hand side of Equation (1). Currently it is a non-self-adjoint operator. Had it been a self-adjoint elliptic operator, its spectrum are all discrete and non-negative (and approaching infinity). We can deduce easily that all positive eigenfunctions decay away as time approaches infinity leaving only the zero eigenfunction which is the stationary solution. Can we somehow transform Equation (1) into one with self-adjoint elliptic operator? More importantly, what is a general theory on the spectrum of this kind of elliptic operators?

There is a sequel to this question.


1 Answer 1


The given PDE (1) can be written as: $$ \partial_t p(t,x) = L^* p(t,x) \tag2 $$ where $L^*$ is the (formal) adjoint of the following operator $$ Lf(x) = - \gamma (x - \theta) f'(x) + \frac{1}{2} k^2 x f''(x) \tag3 $$ Associated to this operator $L$ is a diffusion process which satisfies the SDE: $$ d X = - \gamma (X-\theta) dt + k \sqrt{X} d B $$ where $B$ is a one-dimensional standard Brownian motion. For any $t>0$, the solution of the PDE (1) describes the time evolution of the probability density of $X(t)$ with a point mass initial condition (as given).

The operator $L$ can also be written in conservative form as $$ Lf(x) = \frac{1}{\pi(x)} \partial_x ( M(x) \pi(x) \partial_x f(x) ) \tag4 $$ where we introduced $$ M(x) = \frac{k^2 x}{2} \;, \quad \pi(x) =\frac{1}{x} \exp\left( -\frac{2\gamma (x-\theta \log(x) )}{k^2} \right) \tag5 $$ Let $L^2(\mathbb{R}^+, \pi(x) dx)$ denote the space of functions square integrable with respect to the measure $\pi(x) dx$. Let $\langle \cdot, \cdot \rangle_{\pi}$ denote the $\pi$-weighted $L^2$ inner product defined as: $$ \langle f,g \rangle_{\pi} = \int_{\mathbb{R}^+} f(x) g(x) \pi(x) dx \tag6 $$ for all $f,g \in L^2(\mathbb{R}^+, \pi(x) dx)$. Then $L$ is symmetric with respect to this inner product since $$ \langle f,g \rangle_{\pi} = \langle g,f \rangle_{\pi} $$ for all $f,g \in C_0^{\infty}(\mathbb{R}^+)$. A similar calculation shows that $\pi(x)dx$ is infinitesimally invariant $$ \int_{\mathbb{R}^+} L f(x) \pi(x) dx = 0 $$ for all $f \in C_0^{\infty}(\mathbb{R}^+)$, which implies that $L^* \pi(x)=0$ for all $x \in \mathbb{R}^+$. Furthermore, since the positive function $\pi(x)$ is integrable over $\mathbb{R}^+$, i.e. $Z = \int_{\mathbb{R}^+} \pi(x) dx < \infty$ and the diffusion process corresponding to $L$ is recurrent, then $\pi(x)/Z$ is an invariant probability density for the diffusion process $X$.

For more detail regarding these statements including proofs see, e.g., Sections 4.8 and 5.1 of:

  • Pinsky, Ross G. Positive harmonic functions and diffusion. Vol. 45. Cambridge university press, 1995.
  • $\begingroup$ Thank you very much, Nawaf. I should have realized $L$ could be written in the form of Equation (4) and (5). My further question is whether an elliptic operator in $R^d$ would be able to be written in similar conservative form as in Equation (4) and thus making the multi-dimensional elliptic operator into a symmetric one? $\endgroup$
    – Hans
    Dec 14, 2016 at 23:05
  • $\begingroup$ No. While this trick works for any one-dimensional, elliptic operator, in general this does not work for a $d$-dimensional elliptic operator. $\endgroup$ Dec 14, 2016 at 23:12
  • $\begingroup$ Is there any general theory then to deal with the stationarity (asymptotics) and spectral question regarding the multi-dimensional elliptic operator? $\endgroup$
    – Hans
    Dec 14, 2016 at 23:19
  • $\begingroup$ Off the cuff, I am not aware of any (practical) theory that deals with the spectral question of not necessarily symmetric operators. Here is a practical paper on the asymptotic properties of not necessarily symmetric diffusions by Mattingly, Stuart and Hingham [2002]: sciencedirect.com/science/article/pii/S0304414902001503 See also Chapter 12 of: Da Prato, Giuseppe [2014]. Introduction to stochastic analysis and Malliavin calculus. and, Chapter 4 of: Khasminskii, R. [2011]. Stochastic stability of differential equations. None of these references assume symmetry. $\endgroup$ Dec 14, 2016 at 23:30
  • $\begingroup$ Thank you very much again! Can you take a look at the second part of this question mathoverflow.net/q/257271/32660? $\endgroup$
    – Hans
    Dec 15, 2016 at 0:32

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