# Parabolic PDE Long Time Asymptotics and Elliptic Operator Spectrum

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.

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 $X$.
• 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? – Hans Dec 14 '16 at 23:05
• No. While this trick works for any one-dimensional, elliptic operator, in general this does not work for a $d$-dimensional elliptic operator. – Nawaf Bou-Rabee Dec 14 '16 at 23:12