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This is a follow-up on a previous question. Now the parabolic PDE of $P(t,x,v)$ has two spatial dimensions.
$$ \partial_t P = L^* P \tag1 $$ $$L^*P = \frac12\left(\kappa^2\frac{\partial^2}{\partial v^2}+2\rho\kappa\frac{\partial^2}{\partial x\partial v}+\frac{\partial^2}{\partial x^2}+\frac{\partial}{\partial x}\right)(vP)+\gamma\frac{\partial}{\partial v}((v-\theta)P)$$ where $\kappa,\rho,\theta$ are all positive constants and $\rho\in[0,1]$, with intial-boundary condition $$P(t=0,x,v)= \delta(x)\delta(v-v_0),$$ $$P(t,x,v=0)=P(t,x,v=\infty)=P(t,x=-\infty,v)=P(t,x=\infty,v)=0.$$

Integrating $p$ over $x$ we get the parabolic equation in the previous question.

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?

Currently it is a non-self-adjoint operator. Had it been a self-adjoint elliptic operator, its spectrum would be 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?

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Almost a symmetric diffusion. To see this, note that: $$ L f(x,y) = \operatorname{Trace}\left( M(x,y) D^2 f(x,y) \right) + \mu(x,y) \bullet Df(x,y) \tag{2} $$ where $$ M(x,y) =\frac{1}{2} y \begin{bmatrix} 1 & \rho \kappa \\ \rho \kappa & \kappa^2 \end{bmatrix} \;, \quad \mu(x,y) = - y \begin{bmatrix} \frac{1}{2} \\ \gamma \end{bmatrix} + \begin{bmatrix} 0 \\ \gamma \theta \end{bmatrix} \tag{3} $$ Note also that: $$ \operatorname{div}(M(x,y)) = \frac{1}{2} \begin{bmatrix} \rho \kappa \\ \kappa^2 \end{bmatrix} \quad \text{and} \quad M(x,y)^{-1} \left( \mu(x,y) - \begin{bmatrix} 0 \\ \gamma \theta \end{bmatrix} \right) = \begin{bmatrix} \dfrac{\kappa - 2 \gamma \rho}{\kappa (-1 + \rho^2) } \\ \dfrac{2 \gamma - \kappa \rho}{\kappa^2 (-1 + \rho^2)} \end{bmatrix} $$ which are both constant vectors. Thus, the associated SDE can be written as: $$ d X = - M(x,y) DU(x,y) dt + \begin{bmatrix} 0 \\ \gamma \theta \end{bmatrix} dt +\sqrt{2} B(x,y) dW \tag{4} $$ where $W$ is a two-dimensional standard Brownian motion, $(B B^T)(x,y) = M(x,y)$ and $$ U(x,y) = \dfrac{\kappa - 2 \gamma \rho}{\kappa (-1 + \rho^2) } x + \dfrac{2 \gamma - \kappa \rho}{\kappa^2 (-1 + \rho^2)} y \tag{5} $$ It is almost symmetric in the sense that the differential operator associated to the process $$ X(t) + \operatorname{div}(M(x,y)) t - \begin{bmatrix} 0 \\ \gamma \theta \end{bmatrix} t $$ can be written in conservative form as: $$ \pi(x,y)^{-1} \operatorname{div}( \pi(x,y) M(x,y) Df(x,y) ) \tag{6} $$ where $\pi(x,y) = \exp(-U(x,y))$ with $U(x,y)$ given in (5). In other words, one needs to add a spurious (constant) drift term in order to write $L$ in the form of (6). Looking at (5) it seems like one needs to introduce conditions on $\kappa$, $\gamma$ and $\theta$ to ensure that: for any $\rho \in [0,1]$ the function $\pi(x,y)$ is integrable.

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  • $\begingroup$ The sign in front of $\mu$ in Equation (1) should be +, right? $\endgroup$ – Hans Dec 15 '16 at 2:23
  • $\begingroup$ Yes. I corrected that typo earlier today. $\endgroup$ – Nawaf Bou-Rabee Dec 15 '16 at 19:07
  • $\begingroup$ Thank you, Nawaf. I am going through your derivation. Questions: 1) Does this imply the distribution or the solution of Equation (1) won't converge to the stationary solution? Is it just inconclusive? 2) If it is the latter, is there an example of parabolic equation with only second and first spatial partial derivatives (no zeroth spatial partial derivative) terms that does not converge to its stationary solution? Will it just oscillate or explode? $\endgroup$ – Hans Dec 15 '16 at 21:33
  • $\begingroup$ (1) It is inconclusive. (2) There are many examples of stochastically unstable symmetric and nonsymmetric diffusions which can explode. Symmetry does not imply stability. $\endgroup$ – Nawaf Bou-Rabee Dec 15 '16 at 21:56
  • $\begingroup$ The references given in my answer and comment to your previous question (mathoverflow.net/questions/257239/…) provide sufficient conditions for convergence to a stationary solution. $\endgroup$ – Nawaf Bou-Rabee Dec 15 '16 at 22:24

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