I thought of a possible type of pathwise-defined nonautonomous/stochastic differential equation, and I was wondering if it has been studied before.

Let $(G,\ast)$ be an abelian $C^1$ Lie group. [The main example I have in mind is just $(\mathbb{R}^d,+)$.] Let $U \subset G$ be any non-empty open set. I will define on $U$ a differential equation of the form $$ dx_t \, = \, b(x_t) \, dt \, + \left( \sum_{j=1}^{k_1} \sigma_{1j}(x_t) \circ dW_{1j}(t) \right) \ast \,\ldots\, \ast \left( \sum_{j=1}^{k_n} \sigma_{nj}(x_t) \circ dW_{nj}(t) \right) \ \ \ (\dagger)$$ where

  • $b$, $\sigma_{ij}$ are $C^1$ vector fields on $U$;
  • for each $i \in \{1,\ldots,n\}$, if $k_i>1$ then the vector fields $\{\sigma_{i1},\ldots,\sigma_{ik_i}\}$ commute with each other;
  • $W_{ij} \colon \mathbb{R} \to \mathbb{R}$ are continuous functions.

I will use typical group theory notations $g^{-1}$ and $g^n$, but I am nonetheless picturing the main example of interest as being addition in $\mathbb{R}^d$, not multiplication.

I will use the notational convention that a tangent vector $v \in T_xG$ is an $\mathbb{R}^d$-valued function $v[\cdot]$ on the set of charts $\varphi \colon O \!\ni\! x \to \mathbb{R}^d$ about $x$, satisfying $v[\varphi_2]=(\varphi_2 \circ \varphi_1^{-1})'(\varphi_1(x))v[\varphi_1]$.

For each $i,j$, let $\left(\Phi_{ij}^t\right)_{\! t \in \mathbb{R}}$ be the flow generated by $\sigma_{ij}$. That is: for any $t \in \mathbb{R}$, for any solution $x \colon [\min(0,t),\max(0,t)] \to U$ of the differential equation $\dot{x}=\sigma_{ij}(x)$, we have that $x(0)$ is in the domain of definition of $\Phi_{ij}^t$ and $\Phi_{ij}^t(x(0))=x(t)$.

Definition. Given an open interval $I \subset \mathbb{R}$, a continuous path $(x_t)_{t \in I}$ in $U$ is a solution of $(\dagger)$ if for all $t \in I$, letting $\varphi$ be a chart of $G$ about its identity element, we have \begin{align} \lim_{s \to 0} \, \tfrac{1}{s} &\varphi\!\Bigg( \left( \Phi_{11}^{W_{11}(t+s)-W_{11}(t)} \circ \ldots \circ \Phi_{1k_1}^{W_{1k_1}(t+s)-W_{1k_1}(t)}(x_t) \right)^{\!-1} \ast \\ & \ldots \ast \left( \Phi_{n1}^{W_{n1}(t+s)-W_{n1}(t)} \circ \ldots \circ \Phi_{nk_n}^{W_{nk_n}(t+s)-W_{nk_n}(t)}(x_t) \right)^{\!-1} \ast x_{t+s} \ast x_t^{n-1} \Bigg) = b(x_t)[\varphi]. \end{align}

Has this type of differential equation, or something similar, been studied before?

Admittedly, I haven't yet managed to verify that solutions exist for typical continuous functions $W_{ij}$ (in any sense of the word "typical"); if they don't exist, that would obviously be a good reason for this type of equation not to have been studied!

The physical interpretation is that a system could be subject to different noise influences that are "so separate from each other" that their net effect is not to combine in the usual way as an SDE with multiple diffusion terms $$ \sigma_1(x_t) \circ dW_1(t) \ + \ \ldots \ + \ \sigma_n(x_t) \circ dW_n(t) $$ but instead to superpose as $$ \sigma_1(x_t) \circ dW_1(t) \ \ast \ \ldots \ \ast \ \sigma_n(x_t) \circ dW_n(t) $$ as defined above. Perhaps the most helpful way to picture this is, in the context of $(\mathbb{R}^d,+)$, to see each group of diffusion terms in $(\dagger)$ as representing what the distributional derivative of $x(\cdot)$ would look like locally around $t$ if the other terms were not there, and then add all these approximate local distributions together to get an actual approximate distributional derivative of $x(\cdot)$ locally around $t$. [I don't know if it's possible to make this intuition rigorous.]

Of course, the nice thing about all this -- if solutions exist! -- is that everything is defined sample-pathwise.

UPDATE: I've managed to prove that solutions do not always exist (even for $G=\mathbb{R}^d$ with smooth vector fields). Still, perhaps it could be the case that if $W_{ij}$ are realisations of independent Wiener processes then solutions exist almost surely? (But I'm increasingly pessimistic about this.)

As for my counterexample:

Proposition. Let $(G,\ast)=(\mathbb{R},+)$. Let $W \colon \mathbb{R} \to \mathbb{R}$ be a function with the properties that

  • for all $t$ in a neighbourhood of $0$, there exists $\alpha>\frac{1}{3}$ such that $W$ is $\alpha$-Hölder at $t$;
  • $0$ is in the closure of the set of times $t$ at which $W$ is not $\frac{1}{2}$-Hölder.

Then for every $c \in \mathbb{R} \setminus \{0\}$, the initial value problem $$ \left\{ \begin{array}{r c l} dx_t &=& (x_t \circ dW(t)) \,\ast\, (-x_t \circ dW(t)) \\ x_0 &=& c \end{array} \right. $$ has no solution.

(Almost every sample path of a Wiener process fulfils the conditions: it is locally $\alpha$-Hölder for all $\alpha<\frac{1}{2}$, but the set of times at which it fails to be $\frac{1}{2}$-Hölder has full Lebesgue measure.)

Proof. Suppose for a contradiction that we have a solution $x \colon I \to \mathbb{R}$, where $I$ is a neighbourhood of $0$; assume without loss of generality that $x_t \neq 0$ for all $t \in I$. We have that $$ \lim_{s \to 0} \tfrac{1}{s}\left( -x_t(e^{W(t+s)-W(t)}+e^{W(t)-W(t+s)}-1)+x_{t+s} \right) \ = \ 0. $$ Now for any $r$ sufficiently close to $0$, we have that $$ |e^r - (1+r+\tfrac{1}{2}r^2)| < \tfrac{1}{5.99}|r|^3. $$ Hence, for all $t$ at which $W$ is Hölder with exponent $\,>\!\frac{1}{3}$, we have $$ \lim_{s \to 0} \tfrac{1}{s}\left( -x_t(1+(W(t+s)-W(t))^2)+x_{t+s} \right) \ = \ 0 $$ and therefore $$ \liminf_{s \to 0+} \ \mathrm{sgn}(c)\tfrac{1}{s}(x_{t+s}-x_t) \ = \ \limsup_{s \to 0-} \ \mathrm{sgn}(c)\tfrac{1}{s}(x_{t+s}-x_t) \ = \ 0. $$ This is true for all $t$ in a neighbourhood of $0$, and so it follows (e.g. by a weaker version of the first statement in the accepted answer to this question) that $x_t=c$ for all $t$ in a neighbourhood of $0$. For all $t$ in the interior of this neighbourhood, we have that $$ \lim_{s \to 0} \tfrac{1}{s}\left( c(W(t+s)-W(t))^2 \right) \ = \ 0, $$ and so in particular (since $c$ is non-zero and so can be cancelled) $W$ is $\frac{1}{2}$-Hölder at $t$. But this contradicts the fact that $0$ is in the closure of the set of times at which $W$ is not $\frac{1}{2}$-Hölder. $\ \square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.