## Lecture 13. Linear differential equations driven by rough paths

In this lecture we define solutions of linear differential equations driven by $p$-rough paths, $p \ge 1$ and present the Lyons’ continuity theorem in this setting. Let $x \in \mathbf{\Omega}^p([0,T],\mathbb{R}^d)$ be a $p$-rough path with truncated signature $\sum_{k=0}^{[p]} \int_{\Delta^k [s,t]} dx^{\otimes k},$ and let $x_n \in C^{1-var}([0,T],\mathbb{R}^d)$ be an approximating sequence such that
$\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dx_n^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \to 0.$

Let us consider matrices $M_1,\cdots,M_d \in \mathbb{R}^{n \times n}$. We have the following theorem:

Theorem: Let $y_n:[0,T] \to \mathbb{R}^n$ be the solution of the differential equation
$y_n(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y_n(s)d x^i_n(s).$
Then, when $n \to \infty$, $y_n$ converges in the $p$-variation distance to some $y \in C^{p-var}([0,T],\mathbb{R}^n)$. $y$ is called the solution of the rough differential equation
$y(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y(s)d x^i(s).$

Proof: It is a classical result that the solution of the equation
$y_n(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y_n(s)d x^i_n(s),$
can be expanded as the convergent Volterra series:
$y_n(t)=y_n(s)+ \sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[s,t]}dx_n^{I} \right) y_n(s).$
Therefore, in particular, for $n,m \ge 0$,
$y_n(t)-y_p(t)=\sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[0,t]}dx_n^{I}- \int_{\Delta^{k}[0,t]}dx_p^{I} \right) y(0),$
which implies that
$\| y_n(t)-y_m(t) \| \le \sum^{+\infty}_{k=1}M^k \left\| \int_{\Delta^{k}[0,t]}dx_n^{\otimes k}- \int_{\Delta^{k}[0,t]}dx_m^{\otimes k} \right\| \| y(0) \|$
with $M=\max \{ \| M_1 \| , \cdots , \| M_d \| \}$. From the theorems of the previous lectures, there exists a constant $C \ge 0$ depending only on $p$ and
$\sup_n \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]}$
such that for $k \ge 1$ and $n,m$ big enough:
$\left\| \int_{\Delta^k [0,\cdot]} dx_n^{\otimes k}- \int_{\Delta^k [0,\cdot]} dx_m^{\otimes k} \right\|_{p-var, [0,T]} \le \left( \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j}- \int dx_m^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\left( \frac{k}{p}\right)!}.$
As a consequence, there exists a constant $\tilde{C}$ such that for $n,m$ big enough:
$\| y_n(t)-y_m(t) \| \le \tilde{C} \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j}- \int dx_m^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} .$
This already proves that $y_n$ converges in the supremum topology to some $y$. We now have
$(y_n(t)-y_n(s))-(y_m(t)-y_m(s))$
$=\sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[s,t]}dx_n^{I}y_n(s) -\int_{\Delta^{k}[s,t]}dx_m^{I} y_m(s)\right),$
and we can bound
$\left\| \int_{\Delta^{k}[s,t]}dx_n^{I}y_n(s) -\int_{\Delta^{k}[s,t]}dx_m^{I} y_m(s) \right\|$
$\le \left\| \int_{\Delta^{k}[s,t]}dx_n^{I} \right\| \| y_n(s)-y_m(s) \|+\| y_m(s) \| \left\| \int_{\Delta^{k}[s,t]}dx_n^{I} - \int_{\Delta^{k}[s,t]}dx_m^{I}\right\|$
$\le \left\| \int_{\Delta^{k}[s,t]}dx_n^{I} \right\| \| y_n-y_m \|_{\infty, [0,T]} +\| y_m \|_{\infty, [0,T]} \left\| \int_{\Delta^{k}[s,t]}dx_n^{I} - \int_{\Delta^{k}[s,t]}dx_m^{I}\right\|$
Again, from the theorems of the previous lectures, there exists a constant $C \ge 0$, depending only on $p$ and
$\sup_n \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]}$
such that for $k \ge 1$ and $n,m$ big enough
$\left\| \int_{\Delta^k [s,t]} dx_n^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}, \quad 0 \le s \le t \le T.$
$\left\| \int_{\Delta^k [s,t]} dx_n^{\otimes k}- \int_{\Delta^k [s,t]} dx_m^{\otimes k} \right\| \le \left( \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j}- \int dx_m^{\otimes k} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p} ,$
where $\omega$ is a control such that $\omega(0,T)=1$. Consequently, there is a constant $\tilde{C}$, such that
$\| (y_n(t)-y_n(s))-(y_m(t)-y_m(s)) \|$
$\le \tilde{C} \left( \| y_n-y_m \|_{\infty, [0,T]} + \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j}- \int dx_m^{\otimes k} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \omega(s,t)^{1/p}$
This implies the estimate
$\| y_n -y_m \|_{p-var,[0,T]} \le \tilde{C} \left( \| y_n-y_m \|_{\infty, [0,T]} + \sum_{j=1}^{[p]} \left\| \int dx_n^{\otimes j}- \int dx_m^{\otimes k} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)$
and thus gives the conclusion $\square$

With just a little more work, it is possible to prove the following stronger result whose proof is let to the reader.
Theorem: Let $y_n:[0,T] \to \mathbb{R}^n$ be the solution of the differential equation
$y_n(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y_n(s)d x^i_n(s).$
and $y$ be the solution of the rough differential equation:
$y(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y(s)d x^i(s).$
Then, $y \in \mathbf{\Omega}^p([0,T],\mathbb{R}^d)$ and when $n \to \infty$,
$\sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}- \int dy_n^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \to 0.$

We can get useful estimates for solutions of rough differential equations. For that, we need the following analysis lemma:

Proposition: For $x \ge 0$ and $p \ge 1$,
$\sum_{k=0}^{+\infty} \frac{x^k}{\left( \frac{k}{p} \right)!} \le p e^{x^p}.$

Proof: For $\alpha \ge 0$, we denote
$E_\alpha(x)=\sum_{k=0}^{+\infty} \frac{x^k}{\left( k \alpha \right)!}.$
This is a special function called the Mittag-Leffler function. From the binomial inequality
$E_\alpha(x)^2$
$=\sum_{k=0}^{+\infty} \left( \sum_{j=0}^k \frac{1}{\left( j \alpha \right)!\left( (k-j) \alpha \right)!}\right)x^k$
$\le \frac{1}{\alpha}\sum_{k=0}^{+\infty} 2^{\alpha k} \frac{x^k}{\left( k \alpha \right)!}=\frac{1}{\alpha}E_\alpha(2^\alpha x).$
Thus we proved
$E_\alpha(x)\le\frac{1}{\alpha^{1/2}}E_\alpha(2^\alpha x)^{1/2}.$
Iterating this inequality, $k$ times we obtain
$E_\alpha(x)\le \frac{1}{\alpha^{\sum_{j=1}^k \frac{1}{2^j}}} E_\alpha(2^{\alpha k}x)^{1/(2k)}.$
It is known (and not difficult to prove) that
$E_\alpha(x) \sim_{x \to \infty} \frac{1}{\alpha} e^{x^{1/\alpha}}.$
By letting $k \to \infty$ we conclude
$E_\alpha(x) \le \frac{1}{\alpha} e^{x^{1/\alpha}}.$
$\square$

This estimate provides the following result:

Proposition: Let $y$ be the solution of the rough differential equation:
$y(t)=y(0)+\sum_{i=1}^d \int_0^t M_i y(s)d x^i(s).$
Then, there exists a constant $C$ depending only on $p$ such that for $0 \le t \le T$,
$\| y(t) \| \le p \| y(0)\| e^{ CM \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,t]} \right)^p},$
where $M=\max \{ \| M_1 \|, \cdots, \|M_d\| \}$.

Proof: We have
$y(t)=y(0)+ \sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M_{i_1}\cdots M_{i_k} \left( \int_{\Delta^{k}[0,t]}dx^{I} \right) y(0).$
Thus we obtain
$y(t)\le \left( 1+ \sum^{+\infty}_{k=1}\sum_{I=(i_1,\cdots,i_k)} M^k \left\| \int_{\Delta^{k}[0,t]}dx^{I} \right\| \right) \| y(0) \|$,
and we conclude by using estimates on iterated integrals of rough paths together with the previous lemma $\square$

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