Rough paths theory Fall 2017. Lecture 2

Posted on September 3, 2017 by Fabrice Baudoin

In this lecture we establish the basic existence and uniqueness results concerning differential equations driven by bounded variation paths and prove the continuity in the 1-variation topology of the solution of an equation with respect to the driving signal.

Theorem: Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a Lipschitz continuous map, that is there exists a constant K > 0 such that for every x,y \in \mathbb{R}^e,
\| V(x)-V(y) \| \le K \| x-y \|.
For every y_0 \in \mathbb{R}^e, there is a unique solution to the differential equation:
y(t)=y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le T.
Moreover y \in C^{1-var} ([0,T], \mathbb{R}^e).

Proof: The proof is a classical application of the fixed point theorem. Let 0 < \tau \le T and consider the map \Phi going from the space of continuous functions [0,\tau] \to \mathbb{R}^e into itself, which is defined by
\Phi(y)_t =y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le \tau.
By using estimates on Riemann-Stieltjes integrals, we deduce that
\| \Phi(y^1)-\Phi(y^2) \|_{ \infty, [0,\tau]}
\le \| V(y^1)-V(y^2) \|_{ \infty, [0,\tau]} \| x \|_{1-var,[0,\tau]}
\le K \| y^1-y^2 \|_{ \infty, [0,\tau]} \| x \|_{1-var,[0,\tau]}
If \tau is small enough, then K \| x \|_{1-var,[0,\tau]} < 1, which means that \Phi is a contraction that admits a unique fixed point y. This y is the unique solution to the differential equation:
y(t)=y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le \tau.
By considering then a subdivision
\{ \tau=\tau_1 < \tau_2 < \cdots < \tau_n=T \}
such that K \| x \|_{1-var,[\tau_k,\tau_{k+1}]} < 1, we obtain a unique solution to the differential equation:
y(t)=y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le T
\square

The solution of a differential equation is a continuous function of the initial condition, more precisely we have the following estimate:

Proposition: Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a Lipschitz continuous map such that for every x,y \in \mathbb{R}^e,
\| V(x)-V(y) \| \le K \| x-y \|.
If y^1 and y^2 are the solutions of the differential equations:
y^1(t)=y^1(0)+\int_0^t V(y^1(s)) dx(s), \quad 0\le t \le T,
and
y^2(t)=y^2(0)+\int_0^t V(y^2(s)) dx(s), \quad 0\le t \le T,
then the following estimate holds:
\| y^1 -y^2 \|_{\infty,[0,T]} \le \| y^1(0) -y^2(0) \| \exp \left( K \| x \|_{1-var,[0,T]} \right).

Proof: We have
\| y^1-y^2 \|_{\infty,[0,t]} \le \| y^1(0) -y^2(0) \| +K \int_0^t \| y^1-y^2 \|_{\infty,[0,s]} \| dx(s) \|,
and conclude by Gronwall’s lemma \square

This continuity can be understood in terms of flows. Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a Lipschitz map. Denote by \pi (t,y_0), 0 \le t \le T, y_0 \in \mathbb{R}^e, the unique solution of the equation
y(t)=y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le T.
The previous proposition shows that for a fixed 0 \le t \le T, the map y_0 \to \pi (t,y_0) is Lipschitz continuous. The set \{ \pi (t, \cdot), 0 \le t \le T \} is called the flow of the equation.
Under more regularity assumptions on V, the map y_0 \to \pi (t,y_0) is even C^1 and the Jacobian map solves a linear equation.

Proposition: Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a C^1 Lipschitz continuous map. Let \pi(t,y_0) be the flow of the equation
y(t)=y_0+\int_0^t V(y(s)) dx(s), \quad 0\le t \le T.
Then for every 0\le t \le T, the map y_0 \to \pi (t,y_0) is C^1 and the Jacobian J_t=\frac{\partial \pi(t,y_0)}{\partial y_0} is the unique solution of the matrix linear equation
J_t=Id+ \sum_{i=1}^d\int_0^t DV_i(\pi(s,y_0))J_s dx(s),
where the V_i‘s denote the columns of the matrix V.

We finally turn to the important estimate showing that solutions of differential equations are continuous with respect to the driving path in the 1-variation topology

Theorem: Let x^1,x^2 \in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a Lipschitz and bounded continuous map such that for every x,y \in \mathbb{R}^e,
\| V(x)-V(y) \| \le K \| x-y \|.
If y^1 and y^2 are the solutions of the differential equations:
y^1(t)=y(0)+\int_0^t V(y^1(s)) dx^1(s), \quad 0\le t \le T,
and
y^2(t)=y(0)+\int_0^t V(y^2(s)) dx^2(s), \quad 0\le t \le T,
then the following estimate holds:
\| y^1 -y^2 \|_{1-var,[0,T]} \le \| V \|_\infty \left( 1+ K\| x_1 \|_{1-var,[0,T]} \exp \left( K \| x_1 \|_{1-var,[0,T]} \right) \right) \| x^1 -x^2 \|_{1-var,[0,T]} .

Proof: We first give an estimate in the supremum topology. It is easily seen that the assumptions imply
\| y^1 -y^2 \|_{\infty ,[0,t]} \le K \int_0^t \| y^1 -y^2 \|_{\infty ,[0,s]} \| dx^1(s) \| +\| V \|_\infty \| x^1 -x^2 \|_{1-var,[0,T]}.
From Gronwall’s lemma, we deduce that
\| y^1 -y^2 \|_{\infty ,[0,T]} \le \| V \|_\infty \exp \left( K \| x \|_{1-var,[0,T]} \right) \| x^1 -x^2 \|_{1-var,[0,T]} .
Now, we also have for any 0\le s \le t \le T,
\| y^1(t)-y^2(t)-(y^1(s)-y^2(s))\|\le K \| y^1 -y^2 \|_{\infty ,[0,T]} \| x^1 \|_{1-var,[s,t]} +\| V\|_\infty \| x^1 -x^2 \|_{1-var,[s,t]} .
This implies,
\| y^1 -y^2 \|_{1-var,[0,T]} \le K \| y^1 -y^2 \|_{\infty ,[0,T]} \| x^1 \|_{1-var,[0,T]} +\| V\|_\infty \| x^1 -x^2 \|_{1-var,[0,T]}
and yields the conclusion \square

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