Existence and Uniqueness of Solutions to Stochastic Differential Equations, Strong and Weak Solutions
Definition 1
Let us have a probability space $( \Omega , \mathcal{F} , P)$ and a filtration $\left\{ \mathcal{F}_{t} \right\}_{t \ge 0}$. $$ \begin{align*} f &: [0,T] \times \mathbb{R}^{n} \to \mathbb{R}^{n} \\ g &: [0,T] \times \mathbb{R}^{n} \to \mathbb{R}^{n \times m} \end{align*} $$ Consider the following $n$-dimensional stochastic differential equation for two functions $f$, $g$ and $\mathcal{F}_{t}$-adapted $m$-dimensional Wiener process $W_{t}$: $$ d X_{t} = f \left( t, X_{t} \right) dt + g \left( t, X_{t} \right) d W_{t} $$
- A continuous and $F_{t}$-adapted stochastic process $\left\{ X_{t} \right\}$ that satisfies the equation almost surely at all $t \in [0 , T]$ and is $f \in L^{1} [0,T]$, and $g \in L^{2} [0,T]$ is called the solution of the given equation.
- If the following holds for all other solutions $\left\{ \tilde{X_{t}} \right\}$ than $\left\{ X_{t} \right\}$, this solution is said to be unique: $$ P \left( X_{t} = \tilde{X_{t}} , t \in [0, T] \right) = 1 $$
Theorem: Existence and Uniqueness 2
- (i) Linear growth condition: For some constants $C$, $x \in \mathbb{R}^{n}$, and $t \in [0,T]$ $$ \left| f \left( t , x \right) \right| + \left| g \left( t , x \right) \right| \le C \left( 1 + \left| x \right| \right) $$
- (ii) Uniform Lipschitz condition: For some constants $D$, $x , y \in \mathbb{R}^{n}$, and $t \in [0,T]$ $$ \left| f(t,x) - f(t,y) \right| + \left| g(t,x) - g(t,y) \right| \le D \left| x - y \right| $$
If the above two conditions are satisfied, the following stochastic differential equation $$ d X_{t} = f \left( t, X_{t} \right) dt + g \left( t, X_{t} \right) d W_{t} $$ has a unique solution $X_{t}$ with the following properties: $$ \sup_{t \in [0,T]} E \left[ \left| X_{t} \right|^{2} \right] < \infty $$
- $|g|$ denotes the Frobenius norm $|g|^{2} = \sum_{i,j} \left| g_{ij} \right|^{2}$ of a matrix.
Strong Solution and Weak Solution
The solution $X_{t}$, whose existence is guaranteed by the above theorem, is called a strong solution. $X_{t}$ is seen as the solution obtained under the assumption that we are well aware of the Brownian motion $W_{t}$, that is, there is sufficient information about the given probability space $( \Omega , \mathcal{F} , P)$ so that $W_{t}$ is $\mathcal{F}_{t}$-adapted.
On the other hand, in a situation where only $f$ and $g$ are given, i.e., when there is no information about $W_{t}$, and some $\left( \left( \tilde{X}_{t} , \tilde{W}_{t} \right) , \tilde{F}_{t} \right)$ exists that satisfies the given stochastic differential equation, this is called a weak solution. Specifically, the pair of the solution and $\left( \tilde{X}_{t} , \tilde{W}_{t} \right)$, where $\tilde{W}_{t}$ is a Brownian motion adapted to filtration $\tilde{F}_{t}$ (a Brownian motion that is martingale with respect to $\tilde{F}_{t}$), is the weak solution. Here, it’s not strictly necessary for $\tilde{X}_{t}$ to be $\mathcal{F}_{t}$-adapted.
Of course, a strong solution is also a weak solution, but the converse does not hold. Weak solutions, in a way, might be considered as solutions that are blatantly obvious as solutions, yet cannot be rigorously called solutions in mathematical terms, or solutions that somehow satisfy the equation algebraically.
Example: Tanaka Equation 1
$$ d X_{t} = \operatorname{sign} \left( X_{t} \right) d W_{t} $$ The above-mentioned stochastic differential equation is called the Tanaka Equation. Here, $\operatorname{sign}$ means the sign. In this case, since the diffusion $g \left( t , X_{t} \right) = \operatorname{sign} \left( X_{t} \right)$ does not satisfy the Lipschitz condition near $0$, the existence of a strong solution cannot be guaranteed, and it might even be demonstrated that it does not exist. The rigorous proof of this is omitted as it is not straightforward.
However, if we consider a weak solution, any Brownian motion can be a solution to the Tanaka equation. It essentially has to be Brownian motion, as reflected by the fact that $dX_{t}$, which only affects $dW_{t}$, doesn’t make sense to consider since the probability of $dW_{t}$ being negative or positive is exactly half and half.
Let’s consider any Brownian motion $B_{t}$ and define $X_{t} = B_{t}$ as follows: $$ \tilde{B}_{t} := \int_{0}^{t} \operatorname{sign} \left( B_{s} \right) d B_{s} = \int_{0}^{t} \operatorname{sign} \left( X_{s} \right) d X_{s} $$ Differentiating $t$ gives $$ d \tilde{B}_{t} = \operatorname{sign} \left( X_{t} \right) d X_{t} $$ and multiplying both sides by $\operatorname{sign} \left( X_{t} \right)$ results in $\left( \operatorname{sign} \left( X_{t} \right) \right)^{2} = 1$, thus $$ d X_{t} = \operatorname{sign} \left( X_{t} \right) d \tilde{B}_{t} $$ meaning, $X_{t} = B_{t}$ constitutes a weak solution that satisfies the Tanaka equation.