Theorem 3.2.6 (Blagoveshchenskii-Blagoveshchensk).label Let $\sigma: [0, \infty) \times C([0, \infty); \real^{d}) \to L(\real^{d}; \real^{n})$ and $b: [0, \infty) \times C([0, \infty); \real^{d}) \to \real^{n}$ be previsible path functionals satisfying the Lipschitz condition. If for each $T \ge 0$,

\[\sup_{0 \le s \le T}\norm{\sigma(s, 0)}_{L(\real^d; \real^n)}+ \norm{b(s, 0)}_{\real^n}< \infty\]

then the SDE

\[X_{t} = \xi + \int_{0}^{t} \sigma(s, X)dB_{s} + \int_{0}^{t} b(s, X)ds\]

admits a strong solution

\[F: \real^{n} \times C([0, \infty); \real^{d}) \to C([0, \infty); \real^{n})\]

such that:

  1. (1)

    For each $\theta \in C([0, \infty); \real^{d})$, $F(\cdot, \theta): \real^{d} \to C([0, \infty); \real^{n})$ is continuous.

  2. (2)

    For each solution $X^{y}$ initial condition $y \in \real^{n}$ and $s \ge 0$,

    \[X_{s + t}^{y} = F(X_{s}^{y}, \tau_{-s}B)_{t} \quad \forall t \ge 0\]

    almost surely, where $(\tau_{-s}B)_{r} = B_{r+s}$.

  3. (3)

    For each $y \in \real^{n}$, $\bracs{X_t^y|t \ge 0}$ is a Markov process.