# A solution to stochastic PDE $du(t)= a(t)u(t)\,dt +s(t)\,dz$

Is there a general (integral) solution to $du(t)= -a(t)u(t)\,dt +\sigma(t)\,dz$? Is the following $u(t)=e^{-\int_{t_0}^{t} \alpha(s) \, ds}u(t_0)+\int_{t_0}^t \sigma(v) e^{-\int_v^t a(s) \, ds} \, dz(v)$ correct (which I have seen claimed without a justification)? z(t) is the standard Wiener process. Is there a good reference for it?

• What's $z$? ${}$ – Ian Feb 25 '17 at 4:26
• z is the standard Wienner process. – Adam Feb 25 '17 at 20:57

About the differentiation formula of $u(t)$: Since \begin{align} u(t)&=e^{-\int_{t_0}^ta(s)\mathrm{d}s}u(t_0)+\int_{t_0}^t\sigma(v)e^{-\int_{v}^ta(s)\mathrm{d}s}\mathrm{d}Z(v)\\ &=e^{-\int_{t_0}^ta(s)\mathrm{d}s}\Bigl[u(t_0)+\int_{t_0}^t\sigma(v)e^{\int_{t_0}^va(s)\mathrm{d}s}\mathrm{d}Z(v)\Big]\stackrel{\text{def}}{=}x(t)y(t).\\ \mathrm{d}x(t)&=-a(t)e^{-\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}t=-a(t)x(t)\mathrm{d}t,\\ \mathrm{d}y(t)&=\sigma(t)e^{\int_{t_0}^ta(s)\mathrm{d}s}\mathrm{d}Z(t)=\sigma(t)[x(t)]^{-1}\mathrm{d}Z(t). \end{align} Therefore, $$du(t)=y(t)\mathrm{d}x(t)+x(t)\mathrm{d}y(t) =-a(t)u(t)\mathrm{d}t+\sigma(t)\mathrm{d}Z(t).$$
The $u(t)$ which you given is a solution of your equation. It is suffice using differentiation to verify it is a solution. One way to deduce the general solution is the `method of variation of constant' for ODE. The following book dealt with this topic: B. K. Oksendal, Stochastic Differential Equations, 6 Ed., Sringer Verlag, 2003.
• Differentiating the integral formula, I get $du(t)=e^{\int_{t_0}^{t} a(s)ds}a(t) dt\cdot u(t_0)+\sigma(t)dz$. How do I reconcile it with $du=a(t)u(t)dt+\sigma(t)dz$? (I only see that when t is infinitesimally close to $t_0.$) – Adam Feb 26 '17 at 18:07