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?
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$\begingroup$ What's $z$? ${}$ $\endgroup$– IanCommented Feb 25, 2017 at 4:26
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$\begingroup$ z is the standard Wienner process. $\endgroup$– AdamCommented Feb 25, 2017 at 20:57
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2 Answers
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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, $$ du(t)=y(t)\mathrm{d}x(t)+x(t)\mathrm{d}y(t) =-a(t)u(t)\mathrm{d}t+\sigma(t)\mathrm{d}Z(t). $$
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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.
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$\begingroup$ 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 $t_0.$) $\endgroup$– AdamCommented Feb 26, 2017 at 18:07