# Using an ODE to Solve an SDE

The following is an exercise to solve an SDE using the result of an ODE that has been killing me. I will start with the ODE, which is very simple: $$\begin{cases} \frac{du}{dt} = k + u(t) \\ u(0)=x \end{cases}$$

So, this can be solved formally by:

$u(t)'-u(t)=k$

$u(t)=e^{-\int-1}[\int e^{\int-1}kdt + c]=$

$u(t)=e^t[\int e^{-t}kdt+c]=e^t(-ke^{-t}+c)=ke^t-c$

Applying the initial condition,

$u(0)=x=>x=ke^0-c<=>x=k-c<=>k=x+c$

Now, consider that $f(t,x)$ represents the solution to the above ODE. Find the process $X_t$ that makes $Z_t=f(t,X_t)$ solve the SDE $dZ_t=(c+Z_t)dt+\sigma dW_t$, with $Z_0=x$.

I can find no such process, unless I've botched the ODE.

After researching, this is the progress (I hope) I've made:

Think of our SDE in the form:

$dY_t=(\alpha_t + \beta_tY_t)dt + \nu_tdW_t$, $Y_0=x$

This is a linear differential equation with deterministic coefficients, and thus the solution will be represented by:

$Y_t=e^{\int_0^t \beta_sds}[x+\int_0^t e^{-\int_0^s \beta_udu}\alpha_sds+\int_0^t e^{-\int_0^s \beta_sds}\nu_sdW_s]$

But in our equation we have $\alpha = c,\beta = 1,\nu = \sigma$, thus

$Y_t=e^{\int_0^tds}[x+\int_0^t e^{-\int_0^sdu}cds+\int_0^t e^{-\int_0^sds}\sigma dW_s]$ <=>

$Y_t=e^t[x+c\int_0^te^{-s}ds+\sigma\int_0^te^{-s}dW_s]$

Both remaining integrals = $-e^{-t}+1$ and eventually I get

$Y_t=(x+c+\sigma)e^t-c$, which seems like I'm getting very close (we now have the solution to our ODE, but with the extra $\sigma$ term) but don't see how to tie it all together.

EDIT:

I was going through this once more and now believe I've got it after noticing a mistake. Of course, towards the end when I say both integrals equal $-e^{-t}+1$ I am wrong, since the second one is with respect to dWs, not ds. So, leaving that one in integral form, what you eventually end up with is the solution to the ODE, but with the additional stochastic integral, not just the coefficient sigma.

• The point is that in the ODE situation, if $\frac{du}{dt}-u(t)$ is something independent of $u$ then you can solve the ODE by just integrating $\frac{d}{dt}(e^{-t} u(t))$ and then multiplying both sides of the equation by $e^t$. $\frac{d}{dt} \left ( e^{-t} u(t) \right )$ behaves in a nice way. Thus you might expect that when "$\frac{du}{dt}-u(t)$" involves "$\frac{dW_t}{dt}$" (all formal, since none of these derivatives exist) then $d \left ( e^{-t} u(t) \right )$ will also not explicitly depend on $u$. You then check this with Ito's formula. – Ian May 5 '17 at 21:05
• Have a look at this: math.stackexchange.com/q/592995 – saz May 6 '17 at 5:49
• Thanks guys, these suggestions led me in a good research direction – Archetupon May 6 '17 at 13:30