# Change of variables in stochastic PDE

I have the following stochastic partial differential equation (SPDE):

$$d v = -\mu \frac{\partial v}{\partial x} dt + \frac{1}{2} \frac{\partial^2 v}{\partial x^2} dt - \sqrt{\rho} \frac{\partial v}{\partial x} d M_t$$,

with $$M_t$$ standard Brownian motion and $$\mu, 0 \leq \rho \leq 1$$ real-valued parameters. I read that the solution (without boundary conditions) can be written as the solution of the PDE

$$\frac{\partial u}{\partial t} = \frac{1}{2} (1-\rho) \frac{\partial^2 u}{\partial x^2} - \mu \frac{\partial u}{\partial x}$$

shifted by the current value of the Brownian driver

$$v(t,x) = u(t, x-\sqrt{\rho} M_t)$$

How can I show this explicitly?

NOTE: It might be useful to keep in mind that the SPDE describes the density of the particles

$$d X_t^i = \mu dt + \sqrt{1-\rho} d W_t^i + \sqrt{\rho} d M_t$$, with $$W_t^i$$ independent Brownian motions. $$M_t$$ somehow represents a "common noise".

NOTE: is there maybe a mistake in the SPDE? Should we have $$\frac{1}{2} (1-\rho) \frac{\partial^2 v}{\partial x^2}$$?

• A small clarification: Don't you need $M$ to be a two-parameter object (e.g. a white noise)? Because, if $M$ is just a simple Brownian motion, then for each space variable $x$, you end up with a (linear) stochastic DE. – Sayantan Apr 8 '19 at 15:32
• Not sure what you mean....are you referring to the equation for $v$ or for $u$? In any case, in this simple example, yes, the equations should be linear...what is the problem with that? – Fred G. Apr 8 '19 at 18:03
• By the way, the white noise is going to be (in some sense) the term $\frac{d M_t}{dt}$ – Fred G. Apr 8 '19 at 18:06
• I think you are right: either the spde is given wrong or the pde for $u$ is wrong. You can either add the coefficient $(1-\rho)$ to $\frac{\partial^2 v }{\partial x^2}$ or remove it from the coefficients of $\frac{\partial^2 u }{\partial x^2}$ – Sayantan Apr 9 '19 at 14:40
• Regarding my first comment: I was only trying to point out that since the random noise in your equation is only in the time variable, you may want to look at the SDEs $$\, d v(x,t) = b_x(t)\,d t + \sigma_x(t)\,d B_t$$ where $b_x(t)= -\mu \frac{\partial v}{\partial x} + \frac{1}{2}\frac{\partial^2 v}{\partial x^2}$ and $\sigma_x(t)=\sqrt{\rho} \frac{\partial v}{\partial x}$. But ignore it if it is confusing or not helpful. – Sayantan Apr 9 '19 at 14:46

## 1 Answer

Proving these kind of phase shifts can indeed be difficult as the Ito calculus does not provide a nice chain rule as we have in deterministic calculus. There is however a trick to solve this problem. Choose a smooth test function $$\zeta$$ and define the functional \begin{align} \phi(v_t,M_t)=\langle v_t,\zeta\rangle_{H}=\langle u(t,\cdot),\zeta(\cdot+\sqrt{\rho}M_t)\rangle_{H}, \end{align} where I assume your solution lives in some suitable Hilbert space $$H$$. Now using standard Ito calculus you can derive an SDE for $$\phi$$ (because you know the (S)PDE for $$u$$ and the SDE for $$M_t$$). Then, you can shift the $$\sqrt\rho M_t$$ back to $$u$$ (which results in $$v$$) and then drop the inner product with $$\zeta$$. The resulting SPDE should be equal to your original problem. I hope this answer is still useful for you.