Killing the nonlinearity in SPDE

Question

Consider the SPDE $$dX(t,x) = \left( \frac12 \Delta X(t,x)-\frac{|\nabla X(t,x)|^2}{2}+a(x)\right)\,dt - \nabla X(t,x)\cdot \,dW(t)$$ where $(x,t) \in \mathbb{R}^n \times (0,T)$, $a$ is a differentiable function and $W$ is an $n$-dimensional Brownian motion.

I am interested in making a transformation of $Y= f(X)$ such that we get an equation which doesn't have the $|\nabla X|^2$ part, so that it becomes linear in the new variable and one can study existence/ uniqueness of solution. I noticed that if the original equation had $+ \frac{|\nabla X(t,x)|^2}{2}$ instead of $-\frac{|\nabla X(t,x)|^2}{2}$, then $f(x)= \exp(-x)$ would have worked. Itô's formula would give $$dY(t,x) = \left( -\frac12 \Delta Y(t,x) - a(x)Y(t,x)\right)\,dt- \nabla Y(t,x)\cdot \,dW(t)$$

Answer 1

I think it not possible to find such a map $f$ because if $Y=f(X)$, then the coefficient of $|\nabla Y|^2$ in the Ito's formula will be $-\frac{1}{2f'}$, which cannot be zero always. $$dY= -\nabla Y \cdot\,dW +\left( \frac12 \Delta Y-\frac{1}{2f'} |\nabla Y|^2+a(x)f'\right)\,dt$$

So I think the original SPDE cannot be linearized, if somebody can suggest me how to think about existence and uniqueness of $H^1$ solutions of the original SPDE, it would be a great help.

Answer 2

Well applying Itô's lemma to $Y_t=exp(X_t)$ you get unless mistaken :

$$ \frac {dY_t} {Y_t}=\frac12d<X>_t+dX_t=$$ $$\frac12|\nabla X(t,x)|^2dt+(\frac12.|\Delta X(t,x)|-\frac12.|\nabla X(t,x)|^2+a(x))dt-\nabla X(t,x)dW_t=$$ $$(\frac12|\Delta X(t,x)|+a(x))dt-\nabla X(t,x)dW_t=$$ $$(\frac12|\Delta (ln Y(t,x))|+a(x))dt-\nabla ln(Y(t,x))dW_t$$

as $d<X>_t= |\nabla X(t,x)|^2dt$

Edit : If you take $Y_t=f(X_t)=exp(-X_t)$ then Itô's lemma gives :

$$ \frac {dY_t} {Y_t}=\frac12d<X>_t-dX_t=$$ $$\frac12|\nabla X(t,x)|^2dt-(\frac12.|\Delta X(t,x)|-\frac12.|\nabla X(t,x)|^2+a(x))dt+\nabla X(t,x)dW_t=$$ $$(|\nabla X(t,x)|^2-\frac12|\Delta X(t,x)|-a(x))dt+\nabla X(t,x)dW_t=$$ $$(|\nabla X(t,x)|^2-\frac12|\Delta (ln Y(t,x))|-a(x))dt+\nabla ln(Y(t,x))dW_t$$ So I think you have a problem in your derivation in your question.

Then if you take a general f then comes : $$ dY_t=\frac12f''(X_t) d<X>_t-f'(X_t)dX_t=$$ $$[\frac12.|\nabla X(t,x)|^2.(f''(X_t)-f'(X_t))+f'(X_t).\frac12.|\Delta X(t,x)|+f'(X_t).a(x))]dt+f'(X_t).\nabla X(t,x)dW_t$$

So what you are looking for is a $f$ solving this differential equation : $$f''(x)-f'(x)=0$$ or $$f''(x)=f'(x)$$ the solution of which is in the one dimensional case $f(x)=A.exp(B.x)$ with suitable $A,B$ unless mistaken.