Consider the SPDE $$\frac{\partial}{\partial t}u\_t(x) = \frac{\kappa}{2}\frac{\partial^2}{\partial x^2}u\_t(x) + u\_t(x)(K-u\_t(x)) + \sigma u\_t(x) \xi(t,x),$$ 
where $(t,x)\in {\mathbb R}\_+\times {\mathbb R}$, $\xi$ is a space-time white noise, and $\kappa, K,$ and $\sigma$ are positive constants. Assume the initial data $u\_0(x)$ is measurable, nonnegative, and bounded. Note that the term $u(K-u)$ is not Lip continuous and grows faster than linear. 


I wonder whether the above SPDE has a solution or not? And if a solution exists, is it unique, pathwise or in law? All the references that I know do not treat this special case. Many thanks for your help!



On the existence and uniqueness of solution to SPDE with nonlinear growth coefficients
Call the drift term $b(u) = u (K- u)$ and for simplicity set $K=1$. Consider the following space of functions:
$$C\_{\text{tem}}(\mathbb{R},\mathbb{R}) = \{ f:\mathbb{R} \to \mathbb{R}:\, \sup\_{x \in \mathbb{R}} |f(x)|e^{-\lambda|x|} < \infty \ \forall \lambda >0, \ f \text{ continuous} \} \, . $$
There are results on existence and uniqueness in $C\_{\text{tem}}$ only assuming local Lipschitz conditions and
$$ \big( b(x) - b(y)\big) \cdot (x-y) \leq c(|x-y|^2 + |x-y|) , \ x,y \in \mathbb{R} $$
for some constant $c>0$. See Theorem 4.2 of Iwata, *An infinite dimensional stochastic differential equation with state space* $C(\mathbb{R})$. Note that your $b$ does *not* satisfy this condition. However $b\_1 (u) = b(u) \mathbf{1}(u \geq -1) -2 \mathbf{1}(u< -1) $ does. This allows you to apply Iwata's Theorem. Now you have a solution to the system with $b\_1$ instead of $b$.


Next one can should be able to use Theorem 2.3 of Shiga, *Two contrasting properties of solutions for one-dimensional stochastic partial differential equations*. However, it is only stated assuming *global* Lipschitz conditions. It should also work in the non-global setting. However, one needs to check his proof carefully. This allows you to deduce that your solution is non-negative: $u \geq 0$ and therefore it does not play a role whether you use $b$ or $b\_1$.


