Analytic solution to the following system of PDE

Question

Is there a smooth vector field $u: \mathbb{R}^3\rightarrow \mathbb{R}^3$ such that $$ \dfrac{\partial }{\partial t}u +\lambda\left(u \times \bigtriangledown u\right) = \mu \bigtriangleup u.~~~~~~~~~\lambda\neq 0~~~~~\mu >0 $$ with initial condition $$u(x,0) = u_0(x).$$ The operation $\times$ is the usual cross product in $\mathbb{R}^3$. So equivelently, \begin{equation} \dfrac{\partial }{ \partial t} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} + \lambda\begin{bmatrix} u_2 \dfrac{\partial u_3}{\partial x_3}-u_3 \dfrac{\partial u_2}{\partial x_2} \\ u_3\dfrac{\partial u_1}{\partial x_1} -u_1 \dfrac{\partial u_3}{\partial x_3}\\ u_1\dfrac{\partial u_2}{\partial x_2}- u_2\dfrac{\partial u_1}{\partial x_1} \end{bmatrix} = \mu \bigtriangleup \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} \end{equation} Here $\bigtriangleup $ is the Laplacian operator and $\bigtriangledown u$ is the gradient of $u$ given as $\bigtriangledown u = \left[\dfrac{\partial u_1}{\partial x_1},\dfrac{\partial u_2}{\partial x_2},\dfrac{\partial u_3}{\partial x_3}\right]$. So for example, $\bigtriangleup u_1 = \dfrac{\partial^2 u_1}{\partial x_1^2}+ \dfrac{\partial^2 u_1}{\partial x_2^2}+ \dfrac{\partial^2 u_1}{\partial x_3^2}$.