Hyperbolic flow / vector field - irrotational and divergence-free? My text book on meteorology claims that a hyperbolic flow pattern is both divergence-free and irrotational:


(d) Hyperbolic flow that exhibits both diffluence
  and stretching, but is nondivergent because the
  two terms exactly cancel. Hyperbolic flow also
  exhibits both shear and curvature, but is
  irrotational (i.e., vorticity-free) because the
  two terms exactly cancel.
-- Wallace & Hobbs, Atmospheric Science, 2nd Ed, p 273

In my understanding, that can not be true:


*

*I can obtain a very similar flow pattern from $ grad \ xy $.

*Based on the uniqueness of Helmholtz decomposition, the only divergence-free, irrotational vector field should be $ \vec{f} = \vec{0} $.

*Based on Helmholtz decomposition, any vector field $ \mathbf{u} $ can be represented as $ \mathbf{u} = \mathbf{v} + \mathbf{d} $ with $ \mathbf{v} = \nabla \phi $ and $ \mathbf{d} = \nabla \times \mathbf{A} $. As I understand it, the only divergence-free ( $ \mathbf{v} = 0 $ ) and irrotational ( $ \mathbf{d} = 0 $ ) vector field can be $ \mathbf{u} = 0 $.


Am I missing something or is the text handwaving the math a little too much here?
 A: Based on Hodge-Helmholtz decomposition a vector field $\mathbf{u}$ can be expressed as the sum of an irrotational vector potential $\phi$ and a divergence-free vector field $\mathbf{d}$.
$$\mathbf{u} = \nabla \phi + \mathbf{d}$$
For a flow to be irrotational it has to be able to be derived as the gradient of a vector potential $$\mathbf{u} = \nabla \phi$$ It also follows that is the flow is simply-connected and irrotational then $$\mathbf{u} = \nabla \times \mathbf{d}$$The difference field $\mathbf{u} - \mathbf{d}$ is irrotational therefore it can be resolved as a gradient of a vector potential $\phi$ by taking the divergence of the above equation.  $$\nabla \cdot \mathbf{u} = \nabla^2\phi$$
A vector field satisfying the above equation in which both the lhs and rhs vanish and also satisfies the irrotational condition is both irrotational and divergence-free. Specifically $\nabla \cdot \mathbf{u} =0$  and $\nabla^2\phi = 0$ and $\mathbf{u} = \nabla \phi$.  A possible function $\phi$ for which the vector field $\mathbf{u}$ can be derived from is a harmonic function.
