Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2.
If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations?
Theoretically, it is possible to simulate every fluid flow ever by tracking every single molecule. But direct simulation of turbulence using the Eulerian Navier-Stokes equations requires $Re^{9/4}$ grid points and is thus totally impractical for Reynolds numbers larger than a few thousand. So simulating something with $10^{25}$ things to track is completely impossible.
Let's suppose that the boundary is the x-axis. So along the boundary, the stream function is constant. So, $$\frac{\partial^2\psi}{\partial x^2}=0$$ And from the no-slip boundary condition, $$\frac{\partial \psi}{\partial y}=0$$ This can be used to establish the 2nd order finite difference approximation to the value of the vorticity at the boundary:$$\omega=\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}=\frac{2(\psi(I,1)-\psi(I,0))}{(\Delta y)^2}=\omega(I,0)$$
This is used for the boundary condition on $\omega$.
ADDENDUM
First of all, the stream function is known (and constant) at the solid boundaries (because the solid boundaries are stream lines). So it doesn't have to be solved for. It is determined up to an arbitrary constant, and can thus be taken to be zero at one of the boundaries. At the other boundary, the stream function is equal to the volumetric throughput rate per unit width of channel (which is typically known). So you don't need to solve for the stream function at the solid boundaries.
If the tangential derivative $\partial \psi/\partial x=0$ at all locations along the boundary, it's second partial with respect to x must also be equal to zero. This, of course, all follows from the fact that $\psi$ is constant at the boundary.
To integrate the vorticity equation, you need a boundary condition on the vorticity (or at least a 2nd order finite difference approximation to a boundary condition). Just because $\partial \psi/\partial y=0$ does not mean the the second partial of $\psi$ with respect to y is equal to zero at the boundary; this would imply that the vorticity at the boundary is equal to zero, which we know is not correct.
The variable I in the relationships refers to the I'th x grid point. So, back to the boundary condition on vorticity: We have shown so far that, at the boundary, $$\omega=\frac{\partial^2 \psi}{\partial y^2}$$ subject to the constraint that $\partial \psi/\partial y=0$. If we represent these two conditions in 2nd order finite difference form, we obtain:
$$\omega(I,0)=\frac{\psi(I,1)-2\psi(I,0)+\psi(I,-1)}{(\Delta y)^2}$$and$$\frac{\psi(I,1)-\psi(I,-1)}{2\Delta y}=0$$If we combine these two finite difference equations, we obtain a 2nd order finite difference approximation to the value of the vorticity at the boundary:
$$\omega(I,0)=\frac{2(\psi(I,1)-\psi(I,0))}{(\Delta y)^2}$$
I've successfully used this approach to solving these equations many times.
Best Answer
None of the interesting equations in physics can be derived from simpler principles, because if they could they wouldn't give any new information. That is, those simpler principles would already fully describe the system. Any new equation, whether it's the Navier-Stokes equations, Einstein's equations, the Schrodinger equation, or whatever, must be consistent with the known simpler principles but it has also to incorporate something new.
In this case you appear to have the impression that an attempt to derive the Navier-Stokes equations runs into some impassable hurdle and therefore fails, but this isn't the case. If you search for derivations of the Navier-Stokes equations you will find dozens of such articles, including (as usual) one on Wikipedia. But these are not derivations in the sense that mathematicians will derive theorems from some initial axioms because they require some extra assumptions, for example that the stress tensor is a linear function of the strain rates. I assume this is what Putterman means.
Later:
Phil H takes me to task in a comment, and he's right to do so. My first paragraph considerably overstates the case as the number of equations that introduce a fundamentally new principle are very small.
My answer was aimed at explaining why Putterman says the Navier-Stokes equations can't be derived but actually they can be, as can most equations. Physics is based on reductionism, and while I hesitate to venture into deep philosophical waters physicists basically mean by this that everything can be explained from a small number of basic principles. This is the reason we (some of us) believe that a theory of everything exists. If such a theory does exist then the Navier-Stokes equations could in principle, though not in practice, be derived from it.
Actually the Navier-Stokes equations could in principle be derived from a statistical mechanics treatment of fluids. They don't require any new principles (e.g. relativity or quantum mechanics) that aren't already included in a the theoretical treatment of ideal fluids. In practice they are not derivable because those derivations based on a continuum approach rather than a truly fundamental treatment.