Laser beam in terms of maxwell's equations It is my understanding that a laser creates a beam of light which is contained in a cylinder.
Maxwell's equations speak of the partial derivatives of the electric and magnetic fields. If the electromagnetic wave created by a laser were only present in a cylinder, doesn't this mean that the fields are discontinuous at the boundary of the cylinder? Wouldn't that make the partial derivatives undefined, and then how would Maxwell's equations be interpereted?
How can a laser beam be understood in terms of Maxwell's equations (how can an EM wave propagate only in a cylindrical beam)?
 A: Real, propagating light fields can't have sharp edges as you describe. If you try to impose sharp structure on a field, energy becomes evanescent and cannot propagate. Instead, the energy hangs around the sharp edge as an oscillating store of energy - much like the reactive power in a capacitor or inductor. So sharp edges are stripped off laser beams.
The easiest object of study for understanding your question is a monchromatic field whose electric field is everywhere directed in the $\hat{x}$ direction alone, so that $\vec{E} = \mathcal{E}(x,y,z)\,\exp(-i\,\omega\,t)\,\hat{x}$. You can easily show that this field fulfills Maxwell's equations if $\mathcal{E}$ fulfills Helmholt'z equation:
$$(\nabla^2+k^2)\mathcal{E}=0\tag{1}$$
We can analyze propagation from any transverse (parallel to $\hat{x},\,\hat{y}$, normal to $\hat{z}$) plane to any other transverse plane by representing $\mathcal{E}$ as a superposition of plane waves that each fulfill (1) (since (1) is linear), where we calculate the superposition co-efficients $\tilde{\mathcal{E}}(k_x,k_y)$ through the Fourier transform:
$$\tilde{\mathcal{E}}(k_x,k_y)=\int_{\mathbb{R}^2} \exp\left(-i \left(k_x u + k_y v\right)\right)\,\mathcal{E}
(x,y,0)\,{\rm d} u\, {\rm d} v\tag{2}$$
propagate each separately by witnessing that plane waves $\mathcal{E}(x,\,y,\,0)=\exp\left(i \left(k_x x + k_y y\right)\right)$ propagate by (1) through the formula 
$$\mathcal{E}(x,\,y,\,z)=\exp\left(i \,\left(k_x x + k_y y\right)\right) \exp\left(i \sqrt{k^2 - k_x^2-k_y^2}\, z\right)\tag{3}$$
so that propagation between planes can be calculated by:
$$\mathcal{E}(x,y,z) = \int_{\mathbb{R}^2} \left[\exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \,\sqrt{k^2 - k_x^2-k_y^2}\, z\right)\,\tilde{\mathcal{E}}(k_x,k_y)\right]{\rm d} k_x {\rm d} k_y\tag{4}$$
This is the exact diffraction integral (see further information on its use and restrictions in my answer here) and you can check that it fulfills Helmholtz's equation.
If you try to impose "sharp" structure through edges or subwavelength arrays of radiators, you end up with significant Fourier components in (2) with very high values of $k_x$ and $k_y$: sharp edges need very "wiggly" sinusoids to make them up! From (3) and (4), a plane wave with $k_x^2 + k_y^2>k^2$ will be exponentially decaying with distance, and it cannot transfer energy. It is an Evanescent Field, and it gets stripped off the wave system very swiftly as it propagates.  
A good approximation to (4) when the field is paraxial so that $k_x,\,k_y \ll k$ is the Fresnel diffraction integral:
$$\mathcal{E}(x,y,z) \approx e^{i\,k\,z} \int_{\mathbb{R}^2} \left[\exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(\frac{i}{2\,k} \,(k_x^2+k_y^2)\, z\right)\,\tilde{\mathcal{E}}(k_x,k_y)\right]{\rm d} k_x {\rm d} k_y\tag{5}$$
and it can be shown that the Gaussian Beams diffract according to this approximation, and are therefore paraxially approximate solutions of Maxwell's equations.
Indeed, the Gaussian beams are eigenfunctions of the integral operator in (5), and their eigenvalues all have unit magnitude. This means, physically, that if you bounce a Gaussian beam back and forth in a laser cavity of the right length, it will match up with and reinforce itself after the round trip as it will reproduce itself. This is why Gaussian beams - or at least approximately Gaussian beams - are commonly output by lasers. Any other kind of field doens't match up with itself, so destructive interference swiftly ensures that the Gaussian beams of the appropriate frequency and therfore phase matching condition are selectively amplified by a cavity with gain at the expense of everything else.
