# Green envelopes

Green functions of the Laplacian are bounded in one dimensional domains, a reflection of the fact that square integrability of the derivative implies boundedness in one dimension but not in two or more. So, it’s possible to say something about the envelope ${\sup_y g(x,y)}$ obtained by taking the supremum over source location y.

In order to satisfy the distributional equation ${-\frac{d^2}{dx^2}g(x,y) = \delta_{x=y}}$, Green functions have a corner at y, where the derivative jumps down by 1. Combining this property with the boundary conditions identifies the function. Let’s use the interval (0,1) as the domain.

The Dirichlet boundary condition ${g(0,y) = 0 = g(1,y)}$ leads to the formula ${g(x,y)=\min(x,y)-xy}$:

The upper envelope is the parabola ${x(1-x)}$. This fact becomes more evident if we use more functions.

Mixed Dirichlet-Neumann conditions ${g(0,y) = 0 = g_x'(1,y)}$ yield an even simpler formula ${g(x,y) = \min(x,y)}$. The envelope is boring in this case:

The pure Neumann condition ${g_x'(0,y) = 0 = g_x'(1,y)}$ requires an adjustment to the definition of Green function, since it is incompatible with ${-\frac{d^2}{dx^2}g(x,y) = \delta_{x=y}}$. Instead, let’s ask for ${-\frac{d^2}{dx^2}g(x,y) = \delta_{x=y} - c}$ where the constant c is the reciprocal of the length of the interval. This ensures that the integral of the second derivative is zero. Also, it is more convenient to use the interval (-1,1) here. The Neumann Green function for this interval is ${g(x,y) = (x^2+y^2)/4 - |x-y|/2}$, up to an additive constant.

So the envelope is also determined up to a constant. On the picture above it is ${x^2/2}$.

Finally, let’s consider the periodic condition ${g(0,y) = g(1,y)}$, ${g_x'(0,y) = g_x'(1,y)}$. As with the Neumann condition, the definition is adjusted to allow the second derivative integral be zero. Using the interval (-1, 1) we get ${g(x,y) = (1+x\wedge y)(1-x\vee y)/2 + (x^2+y^2)/4}$ using the notation ${x\wedge y = \min(x,y)}$ and ${x\vee y = \max(x,y)}$ for brevity. Note that the first term, ${(1+x\wedge y)(1-x\vee y)/2}$, is the Dirichlet Green function for this interval. The additional quadratic term brings the integral of the second derivative to zero, which ensures the periodic condition for the first derivative. And the upper envelope is, up to a constant…

a constant. Not exciting but makes perfect sense, considering that we are dealing with the Green function of a circle, where moving the source amounts to shifting the function.