Greens function method for Newtonian potential

Question

this may be a silly question but, well you know when solving for the Poisson equation that gives the Newtonian potential, $\Phi$, (for a point mass, $M$, at the origin) $$\nabla^2 \Phi = 4\pi G M\delta(r)$$ we can use the Green's function method, where the Green's function is given by $$G(x;y)=\frac{1}{|x-y|}$$ In textbooks, I see that the only constraint applied to the resulting $\Phi$ is just that it vanishes at infinity. But why is that so? Doesn't a second order differential equation require 2 constraints?

Thank you.

Answer 1

Addressing an earlier comment: because the equation is inhomogeneous, multiplying the solution by a arbitrary constant will break the DE. In effect the Laplacian will be a Dirac mass of a different size.

What's really going on here is that the general solution to the equation (without the constraint at infinity) is given by a particular solution (such as the usual Newtonian potential) plus a harmonic function. This is just the decomposition of the solution to an inhomogeneous problem into the linear space of homogeneous solutions plus any particular solution.

Now write $\Phi(x) = N(x) + u(x)$ where $N$ is the Newtonian potential and $u$ is harmonic. Require $\Phi$ to vanish at infinity. Since $N$ vanishes at infinity this amounts to requiring that $u$ vanishes at infinity. So now to show $\Phi=N$ you need to show that $u=0$. So you need to show that if a function is harmonic on the whole space and vanishes at infinity then it is zero. This is not difficult to prove using the maximum principle or mean value property.

Answer 2

The constraint at infinity is more powerful than a Cauchy condition. Think of $u''=0$ in $\mathbb{R}$. If you request that $\lim_{x \to \pm\infty} u(x)=0$, then $u=0$ is the only solution.

You should not confuse a Cauchy problem with a boundary value problem: their theories are rather different.

Answer 3

Assume the source is at the origin. The Laplacian is translation invariant; so it suffices to look at the origin and translate that solution to obtain the one for a source at a different point.

The solution for a source at the origin must be radial because of the rotational symmetries of the problem. In $\mathbb{R}^{3}$, the only radial solutions of Laplace's equation have the form $$ \frac{A}{r} + B. $$ There's not really very much you have to choose from.