Property of the Green's Function

Question

I am trying to prove a statement about Green's Functions but can't seem to conclude anything. The statement is this:

Let $G_1$ be the Green's Function of the Laplacian $\Delta$ on $B(0,1)\subset \mathbb R^3$. Let $G_2$ be the Green's Function of the Laplacian $\Delta$ on $B(0,2)\subset \mathbb R^3$. $B(0,1)$ is the Ball centered at the Origin with radius $1$, same analogous statement for $B(0,2)$. Let $x_0\in B(0,1)$. Prove that for all $x\in B(0,1)$ such that $x\neq x_0$, we have:$$G_1(x,x_0)>G_2(x,x_0)$$

So the way I tried to prove this is by considering a region $B(0,1)_\epsilon=B(0,1)\setminus B(0,\epsilon)$ for some $1>\epsilon>0$. Then, $G_1$ is certainly Harmonic on $B(0,1)_\epsilon$. Thus, we can apply the Maximum Property to show that $G_1$ reaches its maximum $M$ on $\partial B(0,1)$. By the minimum Principle, it is also bounded below by $m$. Thus, we see that $m<G_1(x,x_0)<M$.

Now, applying the same region for $B(0,2)$, we get $B(0,2)_\epsilon$. Applying the Minimum Principle and Maximum Principle again (noting that $\partial B(0,1)$ lies within $B(0,2)$, we find that $G_2$ reaches its maximum $N$ and minimum $n$ on $\partial D(0,2)$. Therefore, everywhere on $B(0,2)_\epsilon$, we have that $n<G_2<N$.

I proved beforehand that $G(x,x_0)$ is negative on some domain $\Omega$. I feel like the conclusion to this proof is right in front of me but I can't seem to string these two facts together. If someone has the next step for this proof, I would appreciate it greatly. Thank you!

Answer 1

By definition, for each $x_0 \in B_2$ Green's function $G_2$ is the solution of $$ \begin{cases} \Delta_x G_2 (x,x_0) = \delta_{x_0} & \text{for } x \in B_2, \\ G_2 (x,x_0) = 0 & \text{for } x \in \partial B_2. \end{cases} $$ In the above, $G_2(\cdot,x_0)$ is considered as a function of the first variable and the first equation is meant in distributional sense. The same holds for $G_1$ in $B_1$.

As you already observed, $G_2(x,x_0) < 0$ for $x \in B_2$, $x \neq x_0$. Choose $x_0 \in B_1$ and consider the function $g(x) = G_2(x,x_0) - G_1(x,x_0)$. From what we know about $G_1$ and $G_2$, we conclude that it satisfies $$ \begin{cases} \Delta g(x) = 0 & \text{for } x \in B_1, \\ g(x) < 0 & \text{for } x \in \partial B_1. \end{cases} $$ By maximum principle, $g(x) < 0$ for all $x \in B_1$.

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Note that the above reasoning applies not only to balls, but arbitrary domains $\Omega_1 \subsetneq \Omega_2$.

Answer 2

Going directly from the expression for the Green's function for the ball of radius $R$ centered at $0$ in $\mathbb{R}^{3}$. The expression is as follows:

$$G_{R}(x; y) = \frac{1}{4\pi \left| R\frac{x}{|x|} - |x|\frac{y}{R}\right|} - \frac{1}{4\pi|x - y|}$$

and so from this we see that for $x, x_{0} \in B(0, 1), x \neq x_{0}$ we have

$$G_{2}(x; x_{0}) - G_{1}(x; x_{0}) = \frac{1}{4\pi \left| 2\frac{x}{|x|} - |x|\frac{x_{0}}{2}\right|} - \frac{1}{4\pi \left| \frac{x}{|x|} - |x|x_{0}\right|} < 0$$

since

$$\left| 2\frac{x}{|x|} - |x|\frac{x_{0}}{2}\right| > \left| \frac{x}{|x|} - |x|x_{0}\right|,$$

which isn't hard to check.