1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $x \in B_1$.


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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.