Solution of $\Delta f -\frac{1}{2}hf = 0$ behaves asymptotically as $f(x) = 1 - C/|x|$

Question

Let $f: \mathbb{R}^{3} \to \mathbb{R}$ be the solution of the following PDE: $$\Delta f -\frac{1}{2}h f = 0$$ where $h \in C_{c}^{\infty}(\mathbb{R}^{3})$ (compactly supported an smooth) and $f$ satisfies $\lim_{|x|\to \infty}f(x) = 1$.

I am trying to prove the following: $f$ behaves asymptotically as: $$f(x) = 1 -\frac{C}{|x|}$$ for some constant $C > 0$ and this $C$ is exactly the solution of: $$8 \pi C = \int_{\mathbb{R}^{d}}h(x)f(x).$$

I honestly don't know where to begin.

Answer 1

For the specific question asked, just follow what Giorgio said.

Let $\phi = f-1$, then $\Delta \phi = \frac12 hf$ is a compactly supported function on $\mathbb{R}^3$.

Now let $g$ be the Newton potential of $\frac12 hf$ (so $g = c(\frac12 hf)*\frac{1}{|x|}$ for some appropriate constant $c$), then you also have $\Delta g = \frac12 hf$.

$g$ is well-defined (as $\frac12hf$ has compact support and is continuous), and you can check that it decays to $0$ at infinity.

Thus $\phi - g$ is a harmonic function that vanishes at infinity, and hence by the maximum principle is zero. This shows that $f-1 = g$.

The asymptotics comes straight out of the convolution definition for $g$.

Answer 2

I think you can get the result blowing-down the solution. I'll just sketch the argument since I don't want to be confusing with the details.

Set $u:=1-f$, then $u$ solves $-\Delta u = \frac{h}{2}(1-u)$ and satisfies $\lim_{|x|\to \infty} u(x)=0$. For $r>0$ consider $u_r(x):= ru(rx)$, you can check easily that $u_r$ solves $$-\Delta u_r =\frac{r^3}{2}h(rx)(1-u_r) =: g_r(x)\,. $$ Since $h$ has compact support you have $-\Delta u_r= g_r \to 0$ in $C^2_{\rm loc}(\mathbb{R}^3 \setminus \{0\})$ and moreover $$ \int_{\mathbb{R}^3} g_r \, dx = \frac{r^3}{2} \int_{\mathbb{R}^3} h(rx)(1-u_r(x)) \, dx = \frac{1}{2}\int_{\mathbb{R}^3} h(1-u) = \frac{1}{2}\int_{\mathbb{R}^3} hf =: c_0 \,.$$ Hence $g_r \to c_0\, \delta_{\{x=0\}}$ weakly in the sense of distributions, and from here by elliptic estimates you get that $u_r \to c_0 \, G_3$ where $G_3=C_n/|x|$ is the fundamental solution in $\mathbb{R}^3$. This holds since $G_3$ is the unique solution to $-\Delta G_3=\delta_{\{x=0\}}$ that goes to zero as $|x|\to \infty$. This implies that, for example, in the annulus $A_{1,2}:=B_2(0)\setminus B_1(0)$ there holds $$ \sup_{x\in A_{1,2}} \left|ru(rx) -\frac{c_0 C_n}{|x|} \right| \to 0$$ as $r\to \infty$, or equivalently $$ r \sup_{x \in A_{r,2r}} \left|u(x) -\frac{c_0 C_n}{|x|} \right| \to 0 \,.$$ This implies what you ask.