Regularity of solution of $(-\Delta + w)f = 0$

Question

I am studying the following Schrödinger equation: $$(-\Delta + w)f = 0$$ which represents a quantum state with zero energy. Here $w$ and $f$ are defined on $\mathbb{R}^{3}$. For simplicity, let us assume $w$ is nonnegative, radially symmetric $w(x) = w(|x|) = w(r)$, $r \in (0,\infty)$ and with compact support.

Assume the differential equation above holds in the distribution sense and suppose the solution $f$ is $f \in H^{1}(\mathbb{R}^{3})$. I am actually interested in radially symmetric solutions $f = f(r)$. My question is: how regular the derivative of $f$ can be near the origin? I know I can be the case that $\frac{df}{dr}$ goes to infinity as $r \to 0^{+}$, but is there any condition on, say, $r^{2}\frac{df}{dr}$ or some power $r^{n}\frac{df}{dr}$, $n \ge 1$ in which this product goes to zero as $r \to 0^{+}$?

Edit: The two important cases for me are: $w \in C^{\infty}$ and $w \in L^{3}(\mathbb{R}^{3})$.

Answer 1

As discussed in the comments, I interpret the question as asking about the asymptotics of $f'(r)$, $r\to 0+$, for solutions of $$ -\frac{d^2f}{dr^2} -\frac{2}{r} \frac{df}{dr} + w(r)f(r) = 0 ; \quad\quad\quad\quad (1) $$ the first two terms are the radial part of $-\Delta$.

I'll show that if $w$ is bounded near $r=0$, then $r^2 df/dr$ will be bounded also (and thus $r^{2+\epsilon} df/dr\to 0$). This covers the case $w\in C^{\infty}$; if we only assume $w\in L^3$, the method still has potential, but things get even more tedious.

The substitution $r=1/t$ brings (1) to the form $$ -f'' + (w/t^4)f=0 , $$ and now we are interested in $df/dr = -t^2f'(t)$ for $t\to\infty$.

This looks promising because $w/t^4\in L^1$ if $w$ is continuous, with room to spare, so perturbation methods should work. We can introduce $Y=(f',f)^t$, so $$ Y' = \begin{pmatrix} 0 & w/t^4 \\ 1 & 0 \end{pmatrix} Y . $$ Write $Y=Y_0u$, with $$ Y_0 = \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} $$ chosen as the fundamental matrix of the unperturbed problem. Now another calculation shows that $u'= O(1/t^2) u$, so the coefficients are still in $L^1$, and Gronwall's inequality shows that $u$ is bounded (in fact, Levinson's theorem shows that $u(t)$ converges as $t\to\infty$).

Unwrapping all this, we find that $f'(t)$ stays bounded, and hence so does $r^2 df/dr=-df/dt$.

Answer 2

regarding the case $f\in L^n$ I can give you some suggestion.

Since $L^n$ is contained in the Morrey space $L^{1,n-1}$ and you have Laplace operator you can say that $\nabla f $ belongs to BMO and then locally to any $L^p$ class with $1<p<\infty$.

Reference:

Poisson equations and Morrey spaces J. Math. Anal. Appl. 163, (1992) Theorem 2.7