# Helmholtz Solutions for the Fractional Laplacian and Other Related Operators

Vincent Guan , Mathav Murugan, and Juncheng Wei  Department of Mathematics,
University of British Columbia, Vancouver, BC, V6T 1Z2, Canada;
<EMAIL_ADDRESS>Department of Mathematics, University of British Columbia,
Vancouver, BC, V6T 1Z2, Canada<EMAIL_ADDRESS>Department of Mathematics,
University of British Columbia, Vancouver, BC, V6T 1Z2, Canada;
<EMAIL_ADDRESS>
###### Abstract

We show that the bounded solutions to the fractional Helmholtz equation,
$(-\Delta)^{s}u=u$ for $0<s<1$ in $\mathbb{R}^{n}$, are given by the bounded
solutions to the classical Helmholtz equation $(-\Delta)u=u$ in
$\mathbb{R}^{n}$ for $n\geq 2$ when $u$ is additionally assumed to be
vanishing at $\infty$. When $n=1$, we show that the bounded fractional
Helmholtz solutions are again given by the classical solutions
$A\cos{x}+B\sin{x}$. We show that this classification of fractional Helmholtz
solutions extends for $1<s\leq 2$ and $s\in\mathbb{N}$ when $u\in
C^{\infty}(\mathbb{R}^{n})$. Finally, we prove that the classical solutions
are the unique bounded solutions to the more general equation
$\psi(-\Delta)u=\psi(1)u$ in $\mathbb{R}^{n}$, when $\psi$ is complete
Bernstein and certain regularity conditions are imposed on the associated
weight $a(t)$.

##  1 Introduction

###  1.1 Motivation and Main Results

An elementary result is that the bounded solutions to the equation $-u_{xx}=u$
in $\mathbb{R}$ are given in the form: $u(x)=A\cos{x}+B\sin{x}$. In higher
dimensions, bounded solutions to the classical Helmholtz equation $-\Delta
u=u$ on $\mathbb{R}^{n}$ are expressed in terms of Bessel functions and
spherical harmonics (see Appendix and also [2]). These higher dimensional
solutions are $C^{\infty}$, bounded, and vanishing at $\infty$. Helmholtz
functions arise frequently when solving PDEs such as the heat equation and the
wave equation via separation of variables. They also play fundamental roles in
inverse problems and scattering theory.

In recent years, there is a growing interest in the fractional Laplacian
$(-\Delta)^{s}$, with $0<s<1$, due to various applications involving nonlocal
diffusion [4]. A natural question is to classify solutions to the fractional
Helmholtz equation

$(-\Delta)^{s}u(x)=u(x)\ \ \mbox{in}\ \mathbb{R}^{n}.$ (1.1)

In the first theorem of this paper, we prove that the bounded and vanishing
fractional Helmholtz solutions to (1.1) are the same as the classical
Helmholtz solutions.

######  Theorem 1.1.

Let $u(x)\in L^{\infty}(\mathbb{R}^{n})$. If $n=1$ and $0<s<1$, then all
solutions to $(-\Delta)^{s}u=u$ in $\mathbb{R}$ are $A\cos(x)+B\sin(x)$ for
some constants $A,B\in\mathbb{R}$. If $n\geq 2$ and we assume that $u(x)\to 0$
as $|x|\to\infty$, then $u$ solves $(-\Delta)^{s}u=u$ in $\mathbb{R}^{n}$ for
$0<s<1$ if and only if $(-\Delta)u=u$ in $\mathbb{R}^{n}$.

Remark: The case of $n=1$ was first proved by Fall and Weth in [9].

We will then show that this result can be generalized to powers $1<s\leq 2$ by
reducing this case to the previous case $0<s<1$.

######  Theorem 1.2.

Let $u(x)\in C^{\infty}(\mathbb{R}^{n})\bigcap L^{\infty}(\mathbb{R}^{n})$.
Then $u$ solves $(-\Delta)^{s}u=u$ in $\mathbb{R}^{n}$ for $0<s\leq 2$ if and
only if $(-\Delta)u=u$ in $\mathbb{R}^{n}$.

For $s>2$, we have the following classification result for the polyharmonic
Helmholtz equation.

######  Theorem 1.3.

Let $u(x)\in C^{\infty}(\mathbb{R}^{n})\bigcap L^{\infty}(\mathbb{R}^{n})$ and
$m\in\mathbb{N}$. Then $u$ solves $(-\Delta)^{m}u=u$ in $\mathbb{R}^{n}$ if
and only if $(-\Delta)u=u$ in $\mathbb{R}^{n}$.

Finally, we will consider the Helmholtz equation for complete Bernstein
functions $\psi$ of the Laplace operator. The Bernstein Helmholtz equation is
given by

$\displaystyle\psi(-\Delta)u(x)=u(x)\ \ \mbox{in}\ \mathbb{R}^{n}.$ (1.2)

$\psi(t)$ is said to be a complete Bernstein function if
$\psi(t)=\frac{\mathcal{L}\\{f\\}(t)}{t}$, where $\mathcal{L}$ is the Laplace
transform and $f:[0,\infty)\to[0,\infty)$ is completely monotone, i.e.
$(-1)^{k}f^{(k)}\geq 0$. In particular, $\psi(t)$ generalizes $t^{s}$ with
$0<s<1$.

######  Theorem 1.4.

Let $n\geq 2$. Suppose $u(x)\to 0$ as $|x|\to\infty$ and that the associated
weight $a(t)$ for $\psi(-\Delta)$ in the extension problem (1.5) is $A_{2}$
and obeys $a(t)\sim t^{\alpha}$ for $t\gg 1$ and $-1<\alpha<1$. Then $u\in
L^{\infty}(\mathbb{R}^{n})$ solves $\psi(-\Delta)u=\psi(1)u$ in
$\mathbb{R}^{n}$ if and only if $(-\Delta)u=u$ in $\mathbb{R}^{n}$.

###  1.2 Preliminaries

Recall that $(-\Delta)^{s}$ is defined via the Fourier multiplier
$|\xi|^{2s}$, such that in the distributional sense, we have

$\widehat{(-\Delta)^{s}u}(\xi)=|\xi|^{2s}\hat{u}(\xi).$

We will also use two other definitions. The first is via the singular integral
$(-\Delta)^{s}u(x)=c_{n,s}\textrm{P.V}\int_{\mathbb{R}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+2s}}dy$
where $c_{n,s}$ is a normalization constant. The second definition, made
famous by Caffarelli and Silvestre in [6], characterizes $(-\Delta)^{s}u(x)$
by its harmonic extension $\dot{u}(x,t)$. In particular, $\dot{u}(x,t)\in
H^{1}_{loc}(\mathbb{R}^{n+1}_{+},t^{1-2s})$ weakly solves the harmonic
equation:

$\displaystyle\nabla\cdot(t^{1-2s}\nabla
u)=t^{1-2s}[\Delta_{x}\dot{u}+\frac{1-2s}{t}\dot{u}_{t}+\dot{u}_{tt}]=0\textrm{
on }\mathbb{R}^{n+1}_{+},$ $\displaystyle\lim_{t\to
0}t^{1-2s}\dot{u}_{t}(x,t)=-c_{n,s}(-\Delta)^{s}u\textrm{ on }\mathbb{R}^{n},$
$\displaystyle\dot{u}(x,0)=u(x)\textrm{ on }\mathbb{R}^{n}\times\\{0\\}.$
(1.3)

Similarly, for complete Bernstein functions $\psi(x)$, $\psi(-\Delta)$ is
defined via the Fourier multiplier $\psi(|\xi|^{2})$, such that in the
distributional sense, we have

$\widehat{\psi(-\Delta)u}(\xi)=\psi(|\xi|^{2})\hat{u}(\xi).$

From [12], for any complete Bernstein function $\psi$, we may characterize
$\psi(-\Delta)$ using the harmonic extension $\dot{u}(x,t)\in
H^{1}_{loc}(\mathbb{R}^{n+1}_{+},a(t))$, which solves

$\displaystyle\partial_{s}^{2}\dot{u}(x,s)+A(s)\Delta_{x}\dot{u}(x,s)=0\textrm{
on }\mathbb{R}^{n+1}_{+},$
$\displaystyle\partial_{s}\dot{u}(0,x)=-\psi(-\Delta)u(x)\textrm{ on
}\mathbb{R}^{n},$ $\displaystyle\dot{u}(x,0)=u(x)\textrm{ on
}\mathbb{R}^{n}\times\\{0\\},$ (1.4)

where $A(s)$ is a locally finite measure on some interval $[0,R)$, with $R$
possibly infinite. If $A(s)$ is non-negative and $L^{1}_{\textrm{loc}}$, then
we may apply the change of variables: $ds=(a(t))^{-1}dt$ and $A(s)ds=a(t)dt$.
We call $a(t)$ the associated weight to equation (1.2). This reduces (1.4) to
the familiar form:

$\displaystyle\nabla\cdot(a(t)\nabla\dot{u})=0\textrm{ on
}\mathbb{R}^{n+1}_{+},$ $\displaystyle\lim_{t\to
0}a(t)\dot{u}_{t}(x,t)=-c_{n,a}\psi(-\Delta)u(x)\textrm{ on }\mathbb{R}^{n}.$
(1.5)

For the rest of the paper, the trace condition $\dot{u}(x,0)=u(x)$ will be
implicitly assumed.

###  1.3 Comments about the proofs

Since $(-\Delta)^{s}$ and $\psi(-\Delta)$ are non-local operators, it is
difficult to solve these nonlocal Helmholtz equations directly. We cannot
apply separation of variables as in the $s=1$ case (shown in Appendix), and in
general, it is not easy to compute the integral $(-\Delta)^{s}u$ explicitly.
To prove compatibility of classical solutions with other Helmholtz equations,
it is helpful to consider the equivalent extension problem, i.e. (1.3) for
$(-\Delta)^{s}$ and (1.4), (1.5) for $\psi(-\Delta)$. This way, the problem is
reformulated as a second order elliptical partial differential equation.

Uniqueness of fractional Helmholtz solutions in the class of bounded vanishing
functions for $n\geq 2$ is achieved by working with the extension problem and
using standard energy techniques, following the seminal work of [10]. The
$n=1$ case is handled separately.

Similar but more sophisticated techniques will be used to prove uniqueness in
the general case where we consider the solutions to $\psi(-\Delta)u=u$.
$a(t)\in A_{2}$ is assumed to achieve appropriate Hölder estimates and the
asymptotic behaviour $a(t)\sim t^{\alpha}$ is assumed to apply the estimate
$|\dot{u}_{t}|\leq\frac{C}{t}$ from [5, Prop 4.6].

##  2 Fractional Helmholtz Solutions $0<s<1$

Proof of Theorem 1.1: When $n\geq 2$, Theorem 1.1 is a corollary of Lemma 2.1
and Lemma 5.2. When $n=1$, Theorem 1.1 is a corollary of Lemma 2.1 and Lemma
2.3. ∎

######  Lemma 2.1.

If $u\in L^{\infty}(\mathbb{R}^{n})$ solves $-\Delta u=u$ in $\mathbb{R}^{n}$,
then $(-\Delta)^{s}u=u$ in $\mathbb{R}^{n}$ for $0<s<1$.

Proof: Suppose that $u\in L^{\infty}(\mathbb{R}^{n})$ and $-\Delta u=u$ on
$\mathbb{R}^{n}$. It suffices to show that the extension problem (1.3) for
$(-\Delta)^{s}u=u$ is solvable. Hence, it suffices to show that there is an
extension $\dot{u}(x,t)=u(x)\phi(t)$ with $\phi(0)=1$ such that

$\displaystyle\phi(t)\Delta
u(x)+u(x)\frac{1-2s}{t}\phi^{\prime}(t)+u(x)\phi^{\prime\prime}(t)=0\textrm{
on }\mathbb{R}^{n+1}_{+},$ $\displaystyle\lim\limits_{t\to
0}t^{1-2s}u(x)\phi^{\prime}(t)=-c_{n,s}(-\Delta)^{s}u(x)=-c_{n,s}u(x)\textrm{
on }\mathbb{R}^{n}.$ (2.1)

Since $u$ is a classical Helmholtz solution, (2.1) reduces to

$\displaystyle-\phi(t)+\frac{1-2s}{t}\phi^{\prime}(t)+\phi^{\prime\prime}(t)=0\textrm{
for }t\geq 0,$ $\displaystyle\lim\limits_{t\to
0}t^{1-2s}\phi^{\prime}(t)=-c_{n,s}\in\mathbb{R}.$ (2.2)

It suffices to solve (2.2) with the initial condition $\phi(0)=1$. This ODE is
solved in [6, Section 3.2]. ∎

We now show that [10, Thm 1] can be generalized to non-radial functions in the
following sense:

######  Lemma 2.2.

Let $n\geq 2$, $V(r)\in C^{\infty}(\mathbb{R}^{n})$ be radially non-decreasing
and $u$ be a bounded and vanishing function satisfying
$(-\Delta)^{s}u+V(r)u=0$ on $\mathbb{R}^{n}$. Then the projection of $u$ onto
any spherical harmonical eigenfunction is unique up to a constant factor.

Proof: Let $n\geq 2$ and $u\in L^{\infty}(\mathbb{R}^{n})$ be vanishing and
solve $(-\Delta)^{s}u+V(r)u=0$ on $\mathbb{R}^{n}$ where $V(r)$ is radially
non-decreasing and differentiable. First, we use eigenfunction decomposition
with spherical harmonics to reduce this problem to the case where $u$ is
additionally assumed to be radial. In particular, we consider the extension
$\dot{u}(x,t)$, which solves (1.3). Using spherical harmonics, we obtain the
decomposition

$\displaystyle\dot{u}(r,\theta,t)=\sum_{l}\dot{u_{l}}(r,t)\phi_{l}(\theta),$

where $\Delta_{\mathbb{S}^{n-1}}\phi_{l}(\theta)=-\mu_{l}\phi_{l}(\theta)$ and
$\mu_{l}=l(l+n-2)$. Note that for each fixed eigenvalue $\mu_{l}$, there may
be multiple associated eigenfunctions $\phi_{l}$ included in the sum. Then,
since
$\Delta_{x}u=\partial_{r}^{2}u+\frac{n-1}{r}\partial_{r}u+\frac{\Delta_{\mathbb{S}^{n-1}}u}{r^{2}}$
and $\\{\phi_{l}\\}_{l}$ is an orthogonal family, for each $l$, we may
substitute $\dot{u}=\dot{u_{l}}(r,t)\phi_{l}(\theta)$ into (1.3) to yield

$\displaystyle\dot{u}_{l,rr}+\frac{n-1}{r}\dot{u}_{l,r}+\frac{1-2s}{t}\dot{u}_{l,t}+\dot{u}_{l,tt}-\frac{\mu_{l}\dot{u}_{l}}{r^{2}}=0\textrm{
on }\mathbb{R}^{n+1}_{+},$ $\displaystyle\lim\limits_{t\to
0}t^{1-2s}\dot{u}_{l,t}=c_{n,s}V(r)\dot{u_{l}}(r,0):=c_{n,s}V(r)u_{l}(r)\textrm{
on }\mathbb{R}^{n}.$ (2.3)

Note that $u(x)=\sum\dot{u}_{l}(r,0)\phi_{l}(\theta)=\sum
u_{l}(r)\phi_{l}(\theta)$ by substituting $t=0$. Now, let
$\dot{v}_{l}(r,t)=r^{-l}\dot{u}_{l}(r,t)$ and $v_{l}(r)=r^{-l}u_{l}(r)$. This
yields the system:

$\displaystyle\dot{v}_{l,rr}+\frac{2l+n-1}{r}{\dot{v}}_{l,r}+\frac{1-2s}{t}\dot{v}_{l,t}+{\dot{v}}_{l,tt}=0\textrm{
on }\mathbb{R}^{n+1}_{+},$ $\displaystyle\lim\limits_{t\to
0}t^{1-2s}\dot{v}_{l,t}=c_{n,s}V(r){v}_{l}(r)\textrm{ on }\mathbb{R}^{n}.$
(2.4)

In particular, the last term of the first condition of (2.3) cancels since
$\mu_{l}=l(l+n-2)=l(l-1)+l(n-1)$. Now, note that each $v_{l}$ is bounded,
radial, and vanishing, so we may use the proof of [10, Thm 1] to show that
each $v_{l}$ is unique. To simplify notation, we redefine $u=v_{l}$,
$\dot{u}=\dot{v_{l}}$, and $V(r)=c_{n,s}V(r)$. Following [10, (4.6)], we
define the energy $H(r)$ on $u$ as

$H(r)=\frac{1}{2}\left[\int_{0}^{\infty}t^{1-2s}(\dot{u}_{r}^{2}-\dot{u}_{t}^{2})dt-V(r)u(r)^{2}\right].$
(2.5)

In [10, Prop B.2], it is shown that $H(0)\leq-\frac{1}{2}V(0)u(0)^{2}$ and
$H(\infty)=0$ for any radial $u(r)$ vanishing at infinity. We now show that
$H^{\prime}(r)\leq 0$ for $u(r)=v_{l}(r)$. We compute

$\displaystyle H^{\prime}(r)$
$\displaystyle=\int_{0}^{\infty}\frac{d}{dr}\frac{t^{1-2s}}{2}(\dot{u}_{r}^{2}-\dot{u}_{t}^{2})dt-\frac{1}{2}V^{\prime}(r)u(r)^{2}-V(r)u(r)u^{\prime}(r)$
$\displaystyle=\int_{0}^{\infty}t^{1-2s}\dot{u}_{r}\dot{u}_{rr}dt-\int_{0}^{\infty}t^{1-2s}\dot{u}_{t}\dot{u}_{rt}dt-\frac{1}{2}V^{\prime}(r)u(r)^{2}-V(r)u(r)u^{\prime}(r))$
$\displaystyle=\int_{0}^{\infty}t^{1-2s}\dot{u}_{r}\dot{u}_{rr}dt-t^{1-2s}\dot{u_{t}}\dot{u_{r}}|_{t=0}^{t=\infty}$
$\displaystyle\qquad+\int_{0}^{\infty}t^{1-2s}\dot{u}_{r}\left(\frac{1-2s}{t}\dot{u}_{t}+\dot{u}_{tt}\right)dt-\frac{1}{2}V^{\prime}(r)u(r)^{2}-V(r)u(r)u^{\prime}(r))$
$\displaystyle=\int_{0}^{\infty}t^{1-2s}\dot{u}_{r}\left(\dot{u}_{rr}+\frac{1-2s}{t}\dot{u}_{t}+\dot{u}_{tt}\right)dt-\frac{1}{2}V^{\prime}(r)u(r)^{2})$
$\displaystyle=-\frac{2l+n-1}{r}\int_{0}^{\infty}t^{1-2s}\dot{u}_{r}^{2}dt-\frac{1}{2}V^{\prime}(r)u(r)^{2}\leq
0.$ (2.6)

In the second to last equality, note that
$-t^{1-2s}\dot{u_{t}}\dot{u_{r}}|_{t=0}^{t=\infty}=V(r)u(r)u^{\prime}(r)$
results from applying the boundary condition (2.4) and the estimate
$|\dot{u_{t}}(x,t)|\leq\frac{C}{t}$ from [5, Prop 4.6]. The last equality is
due to the first condition in (2.4) and this expression is non-positive since
$n\geq 2,l\geq 0$, and $V^{\prime}(r)\geq 0$.

If $u(0)=0$, then it is clear that $H(r)=0$ and $\frac{dH}{dr}=0$ given the
properties of $H(r)$ established above. If $n\geq 2$, then we may immediately
conclude that $u_{r}=0$ from (2.6) and hence $u=0$. Equivalently, since (2.4)
is a linear system, $u=v_{l}$ must be the unique bounded and vanishing
solution to (2.4) up to a constant factor. ∎

######  Lemma 2.3.

If $u\in L^{\infty}(\mathbb{R})$ solves $(-\Delta)^{s}u=u$ in $\mathbb{R}$,
then $u^{{}^{\prime\prime}}=u$ in $\mathbb{R}$.

Proof: Recall that the Helmholtz solutions on $\mathbb{R}$ are
$u(x)=A\cos(x)+B\sin(x)$. We may assume $(-\Delta)^{s}u=u$ and $u\in
S^{{}^{\prime}}(\mathbb{R})$, the space of tempered distributions. Then,
taking the Fourier transform on both sides gives
$|\xi|^{2s}\hat{u}(\xi)=\hat{u}(\xi)$ in the weak sense. A quick proof is
shown below. For all $\phi\in S(\mathbb{R})$:

$\langle\hat{u},\phi\rangle=\int{((-\Delta)^{s}u)\hat{\phi}}=\int{u(-\Delta)^{s}\hat{\phi}}=\int{\hat{u}|\xi|^{2s}\phi}=\langle|\xi|^{2s}\hat{u}(\xi),\phi\rangle$
(2.7)

Because the Fourier transform is a bijective map from $S^{{}^{\prime}}\to
S^{{}^{\prime}}$, we note that $\hat{u}(\xi)\in S^{{}^{\prime}}(\mathbb{R})$
and it has support $\\{-1,1\\}$. Thus, we may write
$\hat{u}(\xi)=\sum_{\\{\alpha\leq
N\\}}C_{\alpha}\partial^{\alpha}\delta_{-1}+D_{\alpha}\partial^{\alpha}\delta_{1}$
for some constants $N,C_{\alpha},D_{\alpha}$. Then, taking the inverse
transform, we see that $u(x)=a(x)\sin(x)+b(x)\cos(x)$ for some polynomials
$a(x)$ and $b(x)$. Since $u$ is bounded, $u(x)=A\cos(x)+B\sin(x)$. ∎

##  3 Fractional Helmholtz Solutions $1<s\leq 2$

In this section, we consider fractional Helmholtz solutions with $1<s\leq 2$.
Existence can be proved in a similar way as in Lemma 2.1, by using the
extension problem from [7]. Instead we present a new proof by decomposition of
nonlocal operators. To do so, we use the fact that we can rewrite
$(-\Delta)^{s}u$ with $1<s\leq 2$ in terms of the standard fractional
laplacian with $\frac{1}{2}<s/2\leq 1$ when $u$ has sufficient regularity. We
also use this decomposition to prove uniqueness.

Proof of Theorem 1.2: This is a corollary of Lemma 3.2 and Lemma 3.3. ∎

######  Lemma 3.1.

Let $u\in C^{\infty}(\mathbb{R}^{n})\bigcap L^{\infty}(\mathbb{R}^{n})$. Then
for $1<s\leq 2$, we have $(-\Delta)^{s}u=(-\Delta)^{s/2}[(-\Delta)^{s/2}u]$.

Proof: Let $u\in C^{\infty}(\mathbb{R}^{n})\bigcap L^{\infty}(\mathbb{R}^{n})$
and $1<s\leq 2$. We consider the definition of $(-\Delta)^{s}$ from [1]:

$(-\Delta)^{s}u=L_{2,s}u=c_{n,2,s}\int\frac{u(x-2y)-4u(x-y)+6u(x)-4u(x+y)+u(x+2y)}{|y|^{n+2s}}dy.$

Note also that $L_{1,s/2}u=(-\Delta)^{s/2}u$. Thus, we want to show that
$L_{1,s/2}(L_{1,s/2}u)=L_{2,s}u$. In [1, Thm 1.9] this is obtained for all
$\phi\in C_{c}^{\infty}(\mathbb{R}^{n})$ since

$\mathscr{F}(L_{2,s}\phi)(\xi)=|\xi|^{2s}\mathscr{F}(\phi)(\xi)=|\xi|^{s}\mathscr{F}(L_{1,s/2}\phi)(\xi)=\mathscr{F}(L_{1,s/2}(L_{1,s/2}\phi))(\xi).$
(3.1)

Now, since $u\in L^{\infty}(\mathbb{R}^{n})\bigcap
C^{\infty}(\mathbb{R}^{n})$, we may apply [1, Lemma 1.5], which tells us that
for any $\phi\in C_{c}^{\infty}(\mathbb{R}^{n})$,

$\displaystyle\int_{\mathbb{R}^{n}}u(x)L_{2,s}\phi(x)dx=\int_{\mathbb{R}^{n}}\phi(x)L_{2,s}u(x)dx,$
$\displaystyle\int_{\mathbb{R}^{n}}u(x)L_{1,s/2}\phi(x)dx=\int_{\mathbb{R}^{n}}\phi(x)L_{1,s/2}u(x)dx.$
(3.2)

The desired result then follows from applying (3.1) and (3.2).

$\displaystyle\int_{\mathbb{R}^{n}}\phi(x)L_{2,s}u(x)dx=\int_{\mathbb{R}^{n}}u(x)L_{2,s}\phi(x)dx=\int_{\mathbb{R}^{n}}u(x)L_{1,s/2}(L_{1,s/2}\phi(x))dx$
$\displaystyle=\int_{\mathbb{R}^{n}}L_{1,s/2}u(x)L_{1,s/2}\phi(x)dx=\int_{\mathbb{R}^{n}}\phi(x)L_{1,s/2}(L_{1,s/2}u(x))dx$
(3.3)

∎

######  Lemma 3.2.

If $u$ solves $(-\Delta)u=u$ in $\mathbb{R}^{n}$ and $u\in
C^{\infty}(\mathbb{R}^{n})\bigcap L^{\infty}(\mathbb{R}^{n})$, then $u$ also
solves $(-\Delta)^{s}u=u$ in $\mathbb{R}^{n}$ for $1<s\leq 2$.

Proof: Let $1<s\leq 2$ and $u\in C^{\infty}(\mathbb{R}^{n})\bigcap
L^{\infty}(\mathbb{R}^{n})$ solve $(-\Delta)u=u$ in $\mathbb{R}^{n}$. By
Theorem 1.1, $(-\Delta)^{s/2}u=u$ since $s/2\in(0,1]$. We then apply Lemma 3.1
to yield

$(-\Delta)^{s}u=(-\Delta)^{s/2}[(-\Delta)^{s/2}u]=(-\Delta)^{s/2}u=u.$ (3.4)

∎

To prove uniqueness of these fractional solutions, it suffices to show that
all solutions $(-\Delta)^{s}u=u$, with $1<s\leq 2$, satisfy
$(-\Delta)^{s/2}u=u$, since uniqueness has already been shown for $s\in(0,1]$
with Lemma 2.2 and Lemma 2.3.

######  Lemma 3.3.

Let $1<s\leq 2$. If $u\in C^{\infty}(\mathbb{R}^{n})\bigcap
L^{\infty}(\mathbb{R}^{n})$ solves $(-\Delta)^{s}u=u$ on $\mathbb{R}^{n}$,
then $(-\Delta)^{s/2}u=u$ on $\mathbb{R}^{n}$.

Proof: Let $1<s\leq 2$. and $(-\Delta)^{s}u=u$ on $\mathbb{R}^{n}$. We can
define the system

$\left\\{\begin{array}[]{l}v=(-\Delta)^{s/2}u-u,\\\
(-\Delta)^{s/2}v+v=0.\end{array}\right.$ (3.5)

By Lemma 3.1, it suffices to show that $v=0$. We proceed by using the Maximum
Principle. Let $w(x)=C+|\sqrt{1+x^{2}}|^{\rho}$ with $\rho<s$. We first claim
that we can construct $w$ such that $(-\Delta)^{s/2}w+w>0$. To see this, note
that $|\sqrt{1+x^{2}}|^{\rho}$ asymptotically behaves like $|x|^{\rho}$ while
remaining uniformly bounded for $x$ close to $y$. Hence, we may let
$x=\alpha\tilde{x},y=\alpha\tilde{y}$ and compute

$\displaystyle(-\Delta)^{s/2}w(x)=c_{n,s}\textrm{P.V}\int_{\mathbb{R}^{n}}\frac{|\sqrt{1+x^{2}}|^{\rho}-|\sqrt{1+y^{2}}|^{\rho}}{|x-y|^{n+s}}dy$
$\displaystyle=c_{n,s}\textrm{P.V}\int_{\mathbb{R}^{n}}\frac{|\sqrt{1+\alpha^{2}\tilde{x}^{2}}|^{\rho}-|\sqrt{1+\alpha^{2}\tilde{y}^{2}}|^{\rho}}{\alpha^{n+s}|\tilde{x}-\tilde{y}|^{n+s}}\alpha^{n}d\tilde{y}\sim\alpha^{\rho-s}(-\Delta)^{s}w(\tilde{x}).$
(3.6)

Then, since $\rho-s<0$ and $(-\Delta)^{s}w(\tilde{x})$ is bounded because the
integrand in the last line of (3.6) is $o(|y|^{n})$, we may pick
$\alpha,\rho,C$ such that $(-\Delta)^{s/2}w+w>0$. Now, let
$\tilde{v}=v-\epsilon w$. We immediately see that

$\displaystyle(-\Delta)^{s/2}\tilde{v}+\tilde{v}=(-\Delta)^{s/2}v+v-\epsilon((-\Delta)^{s/2}w+w)$
$\displaystyle=-\epsilon((-\Delta)^{s/2}w+w)<0.$ (3.7)

Note that $v=o(w)=o(|x|^{\rho})$ since we can choose $\rho$ arbitrarily close
to $s$ and $v=o(|x|^{s})$. Therefore, $\tilde{v}(\infty)=-\infty$ and
$\tilde{v}\in C^{2}(\mathbb{R}^{n})$ by Lemma 5.4. From this, we may deduce
that there is some $x_{0}$ such that
$\tilde{v}(x_{0})=\max\limits_{x\in\mathbb{R}^{n}}v(x)$. Hence, we may use the
fact that
$(-\Delta)^{s/2}\tilde{v}(x_{0})=\textrm{P.V}\int\frac{\tilde{v}(x_{0})-\tilde{v}(y)}{|x_{0}-y|^{n+s}}\geq
0$ and (3.7) to obtain the inequalities

$0\leq(-\Delta)^{s/2}\tilde{v}(x_{0})<-\tilde{v}(x_{0})\implies v(x)-\epsilon
w(x)\leq v(x_{0})-\epsilon w(x_{0})\leq 0.$ (3.8)

Taking $\epsilon\to 0$ yields $v(x)\leq 0$. By a symmetric argument, if we let
$\tilde{v}=\epsilon w-v$, then we would have
$(-\Delta)^{s/2}\tilde{v}+\tilde{v}>0$ and $v(x)\geq 0$. This proves $v(x)=0$
as desired. ∎

##  4 Polyharmonic Helmholtz Solutions

In this section, we consider Helmholtz solutions to the equation

$(-\Delta)^{m}u=u\textrm{ in }\mathbb{R}^{n},\textrm{ where }m\in\mathbb{N}.$
(4.1)

Proof of Theorem 1.4: From Lemma 3.2, it is clear that $-\Delta u=u$ will
solve $(-\Delta)^{m}u=u$ for any $m\in\mathbb{N}$ and we have already shown
uniqueness when $m=1,2$. Let us now consider the case $m=3$. To prove
uniqueness of the classical solutions in the class of bounded functions, we
may consider the system

$\displaystyle v=-\Delta u-u,$ $\displaystyle\Delta^{2}v-\Delta v+v=0.$ (4.2)

It suffices to show that $v=0$. To do so, let $\eta$ be a cutoff function
supported in $B_{2R}$. We may multiply the bottom equation by $\eta^{2}v$ and
integrate by parts to yield

$\displaystyle\int(\Delta(\eta v))^{2}+\int|\nabla\eta
v|^{2}+\int\eta^{2}v^{2}=-2\int v\Delta v|\nabla\eta|^{2}+v\eta\nabla
v\cdot\nabla\eta$ $\displaystyle+4\int(\nabla\eta\cdot\nabla
v)^{2}+v\Delta\eta\nabla\eta\cdot\nabla v+\int
v^{2}(\Delta\eta)^{2}-v^{2}\eta\Delta\eta.$ (4.3)

Then, note that we may pick $\eta$ such that
$|\nabla\eta|\lesssim\frac{1}{R}$. Also, from standard elliptical theory,
since $u\in L^{\infty}(\mathbb{R}^{n})$, it follows that all of its
derivatives are uniformly bounded as well. Thus, after using integration by
parts and Cauchy-Schwarz on the right hand side of (4.3), we obtain the
inequality

$\displaystyle\int_{B_{2R}}(\Delta v)^{2}+|\nabla
v|^{2}+v^{2}\lesssim\frac{1}{R^{2}}\int_{B_{2R}}v^{2}+|\nabla v|^{2}.$ (4.4)

Hence, we may iterate the inequality (4.4) on its own right hand side, such
that the bound is scaled by $\frac{1}{R^{2}}$ each time. Taking $R\to\infty$
then yields the desired result $v\equiv 0$.

When $m\geq 4$ we modify the system (4.2) such that $v=-\Delta u-u$ satisfies
$\sum_{j=0}^{m-1}(-\Delta)^{j}v+v=0$. The case of $m\geq 4$ can be proved
similarly as that of $m=3$. We omit the details. This proves Theorem 1.3.

∎

##  5 Complete Bernstein Helmholtz Solutions

We now classify solutions to the Helmholtz equation $\psi(-\Delta)u=u$ in
$\mathbb{R}^{n}$ when $\psi$ is complete Bernstein. Without loss of
generality, we assume that the extension $\dot{u}$ is bounded in
$\mathbb{R}^{n+1}_{+}$.

Proof of Theorem 1.4: Theorem 1.4 is a corollary of Lemma 5.1 and Lemma 5.2. ∎

######  Lemma 5.1.

If $-\Delta u=u$ on $\mathbb{R}^{n}$, then $u$ also solves
$\psi(-\Delta)u=\psi(1)u$ on $\mathbb{R}^{n}$.

Proof: Recall the extension problem for $\psi(-\Delta)u=\psi(1)u$ on
$\mathbb{R}^{n}$ given by

$\displaystyle\partial_{s}^{2}\dot{u}(x,s)+A(s)\Delta_{x}\dot{u}(x,s)=0\textrm{
on }\mathbb{R}^{n+1}_{+},$
$\displaystyle\partial_{s}\dot{u}(0,x)=-\psi(1)u(x)\textrm{ on
}\mathbb{R}^{n}.$ (5.1)

Now, let $\dot{u}(x,s)=\phi(s)u(x)$ such that $\phi(0)=1$. Since $-\Delta
u(x)=u(x)$ on $\mathbb{R}^{n}$, the problem reduces to finding a function
$\phi(s)$ satisfying

$\displaystyle\phi^{\prime\prime}(s)=A(s)\phi(s)\textrm{ for }s\geq 0,$
$\displaystyle\phi^{\prime}(0)=-\psi(1)\in\mathbb{R},$
$\displaystyle\phi(0)=1.$ (5.2)

By [12, Section 3.1], a solution $\phi$ exists. In particular, we may let
$\lambda=1$ in [12, (3.1)] and set $\varphi_{1}=\phi$. ∎

To show uniqueness, we have only been able to provide a proof in the special
case where $a(t)\in A_{2}$ and $a(t)\sim t^{\alpha}$, $|\alpha|<1$ for $t\gg
1$. By scaling the solution, it suffices to consider the case $\psi(1)=1$.

######  Lemma 5.2.

Let $u$ be bounded and vanishing solution to the equation
$\psi(-\Delta)u=\psi(1)u$ on $\mathbb{R}^{n}$ where $n\geq 2$ and the
associated weight $a(t)$ in the extension problem (1.5) is $A_{2}$ and obeys
$a(t)\sim t^{\alpha}$, $|\alpha|<1$ for $t\gg 1$. Then $u$ satisfies $-\Delta
u=u$.

Proof: Let $\psi(-\Delta)u=\psi(1)u$ on $\mathbb{R}^{n}$. We mimic (2.2)-(2.3)
from the proof of Lemma 2.2. Consider the extension $\dot{u}(x,t)$, which
solves (1.5), and decompose it with spherical harmonics to express
$\dot{u}(r,\theta,t)=\sum\dot{u}_{l}(r,t)\phi_{l}(\theta)$. Then, let
$\dot{v}_{l}(r,t)=r^{-l}\dot{u}_{l}(r,t)$ and
$v_{l}=r^{-l}u_{l}(r):=r^{-l}\dot{u}_{l}(r,0)$. For each $l$, $\dot{v}_{l}$
satisfies

$\displaystyle\partial_{t}(a(t)\partial_{r}\dot{v}_{l})+\frac{2l+n-1}{r}a(t)\partial_{r}\dot{v}_{l}+a(t)\partial_{r}^{2}\dot{v}_{l}=0,$
$\displaystyle\lim_{t\to
0}a(t)\partial_{t}\dot{v}_{l}(r,t)=-c_{a,s}\psi(1)v_{l}(r).$ (5.3)

To prove that each $v_{l}$ is unique up to a constant factor in the class of
bounded and vanishing functions, it suffices to show that $v_{l}(0)=0$ implies
$v_{l}\equiv 0$, since (5.3) is a linear system. For simplicity, let
$\dot{v}_{l}=\dot{v}$, $v_{l}=v$, and $-c_{a,s}\psi(1)=c$. Then, multiply the
first equation in (4.3) by $\dot{v}_{r}$ to get

$\displaystyle\partial_{t}(a(t)\partial_{r}\dot{v})\partial_{r}\dot{v}+\frac{2l+n-1}{r}a(t)(\partial_{r}\dot{v})^{2}+a(t)\partial_{r}\dot{v}\partial_{r}^{2}\dot{v}=0.$
(5.4)

Note that the middle term is non-negative. We now integrate the other two
terms over $(0,\infty)\times(0,\infty)$ and show that these integrals are also
non-negative.

$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}(a(t)\dot{v}_{t})_{t}\dot{v}_{r}+a(t)\dot{v}_{r}\dot{v}_{rr}dtdr$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}a(t)\left(\frac{v_{r}^{2}-v_{t}^{2}}{2}\right)_{r}drdt+\int_{0}^{\infty}a(t)\dot{v}_{t}\dot{v}_{r}|_{t=0}^{t=\infty}dr$
$\displaystyle=\int_{0}^{\infty}\frac{a(t)}{2}\left(\dot{v}_{r}^{2}(r,t)-\dot{v}_{t}^{2}(r,t)\right)|_{r=0}^{r=\infty}dt+\int_{0}^{\infty}\lim_{t\to\infty}a(t)\dot{v}_{t}(r,t)\dot{v}_{r}(r,t)+cv(r)\dot{v}_{r}(r,0)dr$
$\displaystyle=\int_{0}^{\infty}\frac{a(t)}{2}\left(\lim_{r\to\infty}\dot{v}_{r}^{2}(r,t)+\dot{v}_{t}^{2}(0,t)\right)dt+c\int_{0}^{\infty}\left(\frac{\dot{v}^{2}(r,0)}{2}\right)_{r}dr$
$\displaystyle=\int_{0}^{\infty}\frac{a(t)}{2}\left(\lim_{r\to\infty}\dot{v}_{r}^{2}(r,t)+\dot{v}_{t}^{2}(0,t)\right)dt-\frac{cv^{2}(0)}{2}.$
(5.5)

Three cancellations were performed to obtain the second to last equality.
First, $\lim\limits_{r\to\infty}a(t)\dot{v}_{t}(r,t)^{2}=0$ due to Parseval’s
Theorem and Lemma 5.3. Second, $a(t)\dot{v}_{r}(0,t)^{2}=0$ since
$\dot{v}_{r}(0,t)=0$ follows from Lemma 5.4. Third, note that
$\lim_{t\to\infty}a(t)\dot{v}_{t}(r,t)\dot{v}_{r}(r,t)=0$ follows from the
estimate $|\dot{v}_{t}(x,t)|\leq\frac{C}{t}$, which can be shown from a
similar rescaling argument done in the proof of [5, Prop 4.6]. This is where
we use the assumption that for sufficiently large $t$, $a(t)\sim t^{\alpha}$
for some $|\alpha|<1$. Then, from (5.4) and (5.5), we may easily compute

$\displaystyle\frac{1}{2}\int_{0}^{\infty}a(t)(\dot{v}_{r}(\infty,t)^{2}+\dot{v}_{t}(0,t)^{2})dt+\int_{0}^{\infty}\int_{0}^{\infty}\frac{2l+n-1}{r}a(t)(\dot{v}_{r})^{2}drdt-\frac{cv^{2}(0)}{2}=0.$
(5.6)

Each of these terms are non-negative since $a(t)\in A_{2}$ implies that
$a(t)>0$ a.e. and $v(0)=0$ is assumed. Thus, if $n\geq 2$, then we may
conclude $\dot{v}_{r}(r,t)=0$ $\forall t\geq 0$ . Since $v(0)=0$, then
$v\equiv 0$ as desired. Hence, each $u_{l}(r)$ is unique up to a constant
factor and $u(r,\theta)=\sum_{l}c_{l}u_{l}(r)\phi_{l}(r)$ for some constants
$c_{l}$. The result follow from Lemma 5.1 and the fact that if
$u(r,\theta)=\sum_{l}u_{l}(r)\phi_{l}(r)$ solves $-\Delta u=u$, then
$\tilde{u}(r,\theta)=\sum_{l}c_{l}u_{l}(r)\phi_{l}(r)$ also solves $-\Delta
u=u$.

∎

######  Lemma 5.3.

Let $u\in L^{\infty}(\mathbb{R}^{n})$ solve (1.2) and $\dot{u}$ solve (1.5)
with $a(t)\in A_{2}$ and $\dot{u}(x,0)$ vanishing at infinity. Then we have
$\lim\limits_{|x|\to\infty}a(t)\dot{u}_{t}(x,t)^{2}=0$.

Proof: Let $\Omega=B_{R}(x_{0})\times[0,R]\subset\mathbb{R}^{n+1}_{+}$. By
expanding and rearranging the first condition of the extension problem (1.5),
$\dot{u}$ solves

$\displaystyle(a(t)\dot{u}_{t})_{t}=-a(t)\Delta_{x}\dot{u}\textrm{ in
}\Omega,$ $\displaystyle a(t)\dot{u}_{t}=u\textrm{ on }\Omega\cap\\{t=0\\}.$
(5.7)

This holds for any fixed $x_{0}$ and $R$. Let $(x,t)\in\Omega$. By the
Fundamental Theorem of Calculus,

$\displaystyle
a(t)\dot{u}_{t}(x,t)-u(x)=-\int_{0}^{t}a(s)\Delta_{x}\dot{u}(x,s)ds.$ (5.8)

Since $a(t)$ is locally integrable, we obtain the estimate

$\displaystyle\|a(t)\dot{u}_{t}(x,t)\|_{L^{\infty}(\Omega)}\leq\|u\|_{L^{\infty}(\Omega)}+\left|\int_{0}^{R}a(t)dt\right|\|\Delta_{x}\dot{u}(x,t)\|_{L^{\infty}(\Omega)}$
$\displaystyle\leq\|u\|_{L^{\infty}(\Omega)}+C\|\Delta_{x}\dot{u}(x,t)\|_{L^{\infty}(\Omega)},$
(5.9)

where $\|u\|_{L^{\infty}(\Omega)}\to 0$ as $x_{0}\to\infty$ since $u(x)$ is
vanishing. The result then follows after applying interpolation inequalities
[11, (6.83), (6.85)] with Lemma 5.4. ∎

######  Lemma 5.4.

If $u\in L^{\infty}(\mathbb{R}^{n})$ solves $\psi(-\Delta)u=u$ on
$\mathbb{R}^{n}$ and $a(t)$, from the extension problem (1.5), is an $A_{2}$
weight, then $\dot{u}(x,\cdot)\in C^{\infty}(\mathbb{R}^{n})$ and hence $u\in
C^{\infty}(\mathbb{R}^{n})$.

Proof: Let $Q_{R}=B_{R}(0)\times(0,R)\subset\mathbb{R}^{n+1}_{+}$ and
$\partial^{\prime}Q_{R}=B_{R}(0)\times\\{0\\}$. The result follows after
adapting [15, Thm 1.2] for $a(t)\in A_{2}$ from the case $a(t)=t^{1-2s}$. That
is, we want to show that if $\dot{u}$ solves (5.7) with $\Omega=Q_{1}$, then

$\displaystyle\sup_{Q_{1/2}}\dot{u}\leq C(\inf_{Q_{1/2}}\dot{u}).$ (5.10)

This is because (5.10) implies that $\|\dot{u}\|_{C^{\beta}(Q_{1})}\leq
C\|\dot{u}\|_{L^{\infty}(Q_{1})}\leq K<\infty$ for some $\beta>0$ [8, Thm
2.3.15]. Below, in line (5.11), we show that when $\beta<1$,
$\dot{u}(x,\cdot)\in C^{\beta}(B_{1})$ implies $\dot{u}(x,\cdot)\in
C^{2\beta}(B_{1})$ with a uniform bound independent of $t$. Hence, we know
that $\dot{u}(x,\cdot)\in C^{1}(B_{1})$ and we may then invoke (5.10) on
$\nabla_{x}\dot{u}$ and repeat the doubling argument to show that
$\dot{u}(x,\cdot)\in C^{2}(B_{1})$. Iterating this process infinitely proves
the result once we rescale for any $R>1$.

Let $\delta>0$ and $v\in\mathbb{R}^{n}$ be a unit vector. We define
$\dot{u}_{\delta,v}(x,\cdot):=\frac{\dot{u}(x+\delta
v,\cdot)-\dot{u}(x,\cdot)}{|\delta|^{\beta}}$. Note that by (5.10),
$\|\dot{u}_{\delta,v}(x,\cdot)\|_{C^{\beta}(B_{1})}\leq K$ for any choice of
$\delta>0$ or $\|v\|=1$. Then, we may compute

$\displaystyle\left|\frac{\dot{u}(x+\delta v,\cdot)+\dot{u}(x-\delta
v,\cdot)-2\dot{u}(x,\cdot)}{\delta^{2\beta}}\right|=\frac{1}{\delta^{\beta}}\left|\frac{\dot{u}(x+\delta
v,\cdot)-\dot{u}(x,\cdot)}{\delta^{\beta}}-\frac{\dot{u}(x,\cdot)-\dot{u}(x-\delta
v,\cdot)}{\delta^{\beta}}\right|$
$\displaystyle=\frac{\left|\dot{u}_{\delta,v}(x,\cdot)-\dot{u}_{\delta,v}(x-\delta
v,\cdot)\right|}{\delta^{\beta}}\leq K.$ (5.11)

By [14], this shows that $\dot{u}(x,\cdot)\in C^{2\beta}(B_{1})$. We can then
double regularity until $2\beta>1$ to show $\dot{u}(x,\cdot)\in C^{1}(B_{1})$.

To prove (5.10), we only need to modify the proof of [15, Lemma 2.3] to hold
for $a(t)\in A_{2}$ since all other steps follow independently. That is, we
need to show that for $f(x,t)\in C^{1}_{c}(Q_{R}\cup\partial^{\prime}Q_{R})$:

$\displaystyle\int_{\partial^{\prime}Q_{R}}|f|^{2}\leq\epsilon\int_{Q_{R}}|\nabla
f|^{2}a(t)+\frac{C(R)}{\epsilon^{\delta}}\int_{Q_{R}}|f|^{2}a(t).$ (5.12)

First, by calculus and Young’s inequality, we may show that

$\displaystyle\int_{\partial^{\prime}Q_{R}}|f|^{p}$
$\displaystyle=-\int_{Q_{R}}\partial_{t}|f|^{p}=-\int_{Q_{R}}p|f|^{p-1}sgn(f)\partial_{t}f$
$\displaystyle\leq\epsilon\int_{Q_{R}}|\nabla
f|^{p}+C\epsilon^{-\frac{1}{p-1}}\int_{Q_{R}}|f|^{p}.$ (5.13)

Next, we claim that for all $0<l<l_{0}$ given some $l_{0}>1$, we have

$\displaystyle\int_{Q_{R}}|f|^{2}a(t)^{-l}\leq C(l)\int_{Q_{R}}|\nabla
f|^{2}a(t).$ (5.14)

In fact, by calculus and the Cauchy-Schwarz inequality, we compute

$\displaystyle
f^{2}(x,t)=\left(\int_{t}^{R}\partial_{t}f(x,s)ds\right)^{2}\leq\left|\int_{t}^{R}a(s)^{-1}ds\right|\left|\int_{t}^{R}a(s)|\partial_{t}f|^{2}ds\right|.$
(5.15)

Since $a(t)\in A_{2}$, then we know that $a(t)\in A_{m}$ for some $1<m<2$ [13,
Lemma 5]. By definition,

$\displaystyle\left(\frac{1}{R}\int_{0}^{R}a(t)\right)\left(\frac{1}{R}\int_{0}^{R}a(t)^{-\frac{m^{\prime}}{m}}\right)^{\frac{m}{m^{\prime}}}\leq
K,$ (5.16)

where $\frac{1}{m}+\frac{1}{m^{\prime}}=1$. Let $b_{0}=\frac{m^{\prime}}{m}$
and note that $b_{0}>1$ since $m<2\implies m^{\prime}>2$. Then, for $R>0$
fixed, we may assume that $\int_{0}^{R}a(t)\geq\delta_{0}>0$. Combining this
fact with (5.16), we get

$\displaystyle\int_{0}^{R}a^{-b_{0}}\leq\tilde{K}\implies\int_{0}^{R}a^{-b}\leq
C:=R^{\frac{b_{0}-b}{b_{0}}}\tilde{K}^{\frac{b}{b_{0}}}\textrm{ }\forall
b<b_{0},$ (5.17)

where the implication is due to Hölder’s inequality. Then, since $1<b_{0}$,
substituting back into (5.15) yields:

$\displaystyle f^{2}(x,t)\leq C\int_{t}^{R}a(s)|\nabla f|^{2}ds.$ (5.18)

Hence, multiplying by $a(t)^{-b}$ and integrating proves (5.14) since

$\displaystyle\int_{Q_{R}}a(t)^{-b}f^{2}\leq\tilde{C}\int_{0}^{R}a(t)^{-b}\int_{B_{R}}\int_{0}^{R}|\nabla
f(x,s)|^{2}a(s)dsdx\leq C\int_{Q_{R}}|\nabla f|^{2}a(s).$

Finally, using (5.13), Hölder’s and Young’s inequalities, and (5.14) applied
on $b=\frac{p}{2-p}$ and $b=1$ in the second last line, we show that

$\displaystyle\int_{\partial^{\prime}Q_{R}}|f|^{2}=\int_{\partial^{\prime}Q_{R}}(|f|^{\frac{2}{p}})^{p}\leq\epsilon\int_{Q_{R}}|\nabla
f^{2/p}|^{p}+C\epsilon^{-\frac{1}{p-1}}\int_{Q_{R}}|f|^{2}$
$\displaystyle=\epsilon(\frac{2}{p})^{p}\int_{Q_{R}}|f|^{2-p}a(t)^{-p/2}|\nabla
f|^{p}a(t)^{p/2}+C\epsilon^{-\frac{1}{p-1}}\int_{Q_{R}}|f|a(t)^{-1/2}|f|a(t)^{1/2}$
$\displaystyle\leq\epsilon(\frac{2}{p})^{p}\left(\int_{Q_{R}}|f|^{2}a(t)^{-\frac{p}{2-p}}\right)^{\frac{2-p}{2}}\left(\int_{Q_{R}}|\nabla
f|^{2}a(t)\right)^{p/2}+$ $\displaystyle
C\epsilon^{-\frac{1}{p-1}}\int_{Q_{R}}\epsilon^{1+\frac{1}{p-1}}|f|^{2}a(t)^{-1}+\epsilon^{-1-\frac{1}{p-1}}|f|^{2}a(t)$
$\displaystyle\leq\epsilon C\int_{Q_{R}}|\nabla
f|^{2}a(t)+\frac{C}{\epsilon^{1+\frac{2}{p-1}}}\int_{Q_{R}}|f|^{2}a(t).$

∎

##  6 Estimates on the Harmonic Extension

In this section, we include some new estimates for the harmonic extension
$\dot{u}(x,t)$ depending on its trace $\dot{u}(x,0)=u(x)$ when $u(x)$ solves
(1.5).

######  Lemma 6.1.

Let $\dot{u}(x,t)\in L^{\infty}(\mathbb{R}^{n+1}_{+})$ solve (1.5) such that
$u(x)=0$, $a(t)\in A_{2}$, and $\psi(-\Delta)u=u$ on $\mathbb{R}^{n}$. Then,
$\dot{u}(x,t)=0$.

Proof:  Let $\tilde{a}(t)$ be the even extension of $a(t)$ so that
$\tilde{a}(t)$ is an $A_{2}$ weight on $\mathbb{R}$. Then, let
$\hat{a}(x,t)=\tilde{a}(t)$. $\hat{a}$ would be an $A_{2}$ weight on
$\mathbb{R}^{n+1}$ by Lemma 6.2. Thus, if we take the odd extension of
$\dot{u}$ across $t=0$, then it will be harmonic in the sense that
$\textrm{div}(\hat{a}\nabla\dot{u})=0$ in $\mathbb{R}^{n+1}$.

Let $Y:=(Y_{s})_{s\geq 0}$ be the diffusion process generated by the equation.
For any open and bounded $A\subset\mathbb{R}^{n+1}$, let $T_{A}$ be the
hitting time $T_{A}=\inf\\{s\geq 0:Y_{s}\in A\\}$ and $\tau_{A}$ be the exit
time $\tau_{A}=\inf\\{s\geq 0:Y_{s}\not\in A\\}$. By the mean value property,
we can express
$\dot{u}(y)=\mathbb{E}_{y}[\dot{u}(Y_{\tau_{A}})]=\mathbb{E}[\dot{u}(Y_{\tau_{A}})|Y_{0}=y]$
for $y=(x,t)\in\mathbb{R}^{n+1}$.

Given $y_{0}\in\mathbb{R}^{n+1}_{+}$, let $x_{0}$ be the projection of $y_{0}$
onto $t=0$. It suffices to show that $\dot{u}(y)=0$ $\forall y\in\Omega$ for
any neighbourhood
$\Omega=B_{R}^{n}(x_{0})\times(0,R)\subset\mathbb{R}^{n+1}_{+}$ where
$B_{R}^{n}(x_{0})\subset\mathbb{R}^{n}$. Then, let
$\Omega_{k}:=B_{2^{k}R}^{n}(x_{0})\times(0,2^{k}R)$, $k\geq 1$. Note that
$\partial\Omega_{k}$ has two parts. The first is
$\Gamma_{1,k}:=\\{(x,t)\in\Omega_{k}:t=0\\}$. Let the rest of the boundary be
$\Gamma_{2,k}$. Note that $\dot{u}(y)=0$ on $\Gamma_{1,k}$ by assumption.
Hence, for any $k\geq 1$ and $y\in\Omega$, applying the mean value property
yields:

$\displaystyle|\dot{u}(y)|=|\mathbb{E}_{y}[\dot{u}(Y_{\tau_{\Omega_{k}}})]|\leq
P_{y}(T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}})\|\dot{u}(x,t)\|_{\infty}.$ (6.1)

Since $\|\dot{u}(x,t)\|_{\infty}\leq M<\infty$, it suffices to show that
$\lim\limits_{k\to\infty}P_{y}(T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}})\to 0$.

Since $\hat{a}(t)\in A_{2}$ and $\textrm{div}(\hat{a}\nabla\dot{u})=0$,
$\dot{u}$ satisfies the elliptic Harnack inequality:
$\operatorname*{ess\,sup}_{B(y,R)}\dot{u}\leq
C_{H}\operatorname*{ess\,inf}_{B(y,R)}\dot{u}$ for any ball
$B(y,R)\subset\mathbb{R}^{n+1}$ [8, Lemma 2.3.5]. Thus, we may apply [3, Lemma
3.7], which shows that for any $\tilde{R}>0$,

$\displaystyle P_{y}(T_{B(y_{1},\tilde{R}/4)}<\tau_{B(y_{0},\tilde{R})})\geq
p_{0}>0\textrm{ }\forall y\in B(y_{0},7\tilde{R}/8),$ (6.2)

for any $y_{1}\in B(y_{0},\tilde{R}/2)\subset\mathbb{R}^{n+1}$ and such that
$p_{0}$ only depends on the constant $C_{H}$ from the Harnack inequality.
Then, for any $k$, we pick $y_{1}^{k}$ as the projection of $y_{0}$ onto
$t=-\frac{2^{k}R}{4}$ and let $\tilde{R}=2^{k}R$. Clearly, $y_{1}^{k}\in
B(y_{0},\frac{2^{k}R}{2})$, which yields

$\displaystyle
P_{y}(T_{B(y_{1}^{k},\frac{2^{k}R}{4})}<\tau_{B(y_{0},2^{k}R)})\geq
p_{0}>0\textrm{ }\forall y\in B\left(y_{0},\frac{7*2^{k}R}{8}\right).$ (6.3)

Given that the diffusion starts at $y\in B(y_{0},\frac{7\cdot 2^{k}R}{8})$,
note that $T_{B(y_{1}^{k},\frac{2^{k}R}{4})}<\tau_{B(y_{0},2^{k}R)}$ implies
that $T_{\Gamma_{1,k}}<T_{\Gamma_{2,k}}$. Thus, for any $k$,

$\displaystyle P_{y}(T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}})\leq 1-p_{0}\textrm{
}\forall y\in B\left(y_{0},\frac{7\cdot 2^{k}R}{8}\right).$ (6.4)

However, since the diffusion is continuous, then
$T_{\Gamma_{2,k+1}}<T_{\Gamma_{1,k+1}}$ implies that
$T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}}$. Since all points on $\Gamma_{2,k}$ are in
$B(y_{0},\frac{7*2^{k+1}R}{8})$, it follows that:

$\displaystyle
P(T_{\Gamma_{2,k+1}}<T_{\Gamma_{1,k+1}}|T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}})\leq
1-p_{0}<1.$ (6.5)

Then, by the strong Markov property, we obtain

$\displaystyle
P_{y}(T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}})=P(T_{\Gamma_{2,k}}<T_{\Gamma_{1,k}}|T_{\Gamma_{2,k-1}}<T_{\Gamma_{1,k-1}})P(T_{\Gamma_{2,k-1}}<T_{\Gamma_{1,k-1}}|T_{\Gamma_{2,k-2}}<T_{\Gamma_{1,k-2}})$
$\displaystyle...P_{y}(T_{\Gamma_{2,1}}<T_{\Gamma_{1,1}})\leq(1-p_{0})^{k}.$
(6.6)

The result therefore follows after taking $k\to\infty$.

∎

######  Lemma 6.2.

If $a(t)$ is a non-negative $A_{2}$ weight on $(0,\infty)$, then it can be
extended to a non-negative $A_{2}$ weight, $\hat{a}(x,t)$ on
$\mathbb{R}^{n+1}$.

Proof: Let $q\geq p\geq 0$. By definition $a(t)\in A_{2}$, $a(t)$ satisfies

$\displaystyle\left(\fint_{p}^{q}a(t)dt\right)\left(\fint_{p}^{q}a(t)^{-1}dt\right)\leq
C\iff\left(\int_{p}^{q}a(t)dt\right)\left(\int_{p}^{q}a(t)^{-1}dt\right)\leq
C|q-p|^{2}.$ (6.7)

Let $\tilde{a}(t)$ be the even extension of $a(t)$. That is,
$\tilde{a}(-t)=\tilde{a}(t)=a(t)$ for $t\geq 0$. Then $\tilde{a}(t)$ is an
$A_{2}$ weight on $\mathbb{R}$. To see why, note that the cases $q,p\geq 0$
and $q,p\leq 0$ immediately reduce to (6.7), so we only need to check the case
when $\hat{p}:=-p>0$ and $q>0$. Let
$M:=\max\\{|p|,|q|\\}=\max\\{\hat{p},q\\}$. We verify that

$\displaystyle\left(\fint_{p}^{q}\tilde{a}(t)dt\right)\left(\fint_{p}^{q}\tilde{a}(t)^{-1}dt\right)$
$\displaystyle=\frac{1}{|q-p|}\left(\int_{p}^{0}\tilde{a}(t)dt+\int_{0}^{q}\tilde{a}(t)dt\right)\frac{1}{|q-p|}\left(\int_{p}^{0}\tilde{a}(t)^{-1}dt+\int_{0}^{q}\tilde{a}(t)^{-1}dt\right)$
$\displaystyle\leq\frac{1}{|q-p|^{2}}\left(\int_{0}^{\hat{p}}a(t)dt+\int_{0}^{q}a(t)dt\right)\left(\int_{0}^{\hat{p}}a(t)^{-1}dt+\int_{0}^{q}a(t)^{-1}dt\right)$
$\displaystyle\leq\frac{1}{|q-p|^{2}}\left(C|p|^{2}+2C|M|^{2}+C|q|^{2}\right)\leq\frac{4C|M|^{2}}{|M|^{2}}=4C.$

In the last line, we have expanded the product and used the non-negativity of
$a(t),\frac{1}{a(t)}$ along with (6.7). Note also that $|q-p|=q+\hat{p}\geq
M$.

Next, we define $\hat{a}(x,t)=\tilde{a}(t)$. Let
$B^{n}(x_{0},R)\subset\mathbb{R}^{n}$ be a ball of radius $R$ centered around
$x_{0}$, and let $t_{0}\in\mathbb{R}$. Then, for any set
$Q_{R}(x_{0},t_{0})=B^{n}(x_{0},R)\times(t_{0}+\frac{R}{2},t_{0}-\frac{R}{2})$,
the $A_{2}$ condition for $\hat{a}(x,t)$ is satisfied on $\mathbb{R}^{n+1}$
since it immediately reduces to the $A_{2}$ condition on $\tilde{a}(t)$. Now,
let $y=(x_{0},t_{0})\in\mathbb{R}^{n+1}$. For any ball
$B(y,R)\subset\mathbb{R}^{n+1}$, note that we may choose $\tilde{R}$ such that
$B(y,R)\subset Q_{\tilde{R}}(x_{0},t_{0})$, and
$|Q_{\tilde{R}}(x_{0},t_{0})|\leq 2|B(y,R)|$. Hence, we have

$\displaystyle\int_{B(y,R)}\hat{a}(y)dy\leq\int_{Q_{\tilde{R}}(x_{0},t_{0})}\hat{a}(x,t)dxdt$
$\displaystyle\leq C|Q_{\tilde{R}}(x_{0},t_{0})|^{2}\leq 4C|B(y,R)|^{2}.$

∎

######  Lemma 6.3.

If $\dot{u}(x,t)$ solves (1.5), such that its trace $\dot{u}(x,0)=u(x)$
vanishes at $\infty$, then
$\lim\limits_{h_{n}\to\infty}\|\dot{u}(x,t)\|_{Q_{R}(x+h_{n})}=0$ where
$Q_{R}(x)=B_{R}(x)\times(0,R)\subset\mathbb{R}^{n+1}_{+}$.

Proof: Fix $x\in\mathbb{R}^{n}$ and let $\\{h_{n}\\}$ be a sequence in
$\mathbb{R}^{n}$ such that $\|h_{n}\|\to\infty$. Define
$\dot{u}_{n}(x,t)=\dot{u}(x+h_{n},t)$. For each $n$, $\dot{u}_{n}(x,t)$ solves
(1.5) with trace $u(x+h_{n})$. By classical elliptic regularity theory, we
know that the limit of solutions
$\dot{u}^{*}=\lim\limits_{n\to\infty}\dot{u}_{n}$ is also a solution to (1.5)
with trace $\lim_{n\to\infty}u(x+h_{n})=0$. Therefore $\dot{u}^{*}$ satisfies
the conditions for Lemma 6.1, which concludes the proof since
$\lim\limits_{n\to\infty}\dot{u}^{*}_{n}=0$.

∎

##  7 Conclusion

In this paper, we have shown that the classical Helmholtz solutions $-\Delta
u=u$ on $\mathbb{R}^{n}$ also solve many other different Helmholtz solutions.
First, we classified these solutions as the bounded and vanishing fractional
Helmholtz solutions for the case $0<s<1$. In the special case $n=1$, we have
proved that the Helmholtz solutions given by $A\sin(x)+B\cos(x)$ are the
bounded fractional solutions using Fourier analysis. This classification
extends to the case $1<s\leq 2$ and the polyharmonic case $s\in\mathbb{N}$,
provided that $u\in C^{\infty}$. The uniqueness proofs use two techniques: the
extension and energy monotonicity of [10], and a decomposition principle. An
open question is that this classification result can be extended for any
$s\in(2,+\infty)\backslash\mathbb{N}$, however the decomposition technique
shown in sections 3 and 4 will not suffice for reducing the case $s>2$ to the
case $0<s\leq 1$. It is also an open question whether or not there exist non-
vanishing fractional Helmholtz solutions.

We have also classified Helmholtz solutions for the more general equation
$\psi(-\Delta)u=\psi(1)u$ when $\psi$ is completely Bernstein. The classical
Helmholtz solutions solve this equation and we have proven that they are
unique in the class of bounded and vanishing functions for $n\geq 2$ when we
impose the regularity conditions $a(t)\in A_{2}$ and $a(t)\sim t^{\alpha}$ for
some $|\alpha|<1$ and $t\gg 1$. These conditions were imposed to obtain
appropriate estimates used in Lemma 5.2 and Lemma 5.4 in order to complete our
uniqueness argument. Further research should be done on whether these
regularity restrictions can be relaxed while preserving the uniqueness of
classical Helmholtz solutions for complete Bernstein Helmholtz solutions.

##  8 Appendix

We briefly classify the solutions to the classical Helmholtz equation and note
their relationship to Bessel functions for dimensions $n\geq 2$. See also [2]
for a complete representation.

Let us first recall that the bounded Bessel function of order $\nu$, denoted
by $J_{\nu}(r)$, satisfies the differential equation
$u_{rr}+\frac{1}{r}u_{r}+(1-\frac{\nu^{2}}{r^{2}})u=0$.

Let $n\geq 2$. Using separation of variables and
$\Delta_{\mathbb{R}^{n}}u=u_{rr}+\frac{n-1}{r}u_{r}+\frac{\Delta_{\mathbb{S}^{n-1}}u}{r^{2}}$,
it is evident that Helmholtz solutions are of the form:

$u(r,\theta)=\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}c_{l,m}r^{\frac{2-n}{2}}J_{n/2+l-1}(r)\phi_{l,m}(\theta)$

where $\phi_{l,m}$ is a Laplacian spherical harmonic of order $l$ and
multiplicity $m$:

$\Delta_{\mathbb{S}^{n-1}}\phi_{l,m}+\mu_{l,m}\phi_{l}=0$, s.t.
$\mu_{l,m}=l(l+n-2)$.

## Acknowledgement

This research is partially supported by NSERC of Canada.

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