# A bespoke multigrid approach for magnetohydrodynamics models of magnetized
plasmas in PETSc

Mark F. Adams and Matthew G. Knepley

###### Abstract

Fully realizing the potential of multigrid solvers often requires custom
algorithms for a given application model, discretizations and even regimes of
interest, despite considerable effort from the applied math community to
develop fully algebraic multigrid (AMG) methods for almost 40 years. Classic
geometric multigrid (GMG) has been effectively applied to challenging, non-
elliptic problems in engineering and scientifically relevant codes, but
application specific algorithms are generally required that do not lend
themselves to deployment in numerical libraries. However, tools in libraries
that support discretizations, distributed mesh management and high performance
computing (HPC) can be used to develop such solvers.

This report develops a magnetohydrodynamics (MHD) code in PETSc (Portable
Extensible Toolkit for Scientific computing) with a fully integrated GMG
solver that is designed to demonstrate the potential of our approach to
providing fast and robust solvers for production applications. These
applications must, however, be able to provide, in addition to the Jacobian
matrix and residual of a pure AMG solver, a hierarchy of meshes and knowledge
of the application’s equations and discretization. An example of a 2D, two
field reduced resistive MHD model, using existing tools in PETSc that is
verified with a “tilt” instability problem that is well documented in the
literature is presented and is an example in the PETSc repository
(src/ts/tutorials/ex48.c). Preliminary CPU-only performance data demonstrates
that the solver can be robust and scalable for the model problem that is
pushed into a regime with highly localized current sheets, which generates
strong, localized non-linearity, that is a challenge for iterative solvers.

PETSc is a widely used numerical library that began in the mid 1990’s as a
sparse linear algebra solver library [BAA+22]. It has since grown to provide a
wide array of numerical tools, such as discretizations and distributed mesh
management, advanced time integrators, nonlinear solvers, support for
downloading and building over 50 third party libraries (solvers, mesh
partitioners, adaptive mesh refinement managers, etc.), as well as
optimization methods and eigensolvers in the tightly integrated Tao and SLEPc
libraries, respectively. PETSc is a fully distributed memory library, built on
the message passing interface (MPI), with support for modern accelerated
architectures (“GPUs”) for the linear solvers, both built-in solvers and third
party solvers, and for GPU sparse matrix assembly. Support for GPU finite
element assembly is currently under development with the libCEED library.
PETSc provides performance portability with two approaches to mitigate risks:
Kokkos and Kokkos Kernels for a portable language and low level numerical
primitives, respectively, and built-in interfaces to vendor low level numeral
primitives. Note, this approach actually yields three back-ends because Kokkos
Kernels can be configured to use vendor “third party libraries” or built-in
methods written in Kokkos.

This paper is intended to demonstrate the potential of a bespoke GMG solver
for existing production codes that require solvers for models and
discretizations that are not supported well with AMG solvers (eg, hyperbolic
problems). Our approach is to use a modular design in a simple code that can
be expanded to support the discretization of a particular application. The
PETSc multigrid framework, and overall support for (matrix) operators,
adaptive mesh refinement and discretizations is designed to input users meshes
and a user’s hierarchy of meshes, and our approach requires that user’s
provide this grid hierarchy. Finally, an application would provide the
algebraic problem (matrix and right hand side, with a possible matrix-free
operator for residual calculations on the fine grid).

This report proceeds by discussing the MHD model and PETSc implementation in
§1, introducing the model problem, the 2D “tilt” instability, showing
verification and robustness with thin current sheets, in §2, presenting small
scale CPU performance results in §3, and concluding in §4.

##  1 The MHD model and PETSc implementation

This project focuses on MHD applications in fusion energy sciences, such as
M3D-C1 and NIMROD. We implement a two field, 2D, incompressible, resistive MHD
code with Poisson brackets and stream functions as per Jardin and Strauss
[Jar04, SL98]. This approach is designed to be extended to full, two-fluid MHD
while reusing all of the terms from the reduced model. To enforce
incompressibility, it is common to introduce stream functions,

$\displaystyle\mathbf{v}$ $\displaystyle=\left(\frac{\partial\phi}{\partial
y},-\frac{\partial\phi}{\partial x}\right),$ (1) $\displaystyle\mathbf{B}$
$\displaystyle=\left(\frac{\partial\psi}{\partial
y},-\frac{\partial\psi}{\partial x}\right),$

where $\mathbf{v}$ is the fluid velocity and $\mathbf{B}$ is the magnetic
field. The two field model uses vorticity $\Omega_{z}$ and magnetic flux
$\psi$, and auxiliary variables potential $\phi$ and (negative) current
density $j_{z}$ according to:

$\displaystyle\partial_{t}\Omega_{z}+\left\\{\Omega_{z},\phi\right\\}-\left\\{j_{z},\psi\right\\}-\mu\nabla^{2}_{\perp}\Omega_{z}$
$\displaystyle=0$ (2)
$\displaystyle\partial_{t}\psi+\left\\{\psi,\phi\right\\}-\eta j_{z}$
$\displaystyle=0$ $\displaystyle\Omega_{z}-\nabla^{2}_{\perp}\phi$
$\displaystyle=0$ $\displaystyle j_{z}-\nabla^{2}_{\perp}\psi$
$\displaystyle=0,$

with Poisson bracket $\left\\{f,g\right\\}=\frac{\partial f}{\partial
x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial
g}{\partial x}$.

The linearization of a bracket $\left\\{f,g\right\\}$, given that it is linear
in both $f$ and $g$, can be expressed with two Frechet derivatives
$D_{f}\left\\{f,g\right\\}+D_{g}\left\\{f,g\right\\}$ where
$D_{f}\left\\{f,g\right\\}=\lim_{\epsilon\to
0}\left(\left\\{f+\epsilon{\hat{f}},g\right\\}-\left\\{f,g\right\\}\right)/\epsilon=\left\\{\hat{f},g\right\\}$,
and
$D\left\\{f,g\right\\}=\left\\{{\hat{f}},g\right\\}+\left\\{f,{\hat{g}}\right\\}$.
Each of these two terms are put into the column corresponding to the
$\hat{\cdot}$ term. The linearized system can be written in matrix form as:

$L\cdot dU=\begin{bmatrix}\frac{1}{\partial
t}-\mu\nabla_{\perp}^{2}+\left\\{\cdot,{\phi}\right\\}&-\left\\{j_{z},\cdot\right\\}&\left\\{{\Omega_{z}},\cdot\right\\}&-\left\\{\cdot,{\psi}\right\\}\\\
.&\frac{1}{\partial
t}+\left\\{\cdot,{\phi}\right\\}&\left\\{{\psi},\cdot\right\\}&-\eta M\\\
M&.&-\nabla_{\perp}^{2}&.\\\
.&-\nabla_{\perp}^{2}&.&M\par\end{bmatrix}\begin{bmatrix}{d\Omega_{z}}\\\
{d\psi}\\\ {d\phi}\\\ {dj_{z}}\end{bmatrix}=\begin{bmatrix}.\\\ .\\\ .\\\ .\\\
.\end{bmatrix}.$ (3)

The PETSc finite element assembly infrastructure processes the finite element
(FE) assembly, calling a user point functions that resembles the strong form.
PETSc supports weak forms with four (hardwired) types of terms with the
standard FE inner product $\left(\cdot,\cdot\right)$ and test functions $v$:

  * •

“g0” mass terms, $M$, of the form $\left(u,v\right)$,

  * •

“g1” first derivative terms $\left(\nabla u,v\right)$,

  * •

“g2” is not used herein but it is used for first derivative terms
$\left(u,\nabla v\right)$,

  * •

“g3” for the weak form of a $-\nabla^{2}u$ term , $\left(\nabla u,\nabla
v\right)$, after integration by parts, which flips the sign.

For scalar problems, the point functions for g0 terms return a scalar, the
point functions for g1 and g2 return a vector and the g3 terms return a second
order tensor (matrix). In our case the g3 functions return scaled identity
tensors (2x2 matrices) and the mass, g0, functions simply return the scale,
1.0. The g1 terms generates a vector that scales the test function that is
dotted with the basis function gradient in the FE integral internal to PETSc.

For instance, $\left\\{f,\hat{g}\right\\}=\frac{\partial f}{\partial
x}\frac{\partial\hat{g}}{\partial y}-\frac{\partial f}{\partial
y}\frac{\partial\hat{g}}{\partial x}=\left(\frac{\partial\hat{g}}{\partial
x},\frac{\partial\hat{g}}{\partial y}\right)\left(-\frac{\partial f}{\partial
y},\frac{\partial f}{\partial
x}\right)^{T}=\nabla\hat{g}\cdot\left(-\frac{\partial f}{\partial
y},\frac{\partial f}{\partial x}\right)^{T}$. Our internal naming convention
uses ‘left’ and ‘right’ to distinguish between two types of linearized bracket
terms and this is a ‘left’ term. Thus the result of the left g1 functions is
$\left(-\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\right)$.
Similarly, the result of the right g1 functions for
$\left\\{\hat{f},g\right\\}$ is $\left(\frac{\partial g}{\partial
y},-\frac{\partial g}{\partial x}\right)$ (this can be seen with the bracket
property that switching the order of the arguments simply flips the sign). For
instance, the point function for the $-\left\\{{j_{z}},\cdot\right\\}$ term
$L(1,2)$ is:

    /* bracket -{ jz , .}, in column of linearized variable .; "n" for negative */
    void g1_njz_left(PetscInt dim, PetscInt Nf, PetscInt NfAux, .... , PetscScalar g1[])
    {
        PetscInt          i,j;
        const PetscScalar *pJzDer   = &u_x[uOff_x[JZ]]; // gradient of jz_z
        for (i = 0; i < dim; ++i)
          for (j = 0; j < dim; ++j)
            g1[j] += -pJzDer[i]*s_K[i][j]; // This should be unrolled, 1/2 no-ops
    }

where s_K is a DxD skew symmetric matrix that encodes the Poisson bracket
(this could be hardwired in future). The result in g1 here is
$\left(\frac{\partial j_{z}}{\partial y},-\frac{\partial j_{z}}{\partial
x}\right)^{T}$. s_K is defined as:

$\begin{bmatrix}0&1\\\ -1&0\\\ \end{bmatrix}.$ (4)

###  1.1 Numerical methods and reproducibility

We discretize (3) with triangular finite elements and use quadratic elements
in this work. Square cells of the cartesian mesh are cut into two triangles.
An initial grid with one cell per processor is generated and PETSc’s command
line option -dm_refine_hierarchy L, resulting in a logically and geometrically
square subdomain on each processor of size $2^{L}\times 2^{L}$. Standard
geometric multigrid preconditioned FGMRES is used as the solver, four
iterations of GMRES with diagonal preconditioning is used as the smoother, and
a direct solver is used for the coarse grid solver. This linear solver is used
in a standard Newton nonlinear solver iteration. An adaptive Runge-Kutta (RK2)
time integrator is use with a maximum time step of $0.025$ (PETSc command line
option -ts_type arkimex -ts_arkimex_type 1bee -ts_adapt_dt_max 0.025). A
tolerance of $10^{-5}$ (-ts_rtol 1e-5) resulted in a minimum time step of
about $0.0005$, which is in line with the time steps reported by Jardin
[Jar04]. Run scripts, data output, and reproducibility information are stored
in the git repository at gitlab.com/markadams4/mg-m3dc1.git.

##  2 Verification with the tilt instability

The tilt instability is an unstable MHD equilibrium that is well documented in
the literature [Jar04, SL98, KPX14]. An initial condition is applied to $\psi$
and is maintained in the boundary condition of the domain $[-2,2]^{2}$
according to:

This is an MHD equilibrium, but it is not stable. One can perturb the initial
condition to aid in developing the instability, but we simply allow the
instability to develop from round-off errors and perhaps the asymmetry of the
mesh (see Figure 2(b)). Strauss provides an image of the initial condition
[SL98], and Jardin provides detailed plots of the evolved fields [Jar04].
Jardin conducts extensive verification studies, and our data is in qualitative
agreement. The simulation time of any point (eg, rotation) in the instability
depends on the initial conditions and numerical errors. To compare with Jardin
we normalize our time by synchronizing with a point close to Jardin’s $t=5$
(Figure 8 (b) and (d) [Jar04]), which fixes our $\hat{t}=10.3$ (in the
repository gitlab.com/markadams4/mg-m3dc1.git data file
src/data/out_064_data_3_64) to Jardin’s $t=0$. Note, these plots where
generated at intervals of $0.1$ and this synchronization is derived from
observation of the plots, and not quantitative data, thus this mapping between
our data and Jardin’s, $t=\hat{t}-10.3$, is approximate.

Figures 1 – 3 show plots of the evolution of the test problem with with a
$512\times 512$ cell grid, with each cell simply divided into two triangles.
With $P2$ elements this is produces a $1025\times 1025$ vertex grid for the
solvers with quadratic shape functions.

(a) $\psi$

(b) $j_{z}$

Figure 1: Tilt instability $t=0$ state

Figure 2 plots at $t=5$ are in good qualitative agreement with Jardin (Figure
8 (d)) and Figure 2(b) shows a detail of the current sheet with some mesh and
visualization effects. Note, Visit is used for these plots and it uses linear
interpolation on a cell and this introduces some of these effects, namely, the
dips in the sheet (upper right in Figure 2(a)). Additionally, the direction of
the split of cells to generate a triangular mesh is not consistent between
processors as is evident in Figures 1(b) and 2(b).

(a) $j_{z}$

(b) $j_{z}$ current sheet detail

Figure 2: Tilt instability at time $t=5$ showing some grid effects and
visualization artifacts

(a) $j_{z}$

(b) $j_{z}$ current sheet detail

Figure 3: Tilt instability at time $t=5.8$ with thin current sheets

Figure 3 plots at the state at $t=5.8$ at a point where the maximum current is
observed, although this data is under-resolved there are multiple local maxima
are observed. Jadin compares growth rates qualitatively with previously
published results that are consistent with our observed growth rates.

##  3 Performance results

Performance is evaluated on Crusher CPU nodes, each with one 64-core AMD EPYC
7A53 “Optimized 3rd Gen EPYC” CPU and two hardware threads per core. One MPI
process is placed on each core, 32 cores are used per node, and the problem is
scaled by a factor of two refinement in each dimension for each case. Two
$64\times 64$ cell subdomains are placed on each core and the node count is
increased by a factor of four for each refinement step, maintaining this same
size local grid for a weak scaling study with about $32K$ vertices per core or
about $1M$ vertices per node, which would be a reasonable if we were using the
accelerators. Note, the finite element device assembly is a work in progress
and not available for this study.

Flexible GMRES is used for the solver, preconditioned by a multigrid F-cycle,
with 4 Jacobi preconditioned GMRES iterations as the smoother, and PETSc’s
built-in algebraic multigrid (AMG) solver as the coarse grid solver. A Newton
nonlinear iteration is used with a relative tolerance of $10^{-9}$. Note,
PETSc’s AMG solver takes care to reduce the number of active processors on its
coarse grids and these solves have a $2\times 2$ cell grid on the coarsest
geometric grid, which is the fine grid for the AMG solver, thus, the AMG
solver reduces the active processors substantially on the first coarse grid in
AMG. The AMG coarse grid is solve with a serial direct solver on one core.

Table 1 presents the time in the linear solver for the first time step of the
model problem (3 nonlinear solves and a total of 9 linear solves).

Table 1: One time step of the three stage adaptive RK2 time integrator with constant subdomain size (weak scaling) Number of nodes: | 2 | 8 | 32
---|---|---|---
Linear solve times (sec) | 4.0 | 5.9 | 8.0
Total number solver iterations | 24 | 31 | 29
Time per iteration (sec) | 0.17 | 0.19 | 0.28
Number of linear solves | 9 | 9 | 7

This data shows that the implementation scaling is fairly good considering
that the multigrid F-cycle applies the coarse grids more often than the
traditional V-cycle. The multigrid F-cycle has the same $O(N)$ work complex,
asymptotically, as V-cycle multigrid but the PRAM complexity, or computational
depth, increases to $O((log^{N})^{2})$ from $O(log^{N})$ and the constants in
the work complexity are a little higher. These are nonlinear problems and the
fluctuation in the iteration count can be caused by new dynamics being
resolved, but the source of the fluctuations in iteration counts in this data
is not clear. This data with the full PETSc output is in the repository
(gitlab.com/markadams4/mg-m3dc1.git src/scale/out_0[02|08|32]_scale_6_32)

##  4 Conclusion and future work

We have presented preliminary results of developing a geometric multigrid
solver for the MHD equations is stream function form, with an implementation
of the 2-field reduced MHD equations in Jardin and Strauss [Jar04, SL98],
which demonstrate the feasibility of using PETSc for developing a tightly
integrated application and multigrid solver for ideal performance. We have
tested the tilt instability problem, and run the analysis out to where thin
current sheets form that go below the resolution of the mesh used in this
study. The dynamics of the system match those reported in Jardin and the
expected physics that we conclude the code is correct. Preliminary performance
results provide only limited data, however the scaling of multigrid methods,
while not perfect, is well understood and as good as any solver for a fully
couple PDE solve of an elliptic/hyperbolic operator. These problems are very
challenging and we have observed some fragility, like failure with large time
steps, which is disappointing but our time steps are in line with Jardin’s
parameters [Jar04].

### Future work:

  * •

add block or Venka like smoothers in PETSc with new batch solvers [Van86,
FKWM19];

  * •

add this block solver as a new low memory plane solver;

  * •

add an anisotropic third dimension and implement semi-coarsening and other
multigrid techniques to ameliorate problems from anisotropy to this MHD code;

  * •

replace the $P2$ finite elements with a mixed FE method with $C^{0}$
$\mathbf{B}$ and $\mathbf{v}$ fields as a better approximation to cubic
$C^{1}$ discretization;

  * •

investigate the dual discretizations for more stable smoothers;

  * •

add device (Kokkos and libCEED) FE device matrix assembly for coarse grid
operators;

  * •

develop a restriction operator for $C^{1}$ FE space to our mixed FE space.

## Acknowledgments

This material is based upon work supported by the U.S. Department of Energy,
Office of Science, Office of Advanced Scientific Computing Research and Office
of Fusion Energy Sciences, Scientific Discovery through Advanced Computing
(SciDAC) program.

## References

  * [BAA+22] Satish Balay, Shrirang Abhyankar, Mark F. Adams, Steven Benson, Jed Brown, Peter Brune, Kris Buschelman, Emil Constantinescu, Lisandro Dalcin, Alp Dener, Victor Eijkhout, Jacob Faibussowitsch, William D. Gropp, Vaclav Hapla, Tobin Isaac, Pierre Jolivet, Dmitry Karpeev, Dinesh Kaushik, Matthew G. Knepley, Fande Kong, Scott Kruger, Dave A. May, Lois Curfman McInnes, Richard Tran Mills, Lawrence Mitchell, Todd Munson, Jose E. Roman, Karl Rupp, Patrick Sanan, Jason Sarich, Barry F. Smith, Stefano Zampini, Hong Zhang, and Junchao Zhang.  PETSc: Portable, Extensible Toolkit for Scientific Computation.  Astrophysics Source Code Library, record ascl:2210.016, October 2022.
  * [FKWM19] P. E. Farrell, M. G. Knepley, F. W., and L. Mitchell.  PCPATCH: software for the topological construction of multigrid relaxation methods.  ACM Trans. Math. Softw., 47:25:1–25:22, 2019.
  * [Jar04] S.C. Jardin.  A triangular finite element with first-derivative continuity applied to fusion mhd applications.  Journal of Computational Physics, 200(1):133–152, 2004.
  * [KPX14] R. Keppens, O. Porth, and C. Xia.  Interacting tilt and kink instabilities in repelling current channels.  The Astrophysical Journal, 795(1):77, oct 2014.
  * [SL98] H.R. Strauss and D.W. Longcope.  An adaptive finite element method for magnetohydrodynamics.  Journal of Computational Physics, 147(2):318–336, 1998.
  * [Van86] S.P Vanka.  Block-implicit multigrid solution of navier-stokes equations in primitive variables.  Journal of Computational Physics, 65(1):138–158, 1986.


# Prescribing oscillation behavior of solutions to the heat equation on
$\mathbb{R}^{n}$ via the initial data and its average integral††thanks:
Mathematics Subject Classification: 35K05, 35K15.

Dong-Ho Tsai

###### Abstract

Motivated by a classical stabilization result for solution to the Cauchy
problem of the heat equation$\ \partial_{t}u=\bigtriangleup u\ $on
$\mathbb{R}^{n}$, we consider its oscillation behavior with radial initial
data $\varphi\left(x\right)=\varphi\left(\left|x\right|\right)\in
C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right).\ $Given four arbitrary finite numbers
$r<\alpha<\beta<s,$ one can construct a radial $\varphi\in
C^{0}\left(\mathbb{R}^{n}\right)\bigcap L^{\infty}\left(\mathbb{R}^{n}\right)$
so that $\varphi\ $together with its corresponding solution$\
u\left(x,t\right)$ satisfy the oscillation behavior:

$\displaystyle\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)$
$\displaystyle=r<\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha$
$\displaystyle<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=s.$

Similarly, given $p<\alpha<\beta<q\ $with $p+q=\alpha+\beta,$ one can find a
radial $\varphi\in C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right)$ so that its average integral
$H\left(\tau\right)\ $and $u\left(x,t\right)$ satisfy

$\displaystyle\liminf_{\tau\rightarrow\infty}H\left(\tau\right)$
$\displaystyle=p<\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha$
$\displaystyle<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=q.$

Here $H\left(\tau\right),\ \tau\in\left(0,\infty\right),\ $is given by

$H\left(\tau\right)=\frac{1}{\left|B\left(0,\tau\right)\right|}\int_{B\left(0,\tau\right)}\varphi\left(y\right)dy,\
\ \ \tau\in\left(0,\infty\right)$

and $B\left(0,\tau\right)$ is the open ball with radius $\tau>0$ centered at
the origin of $\mathbb{R}^{n}$.

##  1 Introduction.

This article is a continuation of our previous ones [TN, CT] and proves
several new interesting results. Consider the initial value problem

$\left\\{\begin{array}[c]{l}u_{t}\left(x,t\right)=\bigtriangleup
u\left(x,t\right),\ \ \ x\in\mathbb{R}^{n},\ \ \ t>0,\vspace{3mm}\\\
u\left(x,0\right)=\varphi\left(x\right),\ \ \
x\in\mathbb{R}^{n},\end{array}\right.$ (1)

where $\varphi:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is a given continuous
function. It is known that if $\varphi:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is
a continuous bounded function, then the function given by convolution integral

$u\left(x,t\right)=\frac{1}{\left(4\pi
t\right)^{n/2}}\int_{\mathbb{R}^{n}}e^{-\frac{\left|x-y\right|^{2}}{4t}}\varphi\left(y\right)dy,\
\ \ x\in\mathbb{R}^{n},\ \ \ t>0$ (2)

is a smooth solution of the heat equation on $\mathbb{R}^{n}\times(0,\infty)\
$with
$\lim_{\left(x,t\right)\rightarrow\left(x_{0},0\right)}u\left(x,t\right)=\varphi\left(x_{0}\right)\
$for any $x_{0}\in\mathbb{R}^{n}.\ $Due to the example by Tychonoff in 1935
(see the book [J], Chapter 7), it is known that the Cauchy problem for the
heat equation (1) has no unique solution on
$\mathbb{R}^{n}\times\left(0,\infty\right)\ $even if the initial data
$\varphi\left(x\right)$ is bounded (unless we impose certain growth condition
of $u\left(x,t\right)$ as $\left|x\right|\rightarrow\infty$ for $t>0$). From
now on, when we say $u\left(x,t\right)$ is ”the solution” of the heat equation
with $u\left(x,0\right)=\varphi\left(x\right),\ x\in\mathbb{R}^{n},$ we always
mean that it is the solution given by the convolution integral (2). As a
consequence, if $\left|\varphi\right|\leq M\ $on $\mathbb{R}^{n}\ $for some
constant $M>0,\ $we also have$\ \left|u\right|\leq M\ $on
$\mathbb{R}^{n}\times[0,\infty).$

In this paper, we shall always assume that $\varphi\in
C^{0}\left(\mathbb{R}^{n}\right)\bigcap L^{\infty}\left(\mathbb{R}^{n}\right)\
$and use$\ H\left(\tau\right),$ $\tau\in[0,\infty),$ to denote the average
integral of$\ \varphi\ $over the open ball
$B\left(0,\tau\right)\subset\mathbb{R}^{n}$ centered at $x=0\ $with radius
$\tau>0,\ $i.e.

$H\left(\tau\right)=\frac{1}{\left|B\left(0,\tau\right)\right|}\int_{B\left(0,\tau\right)}\varphi\left(y\right)dy=\frac{1}{\omega\left(n\right)\tau^{n}}\int_{B\left(0,\tau\right)}\varphi\left(y\right)dy,\
\ \ \tau\in\left(0,\infty\right),$ (3)

where $\omega\left(n\right)=\left|B\left(0,1\right)\right|$ is the volume of
the unit ball in $\mathbb{R}^{n}.$ By continuity, if we define
$H\left(0\right)=\varphi\left(0\right),$ then$\ H\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty).$

The role played by $H\left(\tau\right)$ in the Cauchy problem (1) is the
following beautiful stabilization result:

###### Theorem 1

(See [E, K, RE].) Assume $\varphi\in C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right)$ and $u\left(x,t\right)$ is the solution
of the heat equation (1) given by (2). Then

$\lim_{t\rightarrow\infty}u\left(0,t\right)=0\ \ \ \text{if and only if\ \ \
}\lim_{\tau\rightarrow\infty}\left(\frac{1}{\left|B\left(0,\tau\right)\right|}\int_{B\left(0,\tau\right)}\varphi\left(y\right)dy\right)=0.$
(4)

Moreover,

$\lim_{t\rightarrow\infty}\left(\sup_{x\in\mathbb{R}^{n}}\left|u\left(x,t\right)\right|\right)=0$
(5)

if and only if

$\lim_{\tau\rightarrow\infty}\left(\sup_{x\in\mathbb{R}^{n}}\left|\frac{1}{\left|B\left(x,\tau\right)\right|}\int_{B\left(x,\tau\right)}\varphi\left(y\right)dy\right|\right)=0.$
(6)

Here $\left|B\left(x,\tau\right)\right|$ is the volume of the open ball in
$\mathbb{R}^{n}$ centered at $x\in\mathbb{R}^{n}\ $with radius $\tau>0.$

###### Remark 2

Similar result holds if we replace the limit value $0$ in (4) by any real
number $c.$

###### Remark 3

As demonstrated in [CT], Theorem 1 fails if the initial data $\varphi\ $is
unbounded. In this paper, we will always assume $\varphi\in
C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right)$.

###### Remark 4

Since$\ \left|\varphi\right|\leq M\ $on $\mathbb{R}^{n}\ $for some constant
$M>0,$ the convolution solution $u\left(x,t\right)$ satisfies the following
gradient estimate

$\left|\nabla_{x}u\left(x,t\right)\right|\leq\frac{c\left(n\right)M}{\sqrt{t}},\
\ \ \forall\ \left(x,t\right)\in\mathbb{R}^{n}\times\left(0,\infty\right),$
(7)

where $c\left(n\right)$ is a constant depending only on $n.\ $Hence if we
have$\ \lim_{t\rightarrow\infty}u\left(0,t\right)=0,$ we also have
$\lim_{t\rightarrow\infty}u\left(x,t\right)=0$ for all $x\in\mathbb{R}^{n}$
and the convergence is uniform in $x\in K$ where $K\subset\mathbb{R}^{n}$ is
any compact set.

By Theorem 1 and Remark 2, the
$\lim_{\tau\rightarrow\infty}H\left(\tau\right)$ does not exist if and only if
$\lim_{t\rightarrow\infty}u\left(0,t\right)\ $does not exist. In this case, we
have the following comparison result concerning the oscillation of$\
u\left(0,t\right)$ and$\ H\left(\tau\right)$ as $t\rightarrow\infty\ $and
$\tau\rightarrow\infty\ $(see [CT]), namely

$\liminf_{\tau\rightarrow\infty}H\left(\tau\right)\leq\liminf_{t\rightarrow\infty}u\left(0,t\right)<\limsup_{t\rightarrow\infty}u\left(0,t\right)\leq\limsup_{\tau\rightarrow\infty}H\left(\tau\right).$
(8)

The key point in the proof of (8) is to use the representation formula (see
Lemma 4 in [CT]):

$u\left(0,t\right)=\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}H\left(\sqrt{4t}z\right)dz,\
\ \ \forall\ t\in\left(0,\infty\right)$ (9)

together with the classical Fatou’s Lemma. Same as in [CT], we shall use
$p\leq\alpha<\beta\leq q$ to denote the four finite limit values in (8) in
this paper.

By analogy, can we also compare the oscillation of $u\left(0,t\right),$
$t\in[0,\infty),$ with the oscillation of the initial data
$\varphi\left(x\right)?$ Since $\varphi\left(x\right)\ $is a function defined
on $\mathbb{R}^{n},$ it would be quite tricky unless we assume
$\varphi\left(x\right)=\varphi\left(\tau\right),\
\tau=\left|x\right|\in[0,\infty),$ $x\in\mathbb{R}^{n},$ is a radial function.
With this, both $u\left(0,t\right)$ and $\varphi\left(\tau\right)$ are one-
variable functions defined on $[0,\infty)\ $and we can compare their
oscillations as $t\rightarrow\infty\ $and $\tau\rightarrow\infty$
respectively. From now on, we shall assume that
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ is a radial function.

Clearly, the limit$\ \lim_{\tau\rightarrow\infty}\varphi\left(\tau\right)=0\
$will imply $\lim_{t\rightarrow\infty}u\left(0,t\right)=0,$ but not
conversely. For example, the solution to the one-dimensional heat equation
$u_{t}=u_{xx}$ with radial initial data $\varphi\left(x\right)=\cos x,$
$x\in[0,\infty),$ is given by $u\left(x,t\right)=e^{-t}\cos x,$ which
satisfies$\ \lim_{t\rightarrow\infty}u\left(0,t\right)=0.$ But we have
$\liminf_{x\rightarrow\infty}\varphi\left(x\right)=-1,\
\limsup_{x\rightarrow\infty}\varphi\left(x\right)=1.\ $

For radial $\varphi\in C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty),$ the
identities (2) (at $x=0$) and (3) will become

$u\left(0,t\right)=\frac{n\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n-1}\varphi\left(\sqrt{4t}z\right)dz,\
\ \ \forall\ t\in\left(0,\infty\right)$ (10)

and

$H\left(\tau\right)=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr,\
\ \ \tau\in\left(0,\infty\right),\ \ \ H\left(0\right)=\varphi\left(0\right).$
(11)

In particular, we note the similarity between (10) and (9).

Let $r=\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)\ $and$\
s=\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right).$ We have the
following general inequalities among the six numbers $\alpha,\ \beta,\ p,\ q,\
r,\ s:$

$r\leq p\leq\alpha\leq\beta\leq q\leq s,$ (12)

where the first and last inequalities in (12) can be derived from the
identity$\
H\left(\tau\right)=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr$
and the definition of liminf and limsup. Note that Theorem 1 says that
$\alpha=\beta\ $if and only if $p=q.$ However, $\alpha=\beta$ does not
necessarily imply $r=s\ $(as the above simple example shows), i.e. the
stabilization of $u\left(0,t\right)$ as $t\rightarrow\infty$ does not imply
the stabilization of $\varphi\left(\tau\right)$ as $\tau\rightarrow\infty.$

In this paper we are interested in prescribing the oscillation values
$\alpha<\beta\ $of $u\left(0,t\right)$ as $t\rightarrow\infty$ by choosing
suitable radial $\varphi\left(\tau\right)$ in (10) or by choosing suitable
average integral $H\left(\tau\right)$ in (9) (by (21) below, one can
determine$\ \varphi\left(\tau\right)$ as long as $H\left(\tau\right)$ is
chosen). There are many choices for such $\varphi\left(\tau\right)$ and
$H\left(\tau\right)$. The interesting results here are that one can also
prescribe the oscillation values of $\varphi\left(\tau\right)$ and
$H\left(\tau\right)\ $as $\tau\rightarrow\infty.$

More precisely, the results in this paper are:

  1. 1.

To prescribe, as $\tau\rightarrow\infty$ and $t\rightarrow\infty,$ the
oscillation behavior of $H\left(\tau\right)$ and $u\left(0,t\right)\ $for four
arbitrary finite numbers $p\leq\alpha<\beta\leq q\ $satisfying the symmetry
condition

$p+q=\alpha+\beta.$ (13)

  2. 2.

To prescribe, as $\tau\rightarrow\infty$ and $t\rightarrow\infty,\ $the
oscillation behavior of $\varphi\left(\tau\right)$ and $u\left(0,t\right)$ for
four arbitrary finite numbers $r\leq\alpha\leq\beta\leq s.$

See Theorem 5 and Theorem 8 in the next section.

##  2 Main results and their proofs.

###  2.1 Prescribing $H\left(\tau\right)$ and $u\left(0,t\right).$

The discussion of prescribing $H\left(\tau\right)$ and $u\left(0,t\right)\ $is
motivated by Theorem 1. Denote the four limit values in (8) as $p,\ \alpha,\
\beta,\ q$ respectively. In [CT], with the help of the formula (9), we have
constructed two bounded radial functions
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ so that the following are satisfied:

  1. (a).

$p=-1<\alpha=-\sqrt{A^{2}+B^{2}}<\beta=\sqrt{A^{2}+B^{2}}<q=1.\ $Here $A,\ B$
are the values of certain convergent improper integrals satisfying
$0<A^{2}+B^{2}<1$.

  2. (b).

$p=\alpha<\beta=q\ $for arbitrary two finite numbers $\alpha<\beta.$

In the above two examples, we have the identity

$p+q=\alpha+\beta,\ \ \ \text{where}\ \ \ \alpha<\beta.$ (14)

Our new result in the following is that, as long as the symmetry condition
(14) is satisfied, we can prescribe them. For four numbers
$p\leq\alpha<\beta\leq q$ to satisfy (14), we must have either
$p<\alpha<\beta<q\ $or $p=\alpha<\beta=q.\ $Since the second case has been
done, it suffices to look at the first case. We have:

###### Theorem 5

Let $p<\alpha<\beta<q\ $be four arbitrary finite numbers satisfying the
identity $p+q=\alpha+\beta.\ $One can find a continuous bounded radial
function$\ \varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},\ $so that its average integral $H\left(\tau\right)\ $and
the convolution solution (2) of the problem (1) satisfy

$\liminf_{\tau\rightarrow\infty}H\left(\tau\right)=p<\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=q.$
(15)

More precisely, we can choose the initial data
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ as

$\varphi\left(\tau\right)=\frac{q-p}{2}\sin\left(m\log\left(\tau+1\right)\right)+\frac{q-p}{2}\frac{m\tau}{n\left(\tau+1\right)}\cos\left(m\log\left(\tau+1\right)\right)+\frac{q+p}{2},$
(16)

where$\ \tau\in[0,\infty)$ and$\ m\in\left(0,\infty\right)$ is a number
satisfying the identity

$\displaystyle\left(\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log
z\right)dz\right)^{2}+\left(\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log
z\right)dz\right)^{2}$
$\displaystyle=\left(\frac{\beta-\alpha}{q-p}\right)^{2}\in\left(0,1\right).$
(17)

###### Remark 6

Without the condition $p+q=\alpha+\beta,$ we do not know how to achieve the
result (15) in general. However, we can construct some specific example with
$p+q\neq\alpha+\beta.$ See Section 3.1.

###### Remark 7

Note that if the initial data
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right)$ is radial, the
convolution solution $u\left(x,t\right)\ $of the heat equation will also be
radial in $x\in\mathbb{R}^{n}$ for each fixed time $t>0$.

###  2.2 Prescribing $\varphi\left(\tau\right)$ and $u\left(0,t\right).$

It is due to Theorem 1 that we study the oscillation behavior of$\
H\left(\tau\right)$ and its effect on $u\left(0,t\right).$ Instead, since
$\varphi\left(\tau\right)$ is the initial data, it is perhaps more natural to
study the oscillation relation between$\ \varphi\left(\tau\right)$ and
$u\left(0,t\right).$ However, there is a major difference here. Since we
always assume that $\varphi\left(\tau\right)\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty)$ is continuous and bounded, its average integral function
$H\left(\tau\right)$ will also be continuous and bounded and satisfy
$H^{\prime}\left(\tau\right)=O\left(1/\tau\right)$ as $\tau\rightarrow\infty,$
which we call it a slow oscillation if we have $p<q\ $(see (22) and Remark 14
below). Any such oscillation will cause $u\left(0,t\right)$ to have a slow
oscillation too (by Theorem 1 we have $p<q\ $if and only if $\alpha<\beta$).

On the other hand, $\varphi\left(\tau\right)\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty)$ may not have a slow oscillation (i.e. it may not satisfy
$\varphi^{\prime}\left(\tau\right)=O\left(1/\tau\right)\ $as
$\tau\rightarrow\infty\ $even if it is differentiable) and so its oscillation
(with $r<s$) may not pass to $u\left(0,t\right)$. In fact, for any
$2\pi$-periodic regular oscillation function $\varphi\left(\tau\right),$
$\tau\in[0,\infty),$ the Cauchy problem (1) with$\
u\left(x,0\right)=\varphi\left(\left|x\right|\right),$ $x\in\mathbb{R}^{n},\
$will have the convergence$\
\lim_{t\rightarrow\infty}u\left(0,t\right)=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(\tau\right)d\tau$
(see Lemma 17 below). However, due to the similarity between the two
representation formulas (9) and (10), if $\varphi\left(\tau\right)\ $has a
slow oscillation as $\tau\rightarrow\infty\ $with $r<s,$ then it has the
chance of passing to the function $u\left(0,t\right)\ $as $t\rightarrow\infty$
(see Lemma 18 below).

The major result for prescribing $\varphi\left(\tau\right)$ and
$u\left(0,t\right)\ $is the following:

###### Theorem 8

Let $r<\alpha<\beta<s\ $be four arbitrary finite numbers. One can find a
continuous bounded radial function
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ so that it and the convolution solution (2) of the
problem (1) satisfy

$\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=r<\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=s.$
(18)

More precisely, we can choose the initial data
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ as

$\varphi\left(\tau\right)=\left[C_{1}\sin\left(m\log\left(\tau+1\right)\right)+C_{2}\right]+\psi\left(\tau\right),\
\ \ \tau\in[0,\infty),$ (19)

for some suitable constants $m>0,\ C_{1}>0,\ C_{2}\ $depending on $r,\
\alpha,\ \beta,\ s\ $and some $2\pi$-periodic continuous nonzero function
$\psi\left(\tau\right)$ satisfying$\
\frac{1}{2\pi}\int_{0}^{2\pi}\psi\left(\tau\right)d\tau=0.$

###### Remark 9

Theorem 8 is actually valid for four arbitrary finite numbers$\
r\leq\alpha\leq\beta\leq s.\ $We will deal with the remaining cases in Theorem
20 below.

###### Remark 10

This is a comparison between Theorem 5 and Theorem 8. For any initial data
$\varphi\in C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right),$ no matter it is radial or not, the
function $H\left(\tau\right)$ is always a slow oscillation function as
$\tau\rightarrow\infty$. Therefore, the function space for
$H\left(\tau\right)$ is smaller than the function space for
$\varphi\left(\tau\right),\ $which may explain why we have more general result
in Theorem 8. Note that the function $\varphi\left(\tau\right)$ in (19) does
not have a slow oscillation as $\tau\rightarrow\infty$ due to the term
$\psi\left(\tau\right)$.

####  2.2.1 Proof of Theorem 5.

The proof is an interesting modification of the argument in Section 3.1.1 of
[CT], which can achieve much more general result. Some motivation should be
explained at the beginning. First, we choose
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right)$ to be a radial
function in order to simplify the computation. We think a non-radial function
can also be found as long as we can overcome the computational complexity.
Second, we choose the average integral function $H\left(\tau\right)$ first and
then go back to find its corresponding $\varphi\left(\left|x\right|\right).\
$The function $H\left(\tau\right)\ $has to be oscillatory so that we have$\
\liminf_{\tau\rightarrow\infty}H\left(\tau\right)<\limsup_{\tau\rightarrow\infty}H\left(\tau\right).$
Moreover, since we require $\varphi$ to bounded and continuous (in view of
Theorem 1), its average integral $H\left(\tau\right)$ will satisfy the
derivative estimate$\ H^{\prime}\left(\tau\right)=O\left(1/\tau\right)\ $as$\
\tau\rightarrow\infty\ $(see (22) below). Therefore, a natural choice is
roughly like the function $\sin\left(\log\tau\right),\
\tau\in\left(0,\infty\right).\ $After suitable modification of the function
$\sin\left(\log\tau\right),$ the proof can be achieved.

Given $\varphi\left(x\right)=\varphi\left(\left|x\right|\right)\in
C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty),$ its average integral function

$H\left(\tau\right)=\frac{1}{\omega\left(n\right)\tau^{n}}\int_{B\left(0,\tau\right)}\varphi\left(y\right)dy=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr,\
\ \ \tau\in\left(0,\infty\right),\ \ \ H\left(0\right)=\varphi\left(0\right),$
(20)

is a bounded continuous function on$\ [0,\infty),$ differentiable in
$\left(0,\infty\right),$ and satisfies the identity

$H^{\prime}\left(\tau\right)=-\frac{n}{\tau}H\left(\tau\right)+\frac{n}{\tau}\varphi\left(\tau\right),\
\ \ \forall\ \tau\in\left(0,\infty\right).$ (21)

Hence it has the asymptotic behavior

$H^{\prime}\left(\tau\right)=O\left(\frac{1}{\tau}\right)\ \ \ \text{as\ \ \
}\tau\rightarrow\infty.$ (22)

Moreover, the radial function $\varphi\left(\tau\right)$ can be determined
from$\ H\left(\tau\right)\ $by the identity (21).

Let $m\in\left(0,\infty\right)\ $be a fixed number. Motivated by our principle
stated above, we temporarily choose $H\left(\tau\right)$ (will modify it later
on) to be equal to

$H\left(\tau\right)=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr=\sin\left(m\log\left(\tau+1\right)\right),\
\ \ \tau\in\left(0,\infty\right),\ \ \ H\left(0\right)=0,$ (23)

which, by (21), gives

$\varphi\left(\tau\right)=\sin\left(m\log\left(\tau+1\right)\right)+\frac{m\tau}{n\left(\tau+1\right)}\cos\left(m\log\left(\tau+1\right)\right),\
\ \ \tau\in\left(0,\infty\right).$ (24)

That is, if we choose $\varphi\left(\tau\right)$ to be the function given by
(24) and define $\varphi\left(0\right)=0,\ $then the function
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},\ $will be a radial function defined on $\mathbb{R}^{n},\
$lying in the space $C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right),\ $and its average integral is given
by$\ H\left(\tau\right)=\sin\left(m\log\left(\tau+1\right)\right)$ for all
$\tau\in[0,\infty).\ $In particular, we have $p=-1,\ q=1.$

To find the values of $\alpha,\ \beta,\ $we use the representation formula
(9). We have

$\displaystyle
u\left(0,t\right)=\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log\left(\sqrt{4t}z+1\right)\right)dz$
$\displaystyle=\left\\{\begin{array}[c]{l}\left[\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz\right]\sin\left(m\log\sqrt{4t}\right)\vspace{3mm}\\\
+\left[\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz\right]\cos\left(m\log\sqrt{4t}\right),\
\ \ t\in\left(0,\infty\right)\end{array}\right.\ $ (27)

and the Lebesgue Dominated Convergence Theorem implies

$\displaystyle\lim_{t\rightarrow\infty}\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz$
$\displaystyle=\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log
z\right)dz:=A\left(m\right)\in\left(-1,1\right)$ (28)

and

$\displaystyle\lim_{t\rightarrow\infty}\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz$
$\displaystyle=\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log
z\right)dz:=B\left(m\right)\in\left(-1,1\right),$ (29)

where we can use the identity$\
\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}dz=1\
$to know that $A\left(m\right),\ B\left(m\right)\in\left(-1,1\right).\ $By
Hölder inequality, the constant $A\left(m\right)$ satisfies the estimate

$\displaystyle A^{2}\left(m\right)$
$\displaystyle=\left(\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}\sqrt{e^{-z^{2}}z^{n+1}}\sqrt{e^{-z^{2}}z^{n+1}}\cos\left(m\log
z\right)dz\right)^{2}$
$\displaystyle<\left(\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}dz\right)\left(\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos^{2}\left(m\log
z\right)dz\right)$
$\displaystyle=\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos^{2}\left(m\log
z\right)dz$

and similarly

$B^{2}\left(m\right)<\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin^{2}\left(m\log
z\right)dz$

and we conclude

$0<A^{2}\left(m\right)+B^{2}\left(m\right)<\frac{2\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\left[\cos^{2}\left(m\log
z\right)+\sin^{2}\left(m\log z\right)\right]dz=1.$ (30)

Therefore, we have the convergence

$\lim_{t\rightarrow\infty}\left|u\left(0,t\right)-\left[A\left(m\right)\sin\left(m\log\sqrt{4t}\right)+B\left(m\right)\cos\left(m\log\sqrt{4t}\right)\right]\right|=0,$
(31)

which gives the conclusion

$\displaystyle-1$
$\displaystyle=\liminf_{\tau\rightarrow\infty}H\left(\tau\right)<-\sqrt{A^{2}\left(m\right)+B^{2}\left(m\right)}=\liminf_{t\rightarrow\infty}u\left(0,t\right)$
$\displaystyle<\
\limsup_{t\rightarrow\infty}u\left(0,t\right)=\sqrt{A^{2}\left(m\right)+B^{2}\left(m\right)}<\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=1$
(32)

for fixed $m\in\left(0,\infty\right).$

The idea next is to view$\ m\in\left(0,\infty\right)$ as a variable. We
observe the following:

###### Lemma 11

We have

$\lim_{m\rightarrow\infty}A\left(m\right)=\lim_{m\rightarrow\infty}B\left(m\right)=0,\
\ \ m\in\left(0,\infty\right).$ (33)

Proof. For any $\varepsilon>0,\ $one can find small number$\ \delta>0\ $and
large number$\ M>0,\ $both are independent of $m\in\left(0,\infty\right),$
such that

$\left|\int_{0}^{\delta}e^{-z^{2}}z^{n+1}\cos\left(m\log
z\right)dz\right|+\left|\int_{M}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log
z\right)dz\right|<\varepsilon.$ (34)

On the other hand, by the change of variables, we have

$\int_{\delta}^{M}e^{-z^{2}}z^{n+1}\cos\left(m\log
z\right)dz=\int_{\log\delta}^{\log M}F\left(x\right)\cos\left(mx\right)dx,\ \
\ x=\log z,$

where$\ F\left(x\right)=e^{-e^{2x}+\left(n+2\right)x}.\ $The familiar Riemann-
Lebesgue lemma in analysis (or just use integration by parts) implies

$\lim_{m\rightarrow\infty}\int_{\log\delta}^{\log
M}F\left(x\right)\cos\left(mx\right)dx=0.$ (35)

which, together with (34), implies$\
\lim_{m\rightarrow\infty}A\left(m\right)=0.\ $The proof of
$\lim_{m\rightarrow\infty}B\left(m\right)=0\ $is similar.$\hfill\square$

Since estimate (34) is valid for all $m\in\left(0,\infty\right),\ $the two
improper integrals

$\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\cos\left(m\log z\right)dz,\ \ \
\int_{0}^{\infty}e^{-z^{2}}z^{n+1}\sin\left(m\log z\right)dz$ (36)

converge uniformly with respect to $m\in\left(0,\infty\right).\ $As a
consequence, both $A\left(m\right)\ $and $B\left(m\right)\ $are continuous
functions of $m\in\left(0,\infty\right),\ $with

$\lim_{m\rightarrow\infty}A\left(m\right)=\lim_{m\rightarrow\infty}B\left(m\right)=0,\
\ \ \lim_{m\rightarrow 0}A\left(m\right)=1,\ \ \ \lim_{m\rightarrow
0}B\left(m\right)=0,$ (37)

where the last two limits in (37) are due to the Lebesgue Dominated
Convergence Theorem. Moreover, the intermediate value theorem implies the
existence of a number $m\in\left(0,\infty\right)\ $satisfying

$\sqrt{A^{2}\left(m\right)+B^{2}\left(m\right)}=\frac{\beta-\alpha}{q-p}\in\left(0,1\right).$
(38)

Now we choose $H\left(\tau\right)$ as

$H\left(\tau\right)=\frac{q-p}{2}\sin\left(m\log\left(\tau+1\right)\right)+\frac{q+p}{2},\
\ \ H\left(0\right)=\frac{q+p}{2},\ \ \ \tau\in[0,\infty),$ (39)

where $m\in\left(0,\infty\right)\ $is the number satisfying (38). By (23) and
(24), its corresponding initial bounded radial function
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},$ is given by

$\varphi\left(\tau\right)=\left\\{\begin{array}[c]{l}\frac{q-p}{2}\sin\left(m\log\left(\tau+1\right)\right)\vspace{3mm}\\\
+\frac{q-p}{2}\frac{m\tau}{n\left(\tau+1\right)}\cos\left(m\log\left(\tau+1\right)\right)+\frac{q+p}{2},\
\ \ \varphi\left(0\right)=\frac{q+p}{2},\ \ \
\tau\in[0,\infty).\end{array}\right.$ (40)

Similar to (31), we have

$\lim_{t\rightarrow\infty}\left|u\left(0,t\right)-\left\\{\frac{q-p}{2}\left[A\left(m\right)\sin\left(m\log\sqrt{4t}\right)+B\left(m\right)\cos\left(m\log\sqrt{4t}\right)\right]+\frac{q+p}{2}\right\\}\right|=0,$
(41)

which gives

$\liminf_{t\rightarrow\infty}u\left(0,t\right)=-\frac{q-p}{2}\sqrt{A^{2}\left(m\right)+B^{2}\left(m\right)}+\frac{q+p}{2}=-\frac{\beta-\alpha}{2}+\frac{\beta+\alpha}{2}=\alpha$

and

$\limsup_{t\rightarrow\infty}u\left(0,t\right)=\frac{q-p}{2}\sqrt{A^{2}\left(m\right)+B^{2}\left(m\right)}+\frac{q+p}{2}=\frac{\beta-\alpha}{2}+\frac{\beta+\alpha}{2}=\beta.$

Since we clearly have$\ \liminf_{\tau\rightarrow\infty}H\left(\tau\right)=p,\
\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=q,$ the proof of Theorem 5
is now complete.$\hfill\square$

###### Remark 12

The proof of Theorem 5 also reveals the following interesting observation. For
fixed $q-p>0,\ $if we have small $\beta-\alpha,\ $then by (38) and (33), we
will have large$\ m\in\left(0,\infty\right)\ $and the function
$\varphi\left(\tau\right)$ in (40) will tend to be unbounded due to the term$\
m\tau/\left(n\left(\tau+1\right)\right).$ This may suggest that we cannot find
a bounded radial function$\ \varphi\left(\tau\right)$ satisfying
$p<\alpha=\beta<q.$ This matches with the result in Theorem 1, which implies
that if we have$\ p<\alpha=\beta<q,$ then the initial data $\varphi\in
C^{0}\left(\mathbb{R}^{n}\right)\ $must be unbounded.

###### Remark 13

By (7), for a given initial data $\varphi\in
C^{0}\left(\mathbb{R}^{n}\right)\bigcap
L^{\infty}\left(\mathbb{R}^{n}\right),$ if we have

$\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta,$

then for any fixed $x\in\mathbb{R}^{n}\ $we also have

$\liminf_{t\rightarrow\infty}u\left(x,t\right)=\alpha<\limsup_{t\rightarrow\infty}u\left(x,t\right)=\beta.$
(42)

###### Remark 14

It has been shown in [CT] (see its equations (16) and (37)) that both
functions $H\left(\tau\right)\ $and$\ u\left(0,t\right)$ satisfy

$H^{\prime}\left(\tau\right)=O\left(\frac{1}{\tau}\right)\ \ \text{as\ \
}\tau\rightarrow\infty,\ \ \ \ \
u_{t}\left(0,t\right)=O\left(\frac{1}{t}\right)\ \ \text{as\ \
}t\rightarrow\infty,$ (43)

as long as the initial data $\varphi\ $lies in the space$\
C^{0}\left(\mathbb{R}^{n}\right)\bigcap L^{\infty}\left(\mathbb{R}^{n}\right)$
(no matter it is radial or not). By (43), we may say that any oscillation in$\
H\left(\tau\right)\ $(with $p<q$) or in $u\left(0,t\right)$ (with
$\alpha<\beta$) is a slow oscillation.

As a consequence of Theorem 5, we have the following two corollaries, which
say that we can prescribe the oscillation of $H\left(\tau\right)\ $and
$u\left(0,t\right)$ for three arbitrary different numbers.

###### Corollary 15

For any three different numbers, denoted as $p<\alpha<\beta,\ $one can find a
continuous bounded radial function$\
\varphi\left(x\right)=\varphi\left(\left|x\right|\right),$
$x\in\mathbb{R}^{n},\ $so that its average integral and the solution (2) of
the problem (1) satisfies

$\liminf_{\tau\rightarrow\infty}H\left(\tau\right)=p<\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha<\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=q,$
(44)

where $q=\alpha+\beta-p.\ $Similarly, for any three different numbers, denoted
as $\alpha<\beta<q,\ $one can find a bounded radial function satisfying (44),
where now $p=\alpha+\beta-q.\ $

###### Corollary 16

For any three different numbers, denoted as $p<\alpha<q,\ $with
$\alpha<\left(p+q\right)/2,\ $the same result as in Corollary 15 holds, where
now $\beta=p+q-\alpha.\ $Similarly, for any three different numbers, denoted
as $p<\beta<q,\ $with $\beta>\left(p+q\right)/2,\ $the same result as in
Corollary 15 holds, where now $\alpha=p+q-\beta.\ $

####  2.2.2 Proof of Theorem 8.

To prove Theorem 8, we first need the following:

###### Lemma 17

Assume $\varphi\left(\tau\right)\ $is a $2\pi$-periodic radial function
defined on $\tau\in[0,\infty)$. We have the convergence

$\lim_{t\rightarrow\infty}u\left(0,t\right)=\lim_{\tau\rightarrow\infty}H\left(\tau\right)=\lim_{\tau\rightarrow\infty}\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(\tau\right)d\tau.$
(45)

Proof. Since $\varphi\in C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty),$ by
Remark 2 in Theorem 1, it suffices to prove the identity for
$H\left(\tau\right).\ $The proof is quite straightforward for the case $n=1\
$(see Remark 6 in [TN]), but may need a trick for $n>1.\ $For large $\tau>0,$
we can express it as $\tau=2m\pi+R\ $for some $m\in\mathbb{N}$ and
$R\in[0,2\pi),$ with the understanding that both $m\ $and $R\ $depend on
$\tau$ and as $\tau\rightarrow\infty$ we have $m\rightarrow\infty,$ hence we
obtain

$\displaystyle\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr$
$\displaystyle=\frac{n}{\left(2m\pi+R\right)^{n}}\left(\begin{array}[c]{l}\int_{0}^{2\pi}\varphi\left(r\right)r^{n-1}dr+\int_{2\pi}^{4\pi}\varphi\left(r\right)r^{n-1}dr+\int_{4\pi}^{6\pi}\varphi\left(r\right)r^{n-1}dr\\\
+\cdot\cdot\cdot+\int_{2\left(m-1\right)\pi}^{2m\pi}\varphi\left(r\right)r^{n-1}dr+\int_{2m\pi}^{2m\pi+R}\varphi\left(r\right)r^{n-1}dr\end{array}\right).$
(48)

If we do the change of variables

$\int_{2\pi}^{4\pi}\varphi\left(r\right)r^{n-1}dr=\int_{0}^{2\pi}\varphi\left(s\right)\left(s+2\pi\right)^{n-1}ds,\
\ \ r=s+2\pi,$

and

$\int_{4\pi}^{6\pi}\varphi\left(r\right)r^{n-1}dr=\int_{0}^{2\pi}\varphi\left(s\right)\left(s+4\pi\right)^{n-1}ds,\
\ \ r=s+4\pi,$

$\ ...,\ $etc., (48) becomes

$\displaystyle\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr$
$\displaystyle=\frac{n}{\left(2m\pi+R\right)^{n}}\left(\begin{array}[c]{l}\int_{0}^{2\pi}\varphi\left(s\right)\left[s^{n-1}+\left(s+2\pi\right)^{n-1}+\left(s+4\pi\right)^{n-1}+\cdot\cdot\cdot+\left(s+2\left(m-1\right)\pi\right)^{n-1}\right]ds\vspace{3mm}\\\
+\int_{0}^{R}\varphi\left(s\right)\left(s+2m\pi\right)^{n-1}ds\end{array}\right)$
(51) $\displaystyle:=I+II.$ (52)

For the second term $II\ $in (52), we have

$\lim_{m\rightarrow\infty}\left|II\right|\leq\lim_{m\rightarrow\infty}\left(\frac{n}{\left(2m\pi+R\right)^{n}}\cdot
2\pi\max_{s\in\left[0,2\pi\right]}\left|\varphi\left(s\right)\right|\cdot\left(2\pi+2m\pi\right)^{n-1}\right)=0.$
(53)

To estimate the first term $I\ $in (52), we first look at the integral

$\displaystyle\frac{n}{\left(2m\pi\right)^{n}}\int_{0}^{2\pi}\varphi\left(s\right)\left[s^{n-1}+\left(s+2\pi\right)^{n-1}+\left(s+4\pi\right)^{n-1}+\cdot\cdot\cdot+\left(s+2\left(m-1\right)\pi\right)^{n-1}\right]ds$
$\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(s\right)\underbrace{\frac{n}{m^{n}}\frac{1}{\left(2\pi\right)^{n-1}}\left[s^{n-1}+\left(s+2\pi\right)^{n-1}+\left(s+4\pi\right)^{n-1}+\cdot\cdot\cdot+\left(s+2\left(m-1\right)\pi\right)^{n-1}\right]}ds,$

where the underlined term$\ \underbrace{\cdot\cdot\cdot}$ is equal to

$\displaystyle\underbrace{\cdot\cdot\cdot}$
$\displaystyle=\frac{n}{m^{n}}\left[\left(\frac{s}{2\pi}\right)^{n-1}+\left(\frac{s}{2\pi}+1\right)^{n-1}+\left(\frac{s}{2\pi}+2\right)^{n-1}+\cdot\cdot\cdot+\left(\frac{s}{2\pi}+\left(m-1\right)\right)^{n-1}\right]$

and for $s\in\left[0,2\pi\right]$ we have

$\displaystyle\frac{n}{m^{n}}\left(0^{n-1}+1^{n-1}+2^{n-1}+\cdot\cdot\cdot+\left(m-1\right)^{n-1}\right)$
$\displaystyle\leq\underbrace{\cdot\cdot\cdot}\leq\frac{n}{m^{n}}\left(1^{n-1}+2^{n-1}+3^{n-1}+\cdot\cdot\cdot+m^{n-1}\right).$
(54)

By the inequality

$\displaystyle\frac{n}{m^{n}}\left(0^{n-1}+1^{n-1}+2^{n-1}+\cdot\cdot\cdot+\left(m-1\right)^{n-1}\right)$
$\displaystyle<\frac{n}{m^{n}}\int_{0}^{m}x^{n-1}dx=1<\frac{n}{m^{n}}\left(1^{n-1}+2^{n-1}+3^{n-1}+\cdot\cdot\cdot+m^{n-1}\right)$
(55)

with

$\displaystyle\lim_{m\rightarrow\infty}\left[\frac{n}{m^{n}}\left(1^{n-1}+2^{n-1}+3^{n-1}+\cdot\cdot\cdot+m^{n-1}\right)-\frac{n}{m^{n}}\left(0^{n-1}+1^{n-1}+2^{n-1}+\cdot\cdot\cdot+\left(m-1\right)^{n-1}\right)\right]$
$\displaystyle=\lim_{m\rightarrow\infty}\left(\frac{n}{m^{n}}m^{n-1}\right)=0,$

we must have

$\left\\{\begin{array}[c]{l}\lim_{m\rightarrow\infty}\frac{n}{m^{n}}\left(0^{n-1}+1^{n-1}+2^{n-1}+\cdot\cdot\cdot+\left(m-1\right)^{n-1}\right)=1\vspace{3mm}\\\
\lim_{m\rightarrow\infty}\frac{n}{m^{n}}\left(1^{n-1}+2^{n-1}+3^{n-1}+\cdot\cdot\cdot+m^{n-1}\right)=1\end{array}\right.$

and hence $\lim_{m\rightarrow\infty}\underbrace{\cdot\cdot\cdot}=1,$ which
implies

$\lim_{m\rightarrow\infty}\frac{n}{\left(2m\pi\right)^{n}}\int_{0}^{2m\pi}\varphi\left(r\right)r^{n-1}dr=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(s\right)\left(\lim_{m\rightarrow\infty}\underbrace{\cdot\cdot\cdot}\right)ds=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(s\right)ds.$
(56)

Finally we have

$\lim_{m\rightarrow\infty}I=\lim_{m\rightarrow\infty}\left(\frac{\left(2m\pi\right)^{n}}{\left(2m\pi+R\right)^{n}}\frac{n}{\left(2m\pi\right)^{n}}\int_{0}^{2m\pi}\varphi\left(r\right)r^{n-1}dr\right)=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi\left(s\right)ds.$

The proof is done.$\hfill\square$

As we have said in the paragraph before Theorem 8, if
$\varphi\left(\tau\right)\ $has a slow oscillation as $\tau\rightarrow\infty\
$with $r<s,$ then it has the chance of passing to the function
$u\left(0,t\right)$. The following gives a simple way to find a slow
oscillation function$\ \varphi\left(\tau\right)$ on $[0,\infty)\ $which will
cause $\alpha<\beta.$

###### Lemma 18

Let $g\left(\tau\right)$ be a $2\pi$-periodic non-constant $C^{2}$ function
defined on $[0,\infty)$ and let
$G\left(\tau\right)=g\left(\log\left(\tau+1\right)\right),\
\tau\in[0,\infty).$ Then the radial function$\
\varphi\left(\tau\right)=\frac{\tau}{n}G^{\prime}\left(\tau\right)+G\left(\tau\right),\
\tau\in[0,\infty),$ has a slow oscillation as $\tau\rightarrow\infty,$ lies in
the space $C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty),$ and its average
integral $H\left(\tau\right),\ \tau\in[0,\infty),\ $satisfies$\
H\left(\tau\right)=g\left(\log\left(\tau+1\right)\right),\ \tau\in[0,\infty),\
$with

$p=\liminf_{\tau\rightarrow\infty}g\left(\tau\right)<\limsup_{\tau\rightarrow\infty}g\left(\tau\right)=q,$
(57)

which, by Theorem 1, will also imply $\alpha<\beta.$

Proof. By

$\left\\{\begin{array}[c]{l}\varphi\left(\tau\right)=\frac{\tau}{n}\frac{g^{\prime}\left(\log\left(\tau+1\right)\right)}{\tau+1}+g\left(\log\left(\tau+1\right)\right),\
\ \ \varphi\left(0\right)=g\left(0\right),\vspace{3mm}\\\
\varphi^{\prime}\left(\tau\right)=\left(1+\frac{1}{n}\right)\frac{g^{\prime}\left(\log\left(\tau+1\right)\right)}{\tau+1}+\frac{\tau\left[g^{\prime\prime}\left(\log\left(\tau+1\right)\right)-g^{\prime}\left(\log\left(\tau+1\right)\right)\right]}{n\left(\tau+1\right)^{2}},\
\ \ \tau\in[0,\infty),\end{array}\right.$

we see that$\ \varphi\ $is a slow oscillation function on $[0,\infty),\ $lying
in the space $C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty).\ $Now

$\displaystyle H\left(\tau\right)$
$\displaystyle=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr=\frac{n}{\tau^{n}}\int_{0}^{\tau}\left(\frac{r}{n}G^{\prime}\left(r\right)+G\left(r\right)\right)r^{n-1}dr$
$\displaystyle=\frac{1}{\tau^{n}}\int_{0}^{\tau}\frac{d}{dr}\left(r^{n}G\left(r\right)\right)dr=G\left(\tau\right)=g\left(\log\left(\tau+1\right)\right),\
\ \ \tau\in\left(0,\infty\right)$

and$\ H\left(0\right)=G\left(0\right)=g\left(0\right).\ $Therefore, the
inequality (57) follows.$\hfill\square$

With the help of Lemma 17, we are ready to prove Theorem 8. For any
$r<\alpha<\beta<s,$ we divide the proof into two cases.

Case 1: $r+s=\alpha+\beta,\ r<\alpha<\beta<s.\vspace{3mm}$

In this case we can use the representation formula (10) and, similar to (39),
choose

$\varphi\left(\tau\right)=\frac{s-r}{2}\sin\left(m\log\left(\tau+1\right)\right)+\frac{s+r}{2},\
\ \ \varphi\left(0\right)=\frac{s+r}{2},\ \ \ \tau\in[0,\infty),$ (58)

and perform the same argument as in the proof of Theorem 5 to prescribe them.
More precisely, here we need to choose $m\in\left(0,\infty\right)$ to satisfy
$\sqrt{\tilde{A}^{2}\left(m\right)+\tilde{B}^{2}\left(m\right)}=\frac{\beta-\alpha}{s-r}\in\left(0,1\right),$
where now

$\left\\{\begin{array}[c]{l}\tilde{A}\left(m\right)=\dfrac{n\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n-1}\cos\left(m\log
z\right)dz\in\left(-1,1\right),\vspace{3mm}\\\
\tilde{B}\left(m\right)=\dfrac{n\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n-1}\sin\left(m\log
z\right)dz\in\left(-1,1\right),\end{array}\right.$ (59)

with

$0<\tilde{A}^{2}\left(m\right)+\tilde{B}^{2}\left(m\right)<\frac{n\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n-1}\left[\cos^{2}\left(m\log
z\right)+\sin^{2}\left(m\log z\right)\right]dz=1.$ (60)

Note that in this case the chosen function $\varphi\left(\tau\right),$ like
$H\left(\tau\right)\ $in (39), has a slow oscillation on $[0,\infty).$

Case 2:$\ r+s\neq\alpha+\beta,\ r<\alpha<\beta<s.\vspace{3mm}$

In this case, we may assume $r+s>\alpha+\beta\ $(the treatment for the case
$r+s<\alpha+\beta$ is similar). We will choose$\ \varphi\left(\tau\right)\in
C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty)$ to be the sum of two functions
$\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right),$ where
$\varphi_{1}\left(\tau\right)$ has a slow oscillation, but
$\varphi_{2}\left(\tau\right)$ has a regular oscillation (which is
$2\pi$-periodic).

Let $\lambda\ $be the number satisfying $r+\lambda=\alpha+\beta.\ $Since
$r<\alpha<\beta<s$ and $r+s>\alpha+\beta,$ we have$\ \beta<\lambda<s.$ We can
write $\lambda$ as $\lambda=\varepsilon+\delta$ for some small $\varepsilon>0$
and $\delta\in\mathbb{R}\ $so that we have

$r+\varepsilon<\alpha<\beta<\delta<s.$ (61)

Now we have $\left(r+\varepsilon\right)+\delta=\alpha+\beta\ $and by Case 1,
we can find a radial $\varphi_{1}\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty)\ $such that it and its convolution solution
$u_{1}\left(x,t\right)$ satisfy

$\liminf_{\tau\rightarrow\infty}\varphi_{1}\left(\tau\right)=r+\varepsilon<\liminf_{t\rightarrow\infty}u_{1}\left(0,t\right)=\alpha<\limsup_{t\rightarrow\infty}u_{1}\left(0,t\right)=\beta<\limsup_{\tau\rightarrow\infty}\varphi_{1}\left(\tau\right)=\delta.$
(62)

Next let $\varphi_{2}\left(\tau\right)\ $be a $2\pi$-periodic continuous
function on$\ \tau\in[0,\infty)$ which satisfies

$\frac{1}{2\pi}\int_{0}^{2\pi}\varphi_{2}\left(\tau\right)d\tau=0,\ \ \
\min_{\tau\in\left[0,2\pi\right]}\varphi_{2}\left(\tau\right)=-\varepsilon<0,\
\ \ \max_{\tau\in\left[0,2\pi\right]}\varphi_{2}\left(\tau\right)=s-\delta>0.$
(63)

Such a function $\varphi_{2}\ $clearly exists and, together with its
convolution solution $u_{2}\left(x,t\right),\ $they satisfy

$\liminf_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=-\varepsilon<\lim_{t\rightarrow\infty}u_{2}\left(0,t\right)=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi_{2}\left(\tau\right)d\tau=0<\limsup_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=s-\delta,$
(64)

due to the result in Lemma 17. Finally, we set
$\varphi\left(\tau\right)=\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right),\
\tau\in[0,\infty).\ $We have$\ \varphi\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty)\ $and its convolution solution $u\left(x,t\right)$
satisfy$\ u\left(x,t\right)=u_{1}\left(x,t\right)+u_{2}\left(x,t\right)\ $for
all $\left(x,t\right)\in\mathbb{R}^{n}\times[0,\infty).\ $Since$\
u_{2}\left(0,t\right)\rightarrow 0\ $as $t\rightarrow\infty,$ we have

$\left\\{\begin{array}[c]{l}\liminf_{t\rightarrow\infty}u\left(0,t\right)=\liminf_{t\rightarrow\infty}u_{1}\left(0,t\right)=\alpha\vspace{3mm}\\\
\limsup_{t\rightarrow\infty}u\left(0,t\right)=\limsup_{t\rightarrow\infty}u_{1}\left(0,t\right)=\beta.\end{array}\right.$
(65)

It remains to look at $\varphi\left(\tau\right)$ for $\tau\rightarrow\infty.\
$By (58) we know $\varphi_{1}\left(\tau\right)$ is a slow oscillation function
on $[0,\infty)$ with the form

$\varphi_{1}\left(\tau\right)=\frac{\delta-\left(r+\varepsilon\right)}{2}\sin\left(m\log\left(\tau+1\right)\right)+\frac{\delta+\left(r+\varepsilon\right)}{2},\
\ \ \varphi_{1}\left(0\right)=\frac{\delta+\left(r+\varepsilon\right)}{2},\ \
\ \tau\in[0,\infty)$ (66)

for some fixed$\ m\in\left(0,\infty\right)\ $satisfying$\
\sqrt{\tilde{A}^{2}\left(m\right)+\tilde{B}^{2}\left(m\right)}=\frac{\beta-\alpha}{\delta-\left(r+\varepsilon\right)}\in\left(0,1\right).\
$On the other hand, $\varphi_{2}\left(\tau\right)$ is a regular oscillation
function ($2\pi$-periodic function) on $[0,\infty),$ which oscillates between
$-\varepsilon\ $and $s-\delta.\ $It is not difficult to see that there exists
a sequence $\tau_{j}\rightarrow\infty\ $so that

$\sin\left(m\log\left(\tau_{j}+1\right)\right)\rightarrow-1\ \ \ \text{and\ \
\
}\varphi_{2}\left(\tau_{j}\right)\rightarrow-\varepsilon=\min_{\tau\in\left[0,2\pi\right]}\varphi_{2}\left(\tau\right)<0$
(67)

as $j\rightarrow\infty.\ $ Similarly, there exists a sequence
$\tau_{k}\rightarrow\infty\ $so that

$\sin\left(m\log\left(\tau_{k}+1\right)\right)\rightarrow 1\ \ \ \text{and\ \
\ }\varphi_{2}\left(\tau_{k}\right)\rightarrow
s-\delta=\max_{\tau\in\left[0,2\pi\right]}\varphi_{2}\left(\tau\right)>0$ (68)

as $k\rightarrow\infty.$ By (66), (68), and (67), we have

$\displaystyle\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)$
$\displaystyle=\liminf_{\tau\rightarrow\infty}\left(\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right)\right)$
$\displaystyle=\liminf_{\tau\rightarrow\infty}\varphi_{1}\left(\tau\right)+\liminf_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=\left(r+\varepsilon\right)-\varepsilon=r$
(69)

and

$\displaystyle\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)$
$\displaystyle=\limsup_{\tau\rightarrow\infty}\left(\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right)\right)$
$\displaystyle=\limsup_{\tau\rightarrow\infty}\varphi_{1}\left(\tau\right)+\limsup_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=\delta+\left(s-\delta\right)=s.$
(70)

The proof is done due to (65), (69), and (70).

Combining Case 1 and Case 2, the proof of Theorem 8 is now
complete.$\hfill\square$

###### Remark 19

This is to explain the existence of a sequence $\tau_{k}\rightarrow\infty\ $so
that (68) holds. We first note that, for given $\varepsilon>0,\ $there
exists$\ \delta>0\ $(depending only on $\varepsilon$), such that if
$m\log\left(\tau+1\right)$ (here $m>0$ is fixed) lies in the interval
$\left(2k\pi+\frac{\pi}{2}-\delta,2k\pi+\frac{\pi}{2}+\delta\right)\ $for some
$k\in\mathbb{N},$ then we have$\
\left|\sin\left(m\log\left(\tau+1\right)\right)-1\right|<\varepsilon,$ which
is equivalent for $\tau+1$ to lie in the interval

$\tau+1\in\left(\exp\left(\frac{2k\pi+\frac{\pi}{2}-\delta}{m}\right),\
\exp\left(\frac{2k\pi+\frac{\pi}{2}+\delta}{m}\right)\right).$ (71)

We can choose $k\in\mathbb{N}$ sufficiently large so that the length of the
interval in (71) is sufficiently large. By this observation, we clearly have
(68). The reason for (67) to hold is similar.

###  2.3 The remaining cases not covered by Theorem 8.

For prescribing general$\ r\leq\alpha\leq\beta\leq s,\ $the remaining cases
not covered by Theorem 8 are discussed in the following:

###### Theorem 20

For any four arbitrary finite numbers $r,\ \alpha,\ \beta,\ s\ $in one of the
following cases:

$\left\\{\begin{array}[c]{l}\left(1\right).\ r=\alpha<\beta<s,\ \ \ \ \
\left(2\right).\ r<\alpha=\beta<s,\ \ \ \ \ \left(3\right).\
r<\alpha<\beta=s,\vspace{3mm}\\\ \left(4\right).\ r=\alpha=\beta<s,\ \ \ \ \
\left(5\right).\ r=\alpha<\beta=s,\ \ \ \ \ \left(6\right).\
r<\alpha=\beta=s,\end{array}\right.$ (72)

one can choose a suitable radial initial data
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right)\in
C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty)\ $satisfying

$\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=r,\ \ \
\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=s$ (73)

and its corresponding convolution solution $u\left(x,t\right)$ in (2)
satisfies

$\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha,\ \ \
\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta.$ (74)

###### Remark 21

The case $\ r=\alpha=\beta=s\ $is trivial. Just choose $\varphi\left(x\right)$
to be a constant function.

Proof. $\left(2\right).\ $For $r<\alpha=\beta<s,\ $we can choose
$\varphi\left(\tau\right),\ \tau\in[0,\infty),\ $to be any $2\pi$-periodic
function with average value $\alpha=\beta,$ maximum value $s,\ $minimum value
$r.\ $By Lemma 17, it can be achieved.

$\left(5\right).\ $For $r=\alpha<\beta=s,\ $we choose
$\varphi\left(\tau\right)$ to be the extremely slow oscillation function

$\varphi\left(\tau\right)=\frac{\beta-\alpha}{2}\sin\left[\log\left(\log\left(\tau+2\right)\right)\right]+\frac{\beta+\alpha}{2},\
\ \ \tau\in[0,\infty),$ (75)

which has

$\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=\alpha,\ \ \
\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=\beta.$ (76)

By (10), its corresponding $u\left(0,t\right)$ is given by

$u\left(0,t\right)=\frac{\beta-\alpha}{2}\dfrac{n\omega\left(n\right)}{\pi^{n/2}}\int_{0}^{\infty}e^{-z^{2}}z^{n-1}\sin\left[\log\left(\log\left(\sqrt{4t}z+2\right)\right)\right]dz+\frac{\beta+\alpha}{2},$
(77)

where one can write

$\sin\left[\log\left(\log\left(\sqrt{4t}z+2\right)\right)\right]=\sin\left[\log\left(\log\sqrt{4t}\right)+g\left(t,z\right)\right],$

with

$g\left(t,z\right)=\log\left(1+\frac{1}{\log\sqrt{4t}}\log\left(z+\frac{2}{\sqrt{4t}}\right)\right),\
\ \ t,\ z\in\left(0,\infty\right).$

We see that $\lim_{t\rightarrow\infty}g\left(t,z\right)=0\ $for fixed
$z\in\left(0,\infty\right).$ Hence the Lebesgue Dominated Convergence Theorem
implies

$\lim_{t\rightarrow\infty}\left|u\left(0,t\right)-\left[\frac{\beta-\alpha}{2}\sin\left(\log\left(\log\sqrt{4t}\right)\right)+\frac{\beta+\alpha}{2}\right]\right|=0$
(78)

and so

$\liminf_{t\rightarrow\infty}u\left(0,t\right)=\alpha,\ \ \
\limsup_{t\rightarrow\infty}u\left(0,t\right)=\beta.$ (79)

$\left(4\right).\ $For$\ r=\alpha=\beta<s,$ we choose $\varphi\in
C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty)$ with
$\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=r=\alpha,\
\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=s,$ and for most
$\tau\in[0,\infty)\ $the value of $\varphi\left(\tau\right)$ is equal to
$\alpha\ $and for $\varphi\left(\tau\right)$ not equal to $\alpha,$ it looks
like a thin bump with height $s-\alpha\ $and the supports of these bumps are
spreading further and further apart. By this choice of $\varphi$ we have

$\lim_{\tau\rightarrow\infty}H\left(\tau\right)=\lim_{\tau\rightarrow\infty}\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr=\alpha,$

which implies $\lim_{t\rightarrow\infty}u\left(0,t\right)=\alpha\ $due to
Theorem 1. Hence, the case $r=\alpha=\beta<s$ is achieved.

$\left(6\right).\ $For$\ r<\alpha=\beta=s,\ $the construction of $\varphi$ is
similar to that in $\left(4\right)\ $except that we reverse the role of $r\
$and $s.$

$\left(1\right).\ $For$\ r=\alpha<\beta<s,\ $we first choose
$\varphi_{1}\left(\tau\right)$ to be the function given by (75), which,
together with its corresponding $u_{1}\left(0,t\right),$ will satisfy (76) and
(79). Next, we choose $\varphi_{2}\left(\tau\right).\ $Let $\varepsilon>0\ $be
a fixed small number and for each$\ m\in\mathbb{N},\ $let

$\tau_{m}=\exp\left(\exp\left(2m\pi+\frac{\pi}{2}\right)\right)-2,\ \ \ \
\tilde{\tau}_{m}=\exp\left(\exp\left(2m\pi+\frac{3\pi}{2}\right)\right)-2,$

where we note that $\varphi_{1}\left(\tilde{\tau}_{m}\right)=\alpha,\
\varphi_{1}\left(\tau_{m}\right)=\beta.\ $We require$\
\varphi_{2}\left(\tau\right)$ to be a nonnegative function satisfying

$\varphi_{2}\left(\tau\right)=\left\\{\begin{array}[c]{l}\frac{s-\beta}{\varepsilon}\left(\tau-\left(\tau_{m}-\varepsilon\right)\right),\
\ \ \tau\in\left[\tau_{m}-\varepsilon,\tau_{m}\right]\vspace{3mm}\\\
-\frac{s-\beta}{\varepsilon}\left(\tau-\left(\tau_{m}+\varepsilon\right)\right),\
\ \ \tau\in\left[\tau_{m},\tau_{m}+\varepsilon\right]\vspace{3mm}\\\ 0,\ \ \
\text{otherwise,\ \ \ }\tau\in[0,\infty).\end{array}\right.$

It satisfies $\liminf_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=0\
$and$\ \limsup_{\tau\rightarrow\infty}\varphi_{2}\left(\tau\right)=s-\beta\
$and

$\lim_{\tau\rightarrow\infty}H_{2}\left(\tau\right)=\lim_{\tau\rightarrow\infty}\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi_{2}\left(r\right)r^{n-1}dr=0.$

Hence its corresponding $u_{2}\left(0,t\right)$ satisfies$\
\lim_{t\rightarrow\infty}u_{2}\left(0,t\right)=0\ $due to Theorem 1. Now set
$\varphi=\varphi_{1}+\varphi_{2}\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty).\ $We will have
$u\left(0,t\right)=u_{1}\left(0,t\right)+u_{2}\left(0,t\right)$ with

$\left\\{\begin{array}[c]{l}\liminf_{t\rightarrow\infty}u\left(0,t\right)=\liminf_{t\rightarrow\infty}u_{1}\left(0,t\right)=\alpha\vspace{3mm}\\\
\limsup_{t\rightarrow\infty}u\left(0,t\right)=\limsup_{t\rightarrow\infty}u_{1}\left(0,t\right)=\beta.\end{array}\right.$

Also, by$\
\varphi\left(\tilde{\tau}_{m}\right)=\varphi_{1}\left(\tilde{\tau}_{m}\right)+\varphi_{2}\left(\tilde{\tau}_{m}\right)=\alpha+0=\alpha\
$and$\
\varphi\left(\tau_{m}\right)=\varphi_{1}\left(\tau_{m}\right)+\varphi_{2}\left(\tau_{m}\right)=\beta+\left(s-\beta\right)=s,$
we have

$\left\\{\begin{array}[c]{l}\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=\liminf_{\tau\rightarrow\infty}\left(\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right)\right)=\alpha\vspace{3mm}\\\
\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=\limsup_{\tau\rightarrow\infty}\left(\varphi_{1}\left(\tau\right)+\varphi_{2}\left(\tau\right)\right)=\beta+\left(s-\beta\right)=s.\end{array}\right.$

$\left(3\right)\ $For$\ r<\alpha<\beta=s,\ $the construction of $\varphi$ is
similar to that in $\left(1\right).\vspace{3mm}$

The proof of Theorem 20 is now complete.$\hfill\square$

##  3 Some side issues.

###  3.1 An example of prescribing four different numbers not satisfying (13)
in Theorem 5.

This is related to Theorem 5. Until now, we still do not know how to prescribe
the oscillation of $H\left(\tau\right)\ $and $u\left(0,t\right)$ for four
arbitrary numbers $p<\alpha<\beta<q\ $not satisfying the condition (13). This
will be an interesting problem to explore. However, it is not difficult to
construct a particular example not satisfying (13). We have:

###### Lemma 22

There exists a radial function $\varphi\in C^{0}[0,\infty)\bigcap
L^{\infty}[0,\infty)$ so that its average integral $H\left(\tau\right)\ $and
the convolution solution (2) of the problem (1) satisfy $p<\alpha<\beta<q$
with $p+q\neq\alpha+\beta.$

Proof. The idea of breaking the symmetry, unlike (23), is to choose
$H\left(\tau\right)$ to be equal to the sum of several (at least two)
different slow oscillation functions. For simplicity, we only look at the case
$n=1\ $(the construction for $n>1\ $is similar, but we cannot find numerical
values involving a general variable $n$) and require $H\left(\tau\right)$ to
be equal to the sum of two slow oscillation functions, given by

$\displaystyle H\left(\tau\right)$
$\displaystyle=\frac{n}{\tau^{n}}\int_{0}^{\tau}\varphi\left(r\right)r^{n-1}dr=\frac{1}{\tau}\int_{0}^{\tau}\varphi\left(r\right)dr\
\ \ \left(n=1\right)$
$\displaystyle=\sin\left(\log\left(\tau+1\right)\right)+\sin\left(2\log\left(\tau+1\right)\right),\
\ \ \tau\in\left(0,\infty\right),\ \ \ H\left(0\right)=0.$ (80)

Similar to (24), one can find the corresponding initial radial function
$\varphi\in C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty).$ With the help of
Maple software, we can evaluate

$\left\\{\begin{array}[c]{l}p=\liminf_{\tau\rightarrow\infty}H\left(\tau\right)=\min_{x\in\left[0,2\pi\right]}\left(\sin
x+\sin 2x\right)\approx-1.760172593\vspace{3mm}\\\
q=\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=\max_{x\in\left[0,2\pi\right]}\left(\sin
x+\sin 2x\right)\approx 1.760172593.\end{array}\right.$

By the representation formula
$u\left(0,t\right)=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}e^{-z^{2}}z^{2}H\left(\sqrt{4t}z\right)dz,\
t\in\left(0,\infty\right),\ $we have

$u\left(0,t\right)=\left\\{\begin{array}[c]{l}A\left(t\right)\sin\left(\log\sqrt{4t}\right)+B\left(t\right)\cos\left(\log\sqrt{4t}\right)\vspace{3mm}\\\
+C\left(t\right)\sin\left(2\log\sqrt{4t}\right)+D\left(t\right)\cos\left(2\log\sqrt{4t}\right),\
\ \ t\in\left(0,\infty\right),\end{array}\right.$

where

$\left\\{\begin{array}[c]{l}A\left(t\right)=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}e^{-z^{2}}z^{2}\cos\left(\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz\vspace{3mm}\\\
B\left(t\right)=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}e^{-z^{2}}z^{2}\sin\left(\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz\vspace{3mm}\\\
C\left(t\right)=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}e^{-z^{2}}z^{2}\cos\left(2\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz\vspace{3mm}\\\
D\left(t\right)=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}e^{-z^{2}}z^{2}\sin\left(2\log\left(z+\frac{1}{\sqrt{4t}}\right)\right)dz,\
\ \ t\in\left(0,\infty\right).\end{array}\right.$

As $t\rightarrow\infty,$ we have

$\left\\{\begin{array}[c]{l}\lim_{t\rightarrow\infty}A\left(t\right)=A\approx
0.892253317,\ \ \ \lim_{t\rightarrow\infty}B\left(t\right)=B\approx
0.030945895\vspace{3mm}\\\ \lim_{t\rightarrow\infty}C\left(t\right)=C\approx
0.649173672,\ \ \ \lim_{t\rightarrow\infty}D\left(t\right)=D\approx
0.099535090,\end{array}\right.$

and so

$\displaystyle\lim_{t\rightarrow\infty}\left|u\left(0,t\right)-\left[A\sin\left(\log\sqrt{4t}\right)+B\cos\left(\log\sqrt{4t}\right)+C\sin\left(2\log\sqrt{4t}\right)+D\cos\left(2\log\sqrt{4t}\right)\right]\right|$
$\displaystyle=0,$

which gives

$\left\\{\begin{array}[c]{l}\alpha=\liminf_{t\rightarrow\infty}u\left(0,t\right)=\min_{x\in\left[0,2\pi\right]}\left(A\sin
x+B\cos x+C\sin 2x+D\cos 2x\right)\approx-1.369211837\vspace{3mm}\\\
\beta=\limsup_{t\rightarrow\infty}u\left(0,t\right)=\max_{x\in\left[0,2\pi\right]}\left(A\sin
x+B\cos x+C\sin 2x+D\cos 2x\right)\approx 1.328017886\end{array}\right.$

and we find that $p+q\neq\alpha+\beta.\ \hfill\square$

###### Remark 23

We have used different software to evaluate the above numerical values of $p,\
\alpha,\ \beta,\ q\ $and obtain the same values.

###  3.2 The behavior of $u\left(x,t\right)$ as
$\left|x\right|\rightarrow\infty.$

In Theorem 5 and Theorem 8 we only look at the behavior of $u\left(x,t\right)$
for fixed $x\ $with $t\rightarrow\infty.\ $We may as well look at the behavior
of $u\left(x,t\right)$ for fixed $t\ $with $\left|x\right|\rightarrow\infty.\
$In the following lemma, we discuss such property for a particular radial
initial data. We choose
$\varphi\left(x\right)=\varphi\left(\left|x\right|\right)\in
C^{0}[0,\infty)\bigcap L^{\infty}[0,\infty)$ to be the function in (24) with
$m=1.\ $Its average integral $H\left(\tau\right)\ $is equal to
$\sin\left(\log\left(\tau+1\right)\right),$ $\tau\in[0,\infty),\ $which is our
typical slow oscillation function.

###### Lemma 24

Let $u\left(x,t\right)$ be the solution of the problem (1) given by (2) with
$u\left(x,0\right)=\varphi\left(\left|x\right|\right),$ $x\in\mathbb{R}^{n},\
$where$\ \varphi\left(\tau\right)$ is given by

$\varphi\left(\tau\right)=\sin\left(\log\left(\tau+1\right)\right)+\frac{\tau}{n\left(\tau+1\right)}\cos\left(\log\left(\tau+1\right)\right),\
\ \ \tau\in[0,\infty).$ (81)

Then for each fixed $t>0$ we have

$\lim_{\left|x\right|\rightarrow\infty}\left|u\left(x,t\right)-\left(\sin\left(\log\left|x\right|\right)+\frac{1}{n}\cos\left(\log\left|x\right|\right)\right)\right|=0.$
(82)

###### Remark 25

Note that for fixed $t>0,\ $we have

$\tilde{\alpha}:=\liminf_{\left|x\right|\rightarrow\infty}u\left(x,t\right)=\liminf_{\tau\rightarrow\infty}\varphi\left(\tau\right)=r=-\sqrt{1+\frac{1}{n^{2}}}<\liminf_{\tau\rightarrow\infty}H\left(\tau\right)=p=-1$

and

$\tilde{\beta}:=\limsup_{\left|x\right|\rightarrow\infty}u\left(x,t\right)=\limsup_{\tau\rightarrow\infty}\varphi\left(\tau\right)=s=\sqrt{1+\frac{1}{n^{2}}}>\limsup_{\tau\rightarrow\infty}H\left(\tau\right)=q=1.$

Thus the space oscillation of $u\left(x,t\right)\ $is, in general, not
necessarily bounded by the oscillation of $H\left(\tau\right)\ $(however, it
is always bounded by the oscillation of $\varphi\left(\tau\right)$ due to the
maximum principle). This is different from the behavior for time oscillation,
where we always have$\ r\leq p\leq\alpha\leq\beta\leq q\leq s.$

Proof. By Remark 7, we know that $u\left(x,t\right)\ $is radial in
$x\in\mathbb{R}^{n}\ $for each $t>0.\ $Therefore, it suffices to look at the
oscillation behavior of $u\left(x,t\right)$ for
$\left|x\right|\rightarrow\infty.\ $The representation formulas (9) and (10)
cannot be used here since they are valid only for $u\left(0,t\right).$
Instead, we use the convolution formula (2) to get

$u\left(x,t\right)=\dfrac{1}{\left(4\pi
t\right)^{n/2}}\int_{\mathbb{R}^{n}}e^{-\frac{\left|x-y\right|^{2}}{4t}}\varphi\left(\left|y\right|\right)dy=\dfrac{1}{\pi^{n/2}}\int_{\mathbb{R}^{n}}e^{-\left|z\right|^{2}}\varphi\left(\left|x+\sqrt{4t}z\right|\right)dz,$
(83)

where by (81) we have

$\varphi\left(\left|x+\sqrt{4t}z\right|\right)=\left\\{\begin{array}[c]{l}\sin\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right)\vspace{3mm}\\\
+\frac{\left|x+\sqrt{4t}z\right|}{n\left(\left|x+\sqrt{4t}z\right|+1\right)}\cos\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right).\end{array}\right.$

By the Lebesgue Dominated Convergence Theorem and the identity

$\sin\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right)=\left\\{\begin{array}[c]{l}\sin\left(\log\left|x\right|\right)\cos\left(\log\left(\frac{\left|x+\sqrt{4t}z\right|+1}{\left|x\right|}\right)\right)\vspace{3mm}\\\
+\cos\left(\log\left|x\right|\right)\sin\left(\log\left(\frac{\left|x+\sqrt{4t}z\right|+1}{\left|x\right|}\right)\right),\end{array}\right.$

where

$\lim_{\left|x\right|\rightarrow\infty}\frac{\left|x+\sqrt{4t}z\right|+1}{\left|x\right|}=1,\
\ \ \text{for fixed\ }t\ \text{and }z\text{,}$

we have

$\lim_{\left|x\right|\rightarrow\infty}\left|\sin\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right)-\sin\left(\log\left|x\right|\right)\right|=0$

and so

$\lim_{\left|x\right|\rightarrow\infty}\left|\dfrac{1}{\pi^{n/2}}\int_{\mathbb{R}^{n}}e^{-\left|z\right|^{2}}\sin\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right)dz-\sin\left(\log\left|x\right|\right)\right|=0.$
(84)

Similarly, we have

$\lim_{\left|x\right|\rightarrow\infty}\left|\dfrac{1}{\pi^{n/2}}\int_{\mathbb{R}^{n}}e^{-\left|z\right|^{2}}\left(\tfrac{\left|x+\sqrt{4t}z\right|}{n\left(\left|x+\sqrt{4t}z\right|+1\right)}\cos\left(\log\left(\left|x+\sqrt{4t}z\right|+1\right)\right)\right)dz-\frac{1}{n}\cos\left(\log\left|x\right|\right)\right|=0$
(85)

and (82) follows. The proof is done.$\hfill\square$

Acknowledgement. Research supported by NCTS (National Center for Theoretical
Sciences) and MoST (Ministry of Science and Technology) of Taiwan with grant
number 108-2115-M-007-013-MY2.

## References

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  * [J] F. John,  _Partial Differential Equations, 4th edition,_ Applied Mathematical Sciences, v. 1, Springer-Verlag, 1982.
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Dong-Ho Tsai

Department of Mathematics

National Tsing Hua University

Hsinchu 30013, TAIWAN

E-mail<EMAIL_ADDRESS>


# TweetCOVID: A System for Analyzing Public Sentiments and Discussions about
COVID-19 via Twitter Activities

Jolin Shaynn-Ly Kwan<EMAIL_ADDRESS>Information Systems
Technology and Design PillarSingapore University of Technology and
DesignSingapore and  Kwan Hui Lim<EMAIL_ADDRESS>0000-0002-4569-0901
Information Systems Technology and Design PillarSingapore University of
Technology and DesignSingapore

(2021)

###### Abstract.

The COVID-19 pandemic has created widespread health and economical impacts,
affecting millions around the world. To better understand these impacts, we
present the TweetCOVID system that offers the capability to understand the
public reactions to the COVID-19 pandemic in terms of their sentiments,
emotions, topics of interest and controversial discussions, over a range of
time periods and locations, using public tweets. We also present three example
use cases that illustrates the usefulness of our proposed TweetCOVID system.

COVID-19, Twitter, Sentiment Analysis, Topic Modelling

††journalyear: 2021††copyright: rightsretained††conference: 26th International
Conference on Intelligent User Interfaces; April 14–17, 2021; College Station,
TX, USA††booktitle: 26th International Conference on Intelligent User
Interfaces (IUI ’21 Companion), April 14–17, 2021, College Station, TX,
USA††doi: 10.1145/3397482.3450733††isbn: 978-1-4503-8018-8/21/04††ccs: Human-
centered computing Social networking sites††ccs: Information systems Web and
social media search††ccs: Information systems Social networks

##  1\. Introduction

The COVID-19 pandemic has infected millions and claimed the lives of hundreds
of thousands around the world. In addition, COVID-19 has caused widespread
economic impact, resulting in extreme levels of unemployment across all
industries. In turn, these medical and economical impacts have created immense
emotional stress on the global population. With the ongoing pandemic, it is
important to understand the ever-changing impact of COVID-19 as new
regulations and developments take place over the world.

Towards this effort, we present TweetCOVID, a system for understanding the
impact of COVID-19 on the general public in terms of sentiments, emotions,
topics and controversial discussions. TweetCOVID utilizes publicly available
tweets, alongside sentiment analysis (levels of positivity and negativity),
emotion detection (anger, fear, sadness, disgust, surprise, anticipation,
trust, and joy), topic modelling using Latent dirichlet allocation (LDA),
controversial term tracking and various temporal and spatial analysis.

###  1.1. Related Work

Twitter has been used for various applications, such as online misbehaviour
detection (Fornacciari et al., 2018; Chatzakou et al., 2017), event tracking
(Abdelhaq et al., 2013; Lim et al., 2018; Comito et al., 2019), activity
recommendation (Wang et al., 2018; Madisetty, 2019), etc. While there has been
work using Twitter for COVID-19 related studies, many of these works focus on
a single region (Su et al., 2020; Vicari and Murru, 2020) or for a specific
single purpose, e.g., misinformation (Gruzd and Mai, 2020; Stephens, 2020). In
contrast, our proposed TweetCOVID system provides a comprehensive set of
functionalities to study sentiments, emotions, topics, controversial terms
across various timeframes and locations. This holistic set of capabilities
allow the user to study a flexible range of questions, e.g., what are the
public sentiments in a certain country towards the implementation of a
lockdown and what are the following public discussions?

##  2\. System Architecture and Functionalities

Our proposed system comprises the main components for data
collection/processing, general analytics, sentiment/emotion analysis, topic
modelling, controversy tracking and visualisation. These components are
further elaborated next.

Figure 1. Sample screenshots: Emotions by country and week (top), and overall
sentiment over time (bottom).

  * •

Data Collection/Processing: This component allows the user to either load in
existing Twitter datasets or retrieve datasets via the Twitter API. After
loading/collecting these dataset, the standard processing steps of language
filtering, tokenization, stemming and lowercase conversion are applied to the
dataset before the subsequent analysis.

  * •

General Analytics: This component shows the general statistics and
distribution of the dataset. For example, a user can view the number of tweets
posted in specific time periods and/or at geographic locations to understand
how frequent the discussion about COVID-19 is in specific areas.

  * •

Sentiment/Emotion Analysis: This component shows the general sentiments and
specific emotions associated with COVID-19 related discussions. We use the
AFINN lexicon (Nielsen, 2011) for sentiment analysis that allows us to
determine the positivity and negativity of specific tweets, which we further
aggregate for the overall COVID-19 discussions. For specific emotions, we use
the NRC Word-Emotion Association Lexicon (Mohammad et al., 2013) that allows
us to determine the extent which a tweet displays the eight primary emotions
based on Plutchik theory (Plutchik, 1980). Figure 1 shows two screenshots of
the sentiment and emotion analysis components.

  * •

Topic Modelling: This component displays the various topics that were
discussed over time during the course of the pandemic. We used the popular LDA
(Blei et al., 2003) on the set of weekly tweets to determine the main topics
discussed on a weekly basis. LDA represents tweets as a random mixture over
latent topics and each topic is in turn represented by a set of words. Figure
2 (top) shows a screenshot of this topic modelling component.

  * •

Controversy Tracking: This component tracks and identifies tweets that include
various controversial terms, such as ”wuhan virus”, ”chinese virus”, ”kung
flu”, etc. These tweets are controversial in their content and designed to
incite hatred towards a specific group of people. Figure 2 (bottom) shows a
screenshot of this controversy tracking component.

  * •

Visualisation: This component provides a wide range of visualisation
techniques to best display the processed results from the above-mentioned
components. In addition, a user is able to drill-down and present these
analysis across different time periods and locations.

Figures 1 and 2 shows various example screenshots of our TweetCOVID system.
For a more interactive demonstration of these and additional features, we
refer interested readers to https://youtu.be/ut2EzLeuFwo for a video
demonstration.

Figure 2. Sample screenshots: Discussion topics by week (top), and top words
associated with controversial terms (bottom).

##  3\. Example Use Cases / Demonstration

Using the set of functionalities provided by TweetCOVID, we can utilize the
system for various use cases, which we describe with three examples. For a
more detailed discussion of the main findings, we refer interested readers to
our earlier paper (Kwan and Lim, 2020) that elaborates on various research
questions and findings related to the COVID-19 pandemic.

Use Case 1: Understanding public sentiments and emotions in relation to global
events. Using our system, we observe that public sentiments were noticeably
more negative on 12 March 2020 as compared to the weeks before and after.
Further analysis shows that anxiety and fear levels were heightened after the
WHO Director-General declared COVID-19 as a pandemic on 11 March 2020.
Thereafter, there was a corresponding increase in discussions relating to the
pandemic and the potential impact on the users’ daily life.

Use Case 2: Detecting adjacent topics during the course of the pandemic. Apart
from the COVID-19 pandemic, there were various global incidents that happened
concurrently with the pandemic. Our TweetCOVID system was able to pick up such
incidents in the form of discussion topics. For example, we detected the topic
of the Baltimore Shooting over 12 to 18 March 2020 and how certain users were
linking this incident back to COVID-19.

Use Case 3: Understanding the use of controversial terms during the pandemic.
Our proposed system also tracks the use of controversial terms, such as ”wuhan
virus”, ”chinese virus”, ”kung flu”, and the context they were used in. We
found that more than half of these usage were from the USA, with a majority
from political figures during March 2020. Additionally, the racist nature of
the controversial terms can be seen through the co-occurrence of terms such as
“chinaliedpeopledied”, “racist” and “chinamustexplain”. These terms suggest
misplaced blame upon people of Asian descent for the pandemic and further
perpetuates racism in countries.

##  4\. Conclusion

In this paper, we present a demonstration of the TweetCOVID system for
understanding the impact of COVID-19 on the general public using publicly
available tweets. TweetCOVID offers a range of functionalities, including data
collection/processing, general analytics, sentiment/emotion analysis, topic
modelling, controversy tracking and visualisation. We also demonstrate three
use cases that our system can be used for, including understanding
sentiments/emotions, detecting topics alongside COVID-19 and understanding
controversial terms used. Future work will include exploring other topic
models that are relevant for short text (Jónsson and Stolee, 2015).

###### Acknowledgements.

This research is funded in part by the Singapore University of Technology and
Design under grant SRG-ISTD-2018-140.

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