11institutetext: Shanghai Jiao Tong University, Shanghai, China
11email: {yuncao, heyohai, wnzhang<EMAIL_ADDRESS>
# Unsupervised Diverse Colorization via Generative Adversarial Networks

Yun Cao  Zhiming Zhou  Weinan Zhang  Yong Yu

###### Abstract

Colorization of grayscale images is a hot topic in computer vision. Previous
research mainly focuses on producing a color image to recover the original one
in a supervised learning fashion. However, since many colors share the same
gray value, an input grayscale image could be diversely colorized while
maintaining its reality. In this paper, we design a novel solution for
unsupervised diverse colorization. Specifically, we leverage conditional
generative adversarial networks to model the distribution of real-world item
colors, in which we develop a fully convolutional generator with multi-layer
noise to enhance diversity, with multi-layer condition concatenation to
maintain reality, and with stride 1 to keep spatial information. With such a
novel network architecture, the model yields highly competitive performance on
the open LSUN bedroom dataset. The Turing test on 80 humans further indicates
our generated color schemes are highly convincible.

##  1 Introduction

Image colorization assigns a color to each pixel of a target grayscale image.
Early colorization methods [15, 21] require users to provide considerable
scribbles on the grayscale image, which is apparently time-consuming and
requires expertise. Later research provides more automatic colorization
methods. Those colorization algorithms differ in the ways of how they model
the correspondence between grayscale and color.

Given an input grayscale image, non-parametric methods first define one or
more color reference images (provided by a human or retrieved automatically)
to be used as source data. Then, following the Image Analogies framework [10],
the color is transferred onto the input image from analogous regions of the
reference image(s) [24, 17, 9, 4]. Parametric methods, on the other hand,
learn prediction functions from large datasets of color images in the training
stage, posing the colorization problem as either regression in the continuous
color space [3, 6, 26] or classification of quantized color values [2], which
is a supervised learning fashion.

Whichever seeking the reference images or learning a color prediction model,
all above methods share a common goal, i.e. to provide a color image closer to
the original one. But as we know, many colors share the same gray value.
Purely from a grayscale image, one cannot tell what color of clothes a girl is
wearing or what color a bedroom wall is. Those methods all produce a
deterministic mapping function. Thus when an item could have diverse colors,
their models tend to provide a weighted average brownish color as pointed out
in [14] (See Figure 1 as an example).

Figure 1: Left: The original grayscale image. Middle: Image colorized by non-
adversarial CNNs. Right: Image colorized by human on Reddit. (Figure from
[14])

In this paper, to avoid this sepia-toned colorization, we use conditional
generative adversarial networks (GANs) [8] to generate diverse colorizations
for a single grayscale image while maintaining their reality. GAN is
originally proposed to generate vivid images from some random noise. It is
composed of two adversarial parts: a generative model $G$ that captures the
data distribution, and a discriminative model $D$ that estimates the
probability of whether an image is real or generated by $G$. The generator
part tries to map an input noise to a data distribution closer to the ground
truth data distribution, while the discriminator part tries to distinguish the
generated “fake” data, which comes to an adversarial situation. By careful
designation of both generative and discriminative parts, the generator will
eventually produce results, forming a distribution very close to the ground
truth distribution, and by controlling the input noise we can get various
results of good reality. Thus conditional GAN is a much more suitable
framework to handle diverse colorization than other CNNs. Meanwhile, as the
discriminator only needs the signal of whether a training instance is real or
generated, which is directly provided without any human annotation during the
training phase, the task is in an unsupervised learning fashion.

On the aspect of model designation, unlike many other conditional GANs [12]
using convolution layers as the encoder and deconvolution layers as the
decoder, we build a fully convolutional generator and each convolutional layer
is splinted by a concatenate layer to continuously render the conditional
grayscale information. Additionally, to maintain the spatial information, we
set all convolution stride to 1 to avoid downsizing data. We also concatenate
noise channels to the first half convolutional layers of the generator to
attain more diversity in the color image generation process. As the generator
$G$ would capture the color distribution, we can alter the colorization result
by changing the input noise. Thus we no longer need to train an additional
independent model for each color scheme like [3].

As our goal alters from producing the original colors to producing realistic
diverse colors, we conduct questionnaire surveys as a Turing test instead of
calculating the root mean squared error (RMSE) comparing the original image to
measure our colorization result. The feedback from 80 subjects indicates that
our model successfully produces high-reality color images, yielding more than
$62.6\%$ positive feedback while the rate of ground truth images is 70.0%.
Furthermore, we perform a significance $t$-test to compare the percentages of
human judges as real color images for each test instance (i.e. a real or
generated color image). The resulting $p$-value is $0.1359>0.05$, which
indicates that there is no significant difference between our generated color
images and the real ones. We share the repeatable experiment code for further
research111Experiment code is available at
https://github.com/ccyyatnet/COLORGAN..

##  2 Related Work

###  2.1 Diverse Colorization

The problem of colorization was proposed in the last century, but the research
of diverse colorization was not paid much attention until this decade. In [3],
they used additionally trained model to handle diverse colorization of a scene
image particularly in day and dawn. [26] posed the colorization problem as a
classification task and use class re-balancing at training time to increase
the colorfulness of the result. And in the work of [5], a low dimensional
embedding of color fields using a variational auto-encoder (VAE) is learned.
They constructed loss terms for the VAE decoder that avoid blurry outputs and
take into account the uneven distribution of pixel colors and finally
developed a conditional model for the multi-modal distribution between gray-
level image and the color field embeddings.

Compared with above work, our solution uses conditional generative adversarial
networks to achieve unsupervised diverse colorization in a generic way with
little domain knowledge of the images.

###  2.2 Conditional GAN

Generative adversarial networks (GANs) [8] have attained much attention in
unsupervised learning research during the recent 3 years. Conditional GANs
have been widely used in various computer vision scenarios. [22] used text to
generate images by applying adversarial networks. [12] provided a general-
purpose image-to-image translation model that handles tasks like label to
scene, aerial to map, day to night, edges to photo and also grayscale to
color.

Some of the above works may share a similar goal with us, but our conditional
GAN structure differs a lot from previous work in several architectural
choices mainly for the generator. Unlike other generators which employ an
encoder-like front part consisting of multiple convolution layers and a
decoder-like end part consisting of multiple deconvolution layers, our
generator uses only convolution layers all over the architecture, and does not
downsize data shape by applying convolution stride no more than 1 and no
pooling operation. Additionally, we add multi-layer noise to generate more
diverse colorization, while using multi-layer conditional information to keep
the generated image highly realistic.

##  3 Methods

###  3.1 Problem formulation

GANs are generative models that learn a mapping from random noise vector
$\bm{z}$ to an output color image $\bm{x}$: $G:\bm{z}\rightarrow\bm{x}$.
Compared with GANs, conditional GANs learn a mapping from observed grayscale
image $\bm{y}$ and random noise vector $\bm{z}$, to $\bm{x}$:
$G:\\{\bm{y},\bm{z}\\}\rightarrow\bm{x}$. The generator $G$ is trained to
produce outputs that cannot be distinguished from “real” images by an
adversarially trained discriminator $D$, which is trained with the aim of
detecting the “fake” images produced by the generator. This training procedure
is illustrated in Figure 2.

Figure 2: The illustration of conditional GAN. Generator $G$ is given
conditional information (Grayscale image) together with noise $\bm{z}$, and
produces generated color channels. Discriminator $D$ is trained over the real
color image and the generated color image by $G$. The goal of $D$ is to
distinguish real images from the fake ones. Both nets are trained
adversarially.

The objective of a GAN can be expressed as

$\displaystyle\mathcal{L}_{\text{GAN}}(G,D)=$
$\displaystyle\mathbb{E}_{\bm{x}\sim
P_{\text{data}}{(\bm{x})}}{[\log{D(\bm{x})}]}+$ (1)
$\displaystyle\mathbb{E}_{\bm{z}\sim
P_{\bm{z}}(\bm{z})}{[\log{(1-D(G(\bm{z})))}]},$

while the objective of a conditional GAN is

$\displaystyle\begin{aligned}
\mathcal{L}_{\text{cGAN}}(G,D)=&\mathbb{E}_{\bm{x}\sim
P_{\text{data}}{(\bm{x})}}{[\log{D(\bm{x})}]}+\\\ &\mathbb{E}_{\bm{y}\sim
P_{\text{gray}}(\bm{y}),\bm{z}\sim
P_{\bm{z}}(\bm{z})}{[\log{(1-D(G(\bm{y},\bm{z})))}]},\end{aligned}$ (2)

where $G$ tries to minimize this objective against an adversarial $D$ that
tries to maximize it, i.e.

$\displaystyle G^{*}=\arg\min_{G}\max_{D}{\mathcal{L}_{\text{cGAN}}(G,D)}.$
(3)

Without $\bm{z}$, the generator could still learn a mapping from $\bm{y}$ to
$\bm{x}$, but would produce deterministic outputs. That is why GAN is more
suitable for diverse colorization tasks than other deterministic neural
networks.

###  3.2 Architecture and implementation details

The high-level structure of our conditional GAN is consistent with traditional
ones [12, 5], while the detailed architecture of our generator $G$ differs a
lot.

####  3.2.1 Convolution or deconvolution

Convolution and deconvolution layers are two basic components of image
generators. Convolution layers are mainly used to exact conditional features.
And additionally, many researches [26, 12, 5] use superposition of multiple
convolution layers with stride more than 1 to downsize the data shape, which
works as a data encoder. Deconvolution layers are then used to upsize the data
shape as a decoder of the data representation [18, 12, 5]. While many other
researches share this encoder-decoder structure, we choose to use only
convolution layers in our generator $G$. Firstly, convolution layers are well
capable of feature extraction and transmission. Meanwhile, all the convolution
stride is set to 1 to prevent data shape from downsizing, thus the important
spatial information can be kept along the data flow till the final generation
layer. Some other researches [23, 12] also takes this spatial information into
consideration. They add skip connections between each layer $i$ and layer
$n-i$ to form a “U-Net” structure, where $n$ is the total number of layers.
Each skip connection simply concatenates all channels at layer $i$ with those
at layer $n-i$. Whether adding skip connections or not, the encoder-decoder
structure more tends to extract global features and generate images by this
overall information which is more suitable for global shape transformation
tasks. But in image colorization, we need a very detailed spatial local
guidance to make sure item boundaries will be accurately separated by
different color parts in generated channels. Let alone our modification is
more straightforward and easy to implement. See the structural difference
between “U-Net” and our convolution model in Figure 3.

Figure 3: Structure comparison of auto encoder, U-Net and our generator. Left:
Auto encoder with convolutional encoder part and deconvolutional decoder part.
Middle: U-Net structure [23, 12] with skip connections between layer $i$ and
$n-i$. Right: Our fully convolutional generator with continuous condition
concatenation

.

####  3.2.2 YUV or RGB

A color image can be represented in different forms. The most common
representation is $RGB$ form which splits a color pixel into red, green, blue
three channels. Most computer vision tasks use $RGB$ representation [6, 12,
19] due to its generality. Other kinds of representations are also included
like $YUV$ (or $YCrCb$) [3] and $Lab$ [2, 26, 16]. In colorization tasks, we
have grayscale image as conditional information, thus it is straightforward to
use $YUV$ or $Lab$ representation because the $Y$ and $L$ channel or so called
Luminance channel represents exactly the grayscale information. So while using
$YUV$ representation, we can just predict 2 channels and then concatenate with
the grayscale channel to give a full color image. Additionally, if you use
$RGB$ as image representation, all result channels are predicted, thus to keep
the grayscale of generated color image consistent with the original grayscale
image, we need to add an additional $L1$ loss as a controller to make sure
$Gray(G(\bm{y},\bm{z}))\simeq\bm{y}$:

$\mathcal{L}_{L1}(G)=\mathbb{E}_{\bm{y}\sim P_{\text{gray}}(\bm{y}),\bm{z}\sim
P_{\bm{z}}(\bm{z})}{[\|\bm{y}-Gray(G(\bm{y},\bm{z}))\|]},$ (4)

where for any color image $\bm{x}=(\bm{r},\bm{g},\bm{b})$, the corresponding
grayscale image (or the Luminance channel $Y$) can be calculated by the well-
known psychological formulation:

$\displaystyle Gray(\bm{x})=0.299\bm{r}+0.587\bm{g}+0.114\bm{b}.$ (5)

Note that Eq. (4) can still maintain good colorization diversity, because this
$L1$ loss term only lays a constraint on one dimension out of three channels.
Then the objective function will be modified to:

$G^{*}=\arg\min_{G}\max_{D}{\mathcal{L}_{\text{cGAN}}(G,D)+\lambda\mathcal{L}_{L1}(G)}.$
(6)

Since there is no equality constraint between the recovered grayscale
$Gray(G(\bm{y},\bm{z}))$ and the original one $\bm{y}$, the
$\mathcal{L}_{L1}(G)$ factor will normally be non-zero, which makes the
training unstable due to this additional trade-off. The results will be shown
in Section 4.2 with experimental comparison on both $RGB$ and $YUV$
representations.

####  3.2.3 Multi-layer noise

The authors in [12] mentioned noise ignorance while training the generator. To
handle this problem, they provide noise only in the form of dropout, applied
on several layers of the generator at both training and test time. We also
noticed this problem. Traditional GANs and conditional GANs receive noise
information at the very start layer, during the continuous data transformation
through the network, the noise information is attenuated a lot. To overcome
this problem and make the colorization results more diversified, we
concatenate the noise channel onto the first half of the generator layers (the
first three layers in our case). We conduct experimental comparison on both
one-layer noise and multi-layer noise representations, with results shown in
Section 4.2.

####  3.2.4 Multi-layer conditional information

Other conditional GANs usually add conditional information only in the first
layer, because the layer shape of previous generators changes along their
convolution and deconvolution layers. But due to the consistent layer shape of
our generator, we can apply concatenation of conditional grayscale information
throughout the whole generator layers which can provide sustained conditional
supervision. Though the “U-Net” skip structure of [12] can also help posterior
layers receive conditional information, our model modification is still more
straightforward and convenient.

####  3.2.5 Wasserstein GAN

The recent work of Wasserstein GAN [1] has acquired much attention. The
authors used Wasserstein distance to help getting rid of problems in original
GANs like mode collapse and gradient vanishing and provide a measurable loss
to indicate the progress of GAN training. We also try implementing Wasserstein
GAN modification into our model, but the results are no better than our model.
We make comparison between the results of Wasserstein GAN and our GAN in
Section 4.2.

The illustration of our model structure is shown in Figure 4.

Figure 4: Detailed structure of our conditional GAN. Top: Generator $G$. Each
cubic part represents a Convolution-BatchNorm-ReLU structure. Red connections
represent our modifications of the traditional conditional GANs. Bottom:
Discriminator $D$

.

###  3.3 Training and testing procedure

Algorithm 1 Training phase of our conditional GANs, with the default
parameters $k_{D}=1,k_{G}=1,m=64,s_{z}=100,s=64$.

1: for the number of training iterations do

2: for $k_{D}$ steps do

3: Generate minibatch of $m$ randomly sampled noise
$\\{\bm{z}^{(1)},\dots,\bm{z}^{(m)}\\}$ each of size $[s_{z}]$.

4: Sample minibatch of $m$ grayscale images
$\\{\bm{y}^{(1)},\dots,\bm{y}^{(m)}\\}$ each of shape $[s,s,1]$.

5: Get the corresponding minibatch of $m$ color images
$\\{\bm{x}^{(1)},\dots,\bm{x}^{(m)}\\}$ from data distribution
$p_{\text{data}}(\bm{x})$.

6: Update the discriminator $D$ by:
$\nabla_{\theta_{d}}\frac{1}{m}\sum_{i=1}^{m}[\log{D(\bm{x}^{(i)})}+\log{(1-D(G(\bm{y}^{(i)},\bm{z}^{(i)})))}]$

7: end for

8: for $k_{G}$ steps do

9: Generate minibatch of $m$ randomly sampled noise
$\\{\bm{z}^{(1)},\dots,\bm{z}^{(m)}\\}$

10: Sample minibatch of $m$ grayscale images
$\\{\bm{y}^{(1)},\dots,\bm{y}^{(m)}\\}$

11: Update the generator by:
$\nabla_{\theta_{g}}\frac{1}{m}\sum_{i=1}^{m}\log(1-D(G(\bm{y}^{(i)},\bm{z}^{(i)})))$

12: end for

13: end for

The training phase of our conditional GANs is presented in Algorithm 1. To
assure the BatchNorm layers to work correctly, one cannot feed an image batch
of the same images to test various noise responses. Thus we use multi-round
testing with same batch and rearrange them to test different noise responses
of each image, which is described in Algorithm 2.

Algorithm 2 Testing phase of our conditional GANs with the default parameters
$m=64,s_{z}=100$.

1: Sample minibatch of $m$ grayscale images
$\\{\bm{y}^{(1)},\dots,\bm{y}^{(m)}\\}$

2: for round $i$ in $k_{\text{test}}$ rounds do

3: Generate minibatch of $m$ randomly sampled noise
$\\{\bm{z}^{(1,i)},\dots,\bm{z}^{(m,i)}\\}$ each of size $[s_{z}]$.

4: Generate color image using the trained model $G$:
$\\{\bm{x}^{(1,i)},\dots,\bm{x}^{(m,i)}\\}\leftarrow
G(\\{\bm{y}^{(1)},\dots,\bm{y}^{(m)}\\},\\{\bm{z}^{(1,i)},\dots,\bm{z}^{(m,i)}\\})$

5: end for

6: Rearrange generated results $\bm{x}$
into$\\{\bm{x}^{(1,1)},\dots,\bm{x}^{(1,k_{\text{test}})}\\},\dots,\\{\bm{x}^{(m,1)},\dots,\bm{x}^{(m,k_{\text{test}})}\\}$

##  4 Experiments

###  4.1 Dataset

There are various kinds of color image datasets, and we choose the open LSUN
bedroom dataset222LSUN dataset is available at http://lsun.cs.princeton.edu.
[25] to conduct our experiment. LSUN is a large color image dataset generated
iteratively by human labeling with automatic deep neural classification. It
contains around one million labeled images for each of 10 scene categories and
20 object categories. Among them we choose indoor scene bedroom because it has
enough samples (more than 3 million) and unlike outdoor scenes in which trees
are almost always green and sky is always blue, items in indoor scenes like
bedroom have various colors, as shown in Figure 5. This is exactly what we
need to fully explore the capability of our conditional GAN model. In
experiments, we use $503,900$ bedroom images randomly picked from the LSUN
dataset. We crop a maximum center square out of each image and reshape them
into $64\times 64$ pixel as preprocessing.

Figure 5: Demonstration of LSUN dataset. Top: Outdoor scene (church). Always
blue sky and green trees. Bottom: Indoor scene (bedroom). Various item colors.

###  4.2 Comparison Experiments

####  4.2.1 YUV and RGB

The generated colorization results of a same grayscale image using $YUV$
representation and $RGB$ representation with additional $L1$ loss are shown in
Figure 6. Each group of images consists of $3\times 4$ images generated from a
same grayscale image by each model at a same epoch. Focus on the results in
red boxes, we can see $RGB$ representation suffers from structural miss due to
the additional trade-off between $L1$ loss and the GAN loss. Take the enlarged
image on the top right in Figure 6 as an example, both the wall and the bed on
the left are split by unnaturally white and orange colors, while the results
of $YUV$ setting acquire more smooth transitions. Moreover, the model using
$RGB$ representation shall predict 3 color channels while $YUV$ representation
only predicts 2 channels given the grayscale $Y$ channel as fixed conditional
information, which makes the $YUV$ model training much more stable.

Figure 6: Comparison of different color space representation. Top: training
and testing with $YUV$ representation. Bottom: training with $RGB$
representation and $L1$ loss. Focus on the results in red boxes, $RGB$
representation results lack of item continuity due to the additional trade-off
between $L1$ loss and the GAN loss.

####  4.2.2 Single-layer and multi-layer noise

The generated colorization results of the same grayscale images using single-
layer noise model and multi-layer noise model are shown in Figure 7. Each
group consists of $8\times 8$ images generated from a grayscale image by each
model at the same epoch. We can see from the results that multi-layer noise
possesses our generator $G$ of higher diversity as those results on the right
in Figure 7 are apparently more colorful.

Figure 7: Comparison of single-layer and multi-layer noise model results.
Left: results of single-layer noise model. Right: results of multi-layer noise
model. Apparently multi-layer noise possesses our generator $G$ of higher
diversity. Figure 8: Comparison of single-layer and multi-layer condition
model result. Top: results of single-layer condition model, suffer from
colorization derivation (red box). Bottom: results of multi-layer condition
model, smooth transition.

####  4.2.3 Single-layer and multi-layer condition

The generated colorization results of the same grayscale images using single-
layer conditional model and multi-layer conditional model are shown in Figure
8. We show $2\times 5$ images generated by single-layer condition setting and
multi-layer condition setting at the same epoch. We can see from the results
that the multi-layer condition model makes the generator more structural
information and thus the results of multi-layer condition model are more
stable while the single-layer conditional model suffers from colorization
derivation like those images in red boxes in Figure 8.

####  4.2.4 Wasserstein GAN

Three groups of colorization results of the same grayscale images using GAN
and Wasserstein GAN are shown in Figure 9. We can see from the result that
Wasserstein GAN can produce comparable results as the first two column of
Wasserstein GAN shows, but there are still failed results by Wasserstein GAN
like the last column, even to note that the Wasserstein GAN results(40 epoch)
come after training twice the number of epochs of the GAN results(20 epoch).
This is mainly due to that our training LSUN bedroom dataset is quite large
(503,900 image), the discriminator will not overfit easily, which prevents the
gradient vanishing problem. And also because of the large dataset, the
discriminator needs quite a lot times of optimization to convergence, not to
mention Wasserstein GAN shall not use momentum based optimization strategies
like Adam due to the non-linear parameter value clipping, Wasserstein GAN
needs much longer training to produce comparable results as our model. Since
Wasserstain GAN only helps to improve the stability of GAN training at a price
of much longer training time and we have achieved results of good reality
through our GAN, we did not use Wassserstein GAN structure.

Figure 9: Comparison between the results of GAN and Wasserstein GAN. Each line
consists of the leftmost ground truth color image and three results by GAN,
then three results by Wasserstein GAN. The rightmost images are failed results
by Wasserstein GAN.

More results and discussion of our final model will be shown in the next
section.

##  5 Results and Evaluation

###  5.1 Colorization Results

A variety of image colorization results by our conditional GANs are provided
in Figure 10. Apparently, our fully convolutional (without stride) generator
with multi-layer noise and multi-layer condition concatenation produces
various kinds of colorization schemes while maintaining good reality. Almost
all color parts are kept within correct components without deviation.

###  5.2 Evaluation via Human Study

Previous methods share a common goal to provide a color image closer to the
original one. That is why many of their models [13, 20, 5] take image distance
like RMSE (Root Mean Square Error) and PSNR (Peak Signal-to-Noise Ratio) as
their measurements. And others [11, 12] use additional classifiers to predict
if colorized image can be detected or still correctly classified. But our goal
is to generate diverse colorization schemes, so we cannot take those distance
as our measurements as there exist reasonable colorizations that diverge a lot
from the original color image. Note that some previous work on image
colorization [3, 7, 19, 18] does not provide quantified measurements.

Therefore, just like some previous researches [12, 26], we provide
questionnaire surveys as a Turing test to measure our colorization results. We
ask each of the total 80 participants 20 questions. In each question, we
display 5 color images, one of which is the ground truth image, the others are
our generated colorizations of the grayscale image of the ground truth, and
ask them if any one of them is of poor reality. We add ground truth image
among them as a reference in case of participants do not think any of them is
real. And we arrange all images randomly to avoid any position bias for
participants. The feedback from 80 participants indicates more than 62.6% of
our generated color images are convincible while the rate of ground truth
images is 70.0%. Furthermore, we run significance $t$-test between the ground
truth and the generated images on the percentages of humans rating as real
image for each question. The $p$-value is $0.1359>0.05$, indicating our
generated results have no significant difference with the ground truth images.
Also we calculate the credibility rank within each group of the ground truth
image and the four corresponding generated images. An image gets higher rank
if higher percentage of participants mark it real. And the average credibility
rank of the ground truth images is only 2.5 out of 5, which means at least
$(2.5-1)/(5-1)=37.5\%$ of our generated results are even more convincible than
true images.

##  6 Conclusion

In this paper, we proposed a novel solution to automatically generate diverse
colorization schemes for a grayscale image while maintaining their reality by
exploiting conditional generative adversarial networks which not only solved
the sepia-toned problem of other models but also enhanced the colorization
diversity. We introduced a novel generator architecture which consists of
fully convolutional non-stride structure with multi-layer noise to enhance
diversity and multi-layer condition concatenation to maintain reality. With
this structure, our model successfully generated diversified high-quality
color images for each input grayscale image. We performed a questionnaire
survey as a Turing test to evaluate our colorization result. The feedback from
80 participants indicates our generated colorization results are highly
convincible.

For future work, as so far we have investigated methods to generate color
images by conditional GAN given only corresponding grayscale images, which
provides the model maximum freedom to generate all kinds of colors, we can
also lay additional constraints on the generator to guide the colorization
procedure. Those conditions include but not limited to (i) specified item
color, such as blue bed and white wall etc.; and (ii) global color scheme,
such as warm tone or cool tone etc. And note that given those constraints,
generative adversarial networks shall still produce various vivid
colorizations.

Figure 10: Example results of our conditional GAN on LSUN bedroom data. 20
groups of images, each consists of the leftmost ground truth color image and 4
different colorizations generated by our conditional GANs given the grayscale
version of the ground truth image. One can clearly see that our novel
structure generator produces various colorization schemes while maintaining
good reality.

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#  $L^{2}$-blowup estimates of the plate equation

Ryo Ikehata
Department of Mathematics
Division of Educational Sciences
Graduate School of Humanities and Social Sciences
Hiroshima University
Higashi-Hiroshima 739-8524, Japan<EMAIL_ADDRESS>
###### Abstract

We consider the Cauchy problems in ${\bf R}^{n}$ for the plate equation with a
weighted $L^{1}$-initial data. We derive optimal estimates of the $L^{2}$-norm
of solutions for $n=1,2,3,4$. In particular, such obtained results express
infinite time blowup property in the case when the $0$-th moment of the
initial velocity does not vanish. The idea to derive them is strongly inspired
from an already developed technique [6, 11, 7].

##  1 Introduction

000Keywords and Phrases: Plate equation; weighted $L^{1}$-data; low dimension;
blowup in infinite time; sharp estimates.0002010 Mathematics Subject
Classification. Primary 35L05; Secondary 35B40, 35C20.

This is one more application to the plate and/or beam equation of the method
recently applied in [7] to the wave equation. A similar arugument to these
wave-like estimates to capture a singularity is introduced in [6, 11], and
their related papers are published by the author’s collaborative works (see
[3] and the references therein).
Now, we consider the Cauchy problem of the plate equation:

$\displaystyle u_{tt}+\Delta^{2}u=0,\ \ \ (t,x)\in(0,\infty)\times{\bf
R}^{n},$ (1.1) $\displaystyle u(0,x)=u_{0}(x),\quad u_{t}(0,x)=u_{1}(x),\
x\in{\bf R}^{n}.$ (1.2)

Here, we assume, for the moment, $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times
L^{2}({\bf R}^{n})$.
Concerning the existence of a unique energy solution to problem (1.1)-(1.2),
by the standard semi-group theory one can find that the problem (1.1)-(1.2)
has a unique weak solution

$u\in C([0,\infty);H^{2}({\bf R}^{n}))\cap C^{1}([0,\infty);L^{2}({\bf
R}^{n}))$

satisfying the energy conservation law such that

$E(t)=E(0),\quad t\geq 0,$ (1.3)

where the total energy $E(t)$ for the solution to problem (1.1)-(1.2) can be
defined by

$E(t):=\frac{1}{2}\left(\|u_{t}(t,\cdot)\|^{2}+\|\Delta
u(t,\cdot)\|^{2}\right).$

Here, $\|u\|$ denotes the usual $L^{2}$-norm of $u\in L^{2}({\bf R}^{n})$.

As is frequently appeared in several research papers on the plate and/or beam
models, the assumption $n\geq 5$ is sometimes imposed to treat the problems in
unbounded domains, and for this observation one can cite several typical
papers such as [1], [2], [12], [13], [15], [16] and [14], however, it seems
that there are no any related papers to investigate actively a reason why
$n\geq 5$ (see also [18] for a topic on the $L^{p}$-regularity condition). The
purpose of this paper is to give a partial answer on this kind of problem. In
this connection, in [17, 19] several decay estimates for the quantity
$\|(u_{t}(t,\cdot),\Delta u(t,\cdot))\|_{q}$ can be studied in order to apply
them to nonlinear problems. Our main concern is in observing a singularity
near small frequency region, so we are particular about dealing with the
asymptotic behavior of the quantity $\|u(t,\cdot)\|$. By the way, it should be
mentioned that the equation considered in [13] does not have any singularity
near $0$ frequency region, so a topic taken up in [13] seems to be a little
different from ours.

Before going to introduce our Theorems, we set

$I_{0,n}:=\|u_{0}\|+\|u_{0}\|_{L^{1}({\bf
R}^{n})}+\|u_{1}\|+\|u_{1}\|_{L^{1}({\bf R}^{n})}.$

The following two theorems give a hint about a question why $n\geq 5$ in the
plate equation.

###### Theorem 1.1

Let $n=1,2,3$. Let $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times L^{2}({\bf
R}^{n})$. Then, the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies the
following properties under the additional regularity on the initial data:

$[u_{0},u_{1}]\in L^{1}({\bf R}^{n})\times L^{1}({\bf
R}^{n})\quad\Rightarrow\quad\|u(t,\cdot)\|\leq C_{1}I_{0,n}t^{1-\frac{n}{4}},$
$[u_{0},u_{1}]\in L^{1}({\bf R}^{n})\times L^{1,1}({\bf
R}^{n})\quad\Rightarrow\quad C_{2}\left|\int_{{\bf
R}^{n}}u_{1}(x)dx\right|t^{1-\frac{n}{4}}\leq\|u(t,\cdot)\|$

for $t\gg 1$, where $C_{j}>0$ ($j=1,2$) are constants depending only on $n$.

Our next result is the case of $n=4$.

###### Theorem 1.2

Let $n=4$. Let $[u_{0},u_{1}]\in H^{2}({\bf R}^{4})\times L^{2}({\bf R}^{4})$.
Then, the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies the following
properties under the additional regularity on the initial data:

$[u_{0},u_{1}]\in L^{1}({\bf R}^{4})\times L^{1}({\bf
R}^{4})\quad\Rightarrow\quad\|u(t,\cdot)\|\leq C_{1}I_{0,4}\sqrt{\log t},$
$[u_{0},u_{1}]\in L^{1}({\bf R}^{4})\times L^{1,1}({\bf
R}^{4})\quad\Rightarrow\quad C_{2}\left|\int_{{\bf
R}^{4}}u_{1}(x)dx\right|\sqrt{\log t}\leq\|u(t,\cdot)\|$

for $t\gg 1$, where $C_{j}>0$ ($j=1,2$) are constants depending only on the
space dimension $n=4$.

###### Remark 1.1

One of our advantages in the results is that we never use the compactness
argument as is frequently developed in the wave equation case because of the
non-Kowalewskian property of the equation (1.1). The method will be widely
applicable to the other types of evolution equations. Indeed, it will be
possible to generalize our results to the more general $\sigma$-evolution
equations:

$u_{tt}+(-\Delta)^{\sigma}u=0,$

where $((-\Delta)^{\sigma}f)(x):={\cal F}_{\xi\to
x}^{-1}(|\xi|^{2\sigma}\hat{f}(\xi))(x)$ and $\sigma>0$. A threshold number
$n^{*}$ on the dimension $n$ to divide whether the blow up phenomenon occurs
or not can be defined by $n^{*}=2\sigma$.

###### Remark 1.2

In both Theorems, the so-called infinite time blowup property can be observed
in the case when $n<5$. One has to treat more delicately when we study decay
estimates, asymptotic profiles, nonlinear problems and so on, of the plate
equation considered in unbounded domains, and in particular, in ${\bf R}^{n}$.

Contrary to infinite time blowup results above, in the case when $\int_{{\bf
R}^{n}}u_{1}(x)dx=0$, one has the $L^{2}$-boundedness property (at least) for
$n=3,4$.

###### Theorem 1.3

Let $n=3,4$, and $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times L^{2}({\bf
R}^{n})$. Assume further that $u_{1}\in L^{1,\gamma}({\bf R}^{n})$ with

$n=4\quad\Rightarrow\quad\gamma\in(0,1],$
$n=3\quad\Rightarrow\quad\gamma\in(\frac{1}{2},1].$

If

$\int_{{\bf R}^{n}}u_{1}(x)dx=0,$

then the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies

$\|u(t,\cdot)\|\leq C(\|u_{0}\|+\|u_{1}\|+\|u_{1}\|_{1,\gamma}),$

where $C>0$ is a constant depending on the space dimension $n$ and $\gamma$.

In the case of $n=1,2$, if one assumes an additional vanishing condition of
the $1$th-order moment of the initial velocity $u_{1}$ one can state the
following $L^{2}$-boundedness result.

###### Theorem 1.4

Let $n=1,2$, and $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times L^{2}({\bf
R}^{n})$. Assume further that $u_{1}\in L^{1,\gamma}({\bf R}^{n})$ with

$n=2\quad\Rightarrow\quad\gamma\in(1,2],$
$n=1\quad\Rightarrow\quad\gamma\in(\frac{3}{2},2].$

Under the following two conditions,

$\int_{{\bf R}^{n}}u_{1}(x)dx=0,\qquad\int_{{\bf
R}^{n}}x_{j}u_{1}(x)dx=0\quad(j=1,2,\cdots,n),$ (1.4)

the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies

$\|u(t,\cdot)\|\leq C(\|u_{0}\|+\|u_{1}\|+\|u_{1}\|_{1,\gamma}),$

where $C>0$ is a constant depending on the space dimension $n$ and $\gamma$.

Example. Let $n=1$, and let $g\in C_{0}^{\infty}({\bf R})$ be an odd function,
and choose $u_{1}(x):=g^{\prime}(x)\in C_{0}^{\infty}({\bf R})$. Then, one has

$\int_{{\bf R}}u_{1}(x)dx=0,\quad\int_{{\bf R}}xu_{1}(x)dx=\int_{{\bf
R}}xg^{\prime}(x)dx=-\int_{{\bf R}}g(x)dx=0.$

This expresses an example of the initial velocity satisfying (1.4) for $n=1$.
Similarly, in the case when $n=2$, one can construct an example satsfying
(1.4) by

$u_{1}(x_{1},x_{2}):=g^{\prime}(x_{1})h^{\prime}(x_{2}),$

where $g,h\in C_{0}^{\infty}({\bf R})$.
As a counter part of Theorem 1.4 one can get the infinite time blowup result
in the case when $n=1,2$. For simplicity, we mention only the case of
$\gamma=2$ in order to compare it with Theorem 1.4..

###### Theorem 1.5

(1) Let $n=1$, and $[u_{0},u_{1}]\in H^{2}({\bf R})\times L^{2}({\bf R})$.
Assume further that $u_{1}\in L^{1,2}({\bf R})$ satisfies

$\left|\int_{{\bf R}}u_{1}(x)dx\right|+\left|\int_{{\bf
R}}xu_{1}(x)dx\right|>0.$ (1.5)

Then the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies

$C\left(\left|\int_{{\bf R}}xu_{1}(x)dx\right|t^{\frac{1}{4}}+\left|\int_{{\bf
R}}u_{1}(x)dx\right|t^{\frac{3}{4}}\right)\leq\|u(t,\cdot)\|,\quad t\gg 1,$

where $C>0$ is a constant.
(2) Let $n=2$, and $[u_{0},u_{1}]\in H^{2}({\bf R}^{2})\times L^{2}({\bf
R}^{2})$. Assume further that $u_{1}\in L^{1,2}({\bf R}^{2})$ satisfies

$\left|\int_{{\bf R}^{2}}u_{1}(x)dx\right|+\left|\int_{{\bf
R}^{2}}x_{1}u_{1}(x)dx\right|+\left|\int_{{\bf
R}^{2}}x_{2}u_{1}(x)dx\right|>0.$ (1.6)

Then the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies

$C\left(\left|\int_{{\bf R}^{2}}x_{1}u_{1}(x)dx\right|+\left|\int_{{\bf
R}^{2}}x_{2}u_{1}(x)dx\right|\right)\sqrt{\log t}+C\left|\int_{{\bf
R}^{2}}u_{1}(x)dx\right|\sqrt{t}\leq\|u(t,\cdot)\|,\quad t\gg 1,$

where $C>0$ is a constant.

###### Remark 1.3

As a result of Theorem 1.5, (for examle) in case of $n=1$, even if $\int_{{\bf
R}}u_{1}(x)dx=0$, if $\int_{{\bf R}}xu_{1}(x)dx\neq 0$, then one has the
infinite time blowup result with its rate $t^{1/4}$, however one can know
nothing about the optimality of the blowup rate $t^{1/4}$. A corresponding
result for $n=2$ is true similarly. Of course, one should investigate
(optimal) upper bound estimates of Theorem 1.5, however, this study will be
left to the reader’s interest.

Notation. Throughout this paper, $\|\cdot\|_{q}$ stands for the usual
$L^{q}({\bf R}^{n})$-norm. For simplicity of notation, in particular, we use
$\|\cdot\|$ instead of $\|\cdot\|_{2}$. We also introduce the following
weighted functional spaces.

$L^{1,\gamma}({\bf R}^{n}):=\left\\{f\in L^{1}({\bf
R}^{n})\;\bigm{|}\;\|f\|_{1,\gamma}:=\int_{{\bf
R}^{n}}(1+|x|^{\gamma})|f(x)|dx<+\infty\right\\}.$

One denotes the Fourier transform ${\cal F}_{x\to\xi}(f)(\xi)$ of $f(x)$ by

${\cal F}_{x\to\xi}(f)(\xi)=\hat{f}(\xi):=\displaystyle{\int_{{\bf
R}^{n}}}e^{-ix\cdot\xi}f(x)dx,\quad\xi\in{\bf R}^{n},$

as usual with $i:=\sqrt{-1}$, and ${\cal F}_{\xi\to x}^{-1}$ expresses its
inverse Fourier transform. Finally, we denote the surface area of the
$n$-dimensional unit ball by
$\omega_{n}:=\displaystyle{\int_{|\omega|=1}}d\omega$, and we set
$x=(x_{1},x_{2},\cdots,x_{n})\in{\bf R}^{n}$.

The paper is organized as follows. In Section 2 we try to get $L^{2}$-bound of
solutions via known method due to [9]. In Section 3 we derive the lower bound
estimates of the $L^{2}$-norm of solutions, and in Section 4 we obtain the
upper bound estimates of the $L^{2}$-norm of solutions, and by combining the
results obtained in Sections $3$ and $4$ one can prove Theorems 1.1 and 1.2.
Finally, we prove Theorems 1.3, 1.4 and 1.5 in Section 5.

##  2 $L^{2}$-upper bound estimates: higher dimensional case

In this section, we introduce a device to derive $L^{2}$-estimates of
solutions to problem (1.1)-(1.2) by relying on the modified Morawetz method
developed newly in [9]. For this we prepare the following Hardy type
inequality (see Davies-Hinz [4]). The basic concept in [9] is that the Hardy
type inequality implies the $L^{2}$-upper bound estimate of the solution in
several types of linear evolution equations.

###### Lemma 2.1

Let $n\geq 5$. Then there exists a constant $C^{*}>0$ such that

$\int_{{\bf R}^{n}}\frac{|w(x)|^{2}}{|x|^{4}}dx\leq C^{*}\int_{{\bf
R}^{n}}|\Delta w(x)|^{2}dx$

for all $w\in H^{2}({\bf R}^{n})$.

As in the idea [9], for the solution $u(t,x)$ to problem (1.1)-(1.2) we set

$v(t,x):=\int_{0}^{t}u(s,x)ds.$

Then, the function $v(t,x)$ satisfies the following equation and initial data.

$\displaystyle v_{tt}+\Delta^{2}v=u_{1},\ \ \ (t,x)\in(0,\infty)\times{\bf
R}^{n},$ (2.1) $\displaystyle v(0,x)=0,\quad v_{t}(0,x)=u_{0}(x),\ x\in{\bf
R}^{n}.$ (2.2)

Multiplying the both sides of (2.1)-(2.2) by $v_{t}$, and integrating it over
$[0,t]\times{\bf R}^{n}$ one can get the following energy equality such that

$\frac{1}{2}\|v_{t}(t,\cdot)\|^{2}+\frac{1}{2}\|\Delta
v(t,\cdot)\|^{2}=\frac{1}{2}\|v_{0}\|^{2}+\int_{{\bf R}^{n}}u_{1}(x)v(t,x)dx.$
(2.3)

Now we are in a position to use Lemma 2.1 to proceed the following
computations based on the Schwarz inequality:

$\left|\int_{{\bf R}^{n}}u_{1}(x)v(t,x)dx\right|\leq\int_{{\bf
R}^{n}}|u_{1}(x)||v(t,x)|dx=\int_{{\bf
R}^{n}}\left(|x|^{2}|u_{1}(x)|\right)\left(\frac{|v(t,x)|}{|x|^{2}}\right)dx$
$\leq\left(\int_{{\bf
R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx\right)^{1/2}\left(\int_{{\bf
R}^{n}}\frac{|v(t,x)|^{2}}{|x|^{4}}dx\right)^{1/2}$ $\leq
C_{\varepsilon}\int_{{\bf R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx+\varepsilon\int_{{\bf
R}^{n}}\frac{|v(t,x)|^{2}}{|x|^{4}}dx$ $\leq C_{\varepsilon}\int_{{\bf
R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx+\varepsilon C^{*}\int_{{\bf R}^{n}}|\Delta
v(t,x)|^{2}dx$ (2.4)

with some constant $C_{\varepsilon}>0$, which depends on each $\varepsilon>0$.
Thus, from (2.3) and (2.4) it follows that

$\frac{1}{2}\|v_{t}(t,\cdot)\|^{2}+\frac{1}{2}\|\Delta
v(t,\cdot)\|^{2}\leq\frac{1}{2}\|u_{0}\|^{2}+C_{\varepsilon}\int_{{\bf
R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx+\varepsilon C^{*}\int_{{\bf R}^{n}}|\Delta
v(t,x)|^{2}dx,$

which implies

$\frac{1}{2}\|v_{t}(t,\cdot)\|^{2}+\left(\frac{1}{2}-\varepsilon
C^{*}\right)\|\Delta
v(t,\cdot)\|^{2}\leq\frac{1}{2}\|u_{0}\|^{2}+C_{\varepsilon}\int_{{\bf
R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx$

provided that

$\int_{{\bf R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx<+\infty.$

By choosing $\varepsilon>0$ small enough one can arrive at the crucial
$L^{2}$-estimate by means of the multiplier method only because of
$v_{t}(t,x)=u(t,x)$. The result below compensates the unknown information on
$n\geq 5$.

###### Proposition 2.1

Let $n\geq 5$, and $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times L^{2}({\bf
R}^{n})$. Then, there exists a constant $C>0$ such that the solution $u(t,x)$
to problem (1.1)-(1.2) satisfies

$\|u(t,\cdot)\|^{2}\leq C\left(\|u_{0}\|^{2}+\int_{{\bf
R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx\right)\quad t\geq 0,$

provided additionally that

$\int_{{\bf R}^{n}}|x|^{4}|u_{1}(x)|^{2}dx<+\infty.$ (2.5)

###### Remark 2.1

An essential part of this argument above has already been developed previously
in [2] for the plate equation with damping and the rotational inertia term by
employing the idea due to [9]. This part is just a trial to observe the case
of $n\geq 5$.

###### Remark 2.2

Note that in the case of $4\leq n$, the condition (2.5) together with
$u_{1}\in L^{2}({\bf R}^{n})$ does not necessarily imply $u_{1}\in L^{1}({\bf
R}^{n})$.

So, one can apply the method due to [9] to get the $L^{2}$-bound in the case
when $n\geq 5$, however, this method decisively depends on the existence of
the Hardy type inequality. So, the results for $n=1,2,3,4$ of Theorems 1.1 and
1.2 are quite important since those are independent of any such inequalities.
We have to derive $L^{2}$-bound estimates from the equation itself.
One can proceed the similar argument by using the following inequality in
place of Lemma 2.1 (see [2] for its detail).

###### Lemma 2.2

Let $n\geq 5$. Then there exists a constant $C^{*}>0$ such that

$\|w\|_{L^{\frac{2n}{n-4}}}\leq C^{*}\|\Delta w\|$

for all $w\in H^{2}({\bf R}^{n})$.

##  3 $L^{2}$-lower bound estimates

In this section, we derive the lower bound estimates of the $t$-function
$\|u(t,\cdot)\|$ based on the Plancherel Theorem combined with low frequency
estimates (cf. [6]). For the proof, it suffices to assume $[u_{0},u_{1}]\in
C_{0}^{\infty}({\bf R}^{n})\times C_{0}^{\infty}({\bf R}^{n})$ to have
sufficient regularity of the solution $u(t,x)$ because the density argument
can be applied in the final estimates.
We first set

$L:=\sup_{\theta\neq 0}\left|\frac{\sin\theta}{\theta}\right|<+\infty.$ (3.1)

On the other hand, because of

$\lim_{\theta\to+0}\frac{\sin\theta}{\theta}=1,$

there exists a real number $\delta_{0}\in(0,1)$ such that

$\left|\frac{\sin\theta}{\theta}\right|\geq\frac{1}{2}$ (3.2)

for all $\theta\in(0,\delta_{0}^{2}]$. The following fundamental inequality
will be useful, too:

$|a+b|^{2}\geq\frac{1}{2}|a|^{2}-|b|^{2}$ (3.3)

for all $a,b\in{\bf C}$.
In order to get the lower bound estimate for the quantity $\|u(t,\cdot)\|$, it
suffices to treat $\|w(t,\cdot)\|$ with $w(t,\xi):={\cal
F}_{x\to\xi}(u(t,\cdot))(\xi)=\hat{u}(t,\xi)$ because of the Plancherel
Theorem.

Now we decompose the quantity $\|w(t,\cdot)\|$ as follows: for each $n\geq 1$

$\|w(t,\cdot)\|^{2}=\left(\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}+\int_{|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}\right)|w(t,\xi)|^{2}d\xi=I_{low}^{(n)}(t)+I_{high}^{(n)}(t).$
(3.4)

Here we have just chosen $t>0$ large enough such that

$\frac{\delta_{0}}{\sqrt{t}}\leq 1.$

By the way, in the Fourier space ${\bf R}_{\xi}^{n}$ the problem (1.1)-(1.2)
and its solution $u(t,x)$ can be transformed into the following ODE with
parameter $\xi\in{\bf R}_{\xi}^{n}$:

$\displaystyle w_{tt}+|\xi|^{4}w=0,\ \ \ t>0,\quad\xi\in{\bf R}_{\xi}^{n},$
(3.5) $\displaystyle w(0,\xi)=w_{0}(\xi),\quad w_{t}(0,\xi)=w_{1}(\xi),\ \ \
\xi\in{\bf R}^{n},$ (3.6)

where $w_{0}(\xi):=\hat{u}_{0}(\xi)$ and $w_{1}(\xi):=\hat{u}_{1}(\xi)$. In
addition, one can solve the problem (3.5)-(3.6) as follows:

$w(t,\xi)=\frac{\sin(t|\xi|^{2})}{|\xi|^{2}}w_{1}(\xi)+\cos(t|\xi|^{2})w_{0}(\xi).$
(3.7)

Now, let us derive the lower bound estimates for $I_{low}^{(n)}(t)$ because
one has $\|w(t,\cdot)\|^{2}\geq I_{low}^{(n)}(t)$. One relies on a device
coming from an idea in [6]. Indeed, it follows from (3.7) and (3.3) that

$I_{low}^{(n)}(t)=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\left|\frac{\sin(t|\xi|^{2})}{|\xi|^{2}}w_{1}(\xi)+\cos(t|\xi|^{2})w_{0}(\xi)\right|^{2}d\xi$
$\geq\frac{1}{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi-\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\cos^{2}(t|\xi|^{2})|w_{0}(\xi)|^{2}d\xi$
$=:\frac{1}{2}J_{1}(t)-J_{2}(t).$ (3.8)

Let us first estimate $J_{1}(t)$ by using the decomposition of the initial
data $w_{1}(\xi)$ in the Fourier space:

$w_{1}(\xi)=P+(A(\xi)-iB(\xi)),\quad\xi\in{\bf R}_{\xi}^{n},$

where

$P:=\int_{{\bf R}^{n}}u_{1}(x)dx,$ $A(\xi):=\int_{{\bf
R}^{n}}(\cos(x\xi)-1)u_{1}(x)dx,\quad B(\xi):=\int_{{\bf
R}^{n}}\sin(x\xi)u_{1}(x)dx.$

It is known (see [5]) that there is a constant $M>0$ such that

$|A(\xi)-iB(\xi)|\leq M|\xi|\|u_{1}\|_{L^{1,1}},\quad\xi\in{\bf R}_{\xi}^{n},$
(3.9)

in case of $u_{1}\in L^{1,1}({\bf R}^{n})$. Then, from (3.3) we see that

$J_{1}(t)=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi$
$\geq\frac{P^{2}}{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}d\xi-\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}|A(\xi)-iB(\xi)|^{2}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}d\xi$
$=\frac{P^{2}}{2}K_{1}(t)-K_{2}(t).$ (3.10)

$K_{2}(t)$ can be estimated from above by using and (3.1) and (3.9):

$K_{2}(t)\leq
M^{2}\|u_{1}\|_{L^{1,1}}^{2}\omega_{n}\int_{0}^{\frac{\delta_{0}}{\sqrt{t}}}t^{2}\left(\frac{\sin(t|\xi|^{2})}{t|\xi|^{2}}\right)^{2}|\xi|^{2}d\xi$
$\leq
M^{2}\|u_{1}\|_{L^{1,1}}^{2}L^{2}t^{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}|\xi|^{2}d\xi=M^{2}\|u_{1}\|_{L^{1,1}}^{2}\omega_{n}L^{2}t^{2}\int_{0}^{\frac{\delta_{0}}{\sqrt{t}}}r^{n+1}dr$
$=\frac{L^{2}}{n+2}M^{2}\omega_{n}\delta_{0}^{n+2}\|u_{1}\|_{L^{1,1}}^{2}t^{1-\frac{n}{2}},\quad
t\gg 1.$ (3.11)

On the other hand, one can obtain the lower bound estimate for $K_{1}(t)$
because of (3.2):

$K_{1}(t)=t^{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{(t|\xi|^{2})^{2}}d\xi\geq\frac{t^{2}}{4}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}d\xi$
$=\frac{\omega_{n}\delta_{0}^{n}}{4n}t^{2-\frac{n}{2}},\quad t\gg 1.$ (3.12)

Therefore from (3.10), (3.11) and (3.12) one can get the estimate from below
for $J_{1}(t)$:

$J_{1}(t)\geq\frac{P^{2}}{2}\frac{\omega_{n}\delta_{0}^{n}}{4n}t^{2-\frac{n}{2}}-\frac{L^{2}}{n+2}M^{2}\omega_{n}\delta_{0}^{n+2}\|u_{1}\|_{L^{1,1}}^{2}t^{1-\frac{n}{2}}\quad
t\gg 1.$ (3.13)

Since the upper bound estimate of $J_{2}(t)$ can be easily obtained as
follows:

$J_{2}(t)\leq\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}|w_{0}(\xi)|^{2}d\xi\leq\|u_{0}\|_{L^{1}}^{2}\omega_{n}\int_{0}^{\frac{\delta_{0}}{\sqrt{t}}}r^{n-1}dr=\frac{\omega_{n}\delta_{0}^{n}}{n}\|u_{0}\|_{L^{1}}^{2}t^{-\frac{n}{2}},$

because of (3.13), (3.4) and (3.8) one has just arrived at the following lower
bound estimate for $\|w(t,\cdot)\|$:

$\|w(t,\cdot)\|^{2}\geq
I_{low}^{(n)}(t)\geq\frac{P^{2}}{4}\frac{\omega_{n}\delta_{0}^{n}}{4n}t^{2-\frac{n}{2}}-\frac{M^{2}}{n}\omega_{n}\delta_{0}^{n}\|u_{1}\|_{L^{1,1}}^{2}t^{1-\frac{n}{2}}-\frac{\omega_{n}\delta_{0}^{n}}{n}\|u_{0}\|_{L^{1}}^{2}t^{-\frac{n}{2}}\quad
t\gg 1.$ (3.14)

Therefore, there is a positive real number $t_{0}$ such that

$\|w(t,\cdot)\|^{2}\geq\frac{P^{2}}{32n}\omega_{n}\delta_{0}^{n}t^{2-\frac{n}{2}}$
(3.15)

for all $t\geq t_{0}$ and all $n\in{\bf N}$. It should be mentioned that
$t_{0}>0$ depends on $n$ and the quantities $\|u_{1}\|_{L^{1.1}}$ and
$\|u_{0}\|_{L^{1}}$. From (3.15) with $n\geq 1$ one has the following lemma.

###### Lemma 3.1

Let $n\geq 1$, and $[u_{0},u_{1}]\in L^{1}({\bf R}^{n})\times L^{1,1}({\bf
R}^{n})$. Then, it holds that

$\|w(t,\cdot)\|^{2}\geq CP^{2}t^{2-\frac{n}{2}},\quad t\gg 1.$

Note that the results for $n=4$ in Lemma 3.1 seems to be weak because it shows
just the non-decay property. We expect $\log t$-blowup as in wave equation
case for $n=2$ ([7]). For this purpose we improve the statement for $n=4$ by
relying on a similar argument as in [7] with a trick function $e^{-r^{2}}$.
Indeed, it follows from (3.3) and a similar argument to the previous one that

$\|w(t,\cdot)\|^{2}\geq\frac{1}{2}\int_{{\bf
R}^{4}}\frac{\sin^{2}(tr^{2})}{r^{4}}|w_{1}(\xi)|^{2}d\xi-\int_{{\bf
R}^{4}}\cos^{2}(tr^{2})|w_{0}(\xi)|^{2}d\xi$ $\geq\frac{1}{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}e^{r^{2}}|P+(A(\xi)-iB(\xi))|^{2}d\xi-\|u_{0}\|^{2}$
$\geq\frac{1}{4}P^{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}d\xi-\frac{M^{2}\|u_{1}\|_{L^{1,1}}^{2}}{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}r^{2}d\xi-\|u_{0}\|^{2}$
$\geq\frac{1}{4}P^{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}d\xi-\frac{1}{2}M^{2}\|u_{1}\|_{L^{1,1}}^{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{1}{r^{2}}d\xi-\|u_{0}\|^{2}$
$=:\frac{1}{4}P^{2}U(t)-\frac{1}{4}M^{2}\|u_{1}\|_{L^{1,1}}^{2}\omega_{4}-\|u_{0}\|^{2}.$
(3.16)

Now, we apply an useful idea coming from [11]. For this purpose, we set

$\theta_{j}:=\sqrt{(\frac{1}{4}+j)\frac{\pi}{t}},\quad\tau_{j}:=\sqrt{(\frac{3}{4}+j)\frac{\pi}{t}}\quad(j=0,1,2,\cdots),$

and choose $t>1$ large enough such that $\theta_{0}=\sqrt{\frac{\pi}{4t}}<1$.
Then, because of the fact that

$|\sin(tr^{2})|\geq\frac{1}{\sqrt{2}}$

for $r\in[\theta_{j},\tau_{j}]$ and $j=0,1,2,\cdots$, one can proceed to
estimate:

$U(t)=\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}d\xi\geq\frac{1}{2}\sum_{j=0}^{\infty}\int_{\theta_{j}\leq
r\leq\tau_{j}}\frac{e^{-r^{2}}}{r^{4}}d\xi=\frac{\omega_{4}}{2}\left(\sum_{j=0}^{\infty}\int_{\theta_{j}}^{\tau_{j}}\frac{e^{-r^{2}}}{r}dr\right)$
(3.17)
$\geq\frac{\omega_{4}}{2}\left(\frac{1}{2}\int_{\theta_{0}}^{\infty}\frac{e^{-r^{2}}}{r}dr\right)\geq\frac{\omega_{4}}{4}\int_{\theta_{0}}^{1}\frac{e^{-r^{2}}}{r}dr$
(3.18)
$\geq\omega_{4}\frac{e^{-1}}{4}\int_{\theta_{0}}^{1}\frac{1}{r}dr=\omega_{4}\frac{e^{-1}}{8}(\log
t+\log 4-\log\pi),$ (3.19)

where in the inequality from (3.17) to (3.18) one has just used the monotone
decreasing property of the function $r\mapsto\frac{e^{-r^{2}}}{r}$, and the
fact that the length

$\tau_{j}-\theta_{j}=\frac{\pi}{2t}\frac{1}{\sqrt{(\frac{3}{4}+j)\frac{\pi}{t}}+\sqrt{(\frac{1}{4}+j)\frac{\pi}{t}}}$

is decreasing to $0$ as $j\to\infty$ for each fixed $t>1$. Therefore, by
(3.16) and (3.19) one has the following estimates for $n=4$ for large $t>1$.

###### Lemma 3.2

Let $n=4$, and $[u_{0},u_{1}]\in L^{1}({\bf R}^{n})\times L^{1,1}({\bf
R}^{n})$. Then, it holds that

$\|w(t,\cdot)\|^{2}\geq CP^{2}\log t,\quad t\gg 1.$

Proofs of the lower bound estimates of Theorems 1.1 and 1.2 are direct
consequence of Lemmas 3.1 and 3.2.

##  4 $L^{2}$-upper bound estimates of the solution

In this section, let us derive upper bound estimates of $\|u(t,\cdot)\|$ as
$t\to\infty$ by treating the function $w(t,\xi)$ in both high and low
frequency region.
As in Section 3, one again assumes $[u_{0},u_{1}]\in C_{0}^{\infty}({\bf
R}^{n})\times C_{0}^{\infty}({\bf R}^{n})$ to proceed the proof.

From (3.4) one first derives the upper bound estimate for $I_{low}^{(n)}(t)$
for all $n\geq 1$. Indeed, by (3.8) one has

$I_{low}^{(n)}(t)\leq\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}|\frac{\sin(t|\xi|^{2})}{|\xi|^{2}}w_{1}(\xi)+\cos(t|\xi|^{2})w_{0}(\xi)|^{2}d\xi$
$\leq
2\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+2\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\cos^{2}(t|\xi|^{2})|w_{0}(\xi)|^{2}d\xi$
$=2L_{1}(t)+2L_{2}(t).$ (4.1)

It is easy to obtain the estimate for $L_{2}(t)$ as in the estimate for
$J_{2}(t)$ in Section 2:

$L_{2}(t)\leq\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}|w_{0}(\xi)|^{2}d\xi\leq\|u_{0}\|_{L^{1}}^{2}\omega_{n}\int_{0}^{\frac{\delta_{0}}{\sqrt{t}}}r^{n-1}dr$
$=\frac{\omega_{n}\delta_{0}^{n}}{n}\|u_{0}\|_{L^{1}}^{2}t^{-\frac{n}{2}},$
(4.2)

where one has just used the fat that

$\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}d\xi=\frac{\omega_{n}\delta_{0}^{n}}{n}t^{-\frac{n}{2}},\quad
t\gg 1.$ (4.3)

Let us estimate $L_{1}(t)$. Indeed, from (3.1) and (4.3) one has

$L_{1}(t)=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi\leq\|u_{1}\|_{L^{1}}^{2}t^{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{(t|\xi|^{2})^{2}}d\xi$
$\leq\|u_{1}\|_{L^{1}}^{2}L^{2}t^{2}\frac{\omega_{n}\delta_{0}^{n}}{n}t^{-\frac{n}{2}}=\|u_{1}\|_{L^{1}}^{2}L^{2}\frac{\omega_{n}\delta_{0}^{n}}{n}t^{2-\frac{n}{2}},\quad
t\gg 1.$ (4.4)

Thus, from (4.1), (4.2) and (4.4) one can obtain the low-frequency estimate

$I_{low}^{(n)}(t)\leq
C\left(\|u_{0}\|_{L^{1}}^{2}t^{-\frac{n}{2}}+\|u_{1}\|_{L^{1}}^{2}t^{2-\frac{n}{2}}\right),\quad
t\gg 1,$ (4.5)

with some constant $C>0$.

Next, let us treat $I_{high}^{(n)}(t)$ to get the upper bound estimate.
Indeed, similar to the above estimate one stars with the following
inequalities.

$I_{high}^{(n)}(t)\leq\int_{|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}|\frac{\sin(t|\xi|^{2})}{|\xi|^{2}}w_{1}(\xi)+\cos(t|\xi|^{2})w_{0}(\xi)|^{2}d\xi$
$\leq
2\int_{|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+2\int_{|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}\cos^{2}(t|\xi|^{2})|w_{0}(\xi)|^{2}d\xi$
$=2N_{1}^{(n)}(t)+2N_{2}^{(n)}(t).$ (4.6)

It is easy to treat $N_{2}^{(n)}(t)$ as follows. This can be derived for all
$n\geq 1$:

$N_{2}^{(n)}(t)\leq\int_{|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}|w_{0}(\xi)|^{2}d\xi\leq\|u_{0}\|^{2}.$
(4.7)

Let us estimate $N_{1}^{(n)}(t)$. First, for all $n\geq 1$ one has

$N_{1}^{(n)}(t)=\int_{|\xi|\geq\frac{\delta_{0}}{t^{1/8}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+\int_{\frac{\delta_{0}}{\sqrt{t}}\leq|\xi|\leq\frac{\delta_{0}}{t^{1/8}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi$
$=:R_{1}^{(n)}(t)+R_{2}^{(n)}(t).$ (4.8)

To begin with, we treat $R_{1}^{(n)}(t)$ to get the estimate

$R_{1}^{(n)}(t)\leq\frac{t^{1/2}}{\delta_{0}^{4}}\int_{|\xi|\geq\frac{\delta_{0}}{t^{1/8}}}\sin^{2}(t|\xi|)|w_{1}(\xi)|^{2}d\xi$
$\leq\frac{t^{1/2}}{\delta_{0}^{4}}\int_{|\xi|\geq\frac{\delta_{0}}{t^{1/8}}}|w_{1}(\xi)|^{2}d\xi\leq\frac{\sqrt{t}}{\delta_{0}^{4}}\|u_{1}\|^{2},\quad
t\gg 1.$ (4.9)

On the other hand, in the case of $n\neq 4$, $R_{2}^{(n)}(t)$ can be estimated
as follows:

$R_{2}^{(n)}(t)\leq\|u_{1}\|_{1}^{2}\int_{\frac{\delta_{0}}{\sqrt{t}}\leq|\xi|\leq\frac{\delta_{0}}{t^{1/8}}}\frac{1}{|\xi|^{4}}d\xi$
$=\|u_{1}\|_{1}^{2}\omega_{n}\int_{\frac{\delta_{0}}{\sqrt{t}}}^{\frac{\delta_{0}}{t^{1/8}}}r^{n-5}dr=\frac{\omega_{n}}{(4-n)\delta_{0}^{4-n}}\|u_{1}\|_{1}^{2}(t^{2-\frac{n}{2}}-t^{\frac{1}{2}-\frac{n}{8}}),\quad
t\gg 1.$ (4.10)

Thus, from (4.6), (4.7), (4.8), (4.9) and (4.10) one has the infinite time
blowup estimate for $I_{high}^{(n)}(t)$ in the case of $n=1,2,3$:

$I_{high}^{(n)}(t)=C\left(\|u_{0}\|^{2}+\|u_{1}\|^{2}t^{\frac{1}{2}}+\|u_{1}\|_{L^{1}}^{2}\frac{1}{4-n}(t^{2-\frac{n}{2}}-t^{\frac{1}{2}-\frac{n}{8}})\right),\quad
t\gg 1.$ $\leq
C\left(\|u_{0}\|^{2}+\|u_{1}\|^{2}+\|u_{1}\|_{L^{1}}^{2}\frac{1}{4-n}\right)t^{2-\frac{n}{2}},\quad
t\gg 1.$ (4.11)

Next, let us give sharp estimates for $N_{1}^{(4)}(t)$ in the case of $n=4$ by
using a more delicate computation. For this purpose, by choosing $t>1$
sufficiently large to get the relation $1\leq\log t\leq t\leq t^{2}$ one has a
decomposition of the integrand:

$N_{1}^{(4)}(t)=\int_{|\xi|\geq\frac{\delta_{0}}{(\log
t)^{1/4}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi$
$+\int_{\frac{\delta_{0}}{(\log
t)^{1/4}}\geq|\xi|\geq\frac{\delta_{0}}{t^{1/4}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+\int_{\frac{\delta_{0}}{t^{1/4}}\geq|\xi|\geq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi$
$=:S_{1}(t)+S_{2}(t)+S_{3}(t).$ (4.12)

Let us estimate them in order.
First, one has

$S_{1}(t)\leq\frac{\log
t}{\delta_{0}^{4}}\int_{|\xi|\geq\frac{\delta_{0}}{(\log
t)^{1/4}}}|w_{1}(\xi)|^{2}d\xi$ $\leq\frac{\log
t}{\delta_{0}^{4}}\|u_{1}\|^{2},\quad t\gg 1.$ (4.13)

For $S_{2}(t)$ we see that

$S_{2}(t)\leq\|u_{1}\|_{L^{1}}^{2}\omega_{4}\int_{\frac{\delta_{0}}{t^{1/4}}}^{\frac{\delta_{0}}{(\log
t)^{1/4}}}\frac{1}{r}dr$ $\leq\omega_{4}\|u_{1}\|_{L^{1}}^{2}4^{-1}\left(\log
t-\log(\log t)\right),\quad t\gg 1.$ (4.14)

Finally, let us treat $S_{3}(t)$ to obtain the following estimate:

$S_{3}(t)\leq\|u_{1}\|_{L^{1}}^{2}\int_{\frac{\delta_{0}}{\sqrt{t}}\leq|\xi|\leq\frac{\delta_{0}}{t^{1/4}}}\frac{1}{r^{4}}d\xi$
$\leq\omega_{4}\|u_{1}\|_{L^{1}}^{2}4^{-1}\log t,\quad t\gg 1.$ (4.15)

Thus, it follows from (4.12), (4.13), (4.14) and (4.15) that for large $t\gg
1$

$N_{1}^{(4)}(t)\leq C\left(\|u_{1}\|_{L^{1}}^{2}\log t+\|u_{1}\|^{2}\log
t+\|u_{1}\|_{L^{1}}^{2}(\log t-\log(\log t))\right),\quad t\gg 1$ (4.16)

with some constant $C>0$. Therefore, by combining (4.6), (4.7) and (4.16) one
can get the high-frequency estimate for $n=4$:

$I_{high}^{(4)}(t)\leq
C\left(\|u_{0}\|^{2}+\|u_{1}\|^{2}+\|u_{1}\|_{L^{1}}^{2}\right)\log t,\quad
t\gg 1.$ (4.17)

Finally, it follows from (3.4), (4.5), (4.11) and (4.17) one can get the
crucial upper bound estimates for each $n=1,2,3,4$.

###### Lemma 4.1

Let $n=1,2,3$, and $[u_{0},u_{1}]\in(L^{1}({\bf R}^{n})\cap L^{2}({\bf
R}^{n}))\times(L^{1}({\bf R}^{n})\cap L^{2}({\bf R}^{n}))$. Then, it holds
that

$\|w(t,\cdot)\|^{2}\leq
C(\|u_{0}\|^{2}+\|u_{0}\|_{L^{1}}^{2}+\|u_{1}\|^{2}+\|u_{1}\|_{L^{1}}^{2})t^{2-\frac{n}{2}},\quad
t\gg 1.$

###### Lemma 4.2

Let $n=4$, and $[u_{0},u_{1}]\in(L^{1}({\bf R}^{n})\cap L^{2}({\bf
R}^{n}))\times(L^{1}({\bf R}^{n})\cap L^{2}({\bf R}^{n}))$. Then, it holds
that

$\|w(t,\cdot)\|^{2}\leq
C(\|u_{0}\|^{2}+\|u_{0}\|_{L^{1}}^{2}+\|u_{1}\|^{2}+\|u_{1}\|_{L^{1}}^{2})\log
t,\quad t\gg 1.$

Finally, proofs of the upper bound estimates part of Theorems 1.1 and 1.2 are
direct consequence of Lemmas 4.1 and 4.2, and the Plancherel Theorem.
In connection with these estimates above, one can show the $L^{2}$-upper bound
estimate of the solution itself independently from the Hardy type inequality
(see Proposition 2.1 and Remark 2.2).

###### Proposition 4.1

Let $n\geq 5$, and $[u_{0},u_{1}]\in H^{2}({\bf R}^{n})\times(L^{2}({\bf
R}^{n})\cap L^{1}({\bf R}^{n}))$. Then, there exists a constant $C>0$ such
that the solution $u(t,x)$ to problem (1.1)-(1.2) satisfies

$\|u(t,\cdot)\|\leq C\left(\|u_{0}\|+\|u_{1}\|+\|u_{1}\|_{1}\right)\quad(t\geq
0).$

###### Remark 4.1

A similar $L^{2}$-upper bound estimates can be derived to the solution itself
of the (free) wave equation in the Euclidean space ${\bf R}^{n}$ in the case
of $n\geq 3$, however, the method developed in Proposition 2.1 is widely
applicable to the exterior problem and variavle coefficient cases.

Proof of Proposition 4.1. It suffices to assume the initial data $u_{j}$
($j=0,1$) belongs to $C_{0}^{\infty}({\bf R}^{n})$. Then, one can proceed to
estimate as follows:

$C_{n}\|u(t,\cdot)\|^{2}=\|w(t,\cdot)\|^{2}=\int_{{\bf
R}^{n}}\left|\frac{\sin(t|\xi|^{2})}{|\xi|^{2}}w_{1}(\xi)+\cos(t|\xi|^{2})w_{0}(\xi)\right|^{2}d\xi$
$\leq 2\int_{{\bf
R}^{n}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+2\int_{{\bf
R}^{n}}\cos^{2}(t|\xi|^{2})|w_{0}(\xi)|^{2}d\xi$ $\leq 2\int_{|\xi|\leq
1}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+2\int_{|\xi|\geq
1}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|w_{1}(\xi)|^{2}d\xi+\|u_{0}\|^{2}$
$=2T_{1}(t)+2T_{2}(t)+\|u_{0}\|^{2}.$ (4.18)

Now, about $T_{1}(t)$ one can get

$T_{1}(t)\leq\|w_{1}\|_{\infty}^{2}\int_{|\xi|\leq
1}\frac{1}{|\xi|^{4}}d\xi\leq\omega_{n}\|u_{1}\|_{1}^{2}\int_{0}^{1}r^{n-5}dr$
$=\frac{\omega_{n}}{n-4}\|u_{1}\|_{1}^{2},\quad(t\geq 0).$ (4.19)

For the estimate of $T_{2}(t)$ one can obtain

$T_{2}(t)\leq\int_{|\xi|\geq
1}\sin^{2}(t|\xi|^{2})|w_{1}(\xi)|^{2}d\xi\leq\int_{|\xi|\geq
1}|w_{1}(\xi)|^{2}d\xi$ $\leq\|w_{1}\|^{2}=\|u_{1}\|^{2},\quad(t\geq 0).$
(4.20)

The statement of Proposition 4.1 can be derived because of (4.18)-(4.20).

$\Box$

##  5 Proof of Theorems 1.3, 1.4 and 1.5.

In this section, let us prove Theorems 1.3, 1.4 and 1.5. The first part is
devoted to the proof of Theorem 1.3.
Proof of Theorem 1.3. The whole idea comes from [8]. By applying the spatial
Fourier transform to the both sides of (1.1), again the problem (1.1)-(1.2)
can be reduced to ODE with parameter $\xi$:

$\hat{u}_{tt}(t,\xi)+|\xi|^{4}\hat{u}(t,\xi)=0,\ \ \ \
(t,\xi)\in(0,\infty)\times{\bf R}_{\xi}^{n},$ (5.1)
$\hat{u}(0,\xi)=\hat{u}_{0}(\xi),\quad\hat{u}_{t}(0,\xi)=\hat{u}_{1}(\xi),\,\,\,\,\xi\in{\bf
R}_{\xi}^{n}.$ (5.2)

For the solution $\hat{u}(t,\xi)$ to problem (5.1)-(5.2), one introduces an
auxiliary function

$\hat{w}(t,\xi):=\int^{t}_{0}\hat{u}(s,\xi)ds.$

Then $\hat{w}(t,\xi)$ satisfies

$\hat{w}_{tt}(t,\xi)+|\xi|^{4}\hat{w}(t,\xi)=\hat{u}_{1}(\xi),\ \ \ \
(t,\xi)\in(0,\infty)\times{\bf R}_{\xi}^{n},$ (5.3)
$\hat{w}(0,\xi)=0,\quad\hat{w}_{t}(0,\xi)=\hat{u}_{0}(\xi),\,\,\,\,\xi\in{\bf
R}_{\xi}^{n}.$ (5.4)

Now, we introduce one more auxiliary function $\hat{v}(t,\xi)$ defined on
${\bf R}_{\xi}^{n}\setminus\\{0\\}$ as follows:

$\hat{v}(t,\xi)=\hat{w}(t,\xi)-\frac{\hat{u}_{1}(\xi)}{|\xi|^{4}},\quad\xi\in{\bf
R}_{\xi}^{n}\setminus\\{0\\}.$ (5.5)

Then, the function $\hat{v}(t,\xi)$ satisfies

$\hat{v}_{tt}(t,\xi)+|\xi|^{4}\hat{v}(t,\xi)=0,\ \ \ \ t>0,\quad\xi\in{\bf
R}_{\xi}^{n}\setminus\\{0\\},$ (5.6)
$\hat{v}(0,\xi)=-\frac{\hat{u}_{1}(\xi)}{|\xi|^{4}},\quad\hat{v}_{t}(0,\xi)=\hat{u}_{0}(\xi),\,\,\,\,\xi\in{\bf
R}_{\xi}^{n}\setminus\\{0\\}.$ (5.7)

By multiplying both sided of (5.6) by $\overline{\hat{v}_{t}(t,\xi)}$,
integrating it over $\\{|\xi|\geq\delta\\}$ with small $\delta>0$, and taking
real parts of the resulted equality one can gets

$\frac{d}{dt}\int_{|\xi|\geq\delta}(|\hat{v}_{t}(t,\xi)|^{2}+|\xi|^{4}|\hat{v}(t,\xi)|^{2})d\xi=0,$

so that by integrating it over $[0,t]$ one has

$\int_{|\xi|\geq\delta}(|\hat{v}_{t}(t,\xi)|^{2}+|\xi|^{4}|\hat{v}(t,\xi)|^{2})d\xi=\int_{|\xi|\geq\delta}(|\hat{u}_{0}(\xi)|^{2}+\frac{|\hat{u}_{1}(\xi)|^{2}}{|\xi|^{4}})d\xi,$

where we have just used (5.7). Since
$\hat{v}_{t}(t,\xi)=\hat{w}_{t}(t,\xi)=\hat{u}(t,\xi)$, one can arrive at

$\int_{|\xi|\geq\delta}|\hat{u}(t,\xi)|^{2}d\xi\leq\int_{{\bf
R}_{\xi}^{n}}|\hat{u}_{0}(\xi)|^{2}d\xi+\int_{|\xi|\geq\delta}\frac{|\hat{u}_{1}(\xi)|^{2}}{|\xi|^{4}}d\xi$
$\leq\|u_{0}\|^{2}+\int_{1\geq|\xi|\geq\delta}\frac{|\hat{u}_{1}(\xi)|^{2}}{|\xi|^{4}}d\xi+\int_{1\leq|\xi|}\frac{|\hat{u}_{1}(\xi)|^{2}}{|\xi|^{4}}d\xi$
$\leq\|u_{0}\|^{2}+\int_{1\geq|\xi|\geq\delta}\frac{|\hat{u}_{1}(\xi)|^{2}}{|\xi|^{4}}d\xi+\|u_{1}\|^{2}.$
(5.8)

By applying Lemma 3.1 of [5, page 879] to $\hat{u}_{1}(\xi)$ one has

$|\hat{u}_{1}(\xi)|\leq C|\xi|^{\gamma}\|u_{1}\|_{1,\gamma},$ (5.9)

because of the assumption $P_{1}=\displaystyle{\int_{{\bf
R}^{n}}}u_{1}(x)dx=0$, where $\gamma\in[0,1]$. By substituting (5.9) into
(5.8) one has arrived at the estimate

$\int_{|\xi|\geq\delta}|\hat{u}(t,\xi)|^{2}\leq\|u_{0}\|^{2}+\|u_{1}\|^{2}+C\|u_{1}\|_{1,\gamma}^{2}\int_{1\geq|\xi|\geq\delta}|\xi|^{2\gamma-4}d\xi$
$\leq\|u_{0}\|^{2}+\|u_{1}\|^{2}+C\|u_{1}\|_{1,\gamma}^{2}\frac{\omega_{n}}{2\gamma+n-4},$
(5.10)

provided that $2\gamma+n-4>0$ for $\gamma\in(0,1]$ in the case of $n=4$, and
for $\gamma\in(\frac{1}{2},1]$ in the case of $n=3$. Note that the constant
$C>0$ does not depend on any small $\delta>0$. By letting $\delta\to+0$ in
(5.10), one has the desired estimates via the Plancherel Theorem. $\hfill\Box$

In order to prove Theorem 1.4, we use the following lemma, which is a direct
consequence of Lemma 5.1 with $\gamma\in(1,2]$ of [10].

###### Lemma 5.1

Let $n\geq 1$, and for $\gamma\in(1,2]$, suppose that $f\in L^{1,\gamma}({\bf
R}^{n})$ satisfies

$\int_{{\bf R}^{n}}f(x)dx=0,\qquad\int_{{\bf
R}^{n}}x_{j}f(x)dx=0\quad(j=1,2,\cdots,n).$

Then, it holds that

$|\hat{f}(\xi)|\leq C|\xi|^{\gamma}\|f\|_{1,\gamma},\quad\xi\in{\bf R}^{n}$

with some constant $C=C_{\gamma}>$.

Based on Lemma 5.1, let us prove Theorem 1.4.

Proof of Theorem 1.4. The proof is just a slight modification of those for
Theorem 1.3. Indeed, we use Lemma 5.1 with $f(x):=u_{1}(x)$ and
$\gamma\in(1,2]$ in place of (5.10) with $\gamma\in[0,1]$. Then, (5.10) above
can be interpreted as follows: the positivity $2\gamma+n-4>0$ again holds for
$\gamma\in(1,2]$ in the case of $n=2$, and for $\gamma\in(\frac{3}{2},2]$ in
the case of $n=1$. This implies the desired results. $\hfill\Box$

Finally, let us prove Theorem 1.5 basing on the following equality which is a
direct consequence of [10, Lemma 5.1] with $f(x):=u_{1}(x)$:

$w_{1}(\xi)=P+i\xi P_{1}+E(\xi),$ (5.11)

where

$P:=\int_{{\bf R}}u_{1}(x)dx,\quad P_{1}:=\int_{{\bf R}}xu_{1}(x)dx,$

and the error term $E(\xi)$ satisfies

$|E(\xi)|\leq C|\xi|^{2}\|u_{1}\|_{1,2},\quad\xi\in{\bf R}.$ (5.12)

Proof of (1) of Theorem 1.5. For our purpose it suffices to obtain the lower
bound estimate on $J_{1}(t)$ in (3.8). Indeed, from (5.16) and (3.3) one has

$J_{1}(t)=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|P+i\xi
P_{1}+E(\xi)|^{2}d\xi$
$\geq\frac{1}{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}(P^{2}+|\xi|^{2}P_{1}^{2})d\xi-\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|E(\xi)|^{2}d\xi$
$\geq\frac{P^{2}}{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}d\xi+\frac{P_{1}^{2}}{2}\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{2}}d\xi-\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|E(\xi)|^{2}d\xi$
$=:\frac{P^{2}}{2}K_{1}(t)+\frac{P_{1}^{2}}{2}K_{3}(t)-K_{4}(t),$ (5.13)

where $K_{1}(t)$ is the same function defined in (3.10), and

$K_{3}(t):=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{2}}d\xi,$
$K_{4}(t):=\int_{|\xi|\leq\frac{\delta_{0}}{\sqrt{t}}}\frac{\sin^{2}(t|\xi|^{2})}{|\xi|^{4}}|E(\xi)|^{2}d\xi.$

First, it is easy to see by using (5.17)

$K_{4}(t)\leq C\|u_{1}\|_{1,2}^{2}t^{-\frac{n}{2}},\quad(n=1)$ (5.14)

with some constant $C>0$. While,

$K_{3}(t)\geq\frac{t}{\delta_{0}^{2}}\omega_{1}\int_{0}^{\frac{\delta_{0}}{\sqrt{t}}}\sin^{2}(tr^{2})dr$
$=\left(\frac{\omega_{1}}{\delta_{0}^{2}}\int_{0}^{\delta_{0}}\sin^{2}(\sigma^{2})d\sigma\right)t^{\frac{1}{2}}\quad(t\gg
1).$ (5.15)

On the lower bund of $K_{1}(t)$ one has already derived it in (3.12) wth
$n=1$. Thus, by combining it with (5.18)-(5.20) one has the desired estimates.

$\hfill\Box$

Proof of (2) of Theorem 1.5. Let $n=2$. The following result, once more, is a
direct consequence of [10, Lemma 5.1] with $f(x):=u_{1}(x)$:

$w_{1}(\xi)=P+i\xi\cdot{\bf P}_{1}+E(\xi),$ (5.16)

where

$P:=\int_{{\bf R^{2}}}u_{1}(x)dx,\quad{\bf P}_{1}:=(p_{1},p_{2}),$
$p_{1}:=\int_{{\bf R^{2}}}x_{1}u_{1}(x)dx,\quad p_{2}:=\int_{{\bf
R^{2}}}x_{2}u_{1}(x)dx,$

and the error term $E(\xi)$ satisfies

$|E(\xi)|\leq C|\xi|^{2}\|u_{1}\|_{1,2},\quad\xi\in{\bf R}^{2}.$ (5.17)

Similarly to the estimates developed in (3.16), by using (5.17) one can
proceed the computation as follows:

$\|w(t,\cdot)\|^{2}\geq\frac{1}{2}\int_{{\bf
R}^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}|w_{1}(\xi)|^{2}d\xi-\int_{{\bf
R}^{2}}\cos^{2}(tr^{2})|w_{0}(\xi)|^{2}d\xi$ $\geq\frac{1}{2}\int_{{\bf
R}^{2}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}|P+i\xi\cdot{\bf
P}_{1}+E(\xi)|^{2}d\xi-\|u_{0}\|^{2}$ $\geq\frac{1}{4}P^{2}\int_{{\bf
R}^{2}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}d\xi+\frac{1}{4}K_{5}(t)-C^{2}\|u_{1}\|_{L^{1,2}}^{2}\int_{{\bf
R}^{4}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}r^{4}d\xi-\|u_{0}\|^{2}$
$=:\frac{1}{4}P^{2}U(t)+\frac{1}{4}K_{5}(t)-C\|u_{1}\|_{L^{1,2}}^{2}\omega_{2}-\|u_{0}\|^{2},$
(5.18)

where

$K_{5}(t):=\int_{{\bf
R}^{2}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}|\xi\cdot{\bf P}_{1}|^{2}d\xi.$

First, by similar computations as in (3.17), (3.18) and (3.19) one can get

$U(t)\geq Ct,\quad t\gg 1,$ (5.19)

with some constant $C>0$. So, it suffices to estimate $K_{5}(t)$. In order to
estimate $K_{5}(t)$, we employ an idea coming from [11]. For this, we set

$K:=\left\\{\xi\in{\bf R}^{2}\,:\,\left|\frac{\xi}{|\xi|}\cdot\frac{{\bf
P}_{1}}{|{\bf P}_{1}|}\right|\geq\frac{1}{2}\right\\}.$

Then, one can estimate as follows:

$K_{5}(t):=\int_{{\bf
R}^{2}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{4}}|\xi\cdot{\bf P}_{1}|^{2}d\xi$
$\geq|{\bf
P}_{1}|^{2}\int_{K}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{2}}|\frac{\xi}{|\xi|}\cdot\frac{{\bf
P}_{1}}{|{\bf P}_{1}|}|^{2}d\xi\geq\frac{|{\bf
P}_{1}|^{2}}{4}\int_{K}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r^{2}}d\xi$
$=C\pi\frac{|{\bf
P}_{1}|^{2}}{4}\int_{0}^{\infty}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r}dr=:C\pi\frac{|{\bf
P}_{1}|^{2}}{4}K_{6}(t).$ (5.20)

Once more, as in (3.17), (3.18) and (3.19) one can estimate $K_{6}(t)$:

$K_{6}(t)\geq\sum_{j=0}^{\infty}\int_{\theta_{j}}^{\tau_{j}}e^{-r^{2}}\frac{\sin^{2}(tr^{2})}{r}dr\geq\frac{1}{2}\sum_{j=0}^{\infty}\int_{\theta_{j}}^{\tau_{j}}e^{-r^{2}}r^{-1}dr$
$\geq\frac{1}{4}\int_{\theta_{0}}^{\infty}e^{-r^{2}}r^{-1}dr\geq\frac{e^{-1}}{4}\int_{\theta_{0}}^{1}r^{-1}dr$
$\geq C\log t,\quad t\gg 1.$ (5.21)

The desired estimate can be established from (5.18)-(5.21).

$\hfill\Box$

Acknowledgement. The author would like to thank Wenhui Chen (Shanghai Jiao
Tong University) for his fruitful discussion with me. The work of the author
was supported in part by Grant-in-Aid for Scientific Research (C)20K03682 of
JSPS.

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