1. Introduction
In [1] the following integral equation is of interest;
\begin{equation}
\label{e1}
b(t)=b_0(t)+\int_0^t (t-s)^{\lambda -1}b(s)ds,
\end{equation}
(1)
where \(b_0\) is a smooth functions rapidly decaying with all its derivatives as \(t\to \infty\), \(b_0(t)=0\) if \(t< 0\). We are especially interested in the value \(\lambda=-\frac 1 4\), because of its importance for the Navier-Stokes theory, [
1], Chapter 5, [
2,
3].
The integral in (1) diverges in the classical sense for \(\lambda\le 0\). Our aim is to define this hyper-singular integral. There is a regularization method to define
singular integrals \(J:=\int_{\mathbb{R}}t_+^{\lambda}\phi(t)dt\), \(\lambda< -1\), in distribution theory, [
4]. However, the integral in (1) is a convolution, which is defined in
[
4], p.135, as a direct product
of two distributions. This definition
is not suitable for our purposes because although \(t_+^{\lambda-1}\) for \(\lambda\le 0\), \(\lambda\neq 0,-1,-2,…\) is a distribution
on the space \(C^\infty_0(\mathbb{R}_+)\) of the test functions, but it is not a distribution in the
space \(K=C^\infty_0(\mathbb{R})\) of the test
functions used in [
4]. Indeed, one can find \(\phi\in K\) such that
\(\lim_{n\to \infty}\phi_n=\phi\) in
\(K\), but \(\lim_{n\to \infty}\int_{\mathbb{R}}
t_+^{\lambda-1} \phi(t)dt=\infty\)
for \(\lambda\le 0\), so that
\(t_+^{\lambda-1}\) is not a linear bounded functional in \(K\), i.e., not a distribution. On the other hand, one can check that
\(t_+^{\lambda-1}\) for \(\lambda\in R\)
is a distribution (a bounded linear functional) in the space
\(\mathcal{K}=C^\infty_0(\mathbb{R}_+)\) with
the convergence \(\phi_n\to \phi\) in
\(\mathcal{K}\) defined by the requirements: a) the supports of all \(\phi_n\) belong to an interval \([a,b]\),
\(0< a\le b< \infty\), b) \(\phi_n^{(j)}\to
\phi^{(j)}\) in \(C([a,b])\) for all \(j=0,1,2,….\). Indeed, the functional
\(\int_0^\infty t_+^\lambda\phi(t)dt\) is linear and bounded in
\(\mathcal{K}\):
\[
\left|\int_0^\infty t_+^\lambda\phi_n(t)dt\right|\le (a^\lambda+b^\lambda)
\int_a^b |\phi_n(t)|dt.
\]
A similar estimate holds for the derivatives of \(\phi_n\).
Although \(t_+^{-\frac 5 4}\) is a distribution in \(\mathcal{K}\), the
convolution
\begin{equation}
\label{e2}
h:=\int_0^t(t-s)^{-\frac 5 4}b(s)ds:=
t_+^{-\frac 5 4}\star b
\end{equation}
(2)
cannot be defined similarly to the definition in [
4] because the function
\(\int_0^\infty \phi(u+s) b(s)ds\)
does not, in general, belong to \(\mathcal{K}\) even if
\(\phi\in \mathcal{K}\).
Let us define the convolution \(h\) using the Laplace transform
\[L(b):=\int_0^\infty e^{-pt}b(t)dt, \quad Re p>0.\]
Laplace transform for distributions is studied in [
5]. One has \(L(t_+^{-\frac 5 4}\star b)=L(t_+^{-\frac 5 4})L(b)\).
To define \(L(t^{\lambda-1})\) for \(\lambda\le 0\), note that for Re\(\lambda>0\) the classical definition
\begin{equation}
\label{e3}
\int_0^\infty e^{-pt}t^{\lambda-1}dt=
\frac{\Gamma(\lambda)}{p^\lambda}
\end{equation}
(3)
holds. The right-side of (3)
admits analytic continuation to the
complex plane of \(\lambda\), \(\lambda\neq 0,-1,-2,….\). This allows one to define integral (3)
for any \(\lambda\neq 0,-1,-2,…\).
Recall that the gamma function
\(\Gamma(\lambda)\) has its only singular points, the simple poles, at
\(\lambda=-n\), \(n=0,1,2,…\) with the residue at \(\lambda=-n\) equal to \(\frac{(-1)^n}{n!}\). It is known
that \(\Gamma(z+1)=z\Gamma(z)\), so
\begin{equation}
\label{e3a}
\Gamma(-\frac 1 4)=-4\Gamma(3/4):=-c_1, \quad c_1>0.
\end{equation}
(4)
Therefore, we define
\(h\) by defining \(L(h)\) as follows:
\begin{equation}
\label{e4}
L(h)=-c_1p^{\frac 1 4}L(b), \quad \lambda=-\frac 1 4,
\end{equation}
(5)
and assume that \(L(b)\) can be defined. That \(L(b)\) is well defined in the Navier-Stokes theory follows from the a priori estimates proved in [
1], Chapter 5.
From (5) one gets
\begin{equation}
\label{e4a}
L(b)=-c_1^{-1}p^{-\frac 1 4} L(h).
\end{equation}
(6)
2. Convolution of special functions
Define \(\Phi_\lambda=\frac {t_+^{\lambda-1}}{\Gamma(\lambda)}\).
Lemma 1.
For any \(\lambda, \mu \in \mathbb{R}\) the following formulas hold;
\begin{equation}
\label{e5}
\Phi_\lambda\star \Phi_\mu=\Phi_{\lambda+\mu}, \quad
\Phi_{\lambda +0}\star \Phi_{-\lambda}=\delta(t).
\end{equation}
(7)
Proof.
For Re\(\lambda>0\), Re\(\mu>0\) one has
\begin{equation}
\label{e6}
\Phi_\lambda\star \Phi_\mu=
\frac 1 {\Gamma(\lambda)\Gamma(\mu)}
\int_0^t(t-s)^{\lambda -1}s^{\mu -1}ds
=\frac{t_+^{\lambda+\mu-1}}{\Gamma(\lambda)\Gamma(\mu)}\int_0^1
(1-u)^{\lambda-1}u^{\mu -1}du=\frac{t_+^{\lambda+\mu -1}}{\Gamma(\lambda+\mu)},
\end{equation}
(8)
where we used the known formula for
beta function:
\[B(\lambda, \mu):=\int_0^1u^{\lambda -1}(1-u)^{\mu -1}du=
\frac{\Gamma(\lambda)\Gamma(\mu)}
{\Gamma(\lambda+\mu)}.
\]
Analytic properties of beta function
follow from these of Gamma function.
The function \(\frac 1 {\Gamma(z)}\)
is entire function of \(z\).
Let us now prove the second formula
(7). We have \(\Gamma(\epsilon)\sim \epsilon\) as \(\epsilon \to 0\).
Therefore
\begin{equation}
\label{e6′}
\frac {t_+^{\lambda+\epsilon -\lambda -1}}{\Gamma(\epsilon)}\sim \epsilon t_+^{\epsilon-1}.
\end{equation}
(9)
If \(f\) is any continuous rapidly decaying function then
\begin{equation}
\label{e6a}
\lim_{\epsilon\to 0}\epsilon\int_0^\infty t^{\epsilon-1}f(t)dt=f(0).
\end{equation}
(10)
Indeed, fix a small \(\delta>0\), such that \(f(t)\sim f(0)\)
for \(t\in [0,\delta]\) as \(\delta\to 0\). Then,
as \(\epsilon \to 0\), one has
\begin{equation}
\label{e6b}
\lim_{\epsilon\to +0}\epsilon\int_0^\delta t^{\epsilon-1}f(t)dt=\lim_{\epsilon\to +0} \epsilon f(0)\frac
{t^\epsilon}{\epsilon}|_0^\delta=f(0)\lim_{\epsilon\to +0}\delta^\epsilon=f(0).
\end{equation}
(11)
Note that
\begin{equation}
\label{e6c}
\lim_{\epsilon\to 0} \epsilon\int_\delta^\infty t^{\epsilon-1}f(t)dt=0, \quad \delta>0,
\end{equation}
(12)
because
\(|\int_\delta^\infty t^{\epsilon-1}f(t)dt|\le c\) and \(\epsilon\to 0\).
From (11) and (12)
one obtains (10).
So, the
second formula (7) is proved.
Lemma 1 is proved.
Remark 1.
The first formula (7) of Lemma 1 is proved in [4], pp.150-151. Our proof of the second formula (7) differs from the proof in [4] considerably.
Remark 2.
A different proof of Lemma 1 can be given: \(L(\Phi_{\lambda}\star\Phi_{\mu})=\frac 1 {p^{\lambda+\mu}}\) by formula (3), and
\(L^{-1}(\frac 1 {p^{\lambda+\mu}})=\Phi_{\lambda+\mu}(t)\). If \(\lambda=-\mu\), then \( \frac 1 {p^{\lambda+\mu}}=1\) and \(L^{-1}(1)=\delta(t)\).
3. Integral equation and inequality
Consider equation (1) and the following inequality:
\begin{equation}
\label{e7}
q(t)\le b_0(t)+t_+^{\lambda-1}\star q, \quad q\ge 0.
\end{equation}
(13)
Theorem 1.
Equation (1)
has a unique solution. This solution can be obtained by iterations by solving the Volterra equation
\begin{equation}
\label{e8}
b_{n+1}=-c_1^{-1}\Phi_{1/4}\star b_{n} +c_1^{-1}\Phi_{1/4}\star b_0, \quad
b_{n=0}=
c_1^{-1}\Phi_{1/4}\star b_0, \quad b=\lim_{n\to \infty}b_n.
\end{equation}
(14)
Proof.
Applying to (1) the operator \(\Phi_{1/4}\star\) and
using the second equation (7)
one gets a Volterra equation
\[\Phi_{1/4}\star b=\Phi_{1/4}\star b_0-c_1b,\quad c_1=|\Gamma(-\frac 1 4)|,\]
or
\begin{equation}
\label{e9}
b=-c_1^{-1}\Phi_{1/4}\star b +c_1^{-1}\Phi_{1/4} \star b_0, \quad c_1=4\Gamma(3/4).
\end{equation}
(15)
The operator \(\Phi_\lambda\) with
\(\lambda>0\) is a Volterra-type
equation which can be solved by iterations, see [
1], p.53,
Lemmas 5.10, 5.11. If \(b_0\ge 0\) then
the solution to (1) is non-negative, \(b\ge 0\).
Theorem 1 is proved.
For convenience of the reader let us
prove the results mentioned above.
Lemma 2.
The operator \(Af:=\int_0^t(t-s)^pf(s)ds\) in the space
\(X:=C(0,T)\) for any fixed \(T\in [0,\infty)\) and \(p>-1\) has spectral radius \(r(A)\)
equal to zero, \(r(A)=0\). The equation
\(f=Af+g\) is uniquely solvable in \(X\). Its solution can be obtained by iterations
\begin{equation}
\label{e9a}
f_{n+1}=Af_n+g, \quad f_0=g; \quad
\lim_{n\to \infty}f_n=f,
\end{equation}
(16)
for any \(g\in X\) and the convergence
holds in \(X\).
Proof.
The spectral radius
of a linear operator \(A\) is defined
by the formula \(r(A)=\lim_{n\to \infty}\|A^n\|^{1/n}\). By induction one proves that
\begin{equation}
\label{e9b}
|A^nf|\le t^{n(p+1)}\frac{\Gamma^n(p+1)}{\Gamma(n(p+1)+1)}\|f\|_X, \quad n\ge 1.
\end{equation}
(17)
From this formula and the known asymptotic of the gamma function
the conclusion \(r(A)=0\) follows.
The convergence result (16)
is analogous to the well known
statement for the assumption \(\|A\|< 1\). Lemma 2 is proved.
If \(q\ge 0\) then inequality (13)
implies
\begin{equation}
\label{e9c}
q\le -c_1^{-1}\Phi_{1/4}\star q +c_1^{-1}\Phi_{1/4}\star b_0.
\end{equation}
(18)
Inequality (18) can be solved by iterations with the initial term
\(c_1^{-1}\Phi_{1/4}\star b_0\). This
yields
\begin{equation}
\label{e11}
q\le b,
\end{equation}
(19)
where \(b\) solves (1).
See also [
6,
7].
Conflict of Interests
The author declares no conflict of interest.
