On hyper-singular integrals

1. Introduction

In [1] the following integral equation is of interest;

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 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 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 Therefore, we define \(h\) by defining \(L(h)\) as follows: 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

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;

Proof. For Re\(\lambda>0\), Re\(\mu>0\) one has

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

If \(f\) is any continuous rapidly decaying function then 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 Note that 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:

Theorem 1. Equation (1) has a unique solution. This solution can be obtained by iterations by solving the Volterra equation

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

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

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

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

Inequality (18) can be solved by iterations with the initial term \(c_1^{-1}\Phi_{1/4}\star b_0\). This yields where \(b\) solves (1). See also [6,7].

Conflict of Interests

The author declares no conflict of interest.

References

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