The influence of the coefficients of a system of wave equations coupled by velocities on its stabilization

Abstract

In this work, we consider a system of two wave equations coupled by velocities in a one-dimensional space, with one boundary fractional damping. First, we show that the system is strongly asymptotically stable if and only if the coupling parameter b of the two equations is outside a discrete set of exceptional real values. Next, we show that our system is not uniformly stable. Hence, we look for a polynomial decay rate for smooth initial data. Using a frequency domain approach combined with the multiplier method, we prove that the energy decay rate is greatly influenced by the nature of the coupling parameter b, the arithmetic property of the wave propagation speed a and the order of the fractional damping $\alpha$. Indeed, under the equal speed propagation condition, i.e., $a=1$, we establish an optimal polynomial energy decay rate of type $t^{-\frac{2}{{1-\alpha }}}$ if the coupling parameter $b\notin \pi {\mathbb {Z}}$ and of type $t^{-\frac{2}{{5-\alpha }}}$ if the coupling parameter $b\in \pi {\mathbb {Z}}$. Furthermore, when the wave propagates with different speeds, i.e., $a\not =1$, we prove that, for any rational number $\sqrt{a}$ and almost all irrational numbers $\sqrt{a}$, the energy of our system decays polynomially to zero like as $t^{-\frac{2}{{5-\alpha }}}$. This result still holds if $a\in {\mathbb {Q}}$, $\sqrt{a}\notin {\mathbb {Q}}$ and b small enough.

Appendices

Appendix A: Notions of stability and theorems used

We introduce here the notions of stability that we encounter in this work.

Definition A.1

Assume that A is the generator of a $\hbox {C}_0$-semigroup of contractions $\left( e^{tA}\right) _{t\ge 0}$ on a Hilbert space H. The $C_0$-semigroup $\left( e^{tA}\right) _{t\ge 0}$ is said to be

• 1. strongly stable if

  $$\begin{aligned} \lim _{t\rightarrow +\infty } \Vert e^{tA}x_0\Vert _{H}=0, \quad \forall \ x_0\in H; \end{aligned}$$

• 2. exponentially (or uniformly) stable if there exist two positive constants M and $\epsilon$ such that

  $$\begin{aligned} \Vert e^{tA}x_0\Vert _{H} \le Me^{-\epsilon t}\Vert x_0\Vert _{H}, \quad \forall \ t>0, \ \forall \ x_0\in {H}; \end{aligned}$$

• 3. polynomially stable if there exist two positive constants C and $\alpha$ such that

  $$\begin{aligned} \Vert e^{tA}x_0\Vert _{H}\le C t^{-\alpha }\Vert x_0\Vert _{D(A)}, \quad \forall \ t>0, \ \forall \ x_0\in D\left( A\right) . \end{aligned}$$

  In that case, one says that the solutions of (2.5) decay at a rate $t^{-\alpha }$ (with $\alpha >0$). The $C_0$-semigroup $\left( e^{tA}\right) _{t\ge 0}$ is said to be polynomially stable with optimal decay rate $t^{-\alpha }$ (with $\alpha >0$) if it is polynomially stable with decay rate $t^{-\alpha }$ and for any $\varepsilon >0$ small enough, there exists solutions of (2.5) which do not decay at a rate $t^{-(\alpha +\varepsilon )}$.

$\square$

To obtain strong stability of the $C_0$-semigroup $\left( e^{tA}\right) _{t\ge 0}$, we mention the theorem of Arendt and Batty in [12].

Theorem A.2

(Arendt and Batty in [12]) Assume that A is the generator of a $\hbox {C}_0-$semigroup of contractions $\left( e^{tA}\right) _{t\ge 0}$ on a Hilbert space H. If A has no pure imaginary eigenvalues and $\sigma \left( A\right) \cap i{\mathbb {R}}$ is countable, where $\sigma \left( A\right)$ denotes the spectrum of A, then the $C_0$-semigroup $\left( e^{tA}\right) _{t\ge 0}$ is strongly stable.$\square$

To obtain polynomial stability of the $C_0$-semigroup $\left( e^{tA}\right) _{t\ge 0}$, we mention the theorem of Borichev and Tomilov in [17].

Theorem A.3

(Borichev–Tomilov in [17]) Assume that A is the generator of a strongly continuous semigroup of contractions $\left( e^{tA}\right) _{t\ge 0}$ on H. If $i{\mathbb {R}}\subset \rho (A)$, where $\rho \left( A\right)$ denotes the resolvent set of A, then for a fixed $\ell >0$ the following conditions are equivalent

• 1. ${\sup _{\lambda \in {\mathbb {R}}}\left\| \left( i\lambda I-A\right) ^{-1}\right\| _{{\mathcal {L}}\left( H\right) }=O\left( |\lambda |^\ell \right) ,}$

• 2. ${\Vert e^{tA}x_0\Vert _{{\mathcal {H}}} \le \dfrac{C}{t^{{\frac{1}{\ell }}}} \ \Vert x_0\Vert _{D\left( A\right) } \quad \forall \ t>0,\ x_0\in D\left( A\right) },\$ for some $C>0.$

$\square$

Appendix B: Integrals used

Lemma B.1

Let $0<\alpha <1$ and $\omega (\xi )\in L^2({\mathbb {R}})$. If $|\xi |\omega (\xi )\in L^2({\mathbb {R}})$, then ${\int _{{\mathbb {R}}}|\xi |^{\frac{2\alpha -1}{2}}\omega (\xi )}d\xi$ is well defined.

Proof

We have

$$\begin{aligned} \left| \int _{{\mathbb {R}}}|\xi |^{\frac{2\alpha -1}{2}}\omega (\xi )d\xi \right| \le 2\int _0^{+\infty }\xi ^{\frac{2\alpha -1}{2}}|\omega (\xi )|d\xi \le 2\int _0^{1}\xi ^{\frac{2\alpha -1}{2}}|\omega (\xi )|d\xi +2\int _1^{+\infty }\xi ^{\frac{2\alpha -1}{2}}|\omega (\xi )|d\xi . \end{aligned}$$

Since $0<\alpha <1$ and $\omega (\xi )\in L^2({\mathbb {R}})$, then by using Cauchy-Schwarz inequality, we get

$$\begin{aligned} \int _0^{1}\xi ^{\frac{2\alpha -1}{2}}|\omega (\xi )|d\xi \le \left( \int _0^1 \xi ^{{2\alpha -1}} d\xi \right) ^{\frac{1}{2}} \left( \int _0^1|\omega (\xi )|^2d\xi \right) ^{\frac{1}{2}}\le \frac{1}{2\alpha }\left( \int _0^1|\omega (\xi )|^2d\xi \right) ^{\frac{1}{2}}<+\infty . \end{aligned}$$

On the other hand, since $0<\alpha <1$ and $|\xi |\omega (\xi )\in L^2({\mathbb {R}})$, by using Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} \int _1^{+\infty }\xi ^{\frac{2\alpha -1}{2}}|\omega (\xi )|d\xi= & {} \int _1^{+\infty }\xi ^{\frac{2\alpha -3}{2}}\xi |\omega (\xi )|d\xi \le \left( \int _1^{+\infty } \xi ^{2\alpha -3}d\xi \right) ^{\frac{1}{2}}\left( \int _1^{+\infty }|\xi \omega (\xi )|^2d\xi \right) ^{\frac{1}{2}}\\\le & {} \frac{1}{2(1-\alpha )}\left( \int _1^{+\infty }|\xi \omega (\xi )|^2d\xi \right) ^{\frac{1}{2}}<+\infty . \end{aligned}$$

It follows that ${\int _{{\mathbb {R}}}|\xi |^{\frac{2\alpha -1}{2}}\omega (\xi )}$ is well defined. The proof is thus complete. $\square$

Lemma B.2

(Lemma 2.1 stated in [16]) Let $0<\alpha <1,$ $\eta \ge 0$, and $\lambda \in {\mathbb {C}}$, such that

$$\begin{aligned} \lambda \in D=\left\{ \lambda \in {\mathbb {C}}\ |\ \text {Re}\left\{ \lambda \right\} +\eta >0\right\} \cup \left\{ \lambda \in {\mathbb {C}}\ |\ \text {Im}\left\{ \lambda \right\} \ne 0\right\} , \end{aligned}$$

then

$$\begin{aligned} \kappa (\alpha ) \int _{{\mathbb {R}}}\frac{ \left| \xi \right| ^{2\alpha -1}}{\xi ^2+\eta +\lambda }d\xi = \left( \lambda +\eta \right) ^{\alpha -1}. \end{aligned}$$

$\square$

Lemma B.3

Let $0<\alpha <1,$ $\eta \ge 0$, $f_5\in L^2({\mathbb {R}})$, then the following two integrals

$$\begin{aligned} \mathtt {I}_1(\eta ,\alpha ,f_5)= \gamma \kappa (\alpha ) \int _{{\mathbb {R}}}\frac{|\xi |^{\frac{2\alpha -1}{2}}f_5(\xi )}{1+\eta +\xi ^2}d\xi \ \ \ \text {and}\ \ \ \mathtt {I}_2(\eta ,\alpha )=\gamma \kappa (\alpha ) \int _{{\mathbb {R}}}\frac{|\xi |^{2\alpha -1}}{1+\eta +\xi ^2}d\xi \end{aligned}$$

are well-defined.

Proof

First, $\mathtt {I}_2(\eta ,\alpha )$ can be written as

$$\begin{aligned} \mathtt {I}_2(\eta ,\alpha )=2\gamma \kappa (\alpha )\int _0^1\frac{\xi ^{2\alpha -1}}{1+\eta +\xi ^2}d\xi +2\gamma \kappa (\alpha )\int _1^{+\infty }\frac{\xi ^{2\alpha -1}}{1+\eta +\xi ^2}d\xi . \end{aligned}$$

We have

$$\begin{aligned} \frac{\xi ^{2\alpha -1}}{1+\eta +\xi ^2}\underset{0}{\sim }\frac{\xi ^{2\alpha -1}}{1+\eta }\quad \text {and}\quad \frac{\xi ^{2\alpha -1}}{1+\eta +\xi ^2}\underset{+\infty }{\sim }\frac{1}{\xi ^{3-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$, then $\mathtt {I}_2(\eta ,\alpha )$ is well-defined. Now, for $\mathtt {I}_1(\eta ,\alpha ,f_5)$, using Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} \mathtt {I}_1(\eta ,\alpha ,f_5)\le & {} 2 \gamma \kappa (\alpha )\left( \int _0^{+\infty }\frac{\xi ^{2\alpha -1}}{(1+\eta +\xi ^2)^2}d\xi \right) ^{\frac{1}{2}}\left( \int _{{\mathbb {R}}}\left| f_5(\xi )\right| ^2d\xi \right) ^{\frac{1}{2}} \\\le & {} \sqrt{2\gamma \kappa (\alpha )\mathtt {I}_2(\eta ,\alpha )}\left( \int _{{\mathbb {R}}}\left| f_5(\xi )\right| ^2d\xi \right) ^{\frac{1}{2}}. \end{aligned}$$

Since $\mathtt {I}_2(\eta ,\alpha )$ is well defined and $f_5\in L^2({\mathbb {R}})$, then $\mathtt {I}_1(\eta ,\alpha ,f_5)$ is well-defined. The proof is thus complete. $\square$

Lemma B.4

Let $0<\alpha <1$ and $f_5\in L^2({\mathbb {R}})$. Assume that ($\eta >0$ and $\lambda \in {\mathbb {R}}$) or ($\eta =0$ and $\lambda \in {\mathbb {R}}^{*}$), then the following integrals

$$\begin{aligned} \mathtt {I}_3(\lambda ,\eta ,\alpha )= & {} \gamma \kappa (\alpha )\int _{{\mathbb {R}}} \frac{|\xi |^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}d\xi ,\quad \\ \mathtt {I}_4(\lambda ,\eta ,\alpha )= & {} \gamma \kappa (\alpha ) \int _{{\mathbb {R}}}\frac{|\xi |^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}d\xi , \\ \mathtt {I}_5(\lambda ,\eta ,\alpha ,f_5)= & {} i\lambda \gamma \kappa (\alpha )\int _{{\mathbb {R}}}\frac{f_5(\xi ) |\xi |^{\frac{2\alpha -1}{2}}}{\lambda ^2+(\xi ^2+\eta )^2}d\xi \quad \text {and}\quad \\ \mathtt {I}_6(\lambda ,\eta ,\alpha ,f_5)= & {} -\gamma \kappa (\alpha )\int _{{\mathbb {R}}} \frac{f_5(\xi )|\xi |^{\frac{2\alpha -1}{2}}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}d\xi . \end{aligned}$$

are well-defined.

Proof

First, we have

$$\begin{aligned} \mathtt {I}_3(\lambda ,\eta ,\alpha )= & {} 2\gamma \kappa (\alpha )\int _{0}^{+\infty }\frac{\xi ^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}d\xi =2\gamma \kappa (\alpha )\int _0^1\frac{\xi ^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}d\xi \\&+2\gamma \kappa (\alpha )\int _1^{+\infty }\frac{\xi ^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}d\xi . \end{aligned}$$

We have

$$\begin{aligned} \frac{\xi ^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}\underset{0}{\sim }\frac{\xi ^{2\alpha -1}}{\lambda ^2+\eta ^2}\quad \text {and}\quad \frac{\xi ^{2\alpha -1}}{\lambda ^2+(\xi ^2+\eta )^2}\underset{+\infty }{\sim }\frac{1}{\xi ^{5-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$, then in both cases where ($\eta >0$ and $\lambda \in {\mathbb {R}}$) or ($\eta =0$ and $\lambda \in {\mathbb {R}}^{*}$), we get that $\mathtt {I}_3(\lambda ,\eta ,\alpha )$ is well-defined. Next, we have

$$\begin{aligned} \mathtt {I}_4(\lambda ,\eta ,\alpha )= & {} 2\gamma \kappa (\alpha )\int _0^{+\infty }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}d\xi =2\gamma \kappa (\alpha )\int _0^{1}\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}d\xi \\&+2\gamma \kappa (\alpha )\int _1^{+\infty }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}d\xi . \end{aligned}$$

We have

$$\begin{aligned} \frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}\underset{0}{\sim }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+\eta ^2}\quad \text {and}\quad \frac{\xi ^{2\alpha -1}(\xi ^2+\eta )}{\lambda ^2+(\xi ^2+\eta )^2}\underset{+\infty }{\sim }\frac{1}{\xi ^{3-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$, then in both cases where ($\eta >0$ and $\lambda \in {\mathbb {R}}$) or ($\eta =0$ and $\lambda \in {\mathbb {R}}^{*}$), we get that $\mathtt {I}_4(\lambda ,\eta ,\alpha )$ is well-defined. Now, using Cauchy-Schwarz inequality, we get

$$\begin{aligned} \left| \mathtt {I}_5(\lambda ,\eta ,\alpha ,f_5)\right| \le 2|\lambda |\gamma \kappa (\alpha )\left( \int _0^{+\infty }\frac{\xi ^{2\alpha -1}}{\left( \lambda ^2+(\xi ^2+\eta )^2\right) ^2}d\xi \right) ^{\frac{1}{2}}\left( \int _{{\mathbb {R}}}|f_5(\xi )|^2d\xi \right) ^{\frac{1}{2}}. \end{aligned}$$

We have

$$\begin{aligned} \frac{\xi ^{2\alpha -1}}{\left( \lambda ^2+(\xi ^2+\eta )^2\right) ^2}\underset{0}{\sim }\frac{\xi ^{2\alpha -1}}{\left( \lambda ^2+\eta ^2\right) ^2}\quad \text {and}\quad \frac{\xi ^{2\alpha -1}}{\left( \lambda ^2+(\xi ^2+\eta )^2\right) ^2}\underset{+\infty }{\sim } \frac{1}{\xi ^{9-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$, then

$$\begin{aligned} \int _0^{+\infty }\frac{\xi ^{2\alpha -1}}{(\lambda ^2+(\xi ^2+\eta )^2)^2}d\xi <+\infty . \end{aligned}$$

Thus, using the fact that $f_5\in L^2({\mathbb {R}})$, we get that $\mathtt {I}_5(\lambda ,\eta ,\alpha ,f_5)$ is well-defined. In the same way, using Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} |\mathtt {I}_6(\lambda ,\eta ,\alpha ,f_5)|\le 2\gamma \kappa (\alpha )\left( \int _0^{+\infty }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )^2}{(\lambda ^2+(\xi ^2+\eta )^2)^2}d\xi \right) ^\frac{1}{2}\left( \int _{{\mathbb {R}}}|f_5(\xi )|^2d\xi \right) ^{\frac{1}{2}}. \end{aligned}$$

We have

$$\begin{aligned} \frac{\xi ^{2\alpha -1}(\xi ^2+\eta )^2}{(\lambda ^2+(\xi ^2+\eta )^2)^2}\underset{0}{\sim }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )^2}{\left( \lambda ^2+\eta ^2\right) ^2}\quad \text {and}\quad \frac{\xi ^{2\alpha -1}(\xi ^2+\eta )^2}{(\lambda ^2+(\xi ^2+\eta )^2)^2}\underset{+\infty }{\sim } \frac{1}{\xi ^{5-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$, then

$$\begin{aligned} \int _0^{+\infty }\frac{\xi ^{2\alpha -1}(\xi ^2+\eta )^2}{(\lambda ^2+(\xi ^2+\eta )^2)^2}<+\infty . \end{aligned}$$

Finally, since $f_5\in L^2({\mathbb {R}})$, then $\mathtt {I}_6(\lambda ,\eta ,\alpha ,f_5)$ is well-defined. The proof is thus complete. $\square$

Lemma B.5

Let $0<\alpha <1$, $\eta >0$ and $\lambda >0$, then

$$\begin{aligned} \left\{ \begin{array}{ll} \mathtt {I}_7(\eta ,\alpha )=\gamma \kappa (\alpha )\left( \int _{{\mathbb {R}}}\frac{|\xi |^{2\alpha -1}}{\xi ^2+\eta }d\xi \right) ^{\frac{1}{2}},\quad \text { is well-defined},\\ \mathtt {I}_8(\lambda ,\eta ,\alpha )=\int _{{\mathbb {R}}}\frac{ \left| \xi \right| ^{\alpha +\frac{1}{2}}}{(\lambda +\xi ^2+\eta )^{2}}d\xi =c_1\left( \lambda +\eta \right) ^{\frac{\alpha }{2}-\frac{5}{4}},\\ \mathtt {I}_9(\lambda ,\eta )=\left( \int _{{\mathbb {R}}}\frac{1}{\left( \lambda +\xi ^2+\eta \right) ^2}d\xi \right) ^{\frac{1}{2}}=\sqrt{\frac{\pi }{2}}\frac{1}{\left( \lambda +\eta \right) ^{\frac{3}{4}}},\\ \mathtt {I}_{10}(\lambda ,\eta )=\left( \int _{{\mathbb {R}}}\frac{ \xi ^2}{(\lambda +\xi ^2+\eta )^{4}} d\xi \right) ^{\frac{1}{2}}=\frac{\sqrt{\pi }}{4}\frac{1}{\left( \lambda +\eta \right) ^{\frac{5}{4}}}, \end{array}\right. \end{aligned}$$

where $c_1$ is a positive constant number independent of $\lambda$.

Proof

First, for $\mathtt {I}_7$, we have

$$\begin{aligned} \frac{|\xi |^{2\alpha -1}}{\xi ^2+\eta }\underset{0}{\sim }\frac{\xi ^{2\alpha -1}}{\eta }\quad \text {and}\quad \frac{|\xi |^{2\alpha -1}}{\xi ^2+\eta }\underset{+\infty }{\sim } \frac{1}{\xi ^{3-2\alpha }}. \end{aligned}$$

Since $0<\alpha <1$ and $\eta >0$, then $\mathtt {I}_7$ is well-defined. Next, $\mathtt {I}_8$ can be written as

$$\begin{aligned} \mathtt {I}_8(\lambda ,\eta ,\alpha )=\frac{2}{\left( \lambda +\eta \right) ^{2}}\int _{0}^{\infty }\frac{ \xi ^{\alpha +\frac{1}{2}}}{\left( 1+\frac{\xi ^2}{\lambda +\eta }\right) ^{2}}d\xi . \end{aligned}$$

(B.1)

Thus, equation (B.1) may be simplified by defining a new variable $y=1+\frac{\xi ^2}{\lambda +\eta }$. Substituting $\xi$ by $\left( y-1\right) ^{\frac{1}{2}}\left( \lambda +\eta \right) ^{\frac{1}{2}}$ in (B.1), we get

$$\begin{aligned} \mathtt {I}_8(\lambda ,\eta ,\alpha )=\left( \lambda +\eta \right) ^{\frac{\alpha }{2}-\frac{5}{4}}\int _{1}^{\infty }\frac{\left( y-1\right) ^{\frac{\alpha }{2}-\frac{1}{4}}}{y^2}dy. \end{aligned}$$

Using the fact that $\alpha \in ]0,1[$, it easy to see that $y^{-2}\left( y-1\right) ^{\frac{\alpha }{2}-\frac{1}{4}}\in L^1(1,+\infty )$, therefore there exists $c_1>0$ such that

$$\begin{aligned} \mathtt {I}_8(\lambda ,\eta ,\alpha )=c_1\left( \lambda +\eta \right) ^{\frac{\alpha }{2}-\frac{5}{4}}. \end{aligned}$$

Now, $\mathtt {I}_9(\lambda ,\eta )$ can be written as

$$\begin{aligned} \left( \mathtt {I}_9(\lambda ,\eta )\right) ^2= & {} \frac{2}{(\lambda +\eta )^2}\int _0^{\infty }\frac{1}{\left( 1+\left( \frac{\xi }{\sqrt{\lambda +\eta }}\right) ^2\right) ^2}d\xi =\frac{2}{(\lambda +\eta )^{\frac{3}{2}}}\int _0^{\infty }\frac{1}{\left( 1+s^2\right) ^2}ds \\= & {} \frac{2}{(\lambda +\eta )^{\frac{3}{2}}}\times \frac{\pi }{4}=\frac{\pi }{2(\lambda +\eta )^{\frac{3}{2}}}. \end{aligned}$$

Therefore,  $\mathtt {I}_9(\lambda ,\eta )={\sqrt{\frac{\pi }{2}}\frac{1}{(\lambda +\eta )^{\frac{3}{4}}}}$. Finally, $\mathtt {I}_{10}(\lambda ,\eta )$ can be written as

$$\begin{aligned} \left( \mathtt {I}_{10}(\lambda ,\eta )\right) ^2= & {} \frac{2}{(\lambda +\eta )^4}\int _0^{\infty }\frac{\xi ^2}{\left( 1+\left( \frac{\xi }{\sqrt{\lambda +\eta }}\right) ^2\right) ^4}d\xi \\= & {} \frac{2}{(\lambda +\eta )^{\frac{5}{2}}}\int _0^{\infty }\frac{s^2}{\left( 1+s^2\right) ^4}ds=\frac{2}{(\lambda +\eta )^{\frac{5}{2}}}\times \frac{\pi }{32}. \end{aligned}$$

Then $\mathtt {I}_{10}(\lambda ,\eta )=\frac{\sqrt{\pi }}{4}\frac{1}{(\lambda +\eta )^{\frac{5}{4}}}$. The proof is thus complete. $\square$