$\displaystyle\times\left\\{b^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}\partial_{\overline{b}}h(b,\Phi)dbd\Phi$

Let us explain the meaning of $L$. The function $L$ means the linearization of
the local part associated to $-(-\Delta)^{-1+\frac{\gamma}{2}}\hat{\theta}$
after performing the adapted coordinates and the appropriate scaling from
$\Psi$ to $f$. That is, take the initial stream function (which is a radial
function):

$\hat{\psi}^{0}(re^{i\theta})=-C_{0}r^{1-\alpha},$

and now we use the fact that
$r=(1+\alpha)^{\frac{1}{2}}(-\Psi_{\beta})^{\frac{1}{2}}$, where we recall

$\Psi=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{\alpha-1}{1+\alpha}}(1+f).$

Hence, after the scaling done from $F$ to $\tilde{F}$ in Section 2.4, we
achieve

$\displaystyle L(h)=$
$\displaystyle\beta^{-\frac{\alpha-1}{1+\alpha}}\left\\{-C_{0}^{\frac{-2}{1+\alpha}}(\alpha-1)^{\frac{\alpha-1}{1+\alpha}}\right\\}\partial_{f}\left[-C_{0}(1+\alpha)^{\frac{1-\alpha}{2}}(-\Psi_{\beta})^{\frac{1-\alpha}{2}}\right]h$
$\displaystyle=$
$\displaystyle-\beta^{-\frac{\alpha-1}{1+\alpha}}\left\\{-C_{0}^{\frac{-2}{1+\alpha}}(\alpha-1)^{\frac{\alpha-1}{1+\alpha}}\right\\}C_{0}(1+\alpha)^{\frac{1-\alpha}{2}}\frac{1-\alpha}{2}\left[(-\Psi_{\beta}^{0})^{\frac{1-\alpha}{2}-1}\right]C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\partial_{\beta}(\beta^{\frac{\alpha-1}{1+\alpha}}h).$

Using that
$\partial_{\beta}(\beta^{\frac{\alpha-1}{1+\alpha}}h)=\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{\beta}}h$
together with

$(-\Psi_{\beta}^{0})(\beta)=C_{0}^{\frac{2}{1+\alpha}}\frac{(\alpha-1)^{\frac{2}{1+\alpha}}}{1+\alpha}\beta^{\frac{-2}{1+\alpha}},$

we find

$\displaystyle L(h)=$
$\displaystyle-\beta^{-\frac{\alpha-1}{1+\alpha}}\left\\{-C_{0}^{\frac{-2}{1+\alpha}}(\alpha-1)^{\frac{\alpha-1}{1+\alpha}}\right\\}C_{0}(1+\alpha)^{\frac{1-\alpha}{2}}\frac{1-\alpha}{2}C_{0}^{\frac{1-\alpha}{1+\alpha}}\frac{(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}}{(1+\alpha)^{\frac{1-\alpha}{2}}}\beta^{\frac{\alpha-1}{1+\alpha}}C_{0}^{\frac{-2}{1+\alpha}}\frac{(\alpha-1)^{\frac{-2}{1+\alpha}}}{(1+\alpha)^{-1}}\beta^{\frac{2}{1+\alpha}}$
$\displaystyle\times
C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{\beta}}h$
$\displaystyle=$
$\displaystyle-\frac{(1+\alpha)}{2}\partial_{\overline{\beta}}h$

Using now the expression of $\partial_{\overline{\beta}}h$ we finally get

$L(h)=-\frac{\alpha-1}{2}h-\frac{(1+\alpha)}{2}\beta h_{\beta}$

Hence, we find

$\displaystyle\partial_{f}\tilde{F}(0,1)h(\beta,\phi)=$
$\displaystyle\frac{(1+\alpha)}{2}\partial_{\beta}(\beta h(\beta,\phi))$
$\displaystyle-\frac{C_{\gamma}}{4\pi
C_{0}}\frac{1+\alpha}{\alpha-1}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{b\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\frac{1}{\alpha-1}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\partial_{\overline{\varPsi}}h(b,\Phi)dbd\Phi}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}},$
$\displaystyle+\frac{\gamma}{2}\frac{C_{\gamma}}{2\pi
C_{0}}\frac{1}{1+\alpha}\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\beta^{-\frac{\gamma}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}+1}}$
$\displaystyle\times\left\\{b^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}\partial_{\overline{b}}h(b,\Phi)dbd\Phi.$

∎

######  Remark 4.2.

The denominator $D(0)$ agrees with

$\displaystyle
D(0)(\beta,\phi,b,\Phi)=\left(\frac{\alpha-1}{1+\alpha}\right)\left\\{(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})^{2}+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\phi+\beta-\Phi-b)\right\\}.$

In the following proposition, we show an extra cancellation of the linearized
operator amounting to a more compact expression of it.

######  Proposition 4.3.

The linearized operator in Proposition 4.1 can be written as

$\displaystyle\partial_{f}\tilde{F}(0,1)h(\beta,\phi)=\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h(\beta,\phi))$ $\displaystyle-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varPsi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}.$
(85)

###### Proof.

To obtain such expression, we will do an integration by parts in the last term
of $I$ in (4.1).

First, note that using $\partial_{\varPsi}=\partial_{\Phi}-\partial_{b}$ we
get

$\displaystyle\partial_{\varPsi}\frac{1}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}=$
$\displaystyle-\frac{\gamma}{2}\frac{2}{1+\alpha}\left(\frac{\alpha-1}{1+\alpha}\right)b^{-1}\frac{b^{\frac{-2}{1+\alpha}}-b^{\frac{-1}{1+\alpha}}\beta^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}+1}}.$

Then

$\displaystyle I_{3}=$ $\displaystyle-\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{\varPsi}\frac{1}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle=$ $\displaystyle-\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{\Phi}\frac{1}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle+\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{\gamma}{2}}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{b}\frac{1}{D(0)(\beta,\phi,b,\Phi)^{\frac{\gamma}{2}}}dbd\Phi.$

Taking into account the definition of $\varPsi$, we can integrate by parts
there

Let us start with $\partial_{\Phi}$. Note that

$\displaystyle\int_{0}^{2\pi}$ $\displaystyle
b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{\Phi}\frac{1}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}d\Phi$
$\displaystyle=-\int_{0}^{2\pi}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\Phi}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}d\Phi,$

since the boundary terms vanish due to the periodicity of the function.
However, the boundary terms with respect to the integration in $b$ will depend
on the choice of the parameters:

$\displaystyle\int_{0}^{\infty}$ $\displaystyle
b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{b}\frac{1}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}db$
$\displaystyle=$
$\displaystyle\lim_{b\rightarrow+\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle-\lim_{b\rightarrow
0}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle-\int_{0}^{\infty}\frac{\frac{-\gamma-1+\alpha}{1+\alpha}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)+b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{b}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}db.$

Since $\partial_{\overline{b}}h\in\mathcal{X}^{\sigma,\delta}$, we have that

$\displaystyle\lim_{b\rightarrow+\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}=\lim_{b\rightarrow+\infty}\beta^{\frac{\gamma}{1+\alpha}}b^{\frac{-\gamma}{1+\alpha}}b^{\frac{\alpha-1}{1+\alpha}}b^{-\sigma}=0,$

since $\gamma>\alpha-1$ by (56). On the other hand

$\displaystyle\lim_{b\rightarrow
0}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}=\lim_{b\rightarrow
0}b^{\frac{\gamma}{1+\alpha}}b^{\frac{-\gamma}{1+\alpha}}b^{\frac{\alpha-1}{1+\alpha}}b^{\sigma}=0,$

since $\alpha>1$. Hence, in this term the boundary terms also vanish obtaining

$\displaystyle\int_{0}^{\infty}$ $\displaystyle
b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\overline{b}}h(b,\Phi)\partial_{b}\frac{1}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}db$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}\frac{\frac{-\gamma-2+\alpha}{1+\alpha}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)+b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{b}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}db.$

By putting everything together we find

$\displaystyle\mathcal{I}_{3}=$ $\displaystyle\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\Phi}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi,$
$\displaystyle-\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\frac{-\gamma-2+\alpha}{1+\alpha}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)+b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{b}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle=$ $\displaystyle\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b\partial_{\varPsi}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi,$
$\displaystyle-\frac{1/\mu-2}{1+\alpha}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$

Note now

$b\partial_{\varPsi}\partial_{\overline{b}}h=\partial_{\overline{b\varPsi}}h-\frac{2}{1+\alpha}\partial_{\overline{b}}h,$

implying

$\displaystyle\mathcal{I}_{3}=$ $\displaystyle\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b\varPsi}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi,$
$\displaystyle-\frac{2}{1+\alpha}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi,$
$\displaystyle-\frac{1/\mu-2}{1+\alpha}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$ $\displaystyle\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b\varPsi}}h(b,\Phi)}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi,$
$\displaystyle-\frac{1/\mu}{1+\alpha}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1+\alpha}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\partial_{\overline{b}}h(b,\Phi)dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}.$

Joining it with $I_{1}$ and $I_{2}$ we find

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h(\beta,\phi))$ $\displaystyle-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varPsi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}.$

That concludes the proof. ∎

###  4.2. Fredholmness at the trivial solution

By using the expression from Proposition 4.3 we can prove that the linear
operator is Fredholm of zero index in the following proposition. Since our
goal is the use of the Implicit Function theorem, and then, proving that
$\partial_{f}F(0,1)$ is an isomorphism we need to study its kernel and range.
However, by the general theory on Fredholm operators we can restrict ourselves
to check the kernel.

######  Proposition 4.4.

Let $\gamma,\sigma\in(0,1)$ and $\alpha\in(1,2)$ satisfying (56). Then the
linear operator $\partial_{f}\tilde{F}(0,1):X_{1}^{\sigma,\delta}\rightarrow
Y^{\sigma,\delta}$ is Fredholm of zero index, for any
$\delta<\min\\{1-\gamma,\sigma\\}$.

###### Proof.

We can decompose the linear operator as follows

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h(\beta,\phi))$ (86) $\displaystyle-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varphi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
(87) $\displaystyle=$
$\displaystyle\frac{(1+\alpha)}{2}\mathcal{L}_{1}-\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\mathcal{L}_{2},$ (88)

where

$\mathcal{L}_{1}(h):=\partial_{\beta}(\beta h(\beta,\phi)),$

and

$\mathcal{L}_{2}(h):=\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varphi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}.$

The idea of the proof is to show that
$\mathcal{L}_{1}:X_{1}^{\sigma,\delta}\rightarrow Y^{\sigma,\delta}$ is an
isomorphism, whereas $\mathcal{L}_{2}:X_{1}^{\sigma,\delta}\rightarrow
Y^{\sigma,\delta}$ is compact. Then, since compact perturbations of Fredholm
operators remain being Fredholm of zero index, we can conclude the proof.

$\bullet$ $\mathcal{L}_{1}:X_{1}^{\sigma,\delta}\rightarrow Y^{\sigma,\delta}$
is an isomorphism.

Take $g\in Y^{\sigma,\delta}$, we wish to solve $\mathcal{L}_{1}(h)=g$. We can
explicitly solve such differential equation as

$h(\beta,\phi)=\frac{K}{\beta}+\frac{1}{\beta}\int_{0}^{\beta}g(s,\phi)ds.$

for any $K$. However, $h$ must lie in $X_{1}^{\sigma,\delta}$. In particular,
$h$ vanishes at $0$ and $K=0$:

$h(\beta,\phi)=\frac{1}{\beta}\int_{0}^{\beta}g(s,\phi)ds=\int_{0}^{1}g(s\beta,\phi).$
(89)

Let us now check that if $g\in Y^{\sigma,\delta}$, then $h\in
X_{1}^{\sigma,\delta}$, where the norm of $X_{1}^{\sigma,\delta}$ is given in
(46). Let us start showing that

$||h||_{\mathcal{X}^{\sigma,\delta}}=||h||_{L^{\infty}_{\sigma}}+||h||_{C^{\delta}_{\beta}}<C.$

We find that

$\displaystyle||h||_{L^{\infty}_{\sigma}}=\sup_{\beta\in[0,+\infty)}\frac{(1+\beta)^{2\sigma}}{\beta^{\sigma}}\int_{0}^{1}g(s\beta,\phi)ds\leq||g||_{L^{\infty}_{\sigma}}\sup_{\beta\in[0,+\infty)}(1+\beta)^{2\sigma}\int_{0}^{1}\frac{s^{\sigma}}{(1+s\beta)^{2\sigma}}ds<C,$

since $\sigma<1$. Let us now show that $||h||_{C^{\delta}_{\beta}}<C$:

$\displaystyle|h(\beta,\phi)-h(b,\phi)|\leq\int_{0}^{1}|g(s\beta,\phi)-g(sb,\phi)|ds\leq
C||g||_{C^{\delta}_{\beta}}\int_{0}^{1}\frac{|s\beta-
sb|^{\alpha}}{|s\beta+sb|^{\alpha}}ds\leq
C||g||_{C^{\delta}_{\beta}}\frac{|\beta-b|^{\delta}}{|\beta+b|^{\delta}},$

implying that $||h||_{C^{\delta}_{\beta}}<C||g||_{C^{\delta}_{\beta}}$. Let us
show that $||h||_{\textnormal{Lip}_{\phi}}<C$:

$|h(\beta,\phi)-h(\beta,\Phi)|\leq\int_{0}^{1}|g(s\beta,\phi)-g(s\beta,\Phi)|ds\leq||g||_{\textnormal{Lip}_{\phi}}\sin((\Phi-\phi)/2),$

implying that

$||h||_{\textnormal{Lip}_{\phi}}<||g||_{\textnormal{Lip}_{\phi}}.$

Let us move to $\beta h_{\beta}$ and note that

$\beta h_{\beta}(\beta,\phi)=-h(\beta,\phi)+g(\beta,\phi),$

and then

$\displaystyle||\beta h_{\beta}||_{\mathcal{X}^{\sigma,\delta}}+||\beta
h_{\beta}||_{\textnormal{Lip}_{\phi}}\leq$
$\displaystyle||h||_{\mathcal{X}^{\sigma,\delta}}+||h||_{\textnormal{Lip}_{\phi}}$
$\displaystyle+||g||_{\mathcal{X}^{\sigma,\delta}}+||g||_{\textnormal{Lip}_{\phi}}$
$\displaystyle\leq$ $\displaystyle C.$

Next, we will show that $\beta h_{\varphi}\in\mathcal{X}^{\sigma,\delta}$ and
we will separate in $\beta\in[0,\beta_{0})$ and $\beta\in[\beta_{0},+\infty)$.
For the first case, it can be written as

$\displaystyle h_{\varphi}=$
$\displaystyle\frac{1}{\beta}\int_{0}^{\beta}g_{\phi}(s,\phi)ds+\frac{1}{\beta}h(\beta,\phi)-\frac{1}{\beta}g(\beta,\phi)$
$\displaystyle=$
$\displaystyle\frac{1}{\beta}\int_{0}^{\beta}g_{\varphi}(s,\phi)ds+\frac{1}{\beta}\int_{0}^{\beta}g_{s}(s,\phi)ds+\frac{1}{\beta}h(\beta,\phi)-\frac{1}{\beta}g(\beta,\phi)$
$\displaystyle=$
$\displaystyle\frac{1}{\beta}\int_{0}^{\beta}g_{\varphi}(s,\phi)ds+\frac{1}{\beta}g(\beta,\phi)ds+\frac{1}{\beta}h(\beta,\phi)-\frac{1}{\beta}g(\beta,\phi),$

and then

$\beta h_{\varphi}=\int_{0}^{\beta}g_{\varphi}(s,\phi)ds+h(\beta,\phi).$ (90)

Since we have showed that $h\in\mathcal{X}^{\sigma,\delta}$, it remains only
to check the first term:

$\displaystyle\sup_{\beta\in[0,\beta_{0})}\frac{(1+\beta)^{2\sigma}}{\beta^{\sigma}}\int_{0}^{\beta}g_{\varphi}(s,\phi)ds\leq$
$\displaystyle||\beta
g_{\varphi}||_{L^{\infty}_{\sigma}}\sup_{\beta\in[0,\beta_{0})}\frac{1}{\beta^{\sigma}}\int_{0}^{\beta}\frac{1}{s}\frac{s^{\sigma}}{(1+s)^{2\sigma}}ds$
$\displaystyle\leq$ $\displaystyle||\beta
g_{\varphi}||_{L^{\infty}_{\sigma}}\sup_{\beta\in[0,\beta_{0})}\int_{0}^{1}s^{\sigma-1}ds$
$\displaystyle\leq$ $\displaystyle||\beta
g_{\varphi}||_{L^{\infty}_{\sigma}}.$

In the case $\beta\in[\beta_{0},+\infty)$ we write $h$ as

$h(\beta,\phi)=\frac{1}{\beta}\int_{\infty}^{\beta}g(s,\phi)ds,$

instead of (89) by using that $h$ must vanish at $0$ and $+\infty$. In that
case, we have

$\beta h_{\varphi}=\int_{\infty}^{\beta}g_{\varphi}(s,\phi)ds+h(\beta,\phi),$
(91)

and hence it remains to check the first term:

$\displaystyle\sup_{\beta\in[\beta_{0},+\infty)}\frac{(1+\beta)^{2\sigma}}{\beta^{\sigma}}\int_{\infty}^{\beta}g_{\varphi}(s,\phi)ds\leq$
$\displaystyle||\beta
g_{\varphi}||_{L^{\infty}_{\sigma}}\sup_{\beta\in[\beta_{0},+\infty)}\beta^{\sigma}\int_{\infty}^{\beta}\frac{1}{s^{1+\sigma}}ds$
$\displaystyle\lesssim$ $\displaystyle||\beta
g_{\varphi}||_{L^{\infty}_{\sigma}},$

since $\sigma\in(0,1)$. In that way, we have showed that

$||\beta h_{\varphi}||_{L^{\infty}_{\sigma}}\leq C||\beta g_{\varphi}||.$

Let us now check that $||\beta h_{\varphi}||_{C^{\delta}_{\beta}}<C$, which
amounts to check such property to the first term in (90), that is,

$\displaystyle\left|\int_{\beta}^{b}g_{\varphi}(s,\phi)ds\right|\leq
C\frac{|\beta-b|^{\delta}}{|\beta+b|^{\delta}}.$

Take first $|\beta+b|<\beta_{0}<1$, and then

$\displaystyle\left|\int_{\beta}^{b}g_{\varphi}(s,\phi)ds\right|\leq
C||g_{\varphi}||_{\mathcal{X}^{\sigma,\delta}}\int_{\beta}^{b}s^{\sigma-1}ds\leq
C|\beta-b|^{\sigma}\leq C\frac{|\beta-b|^{\delta}}{|\beta+b|^{\delta}},$

if $\delta<\sigma<1$. In the case $||\beta+b||>\beta_{0}$ we use the
expression of $\beta h_{\varphi}$ we find

$\displaystyle\left|\int_{\beta}^{b}g_{\varphi}(s,\phi)ds\right|\leq
C||g_{\varphi}||_{\mathcal{X}^{\sigma,\delta}}\int_{\beta}^{b}s^{-\sigma-1}ds\leq
C\left|\frac{1}{\beta^{\sigma}}-\frac{1}{b^{\sigma}}\right|\leq
C\frac{|\beta-b|^{\sigma}}{|\beta+b|^{2\sigma}}\leq
C\frac{|\beta-b|^{\sigma}}{|\beta+b|^{\sigma}}\leq
C\frac{|\beta-b|^{\delta}}{|\beta+b|^{\delta}}.$

Let us finish with $\beta^{2}h_{\beta,\varphi}$ noting that

$\displaystyle h_{\beta,\varphi}(\beta,\phi)=$
$\displaystyle-\frac{1}{\beta}\frac{1}{\beta}\int_{0}^{\beta}g_{\varphi}(s,\phi)ds+\frac{1}{\beta}g_{\varphi}(\beta,\phi)+\frac{1}{\beta}h_{\beta}(\beta,\phi)-\frac{1}{\beta^{2}}h(\beta,\phi)$
$\displaystyle=-\frac{1}{\beta}(h_{\varphi}(\beta,\phi)-\frac{1}{\beta}h(\beta,\phi))+\frac{1}{\beta}g_{\varphi}(\beta,\phi)+\frac{1}{\beta}h_{\beta}(\beta,\phi)-\frac{1}{\beta^{2}}h(\beta,\phi),$

and then

$\beta^{2}h_{\beta,\varphi}(\beta,\phi)=-\beta h_{\varphi}(\beta,\phi)+\beta
g_{\varphi}(\beta,\phi)+\beta h_{\beta}(\beta,\phi),$

implying then that $\beta^{2}h_{\beta,\varphi}\in\mathcal{X_{\sigma,\alpha}}.$

$\bullet$ $\mathcal{L}_{2}:X_{1}^{\sigma,\delta}\rightarrow Y^{\sigma,\delta}$
is compact. From Proposition 3.10 we have that $\mathcal{L}_{2}\in
Y^{\sigma,\delta}$. To prove that $\mathcal{L}_{2}$ is compact, we will prove
that $\mathcal{L}_{2}\in Y^{\sigma,\delta^{\prime}}$, with
$\delta^{\prime}>\delta$. Then, since the embedding
$Y^{\sigma,\delta^{\prime}}\subset Y^{\sigma,\delta}$ is compact (we refer to
[65, Lemma 2.2]) we have that
$\mathcal{L}_{2}:X_{1}^{\sigma,\delta}\rightarrow Y^{\sigma,\delta}$ is a
compact operator.

Take $\mathcal{L}_{2}$ and let us prove that
$\beta\partial_{\varphi}\mathcal{L}_{2}\in C^{\delta^{\prime}}_{\beta}$ with
$\delta^{\prime}>\delta$. For that, we compute
$\partial_{\varphi}\mathcal{L}_{2}$:

$\displaystyle\partial_{\varphi}\mathcal{L}_{2}(h)=C\beta^{-1}\mathcal{L}_{2}(h)$
$\displaystyle+C\beta^{-1}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varphi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))\left\\{\beta^{\frac{-2}{1+\alpha}}-\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\right\\}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{2+\gamma}{2}}},$

for some constant $C$. Then

$\displaystyle\beta\partial_{\varphi}\mathcal{L}_{2}(h)=C\mathcal{L}_{2}(h)$
$\displaystyle+C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varphi}}h(b,\Phi)+\partial_{\overline{b}}h(b,\Phi))\left\\{\beta^{\frac{-1}{1+\alpha}}(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\phi+\beta-\Phi-b))\right\\}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{2+\gamma}{2}}}.$

Note that in the previous expression the singularity is critical. To solve
this problem we should integrate by parts somehow but we can not differentiate
$\partial_{\overline{\varphi}}h$ with respect to $\partial_{\varphi}$ and then
we have to do the following decomposition:

$\displaystyle\beta\partial_{\varphi}\mathcal{L}_{2}(h)=C\mathcal{L}_{2}(h)$
$\displaystyle+C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{\varphi}}h(b,\Phi)-\partial_{\overline{\varphi}}h(\beta,\phi))\left\\{\beta^{\frac{-1}{1+\alpha}}(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\phi+\beta-\Phi-b))\right\\}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{2+\gamma}{2}}}$
$\displaystyle+C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))\left\\{\beta^{\frac{-1}{1+\alpha}}(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\phi+\beta-\Phi-b))\right\\}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{2+\gamma}{2}}}$
$\displaystyle+C\partial_{\overline{\varphi}}h(\beta,\phi)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{\beta^{\frac{-1}{1+\alpha}}(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\phi+\beta-\Phi-b))\right\\}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{2+\gamma}{2}}}$
$\displaystyle=:$ $\displaystyle
C\mathcal{L}_{2}(h)+L_{2,1}(h)+L_{2,2}(h)+L_{2,3}(h).$

Let us show the details for one of the terms and we have chosen $L_{2,2}$ due
to its difficulty. Note that the singularity coming from the denominator is
not a priori integrable and we need to use some extra cancellation. To achieve
that, we need to integrate by parts and then we have to add and subtract some
appropriate function in the denominator. Indeed, denote

$J(\beta,\phi,b,\Phi):=\\{\beta_{1}^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta_{1}^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta_{1}-\Phi-b)\\}.$

Notice that $J$ is indeed the denominator at the trivial solution and then
$J=CD(0)$, for some constant $C$, where $D(f)$ is defined in (51). In
particular, following Proposition 3.7 and the particular expression for $J$
one finds

$\displaystyle|J(\beta,\phi,b,\Phi)|\geq
C(\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}})\left\\{\frac{|\beta-b|^{2}}{|\beta+b|^{2}}+\sin^{2}((\beta-b+\phi-\Phi)/2)\right\\}.$
(92)

By using the expression of $J$ we also find

$\displaystyle\partial_{\varphi}J(\beta,\phi,b,\Phi)=\frac{2}{1+\alpha}\beta^{-1}\left\\{\beta^{\frac{-1}{1+\alpha}}(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})+\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\beta+\phi-b-\Phi)\right\\},$

and then

$\displaystyle\beta\partial_{\varphi}J(\beta,\phi,b,\Phi)+b\partial_{\varPsi}J(\beta,\phi,b,\Phi)=$
$\displaystyle\frac{2}{1+\alpha}\left\\{(\beta^{\frac{-1}{1+\alpha}}-b^{\frac{-1}{1+\alpha}})^{2}+2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}(1-\cos(\beta+\phi-b-\Phi)\right\\}$
$\displaystyle=$ $\displaystyle\frac{2}{1+\alpha}J(\beta,\phi,b,\Phi).$

Motivated by the previous computation, we decompose $L_{2,2}$ as follows

$\displaystyle L_{2,2}(h)(\beta)=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle+\frac{2}{\gamma}C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}bb^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))$
$\displaystyle\partial_{\varPsi}\frac{1}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi.$

Integrating by parts in the last term we have that the boundary terms vanish
since $h\in X_{1}^{\sigma,\delta}$ (we refer to the proof of Proposition 4.3)
and then

$\displaystyle L_{2,2}(h)(\beta)=$ $\displaystyle
C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle-\frac{2}{\gamma}C\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{bb^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b},\varPsi}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle-\frac{2}{\gamma}C\frac{1+\gamma-\alpha}{1+\alpha}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle=:$ $\displaystyle L_{2,2,1}+L_{2,2,2}+L_{2,2,3}.$

Take now $\beta_{1}<\beta_{2}$ implying

$\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}<\beta_{1}^{-\frac{\alpha-1}{1+\alpha}},$
(93)

and let us show the Hölder regularity for $L_{2,2}$. By implementing the
difference of $L_{2,2}$ in the two points we find

$\displaystyle|L_{2,2}(h)(\beta_{1})-$ $\displaystyle
L_{2,2}(h)(\beta_{2})|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta^{\prime}}}{|\beta_{1}+\beta_{2}|^{\delta^{\prime}}}(\beta_{1}^{-\frac{\alpha-1}{1+\alpha}}+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}})\Big{\\{}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}dbd\Phi$
$\displaystyle+\int_{0}^{2\pi}\int_{0}^{\infty}\frac{bb^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b},\varPsi}h(b,\Phi))dbd\Phi}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}$
$\displaystyle+\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))dbd\Phi}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}}\Big{\\}}$
$\displaystyle+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|}{J(\beta_{1},\phi,b,\Phi)^{\frac{2-\gamma}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{2-\gamma}{2}}}dbd\Phi$
$\displaystyle+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{bb^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b},\varPsi}h(b,\Phi))}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|}{J(\beta_{1},\phi,b,\Phi)^{\frac{2-\gamma}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{2-\gamma}{2}}}dbd\Phi$
$\displaystyle+\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}(\partial_{\overline{b}}h(b,\Phi))}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|}{J(\beta_{1},\phi,b,\Phi)^{\frac{2-\gamma}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{2-\gamma}{2}}}dbd\Phi,$

where $\delta^{\prime}$ is any number in $(0,1)$. Taking into account (93) the
first three terms can be related to the operator $\mathcal{H}$ defined in
(55), and for the last three we use that $h\in X_{1}^{\sigma,\delta}$
obtaining

$\displaystyle|L_{2,2}(h)(\beta_{1})-L_{2,2}(h)(\beta_{2})|\leq
C\frac{|\beta_{1}-\beta_{2}|^{\delta^{\prime}}}{|\beta_{1}+\beta_{2}|^{\delta^{\prime}}}\Big{\\{}||\mathcal{H}_{0}(\partial_{\overline{b}}h(b,\Phi),1)||_{L^{\infty}}+||\mathcal{H}_{0}(b(\partial_{\overline{b},\varPsi}h(b,\Phi)),1)||_{L^{\infty}}\Big{\\}}$
$\displaystyle+C||h||_{X_{1}^{\sigma,\delta}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|}{J(\beta_{1},\phi,b,\Phi)^{\frac{2-\gamma}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{2-\gamma}{2}}}dbd\Phi.$

By Proposition 3.9 we have that

$||\mathcal{H}_{0}(\partial_{\overline{b}}h(b,\Phi),1)||_{L^{\infty}}+||\mathcal{H}_{0}(b(\partial_{\overline{b},\varPsi}h(b,\Phi)),1)||_{L^{\infty}}\leq
C||h||_{X_{1}^{\sigma,\delta}}$

Next, we need to estimate the difference of $J$ in these two points:

$\displaystyle|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|\leq
C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})(J(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}})$
$\displaystyle+C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}\beta_{2}^{\frac{-1}{1+\alpha}}|\beta_{1}+\beta_{2}|(J(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}})$
(94) $\displaystyle\leq$ $\displaystyle
C\frac{|\beta_{1}-\beta_{2}|}{|\beta_{1}+\beta_{2}|}(1+|\beta_{1}+\beta_{2}|)(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})(J(\beta_{1},\phi,b,\Phi)^{\frac{1}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{1}{2}})$

and interpolating with

$|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|\leq
C|J(\beta_{1},\phi,b,\Phi)+J(\beta_{2},\phi,b,\Phi)|,$

we achieve

$\displaystyle|J(\beta_{1},\phi,b,\Phi)-J(\beta_{2},\phi,b,\Phi)|\leq$
$\displaystyle
C\frac{|\beta_{1}-\beta_{2}|^{\delta^{\prime}}}{|\beta_{1}+\beta_{2}|^{\delta^{\prime}}}(1+|\beta_{1}+\beta_{2}|^{\delta^{\prime}})(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})^{\delta^{\prime}}$
(96)
$\displaystyle\times(J(\beta_{1},\phi,b,\Phi)^{1-\frac{\delta^{\prime}}{2}}+J(\beta_{2},\phi,b,\Phi)^{1-\frac{\delta^{\prime}}{2}}),$
(97)

where $\delta^{\prime}\in(0,1)$. Finally,

$\displaystyle|L_{2,2}(h)(\beta_{1})-L_{2,2}(h)(\beta_{2})|\leq
C||h||_{X_{1}^{\sigma,\delta}}\frac{|\beta_{1}-\beta_{2}|^{\delta^{\prime}}}{|\beta_{1}+\beta_{2}|^{\delta^{\prime}}}\Big{[}1$
$\displaystyle+(1+|\beta_{1}+\beta_{2}|^{\delta^{\prime}})(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})^{\delta^{\prime}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}}{J(\beta_{1},\phi,b,\Phi)^{\frac{\gamma}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\frac{J(\beta_{1},\phi,b,\Phi)^{-\frac{\delta^{\prime}}{2}}+J(\beta_{2},\phi,b,\Phi)^{-\frac{\delta^{\prime}}{2}}}{J(\beta_{1},\phi,b,\Phi)^{\frac{-\gamma}{2}}+J(\beta_{2},\phi,b,\Phi)^{\frac{-\gamma}{2}}}dbd\Phi\Big{]}$
$\displaystyle\leq
C||h||_{X_{1}^{\sigma,\delta}}\frac{|\beta_{1}-\beta_{2}|^{\delta^{\prime}}}{|\beta_{1}+\beta_{2}|^{\delta^{\prime}}}\Big{[}1$
$\displaystyle+(1+|\beta_{1}+\beta_{2}|^{\delta^{\prime}})(\beta_{1}^{\frac{-1}{1+\alpha}}+\beta_{2}^{\frac{-1}{1+\alpha}})^{\delta^{\prime}}\beta_{2}^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}b^{\sigma}/(1+b)^{2\sigma}}$
$\displaystyle\times\left(\frac{1}{J(\beta_{1},\phi,b,\Phi)^{\frac{\delta^{\prime}}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}+\frac{1}{J(\beta_{1},\phi,b,\Phi)^{\frac{\delta^{\prime}}{2}}J(\beta_{2},\phi,b,\Phi)^{\frac{\gamma}{2}}}\right)dbd\Phi\Big{]}$

The last integral is similar to the one in (3.3) and then we find that it is
bounded if $\delta^{\prime}<\sigma$. Furthermore, since $\delta^{\prime}$ is
arbitrary we can choose $\delta^{\prime}>\delta$ to have the compactness. ∎

###  4.3. Expression of the linearized operator in Fourier series

In the next proposition, we show that the linear operator can be computed in
Fourier modes depending on the Fourier coefficients of $h$. Assume then that
$h$ takes the form

$h(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}h_{k}(\beta)e^{ik\phi}.$
(98)

######  Proposition 4.5.

Let $h$ be expressed as (98). Then

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}e^{ik\phi}\left\\{\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h_{k}(\beta))\right.$
$\displaystyle\left.-ik\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)e^{ik\beta}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{e^{-ikb}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)\cos(k\eta)dbd\eta}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}\right\\}.$

###### Proof.

Let us assume that
$h(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}h_{k}(\beta)e^{ik\phi}$ in
the expression of Proposition 4.3. Using the definition of
$\partial_{\overline{\beta}}$ and $\partial_{\overline{\varphi}}$ we find

$\partial_{\overline{\beta}}h(\beta,\phi)=\frac{\alpha-1}{1+\alpha}h(\beta,\phi)+\beta\partial_{\beta}h(\beta,\phi)$

and

$\partial_{\overline{\varphi}}h(\beta,\phi)=-\frac{\alpha-1}{1+\alpha}h(\beta,\phi)+\beta\partial_{\varphi}h(\beta,\phi)$

obtaining

$\partial_{\overline{\varphi}}h(\beta,\phi)+\partial_{\overline{\beta}}h(\beta,\phi)=\beta(\partial_{\varphi}h(\beta,\phi)+\partial_{\beta}h(\beta,\phi))=\beta\partial_{\phi}h(\beta,\phi),$

taking into account that
$\partial_{\varphi}=\partial_{\phi}-\partial_{\beta}$. Hence, we can compute
the Fourier series of the above term as

$\partial_{\overline{\varphi}}h(\beta,\phi)+\partial_{\overline{\beta}}h(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}ik\beta
h_{k}(\beta)e^{ik\phi}.$ (99)

Moreover

$\partial_{\beta}(\beta
h(\beta,\phi))=\sum_{k\in\operatorname{\mathbb{Z}}}\partial_{\beta}(\beta
h_{k}(\beta))e^{ik\phi}.$ (100)

Inserting (99) and (100) into (4.3) we get

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}e^{ik\phi}\left\\{\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h_{k}(\beta))\right.$ (101)
$\displaystyle\left.-ike^{-ik\phi}\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)e^{ik\Phi}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}\right\\}.$

Note that we can write the integral as follows

$\displaystyle\int_{0}^{2\pi}\int_{0}^{\infty}$
$\displaystyle\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)e^{ik\Phi}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle=e^{ik\phi}e^{ik\beta}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{e^{-ikb}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)e^{ik(\Phi+b-\phi-\beta)}dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\phi+\beta-\Phi-b)\\}^{\frac{\gamma}{2}}}$
$\displaystyle=e^{ik\phi}e^{ik\beta}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{e^{-ikb}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)\cos(k\eta)dbd\eta}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}$

Hence, coming back to (101) we find

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}e^{ik\phi}\left\\{\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h_{k}(\beta))\right.$
$\displaystyle\left.-ik\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)e^{ik\beta}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{e^{-ikb}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)\cos(k\eta)dbd\eta}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}\right\\}.$

∎

##  5\. Invertibility of the linear operator

In this section, we shall study the invertibility of the linearized operator
at the trivial solution in order to later apply the infinite dimensional
Implicit Function theorem. From Proposition 4.4 we know that the linear
operator is a compact perturbation of a isomorphism and thus it is a Fredholm
operator of zero index. In that way, it reduces to check that the kernel is
trivial. In order to study the kernel equation, we relate this problem with
the moment problem by using the Mellin and Laplace transform. More
specifically, once we apply the Mellin transform to the kernel equation, we
arrive to a recurrence equation with singular coefficients. However, we can
overcome this by studying the special structure of such recurrence equation.
Later, we shall relate the solution of the recurrence equation with the moment
problem by using the Laplace transform, concluding that the solution is
singular and thus it does not belong to our function space. That will amount
to a trivial kernel.

###  5.1. Kernel study

Note that the kernel study reduces to the following problem.

$\partial_{f}F(0,1)h=0,h\in X_{1}^{\sigma,\delta}\Rightarrow h=0.$ (102)

Since we can restrict ourselves to work with the linearized operator as a
Fourier series, hence we can directly check that the linear operator at the
mode $k$ is trivial, for any $k\in\operatorname{\mathbb{Z}}$. The linearized
operator is given by Proposition 4.5 and can be written as

$\displaystyle\partial_{f}\tilde{F}(0,1)(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}e^{ik\phi}\left\\{\frac{(1+\alpha)}{2}\partial_{\beta}(\beta
h_{k}(\beta))\right.$
$\displaystyle\left.-ik\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)e^{ik\beta}\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{e^{-ikb}b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bh_{k}(b)\cos(k\eta)dbd\eta}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}\right\\}.$
$\displaystyle=$
$\displaystyle\sum_{k\in\operatorname{\mathbb{Z}}}e^{ik\phi}e^{ik\beta}\left\\{\frac{(1+\alpha)}{2}e^{-ik\beta}\partial_{\beta}(\beta
h_{k}(\beta))-ikJ_{k}(e^{-ikb}h_{k})(\beta)\right\\},$

where

$J_{k}(g_{k})(\beta):=\frac{1}{\mu}\frac{C_{\gamma}}{4\pi
C_{0}}\left(\frac{1}{\alpha-1}\right)\beta^{-\frac{\alpha-1}{1+\alpha}}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}bg_{k}(b)\cos(k\eta)dbd\Phi}{\\{\beta^{\frac{-2}{1+\alpha}}+b^{\frac{-2}{1+\alpha}}-2\beta^{\frac{-1}{1+\alpha}}b^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}},$

and

$g_{k}(\beta):=e^{-ik\beta}h_{k}(\beta).$

Define also

$\mathcal{L}_{k}(h)(\beta):=\frac{(1+\alpha)}{2}e^{-ik\beta}\partial_{\beta}(\beta
h_{k}(\beta))-ikJ_{k}(e^{-ikb}h_{k})(\beta),$ (103)

and we wish to prove that $\mathcal{L}_{k}$ has a trivial kernel for each
$k\in\operatorname{\mathbb{Z}}$. For $h\in X_{1}^{\sigma,\delta}$ one has that
$h_{k}\in\mathcal{X}_{0}$ defined in (49) where

$\displaystyle\mathcal{X}_{0}=$
$\displaystyle\left\\{f:\beta\in[0,\infty)\rightarrow\operatorname{\mathbb{R}},\quad
f,\beta f^{\prime}\in\mathcal{X}^{\sigma,\delta}\right\\}.$

Hence, we will check that the kernel of $\mathcal{L}_{k}$ is trivial in the
larger space $\mathcal{X}_{0}$ for any $k\in\operatorname{\mathbb{Z}}$, which
implies (102).

The case $k=0$ is straightforward since

$\mathcal{L}_{0}(h)(\beta):=\frac{(1+\alpha)}{2}e^{-ik\beta}\partial_{\beta}(\beta
h_{k}(\beta)),$

which has a trivial kernel.

In the following proposition we use an auxiliary function $g_{k}$ to work with
(103).

######  Proposition 5.1.

Let $g_{k}(\beta)=e^{-ik\beta}h_{k}(\beta)$ then $\mathcal{L}_{k}=0$, defined
in (103), agrees with

$\displaystyle\tilde{\mathcal{L}}_{k}(g_{k}):=\frac{1+\alpha}{2}g_{k}(\beta)+ik\frac{1+\alpha}{2}\beta
g_{k}(\beta)+\frac{1+\alpha}{2}\beta
g_{k}^{\prime}(\beta)-ikJ_{k}(g_{k})(\beta)=0,$

where

$J_{k}(\beta)=\int_{0}^{\infty}g_{k}(b)K_{k}(\beta/b)db,$ (104)

and

$K_{k}(z):=\frac{C_{\gamma}}{\mu 4\pi
C_{0}(\alpha-1)}\int_{0}^{2\pi}\frac{z^{-\frac{\alpha-1}{1+\alpha}}\cos(k\eta)d\eta}{\\{z^{\frac{-2}{1+\alpha}}+1-2z^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}.$

###### Proof.

Just note that

$e^{ik\beta}\beta h_{k}^{\prime}(\beta)=\beta g_{k}^{\prime}(\beta)+ik\beta
g_{k}(\beta),$

and use the expression of $J_{k}$. ∎

Note that $K_{-k}=K_{k}$ and hence although we will work with
$k\in\operatorname{\mathbb{N}}^{\star}$, we can always exchange $k$ by $-k$.
In the following, we give the monotonicity of $K_{k}$ with respect to $k$.

######  Proposition 5.2.

For any $z\in(0,\infty)$, the sequence
$k\in\operatorname{\mathbb{N}}^{\star}\mapsto K_{k}(z)$ is strictly
decreasing.

###### Proof.

Let us define the auxiliary function

$T_{k}(z):=\int_{0}^{2\pi}\frac{\cos(k\eta)d\eta}{\\{z^{2}+1-2z\cos(\eta)\\}^{\frac{\gamma}{2}}},$
(105)

and then $K_{k}$ can be written as

$K_{k}(z)=\frac{C_{\gamma}}{\mu 4\pi
C_{0}(\alpha-1)}z^{-\frac{\alpha-1}{1+\alpha}}T_{k}(z^{\frac{-1}{1+\alpha}}).$

Hence, in order to study the monotonicity of $k\mapsto K_{k}(z)$ with
$z\in(0,\infty)$, it reduces to study the monotonicity of $k\mapsto T_{k}(z)$
with $z\in(0,\infty)$. Moreover, by using Lemma B.1 we can write $T_{k}$ in
terms of the Gauss Hypergeometric function as follows

$T_{k}(z)=\frac{2\pi\left(\frac{\gamma}{2}\right)_{k}2^{k}\left(\frac{1}{2}\right)_{k}}{(2k)!}\frac{1}{(2z)^{\frac{\gamma}{2}}}\left(\frac{2z}{(1+z)^{2}}\right)^{k+\frac{\gamma}{2}}{}_{2}F_{1}\left(k+\frac{\gamma}{2},k+\frac{1}{2},2k+1,\frac{4z}{(1+z)^{2}}\right).$
(106)

Denoting

$x:=\frac{4z}{(1+z)^{2}},$

where note that $x\in[0,1]$, and by virtue of the integral representation of
the Gauss Hypergeometric function (124) we obtain

$\displaystyle T_{k}(z)=$
$\displaystyle\frac{2\pi\left(\frac{\gamma}{2}\right)_{k}2^{k}\left(\frac{1}{2}\right)_{k}}{(2k)!2^{k+\frac{\gamma}{2}}}\frac{1}{(2z)^{\frac{\gamma}{2}}}x^{k+\frac{\gamma}{2}}{}_{2}F_{1}\left(k+\frac{\gamma}{2},k+\frac{1}{2},2k+1,x\right)$
$\displaystyle=$
$\displaystyle\frac{2\pi\left(\frac{\gamma}{2}\right)_{k}2^{k}\left(\frac{1}{2}\right)_{k}}{(2k)!2^{k+\frac{\gamma}{2}}}\frac{1}{(2z)^{\frac{\gamma}{2}}}x^{k+\frac{\gamma}{2}}\frac{(2k)!}{\Gamma(k+\frac{\gamma}{2})\Gamma(k+1-\frac{\gamma}{2})}\int_{0}^{1}t^{k+\frac{\gamma}{2}-1}(1-t)^{k-\frac{\gamma}{2}}(1-xt)^{-k-\frac{1}{2}}dt$
$\displaystyle=$
$\displaystyle\frac{2\pi}{\Gamma(\frac{\gamma}{2})\Gamma(\frac{1}{2})2^{\gamma}z^{\frac{\gamma}{2}}}\frac{\Gamma(\frac{1}{2}+k)}{\Gamma(k+1-\frac{\gamma}{2})}x^{k+\frac{\gamma}{2}}\int_{0}^{1}t^{k+\frac{\gamma}{2}-1}(1-t)^{k-\frac{\gamma}{2}}(1-xt)^{-k-\frac{1}{2}}dt$
$\displaystyle=:$
$\displaystyle\frac{2\pi}{\Gamma(\frac{\gamma}{2})\Gamma(\frac{1}{2})2^{\gamma}z^{\frac{\gamma}{2}}}T^{1}_{k}T^{2}_{k}(x)T^{3}_{k}(x),$

with

$\displaystyle T^{1}_{k}:=$
$\displaystyle\frac{\Gamma(\frac{1}{2}+k)}{\Gamma(k+1-\frac{\gamma}{2})},$
$\displaystyle T^{2}_{k}(x):=$ $\displaystyle x^{k+\frac{\gamma}{2}},$
$\displaystyle T^{3}_{k}(x):=$
$\displaystyle\int_{0}^{1}t^{k+\frac{\gamma}{2}-1}(1-t)^{k-\frac{\gamma}{2}}(1-xt)^{-k-\frac{1}{2}}dt.$

Note that $T^{1}_{k},T^{2}_{k},T^{3}_{k}$ are positive functions for
$x\in[0,1]$. Moreover, $k\mapsto T^{2}_{k}(x)$ is strictly decreasing for
$x\in(0,1)$. The last function $T^{3}_{k}$ can be written as

$T^{3}_{k}(x)=\int_{0}^{1}t^{\frac{\gamma}{2}-1}(1-t)^{-\frac{\gamma}{2}}(1-xt)^{-\frac{1}{2}}\left(\frac{t(1-t)}{1-xt}\right)^{k}dt,$

obtaining that $k\mapsto T^{3}_{k}(x)$ is strictly decreasing for $x\in(0,1)$
since $\frac{t(1-t)}{1-xt}\in(0,1)$. For the first term $T^{1}_{k}$ let us
differentiate in $k$ obtaining

$\partial_{k}T^{1}_{k}=\frac{\Gamma^{\prime}(\frac{1}{2}+k)\Gamma(k+1-\frac{\gamma}{2})-\Gamma(\frac{1}{2}+k)\Gamma^{\prime}(k+1-\frac{\gamma}{2})}{\Gamma(k+1-\frac{\gamma}{2})^{2}}.$

We claim that

$\partial_{k}T^{1}_{k}<0.$ (107)

Indeed, that condition agrees with

$\Gamma^{\prime}(\frac{1}{2}+k)\Gamma(k+1-\frac{\gamma}{2})<\Gamma(\frac{1}{2}+k)\Gamma^{\prime}(k+1-\frac{\gamma}{2}),$

which agrees with

$\psi^{0}(\frac{1}{2}+k)=\frac{\Gamma^{\prime}(\frac{1}{2}+k)}{\Gamma(\frac{1}{2}+k)}<\frac{\Gamma^{\prime}(k+1-\frac{\gamma}{2})}{\Gamma(k+1-\frac{\gamma}{2})}=\psi^{0}(k+1-\frac{\gamma}{2}),$

where $\psi^{0}$ is the digamma function which is increasing. Hence, our claim
(107) holds true. As a consequence, we get that $k\mapsto
T^{1}_{k}T^{2}_{k}(x)T^{3}_{k}(x)$ strictly decreases for $x\in(0,1)$,
obtaining the announced result. ∎

###  5.2. Mellin transform

Mellin transforms are very useful when dealing with integral differential
equations where the integral operator is of the type (104) due to the product
property (126):

$M[J_{k}](s)=M[K](s)M[g_{k}](s+1).$ (108)

Hence, applying the Mellin transform to the expression given in Proposition
5.1 we obtain the following equivalent equation.

######  Proposition 5.3.

The equation $\tilde{\mathcal{L}}_{k}(g_{k})=0$ agrees with

$\displaystyle
M[\tilde{\mathcal{L}}_{k}(g_{k})](s)=\frac{1+\alpha}{2}\left(1-s\right)G_{k}(s)-ik\left(\tilde{K}_{k}(s)-\frac{1+\alpha}{2}\right)G_{k}(s+1)=0,$
(109)

where $G_{k}=M[g_{k}]$ and $\tilde{K}_{k}=M[K_{k}]$.

###### Proof.

The proof is based on the properties (126)–(128) together with (108). ∎

######  Remark 5.4.

Given $h_{k}\in\mathcal{X}_{0}^{\sigma,\delta}$, then $M[g_{k}]$ is well
defined in

$\\{s\in\operatorname{\mathbb{C}},\quad-\sigma<\textnormal{Re}(s)<\sigma\\}.$

######  Remark 5.5.

Let us again remark that the kernel $K_{k}$ is symmetric with respect to $k$:
$K_{-k}=K_{k}$. Hence, in order to work with
$k\in\operatorname{\mathbb{Z}}/\operatorname{\mathbb{N}}$ we can do the change
$k\mapsto-k$. In that way, the kernel equation (109) reads as

$\displaystyle
M[\tilde{\mathcal{L}}_{k}(g_{k})](s)=\frac{1+\alpha}{2}\left(1-s\right)G_{k}(s)+ik\left(\tilde{K}_{k}(s)-\frac{1+\alpha}{2}\right)G_{k}(s+1)=0,$

with $k\in\operatorname{\mathbb{N}}^{\star}$.

In that way, Equation (109) can be written as

$G_{k}(s+1)=\frac{1}{k}F_{k}(s)G_{k}(s),$ (110)

where

$F_{k}(s):=i\frac{1-s}{(1-\frac{2}{1+\alpha}\tilde{K}_{k}(s))}.$

In the following, let us give a better expression for $F_{k}$ together with
some properties. As a consequence, we shall look for the values of $s$ such
that (110) is well-defined.

######  Proposition 5.6.

The Mellin transform of $K_{k}$ can be written as follows

$\tilde{K}_{k}(s)=\frac{C_{\gamma}(1+\alpha)}{\mu 4\pi
C_{0}(\alpha-1)}M[T_{k}(z)](-(1+\alpha)s+\alpha-1),$

for

$\frac{\alpha-1-k-\gamma}{1+\alpha}<s<\frac{\alpha-1+k}{1+\alpha},$

and where $T_{k}$ is defined in (105). Moreover,

$\lim_{s\rightarrow\frac{\alpha-1-k-\gamma}{1+\alpha}}\tilde{K}_{k}(s)=\lim_{s\rightarrow\frac{\alpha-1+k}{1+\alpha}}\tilde{K}_{k}(s)=+\infty,$

for $k\in\operatorname{\mathbb{N}}^{\star}$.

###### Proof.

Recall the expression for $K_{k}$:

$K_{k}(z)=\frac{C_{\gamma}}{\mu 4\pi
C_{0}(\alpha-1)}\int_{0}^{2\pi}\frac{z^{-\frac{\alpha-1}{1+\alpha}}\cos(k\eta)d\eta}{\\{z^{\frac{-2}{1+\alpha}}+1-2z^{\frac{-1}{1+\alpha}}\cos(\eta)\\}^{\frac{\gamma}{2}}}.$

Using $T_{k}$ in (105) we have

$K_{k}(z)=\frac{C_{\gamma}}{\mu 4\pi
C_{0}(\alpha-1)}z^{-\frac{\alpha-1}{1+\alpha}}T_{k}(z^{\frac{-1}{1+\alpha}}).$

Now, let us use some properties about the Mellin transform described in
Appendix B. By (127) we get

$M[K_{k}](s)=\tilde{K}_{k}(s)=\frac{C_{\gamma}}{\mu 4\pi
C_{0}(\alpha-1)}M[T_{k}(z^{\frac{-1}{1+\alpha}})]\left(s-\frac{\alpha-1}{1+\alpha}\right).$

Moreover, by (130) one finds

$M[T_{k}(z^{\frac{-1}{1+\alpha}})](s)=(1+\alpha)M[T_{k}(z)](-(1+\alpha)s),$

and hence

$M[T_{k}(z^{\frac{-1}{1+\alpha}})](s-\frac{\alpha-1}{1+\alpha})=(1+\alpha)M[T_{k}(z)](-(1+\alpha)s+\alpha-1),$

which implies

$\tilde{K}_{k}(s)=\frac{C_{\gamma}(1+\alpha)}{\mu 4\pi
C_{0}(\alpha-1)}M[T_{k}(z)](-(1+\alpha)s+\alpha-1).$ (111)

Let us check where $M[T_{k}(z)](s)$ is well-defined. Since $T_{k}$ can be
related to the Gauss Hypergeometric function by using Lemma B.1, we get

$\displaystyle M[T_{k}](s)=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{2\pi}\frac{z^{s-1}\cos(k\eta)d\eta
dz}{\left\\{z^{2}+1-2z\cos(\eta)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$ $\displaystyle\frac{2\pi
2^{2k}\left(\frac{\gamma}{2}\right)_{k}\left(\frac{1}{2}\right)_{k}}{(2k)!}\int_{0}^{\infty}\frac{z^{s-1+k}}{(1+z)^{\gamma+2k}}F\left(k+\frac{\gamma}{2},k+\frac{1}{2},2k+1,\frac{4z}{(1+z)^{2}}\right)dz.$

Since the parameters of the Gauss Hypergeometric function are positive, this
function increases and then for any $k\in\operatorname{\mathbb{N}}^{\star}$,
$\gamma\in(0,1)$ and $z\in(0+\infty)$ one has

$1<F\left(k+\frac{\gamma}{2},k+\frac{1}{2},2k+1,\frac{4z}{(1+z)^{2}}\right)<F\left(k+\frac{\gamma}{2},k+\frac{1}{2},2k+1,1\right)<+\infty.$

Hence $M[T_{k}](s)$ is well-defined for $s\in(-k,k+\gamma)$, where that
condition comes form the integrability of the integral. Moreover

$\displaystyle M[T_{k}](s)\geq\frac{2\pi
2^{2k}\left(\frac{\gamma}{2}\right)_{k}\left(\frac{1}{2}\right)_{k}}{(2k)!}\int_{0}^{\infty}\frac{z^{s-1+k}}{(1+z)^{\gamma+2k}}dz,$
(112)

implying

$\lim_{s\rightarrow-k}T_{k}(s)=\lim_{s\rightarrow k+\gamma}T_{k}(s)=+\infty.$

Coming back to $\tilde{K}$ through (111), we get that $\tilde{K}$ is well-
defined if

$\frac{\alpha-1-k-\gamma}{1+\alpha}<s<\frac{\alpha-1+k}{1+\alpha},$

and

$\lim_{s\rightarrow\frac{\alpha-1-k-\gamma}{1+\alpha}}\tilde{K}(s)=\lim_{s\rightarrow\frac{\alpha-1+k}{1+\alpha}}\tilde{K}(s)=+\infty.$

∎

As a consequence of Proposition 5.2 we get the monotonicity of the Mellin
transform of $K$, that is, $\tilde{K}$:

######  Corollary 5.7.

For any $s\in[0,\frac{\alpha-1+k}{1+\alpha})$, the sequence
$k\in\operatorname{\mathbb{N}}^{\star}\mapsto\tilde{K}_{k}(s)$ is strictly
decreasing.

In the following proposition, we give the monotonicity of $\tilde{K}_{k}$ with
respect to $s$.

######  Proposition 5.8.

The function $s\in[0,\frac{\alpha-1+k}{1+\alpha})\mapsto\tilde{K}_{k}(s)$
increases, for any $k\in\operatorname{\mathbb{N}}^{\star}$. Moreover,
$\frac{2}{1+\alpha}\tilde{K}_{k}(0)<1$.

###### Proof.

By the expression of $\tilde{K}$ in Proposition 5.6, we can relate
$\tilde{K}_{k}$ with $M[T_{k}]$. Hence if $s\in(-k,\alpha-1]\mapsto
M[T_{k}](s)$ decreases, we achieve that
$s\in[0,\frac{\alpha-1+k}{1+\alpha})\mapsto\tilde{K}_{k}(s)$ increases.
Moreover, using the explicit expression of $T_{k}$ we get that
$T_{k}(z^{-1})=z^{\gamma}T_{k}(z)$. In that way, $M[T_{k}]$ can be written as
follows

$\displaystyle M[T_{k}](s)=$ $\displaystyle\int_{0}^{\infty}z^{s-1}T_{k}(z)dz$
$\displaystyle=$
$\displaystyle\int_{0}^{1}z^{s-1}T_{k}(z)dz+\int_{1}^{\infty}z^{s-1}T_{k}(z)dz$
$\displaystyle=$
$\displaystyle\int_{0}^{1}T_{k}(z)(z^{s-1}+z^{1-s}z^{\gamma}z^{-2})dz.$

Then, computing its derivative we observe

$\displaystyle M[T_{k}]^{\prime}(s)=$
$\displaystyle\int_{0}^{1}\ln(z)T_{k}(z)(z^{s-1}-z^{1-s}z^{\gamma}z^{-2})dz$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\ln(z)T_{k}(z)z^{s-1}(1-z^{\gamma-2s})dz.$

Since $s\in(-k,\alpha-1)$ and $\alpha$ satisfies (56), we get that
$\gamma-2s>0$ and hence $M[T_{k}]^{\prime}(s)<0$. As a consequence $M[T_{k}]$
is strictly decreasing.

About the second point of the statement and by virtue of Corollary 5.7, it is
enough to prove that $\frac{2}{1+\alpha}\tilde{K}_{1}(0)<1$. Moreover, let us
check that $\alpha\mapsto\frac{2}{1+\alpha}\tilde{K}_{1}(0)$ strictly
decreases and hence we can restrict ourselves to $\alpha=1$. From Proposition
5.1 we get

$\displaystyle\frac{2}{1+\alpha}\tilde{K}_{1}(0)=\frac{C_{\gamma}}{\mu 2\pi
C_{0}(\alpha-1)}M[T_{1}](\alpha-1).$

Moreover, from Proposition A.3 and the expression of $\mu$ in (16) we get

$\displaystyle\frac{2}{1+\alpha}\tilde{K}_{1}(0)=\frac{(1+\alpha-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](\alpha-1).$

By the previous estimates concerning the monotonicity of $s\mapsto
M[T_{k}](s)$, we have that $\alpha\in(1,1+\gamma)\mapsto M[T_{1}](\alpha-1)$
is strictly decreasing. Moreover, one has that
$M[T_{0}]^{\prime}(1-\alpha+\gamma)<0$ by similar arguments. Hence, we claim
that

$g(\alpha):=\frac{(1+\alpha-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)},$

is positive and strictly decreasing. The positivity comes from the fact that
$M[T_{0}]$ is positive. In order to prove the monotonicity, let us
differentiate:

$\displaystyle g^{\prime}(\alpha)=$
$\displaystyle\frac{(\alpha-1)M[T_{0}](1-\alpha+\gamma)-(1+\alpha-\gamma)M[T_{0}](1-\alpha+\gamma)+(\alpha-1)(1+\alpha-\gamma)M[T_{0}]^{\prime}(1-\alpha+\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)^{2}}$
$\displaystyle=$
$\displaystyle\frac{-(2-\gamma)M[T_{0}](1-\alpha+\gamma)+(\alpha-1)(1+\alpha-\gamma)M[T_{0}]^{\prime}(1-\alpha+\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)^{2}},$

which is negative since $\gamma<1$. Hence, we have proved our claim. Then

$\frac{2}{1+\alpha}\tilde{K}_{1}(0)\leq\lim_{\alpha\rightarrow
1^{+}}\frac{(1+\alpha-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](\alpha-1)=\lim_{\alpha\rightarrow
1^{+}}\frac{(2-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](0).$

Note that

$\lim_{\alpha\rightarrow 1^{+}}M[T_{0}](1-\alpha+\gamma)=+\infty,$

and hence there is a competition in the denominator that we will have to
solve. Note that we can again relate $M[T_{0}]$ with the Gauss Hypergeometric
function:

$\displaystyle M[T_{0}](1-\alpha+\gamma)=$ $\displaystyle
2\pi\int_{0}^{\infty}\frac{z^{\gamma-\alpha}}{(1+z)^{\gamma}}{}_{2}F_{1}\left(\frac{\gamma}{2},\frac{1}{2},1,\frac{4z}{(1+z)^{2}}\right)dz$
$\displaystyle>$ $\displaystyle
2\pi\int_{0}^{\infty}\frac{z^{\gamma-\alpha}}{(1+z)^{\gamma}}dz$
$\displaystyle=$ $\displaystyle
2\pi\frac{\Gamma(1+\gamma-\alpha)\Gamma(\alpha-1)}{\Gamma(\gamma)}.$

Since $x\Gamma(x)=\Gamma(x+1)$ we get

$\lim_{\alpha\rightarrow 1}(\alpha-1)\Gamma(\alpha-1)=1,$

and hence

$\lim_{\alpha\rightarrow
1^{+}}\frac{(2-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](0)<\frac{(2-\gamma)}{2\pi}M[T_{1}](0).$

Using now the expression of $T_{1}$ in terms of the Gauss Hypergeometric
function (106) we find

$\displaystyle
M[T_{1}](0)=\pi\gamma\int_{0}^{\infty}\frac{1}{(1+z)^{2+\gamma}}{}_{2}F_{1}\left(1+\frac{\gamma}{2},\frac{3}{2},3,\frac{4z}{(1+z)^{2}}\right)dz.$

By writing the Gauss Hypergeometric function in series, integrating and
summing again we find the following expression

$\displaystyle
M[T_{1}](0)=\pi\gamma\frac{\Gamma(-\frac{\gamma}{2})\Gamma(\frac{\gamma-1}{2})}{\Gamma(\frac{1-\gamma}{2})\Gamma(\frac{\gamma}{2})}.$

Using that

$\Gamma(z)=-\frac{\Gamma(-z)\Gamma(1+z)}{\Gamma(1-z)},$

we find

$\displaystyle
M[T_{1}](0)=\pi\gamma\frac{\Gamma(1-\frac{\gamma}{2})\Gamma(\frac{\gamma+1}{2})}{\Gamma(1+\frac{\gamma}{2})\Gamma(\frac{3}{2}-\frac{\gamma}{2})},$

and hence

$\lim_{\alpha\rightarrow
1^{+}}\frac{(2-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](0)<\frac{(2-\gamma)\gamma}{2}\frac{\Gamma(1-\frac{\gamma}{2})\Gamma(\frac{\gamma+1}{2})}{\Gamma(1+\frac{\gamma}{2})\Gamma(\frac{3}{2}-\frac{\gamma}{2})}=:p(\gamma).$

Note that $p(1)=1$ and if we check that $p$ is strictly increasing, we can
arrive to the announced result given that:

$\lim_{\alpha\rightarrow
1^{+}}\frac{(2-\gamma)}{(\alpha-1)M[T_{0}](1-\alpha+\gamma)}M[T_{1}](0)<\frac{(2-\gamma)\gamma}{2}\frac{\Gamma(1-\frac{\gamma}{2})\Gamma(\frac{\gamma+1}{2})}{\Gamma(1+\frac{\gamma}{2})\Gamma(\frac{3}{2}-\frac{\gamma}{2})}<p(1)=1,$

for any $\gamma\in(0,1)$.

Hence, it remains to check that $p$ is strictly increasing. Note that $p$ can
be written as

$p(\gamma)=2\frac{\Gamma(2-\frac{\gamma}{2})\Gamma(\frac{1+\gamma}{2})}{\Gamma(\frac{3-\gamma}{2})\Gamma(\frac{\gamma}{2})},$

and whose derivative reads as follows

$p^{\prime}(\gamma)=\frac{\Gamma(2-\frac{\gamma}{2})\Gamma(\frac{1+\gamma}{2})}{\Gamma(\frac{3-\gamma}{2})\Gamma(\frac{\gamma}{2})}\left\\{\psi^{0}(\frac{3-\gamma}{2})-\psi^{0}(2-\frac{\gamma}{2})-\psi^{0}(\frac{\gamma}{2})+\psi^{0}(\frac{1+\gamma}{2})\right\\},$

where $\psi^{0}$ is the digamma function. Now, using the integral
representation of the digamma function (125) we find

$\psi^{0}(\frac{3-\gamma}{2})-\psi^{0}(2-\frac{\gamma}{2})-\psi^{0}(\frac{\gamma}{2})+\psi^{0}(\frac{1+\gamma}{2})=\int_{0}^{1}\frac{-t^{\frac{1-\gamma}{2}}-t^{-\frac{1-\gamma}{2}}+t^{\frac{2-\gamma}{2}}+t^{-\frac{2-\gamma}{2}}}{1-t}dt,$

where we can prove that the function inside the integral is positive for any
$t\in(0,1)$ since $\gamma\in(0,1)$. Hence we obtain $p^{\prime}(\gamma)>0$,
for any $\gamma\in(0,1)$, and the claim holds true. That concludes the proof.
∎

As a consequence of Proposition 5.6 and Proposition 5.8, and using the
intermediate theorem, we get the following corollary:

######  Corollary 5.9.

For any $k\in\operatorname{\mathbb{N}}^{\star}$, there exists a unique
$s_{0}\in[0,\frac{\alpha-1+k}{1+\alpha})$ such that $\tilde{K}_{k}(s_{0})=1$.

Hence, using the expression of $\tilde{K}$ in Proposition 5.6 we find

$(1-s)G_{k}(s)+ik\left(1-\frac{C_{\gamma}}{\mu 2\pi
C_{0}(\alpha-1)}M[T_{k}(z)](-(1+\alpha)s+\alpha-1)\right)G_{k}(s+1)=0,$

amounting to

$G_{k}(s+1)=\frac{i}{k}\frac{1-s}{1-\frac{C_{\gamma}}{\mu 2\pi
C_{0}(\alpha-1)}M[T_{k}(z)](-(1+\alpha)s+\alpha-1)}G_{k}(s).$ (113)

Note that by Proposition 5.6 $M[T_{k}(z)](-(1+\alpha)s+\alpha-1)$ is well-
defined for

$\frac{\alpha-1-k-\gamma}{1+\alpha}<s<\frac{\alpha-1+k}{1+\alpha},$

Using $F_{k}$ one has

$F_{k}(s)=i\frac{1-s}{1-\frac{C_{\gamma}}{\mu 2\pi
C_{0}(\alpha-1)}M[T_{k}(z)](-(1+\alpha)s+\alpha-1)}.$

Moreover, due to the limits of $\tilde{K}$ in Proposition 5.6, we get that

$\lim_{s\rightarrow\frac{\alpha-1-k-\gamma}{1+\alpha}}F_{k}(s)=\lim_{s\rightarrow\frac{\alpha-1+k}{1+\alpha}}F_{k}(s)=0,$

and hence we can extend it to zero outside
$(\frac{\alpha-1-k-\gamma}{1+\alpha},\frac{\alpha-1+k}{1+\alpha})$. Moreover,
due to (56) we have that

$\frac{\alpha-1-\gamma-k}{1+\alpha}<\frac{\alpha-1-\gamma}{1+\alpha}<-\sigma<s<\sigma<\frac{\alpha-1}{1+\alpha}<\frac{\alpha-1+k}{1+\alpha}.$

Since a priori $G_{k}$ is well-defined for $s\in(-\sigma,\sigma)$, one obtains
that (113) is well-defined for $s\in(-\sigma,\sigma)$, and then $G_{k}$ is
well-defined for $s\in(-\sigma+1,\sigma+1)$.

Finally, in the following proposition we achieve that the kernel is trivial.
For this goal, we observe (113) as a recurrence equation and we relate the
solutions to the moment problem. In that way, using the Laplace transform we
achieve that the solution $h_{k}$ is singular and thus it does not belong to
our function spaces, concluding that the kernel is trivial.

######  Proposition 5.10.

If $G_{k}$ solves (113), with $g_{k}(\beta)=e^{-ik\beta}h_{k}(\beta)$ and
$h_{k}\in\mathcal{X}_{0}^{\sigma,\delta}$, hence $h_{k}=0$.

###### Proof.

Taking $s=M\in\operatorname{\mathbb{N}}^{\star}$, we can solve the recurrence
equation (113) as

$G_{k}(M)=\frac{1}{k}G_{k}(0)\prod_{n=1}^{M-1}F_{k}(n).$

Note that $G_{k}(0)$ is well-defined and then

$G_{k}(1)=\frac{1}{k}F_{k}(0)G_{k}(0),$

which is finite due to Proposition 5.8.

First, assume that $\frac{2}{1+\alpha}\tilde{K}_{k}(N)\neq 1$, for any
$N\in\operatorname{\mathbb{N}}$ and $N\leq\frac{\alpha-1+k}{1+\alpha}$. In
that case, since $F_{k}(1)=0$ we get $G_{k}(M)=0$ for any $M\geq 2$. Let
$\mathcal{L}$ be the Laplace Transform operator, and thus

$\displaystyle\mathcal{L}[g_{k}](t)=$
$\displaystyle\int_{0}^{\infty}e^{-tz}g(z)dz$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}\int_{0}^{\infty}z^{n}g_{(}z)dz$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}G_{k}(n+1)$
$\displaystyle=$ $\displaystyle G_{k}(1).$

Since the inverse Laplace transform of a constant is a dirac mass and $g_{k}$
is continuous, we get that necessarily $g_{k}=0.$

Second, assume that there exists $N_{0}\in\operatorname{\mathbb{N}}$ with
$N_{0}\in[1,\frac{\alpha-1+k}{1+\alpha})$ such that

$\frac{2}{1+\alpha}\tilde{K}_{k}(N_{0})=1.$

In that case $\frac{2}{1+\alpha}\tilde{K}_{k}$ has a simple root at $N_{0}$.
On the one hand, if $N_{0}=1$ we have that

$\lim_{s\rightarrow
1}\frac{s-1}{\frac{2}{1+\alpha}\tilde{K}_{k}(s)-1}\in\operatorname{\mathbb{R}},$

since the root is simple. In that case $G_{k}(2)$ is well-defined and we can
write

$G_{k}(M)=\frac{1}{k}G_{k}(0)\prod_{n=1}^{M-1}F_{k}(M).$

Moreover, note that for $M\geq\frac{\alpha-1+k}{1+\alpha}$ we have that
$F_{k}(M)=0$, implying that $G_{k}(M)=0$ for
$M\geq\frac{\alpha-1+k}{1+\alpha}$. Hence the Laplace transform of $g_{k}$ is
a polynomial. On the other hand, if $N_{0}\geq 2$, we have

$\lim_{s\rightarrow N_{0}}F_{k}(1)F_{k}(s)\in\operatorname{\mathbb{R}},$

since $F_{k}(1)=0$ and the pole of $F_{k}$ at $N_{0}$ is simple. In that case,
we find again that the Laplace transform of $g_{k}$ is a polynomial. Now, by
using the expression of the Laplace transform we find

$\mathcal{L}_{g_{k}}(t)=\int_{0}^{\infty}e^{-tz}g_{k}(z)dz.$

Since $g_{k}\in L^{\infty}([0,\infty))$ we get that

$\lim_{t\rightarrow+\infty}|\mathcal{L}_{g_{k}}(t)|=0,$

which contradicts the fact that the Laplace transform is a polynomial. Hence
$g_{k}=0$. ∎

##  6\. $m$-fold symmetry and main theorem in the variables $(\beta,\phi)$

In this section, we introduce the $m$-fold symmetry to the function spaces,
and then we will work with the extra parameter $m$. Moreover, we will give the
main result in the adapted coordinates which is a consequence of the infinite
dimensional Implicit Function theorem applied to $\tilde{F}$.

###  6.1. $m$-fold symmetry in the function spaces

In the following, we introduce the $m$-fold symmetry in the function spaces
(45), (47) and (48):

$\displaystyle X_{1,m}^{\sigma,\delta}:=$ $\displaystyle\left\\{f\in
X_{1}^{\sigma,\delta},\quad
f\left(\beta,\phi+\frac{2\pi}{m}\right)=f(\beta,\phi),\,(\beta,\phi)\in[0,+\infty)\times[0,2\pi]\right\\},$
$\displaystyle X_{2,m}^{p}:=$ $\displaystyle\left\\{f\in X_{2}^{p},\quad
f\left(\phi+\frac{2\pi}{m}\right)=f(\phi),\,\phi\in[0,2\pi]\right\\},$
$\displaystyle Y_{m}^{\sigma,\delta}:=$ $\displaystyle\left\\{f\in
Y^{\sigma,\delta},\quad
f\left(\beta,\phi+\frac{2\pi}{m}\right)=f(\beta,\phi),\,(\beta,\phi)\in[0,+\infty)\times[0,2\pi]\right\\}.$

Note that $f\in X_{1,m}^{\sigma,\delta}$ can be written in Fourier series as

$f(\beta,\phi)=\sum_{k\in\operatorname{\mathbb{Z}}}f_{k}(\beta)e^{ikm\phi}.$

Since the only variation that we have introduced in the function spaces is the
$m$-fold symmetry, all the previous analysis also works here. We only need to
check the symmetry persistence of the nonlinear functional.

######  Proposition 6.1.

If $f\left(\beta,\phi+\frac{2\pi}{m}\right)=f(\beta,\phi)$ and
$\Omega\left(\phi+\frac{2\pi}{m}\right)=\Omega(\phi)$ then
$\tilde{F}(f,\Omega)\left(\beta,\phi+\frac{2\pi}{m}\right)=\tilde{F}(f,\Omega)(\beta,\phi)$.

###### Proof.

Let us write $\tilde{F}(f,\Omega)(\beta,\phi+\frac{2\pi}{m})$ using (2.4):

$\displaystyle\tilde{F}(f,\Omega)(\beta,\phi+\frac{2\pi}{m})=1+f(\beta,\phi+\frac{2\pi}{m})$
$\displaystyle-(-1)^{\frac{-1}{2\mu}}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi)\right\\}^{-\frac{1}{2\mu}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(b,\Phi)\right\\}\tilde{\Omega}(\Phi)dbd\Phi}{D(f)(\beta,\phi+\frac{2\pi}{m},b,\Phi)^{\frac{\gamma}{2}}}.$

Doing the change of variables $\Phi\mapsto\Phi+\frac{2\pi}{m}$ we get

$\displaystyle\tilde{F}(f,\Omega)(\beta,\phi+\frac{2\pi}{m})=1+f(\beta,\phi+\frac{2\pi}{m})$
$\displaystyle-(-1)^{\frac{-1}{2\mu}}\frac{C_{\gamma}(\alpha-1)^{\frac{\alpha-1}{2}}}{4\pi
C_{0}(1+\alpha)^{\frac{\alpha-1}{2}}}\beta^{-\frac{\alpha-1}{1+\alpha}}$
$\displaystyle\times\int_{0}^{2\pi}\int_{0}^{\infty}\frac{b^{-\frac{\gamma}{1+\alpha}}b^{\frac{-2}{1+\alpha}}\left\\{-\frac{\alpha-1}{1+\alpha}+\partial_{\overline{\varphi}}f(b,\Phi+\frac{2\pi}{m})\right\\}^{-\frac{1}{2\mu}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(b,\Phi+\frac{2\pi}{m})\right\\}\tilde{\Omega}(\Phi+\frac{2\pi}{m})dbd\Phi}{D(f)(\beta,\phi+\frac{2\pi}{m},b,\Phi+\frac{2\pi}{m})^{\frac{\gamma}{2}}}.$

Using that $f\left(\beta,\phi+\frac{2\pi}{m}\right)=f(\beta,\phi)$ and
$\Omega\left(\phi+\frac{2\pi}{m}\right)=\Omega(\phi)$, and the expression of
$D(f)$ in (40) we achieve
that$\tilde{F}(f,\Omega)\left(\beta,\phi+\frac{2\pi}{m}\right)=\tilde{F}(f,\Omega)(\beta,\phi)$,
concluding the proof. ∎

Let us remark that the spectral study developed in Section 5 follows
similarly, one just have to change $k\mapsto km$.

###  6.2. Main theorem in the adapted coordinates

Finally, let us state the main theorem in the adapted coordinates, which is a
consequence of the infinite dimensional Implicit Function theorem.

######  Theorem 6.2.

Let $\gamma,\sigma\in(0,1)$ and $\alpha\in(1,1+\gamma)$ satisfying (56). Then,
for each $m\geq 1$, there exists $\varepsilon>0$ and a $C^{1}$ function
$\tilde{\Omega}\in
B_{X_{2,m}^{p}}(1,\varepsilon)\mapsto\tilde{f}(\tilde{\Omega})\in
X_{1,m}^{\sigma,\delta}$ such that

$\tilde{F}(f,\tilde{\Omega})=0,\,(f,\tilde{\Omega})\in
B_{X_{1,m}^{\sigma,\delta}}(0,\varepsilon)\times
B_{X_{2,m}^{p}}(1,\varepsilon)\Longleftrightarrow f=\tilde{f}(\Omega),$

for $\delta<\min\\{1-\gamma,\sigma\\}$ and $p>\frac{1}{1-\gamma}$.

###### Proof.

From Proposition 3.10 and Proposition 6.1 we get that
$\tilde{F}:B_{X_{1,m}^{\sigma,\delta}}(0,\varepsilon)\times
X_{2,m}^{p}(1,\varepsilon)\rightarrow Y_{m}^{\sigma,\delta}$ is well-defined
and $C^{1}$. Moreover, by Proposition 2.5 we have that $\tilde{F}(0,1)=0$.
Moreover, $\partial_{f}\tilde{F}(0,1)$ is Fredholm of zero index by
Proposition 4.4 and has a trivial kernel by (5.10). Hence,
$\partial_{f}\tilde{F}(0,1)$ is a isomorphism. The infinite dimensional
Implicit Function theorem to $\tilde{F}$ concludes the proof. ∎

######  Remark 6.3.

A priori $\Psi$ and $\Omega$ are complex functions due to the function spaces,
however we are interested in real functions. That is not a problem since
assuming in the function space that $f$ and $\tilde{\Omega}$ are real
functions we obtain that the nonlinear operator is also a real function.

######  Remark 6.4.

Taking $\tilde{f}$ constructed in Theorem 6.2, we find

$\displaystyle\Psi(\beta,\phi)=$ $\displaystyle-
C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{\alpha-1}{1+\alpha}}(1+\tilde{f}(\tilde{\Omega})(\beta,\phi)),$
$\displaystyle\Omega(\phi)=$
$\displaystyle(1+\alpha)^{-\frac{1}{2\mu}}\tilde{\Omega}(\phi),$

satisfying $F(\Psi,\Omega)=0$. Moreover, we can come back to $\Theta$ and find

$\displaystyle\Theta(\beta,\phi)=(\Psi_{\varphi})^{-\frac{1+\alpha-\gamma}{2}}\Omega(\phi),$

which together with $\Psi$ verifies (25), meaning the vorticity equation in
the adapted coordinates $(\beta,\phi)$. Note that $\Theta$ is just a
perturbation of the trivial one. Indeed, we can compute $\Psi_{\varphi}$ as

$\displaystyle\Psi_{\varphi}(\beta,\phi)=$ $\displaystyle-
C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\left(-\frac{\alpha-1}{1+\alpha}-\frac{\alpha-1}{1+\alpha}\tilde{f}(\tilde{\Omega})(\beta,\phi)+\beta\tilde{f}_{\varphi}(\tilde{\Omega})(\beta,\phi)\right)$
$\displaystyle=$ $\displaystyle-
C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{-2}{1+\alpha}}\left(-\frac{\alpha-1}{1+\alpha}+\tilde{f}_{\overline{\varphi}}(\tilde{\Omega})(\beta,\phi)\right),$
(114)

where recall that $\tilde{f}_{\varphi}(\tilde{\Omega})$ decays at $0$ and
$\infty$ in $\beta$. Hence $\Psi_{\varphi}$ is singular at $\beta=0$ as the
trivial solution, and decays at $\beta\rightarrow+\infty$ as the trivial
solution. Coming back to $\Theta$ we get

$\Theta(\beta,\phi)=(1+\alpha)^{-\frac{1}{2\mu}}\left[C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\right]^{-\frac{1+\alpha-\gamma}{2}}\beta^{\frac{1+\alpha-\gamma}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}-\tilde{f}_{\overline{\varphi}}(\tilde{\Omega})(\beta,\phi)\right)^{-\frac{1+\alpha-\gamma}{2}}\tilde{\Omega}(\phi).$

Note that $1+\alpha-\gamma>0$ and hence $\Theta(0,\phi)=0$, for any
$\phi\in[0,2\pi])$. In particular, we find

$\Theta(\beta,\phi)\sim\beta^{\frac{1+\alpha-\gamma}{1+\alpha}},\quad\beta\sim
0,\infty.$

##  7\. Recovering the solution in the original coordinates

This section is devoted to recover the solution in the physical variables
$x\in\operatorname{\mathbb{R}}^{2}$ and hence give the associated Theorem 6.2
in those variables. First, we study the invertibility of the change of
coordinates, which is a nonlinear implicit change of variables, and later we
state the main result.

###  7.1. Invertibility of the change of coordinates

In order to come back to the physical variable
$z\in\operatorname{\mathbb{R}}^{2}$ we should invert the adapted coordinates.
From Proposition 2.2 we have that

$r(\beta,\phi)=(-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}},\quad\vartheta(\beta,\phi)=\beta+\phi.$
(115)

In the following proposition, let us study the regularity of $r(\beta,\phi)$
and $\vartheta(\beta,\phi)$. For that aim, let us define the following
function space:

$L^{\infty}_{0,\sigma}:=\left\\{g=g(0)+\tilde{g},\quad\tilde{g}\in
L^{\infty}_{\sigma}\right\\}.$

######  Proposition 7.1.

Let
$(r,\vartheta):[0,\infty)\times[0,2\pi]\rightarrow[0,\infty)\times[0,2\pi]$
defined as (115). There exists $\varepsilon<1$ such that if $f\in
B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$ then $\vartheta\in
C^{\infty}([0,\infty)\times[0,2\pi])$ and

$\beta^{\frac{-1}{1+\alpha}}r\in L^{\infty}_{0,\sigma}\cap
C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi},$

and

$\beta^{\frac{-1}{1+\alpha}}\beta r_{\varphi}\in L^{\infty}_{\sigma}\cap
C^{\delta}_{\beta}.$

###### Proof.

First, note that $\vartheta\in C^{\infty}([0,+\infty)\times[0,2\pi])$. Next,
from (6.4) we find

$(-\Psi_{\beta})^{\frac{1}{2}}(\beta,\phi)=C_{0}^{\frac{1}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{2(1+\alpha)}}\beta^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta,\phi)\right)^{\frac{1}{2}}.$
(116)

Let us check the regularity of such function, which will give us the
regularity of $r$ due to (115). Denote

$g(\beta,\phi):=\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta,\phi)\right)^{\frac{1}{2}}=\left(\frac{\alpha-1}{1+\alpha}+\frac{\alpha-1}{1+\alpha}f(\beta,\phi)+\beta{f}_{{\beta}}(\beta,\phi)\right)^{\frac{1}{2}}.$

Since $f\in X_{1}^{\sigma,\delta}$, we have that $f,\beta f_{\beta}\in
L^{\infty}_{\sigma}\cap C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi}$. Hence,
denoting

$g_{0}=g(0,\phi)=\left(\frac{\alpha-1}{1+\alpha}\right)^{\frac{1}{2}},$

we get that

$g-g_{0}\in L^{\infty}_{\sigma},\quad g\in
C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi},$

and then

$\beta^{\frac{-1}{1+\alpha}}r\in L^{\infty}_{0,\sigma}\cap
C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi}.$

Hence, we find the following asymptotic for $g$:

$g\sim_{0,\infty}g_{0}+\frac{\beta^{\sigma}}{(1+\beta)^{2\sigma}}.$

Coming back to $(\Psi_{\beta})^{\frac{1}{2}}$ we get that

$\beta^{\frac{1}{1+\alpha}}(\Psi_{\beta})^{\frac{1}{2}}\in
L^{\infty}_{0,\sigma}\cap C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi},$

with the following asymptotics:

$(\Psi_{\beta})^{\frac{1}{2}}\sim_{0,\infty}\beta^{\frac{-1}{1+\alpha}}\left(1+\frac{\beta^{\sigma}}{(1+\beta)^{2\sigma}}\right),$

and then

$(\Psi_{\beta})^{\frac{1}{2}}\sim_{0,\infty}\beta^{\frac{-1}{1+\alpha}}.$

We can also compute $\partial_{\varphi}g$ as

$\partial_{\varphi}g(\beta,\phi)=\frac{\frac{\alpha-1}{1+\alpha}f_{\varphi}(\beta,\phi)+\beta
f_{\beta\varphi}(\beta,\phi)}{2\left(\frac{\alpha-1}{1+\alpha}+\frac{\alpha-1}{1+\alpha}f(\beta,\phi)+\beta{f}_{{\beta}}(\beta,\phi)\right)^{\frac{1}{2}}}.$

Since $\beta f_{\varphi},\beta^{2}f_{\beta,\varphi}\in L^{\infty}_{\sigma}\cap
C^{\delta}_{\beta}$ we obtain

$\beta\partial_{\varphi}g\in L^{\infty}_{\sigma}\cap C^{\delta}_{\beta},$

implying

$\beta^{\frac{-1}{1+\alpha}}\beta\partial_{\varphi}(\Psi_{\beta})^{\frac{1}{2}}\in
L^{\infty}_{\sigma}\cap C^{\delta}_{\beta}.$

Finally, one finds that

$\beta^{\frac{-1}{1+\alpha}}r\in L^{\infty}_{0,\sigma}\cap
C^{\delta}_{\beta}\cap\textnormal{Lip}_{\phi}$

and

$\beta^{\frac{-1}{1+\alpha}}\beta r_{\varphi}\in L^{\infty}_{\sigma}\cap
C^{\delta}_{\beta}.$

∎

In the next proposition, we show that if $f\in
B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$ hence the jacobian of the
transformation $(\beta,\phi)\mapsto(z_{1},z_{2})$ is non vanishing except from
$\beta=+\infty$.

######  Proposition 7.2.

There exists $\varepsilon<1$ such that if $f\in
B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$ hence the jacobian $|J|$ of the
transformation $(\beta,\phi)\mapsto(z_{1},z_{2})$ does not vanish for
$(\beta,\phi)\in(0,\infty)\times[0,2\pi]$.

###### Proof.

From Proposition 2.3 we have that

$|J|=\frac{1+\alpha}{2}\Psi_{\beta\varphi}.$

Writing $\Psi$ in terms of $f$ and using (6.4) we find

$\Psi_{\beta\varphi}=-C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{-1}\beta^{\frac{-2}{1+\alpha}}\left\\{\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(\beta,\phi)\right\\}.$

Note that

$\frac{2(\alpha-1)}{(1+\alpha)^{2}}+\partial_{\overline{\beta\varphi}}f(\beta,\phi)\geq\frac{2(\alpha-1)}{(1+\alpha)^{2}}-\partial_{\overline{\beta\varphi}}f(\beta,\phi)\geq\frac{2(\alpha-1)}{(1+\alpha)^{2}}-||f||_{B_{X_{1}^{\sigma,\delta}}}\geq\varepsilon_{0}>0,$

which, together with $\beta\in(0,\infty)$ we find that $|J|>0$. Moreover,
notice that

$\lim_{\beta\rightarrow
0}|J|=+\infty,\quad\lim_{\beta\rightarrow+\infty}|J|=0.$

∎

By the previous Proposition 7.2 we have that the jacobian is not vanishing and
then we can invert the change of coordinates and find
$(\beta(z_{1},z_{2}),\phi(z_{1},z_{2}))$, for
$(\beta,\phi)\in(0,\infty)\times[0,2\pi]$. Since the jacobian of
$(z_{1},z_{2})\mapsto(\beta,\phi))$ is $|J|^{-1}$ which is vanishing at
$\beta=0$, we will identify such point with $(z_{1},z_{2})=(0,0)$.

Indeed, we shall first use the polar coordinates and then the main task is to
invert

$(r(\beta,\phi),\vartheta(\beta,\phi))=(\tilde{r},\tilde{\vartheta}),$

finding $\beta=\beta(\tilde{r},\tilde{\vartheta})$ and
$\phi=\phi(\tilde{r},\tilde{\vartheta})$, which is possible due to Proposition
7.2 by using the Inverse Function theorem. More precisely, in the next
proposition we understand better the functions $\beta(r,\vartheta)$ and
$\phi(r,\vartheta)$. For that, let us define $Z^{\sigma,\alpha}$ as

$Z^{\sigma,\alpha}:=\left\\{\tilde{\beta}\in
C_{b}([0,\infty)\times[0,2\pi]),\quad\tilde{\beta},r\partial_{r}\tilde{\beta}\in
L^{\infty}_{\sigma(1+\alpha)}\right\\}.$

######  Proposition 7.3.

Consider
$(r,\vartheta):[0,\infty)\times[0,2\pi]\rightarrow[0,\infty)\times[0,2\pi]$ in
(115). There exists $\varepsilon>0$ such that for $f\in
B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$ the following is satisfied. There
exists a function $\tilde{\beta}\in Z^{\sigma,\alpha}$ such that defining

$\beta(\tilde{r},\tilde{\vartheta})=r^{-1-\alpha}(1+\tilde{\beta}(\tilde{r},\tilde{\vartheta})),\quad\phi(\tilde{r},\tilde{\vartheta})=\tilde{\vartheta}-\beta(\tilde{r},\tilde{\vartheta}),$

we find

$(r(\beta(\tilde{r},\tilde{\vartheta}),\phi(\tilde{r},\tilde{\vartheta})),\vartheta(\beta(\tilde{r},\tilde{\vartheta}),\phi(\tilde{r},\tilde{\vartheta})))=(\tilde{r},\tilde{\vartheta}),$

for any $(\tilde{r},\tilde{\vartheta})\in[0,\infty)\times[0,2\pi]$.

###### Proof.

Using (115), our main task is to invert

$\left((-(1+\alpha)\Psi_{\beta})^{\frac{1}{2}},\beta+\phi\right)=(r,\vartheta),$

for any $(r,\vartheta)\in[0,+\infty)\times[0,2\pi]$. Hence, we want to find
two functions
$\beta,\phi:[0,+\infty)\times[0,2\pi]\rightarrow[0,+\infty)\times[0,2\pi]$
such that

$\displaystyle(-(1+\alpha)\Psi_{\beta}(\beta(r,\vartheta),\phi(r,\vartheta))^{\frac{1}{2}}=$
$\displaystyle r,$ $\displaystyle\beta(r,\vartheta)+\phi(r,\vartheta)=$
$\displaystyle\vartheta.$

From the second equation we find

$\phi(r,\vartheta)=\vartheta-\beta(r,\vartheta),$ (117)

and inserting it in the first equation we shall solve

$(-(1+\alpha)\Psi_{\beta}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))^{\frac{1}{2}}-r=0,$
(118)

for any $(r,\vartheta)\in[0,+\infty)\times[0,2\pi]$. Let us recall that $\Psi$
is indeed a perturbation of the trivial one, and by (116) we find

$(-\Psi_{\beta})^{\frac{1}{2}}(\beta,\phi)=C_{0}^{\frac{1}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{2(1+\alpha)}}\beta^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta,\phi)\right)^{\frac{1}{2}},$

for $f\in B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$. Hence, (118) agrees with

$(1+\alpha)^{\frac{1}{2}}C_{0}^{\frac{1}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{2(1+\alpha)}}\beta^{\frac{-1}{1+\alpha}}(r,\vartheta)\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)^{\frac{1}{2}}-r=0.$
(119)

Define now

$\beta_{0}(r)=C_{0}(\alpha-1)r^{-1-\alpha},$

agreeing with the change of coordinates for the trivial solution, which was
given in Proposition 2.5. Hence, decompose $\beta$ as

$\beta=\beta_{0}(1+\tilde{\beta}),$

and hence we can write (119) as $G(\tilde{\beta},f)(r,\vartheta)=0$ for any
$(r,\vartheta)\in(0,+\infty)\times[0,2\pi]$, where

$G(\tilde{\beta},f)(r,\vartheta):=\frac{(1+\alpha)^{\frac{1}{2}}}{(\alpha-1)^{\frac{1}{2}}}(1+\tilde{\beta}(r,\vartheta))^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)^{\frac{1}{2}}-1.$
(120)

We claim that there exists $\varepsilon<1$ such that $G:\times
B_{Z^{\sigma,\alpha}}(0,\varepsilon)\times
B_{X_{1}^{\sigma,\delta}(0,\varepsilon)}\rightarrow Z^{\sigma,\alpha}$ is
well-defined and $C^{1}$. Hence, we would like to find nontrivial
$(\tilde{\beta},f)$ such that $G(\tilde{\beta},f)(r)=0$ for any
$r\in[0,+\infty)$. We have the trivial solution given by

$G(0,0)(r,\vartheta)=0,\quad\forall(r,\vartheta)\in[0,+\infty)\times[0,2\pi].$

We wish to implement the Implicit Function theorem there and thus we need to
compute the linearized operator:

$\displaystyle\partial_{\tilde{\beta}}G(0,0)h(r,\vartheta)=-\frac{1}{1+\alpha}h(r,\vartheta),$

which is clearly an isomorphism from $Z^{\sigma,\alpha}$ into
$Z^{\sigma,\alpha}$.

Finally, it remains to check the well-definition of $G$. For that, let us
write $G$ as follows

$\displaystyle G(\tilde{\beta},f)(r)=$
$\displaystyle\frac{(1+\alpha)^{\frac{1}{2}}}{(\alpha-1)^{\frac{1}{2}}}\frac{(1+\tilde{\beta}(r,\vartheta))^{\frac{-2}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)-\frac{(\alpha-1)}{(1+\alpha)}}{(1+\tilde{\beta}(r,\vartheta))^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)^{\frac{1}{2}}+\frac{(\alpha-1)^{\frac{1}{2}}}{(1+\alpha)^{\frac{1}{2}}}}$
$\displaystyle=$
$\displaystyle\frac{(1+\alpha)^{\frac{1}{2}}}{(\alpha-1)^{\frac{1}{2}}}\frac{\left[(1+\tilde{\beta}(r,\vartheta))^{\frac{-2}{1+\alpha}}-1\right]\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)}{(1+\tilde{\beta}(r,\vartheta))^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)^{\frac{1}{2}}+\frac{(\alpha-1)^{\frac{1}{2}}}{(1+\alpha)^{\frac{1}{2}}}}$
$\displaystyle+\frac{(1+\alpha)^{\frac{1}{2}}}{(\alpha-1)^{\frac{1}{2}}}\frac{{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))}{(1+\tilde{\beta}(r,\vartheta))^{\frac{-1}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}+{f}_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\right)^{\frac{1}{2}}+\frac{(\alpha-1)^{\frac{1}{2}}}{(1+\alpha)^{\frac{1}{2}}}}.$

Keep in mind that $f\in X_{1}^{\sigma,\delta}$. First, note that
$f_{\overline{\beta}}\in L^{\infty}_{\sigma}$. Assuming that $\tilde{\beta}\in
L^{\infty}$, we have that

$\beta(r,\vartheta)\sim_{0,\infty}r^{-1-\alpha},$

and hence

$f_{\overline{\beta}}(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\sim_{0,\infty}\frac{r^{\sigma(1+\alpha)}}{(1+r)^{2\sigma(1+\alpha)}}.$
(121)

Hence, it is natural to assume that $\tilde{\beta}\in
L^{\infty}_{\sigma(1+\alpha)}$. Moreover, for $\varepsilon$ small and
$\tilde{\beta}\in B_{L^{\infty}_{\sigma(1+\alpha)}}(0,\varepsilon)$ we have
that all the denominator in the previous expression of $G$ are bounded. Hence,
from (121) we find that $G(\tilde{\beta},f)\in L^{\infty}_{\sigma(1+\alpha)}$.

Since in the expression of $G$ we have
$\partial_{\overline{\beta}}f(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))$,
it suggests us that we can only take one derivative in $r$. Indeed

$\partial_{r}(\partial_{\overline{\beta}}f(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta)))=(\partial_{\varphi}\partial_{\overline{\beta}}f)(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta))\partial_{r}\beta(r,\vartheta).$

Since $\beta(\partial_{\varphi}\partial_{\overline{\beta}}f)\in
L^{\infty}_{\sigma}$ we find that

$r\partial_{r}(\partial_{\overline{\beta}}f(\beta(r,\vartheta),\vartheta-\beta(r,\vartheta)))\sim_{0,\infty}\frac{r^{\sigma(1+\alpha)}}{(1+r)^{2\sigma(1+\alpha)}}.$

That motivates us to include $r\partial_{r}\tilde{\beta}\in
L^{\infty}_{\sigma(1+\alpha)}$ in the space $Z^{\sigma,\alpha}$. With such
space, we have the well-definition of $G$ and then we conclude the proof by
means of the Implicit Function theorem.

By (117) we find that $\phi\in Z_{0}^{\sigma,\alpha}$, where

$Z_{0}^{\sigma,\alpha}:=\left\\{\tilde{\beta}\in
C_{b}([0,\infty)\times[0,2\pi]),\quad\tilde{\beta},r\partial_{r}\tilde{\beta}\in
L^{\infty}_{0,\sigma(1+\alpha)}\right\\}.$

∎

###  7.2. Solution in the physical variables

From Theorem 6.2 together with Remark 6.4 we have constructed solutions
$(\Psi,\Theta)$ to (25), which is nothing but the vorticity equation in the
adapted coordinates, and (2.2) which refers to the elliptic relation between
$\Psi$ and $\Theta$. In particular, we find $(\Psi,\Theta)$ as

$\displaystyle\Psi(\beta,\phi)=$ $\displaystyle-
C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\beta^{\frac{\alpha-1}{1+\alpha}}(1+\tilde{f}(\tilde{\Omega})(\beta,\phi)),$
$\displaystyle\Theta(\beta,\phi)=$
$\displaystyle(1+\alpha)^{-\frac{1}{2\mu}}\left[C_{0}^{\frac{2}{1+\alpha}}(\alpha-1)^{\frac{1-\alpha}{1+\alpha}}\right]^{-\frac{1+\alpha-\gamma}{2}}\beta^{\frac{1+\alpha-\gamma}{1+\alpha}}\left(\frac{\alpha-1}{1+\alpha}-\tilde{f}_{\overline{\varphi}}(\tilde{\Omega})(\beta,\phi)\right)^{-\frac{1+\alpha-\gamma}{2}}\tilde{\Omega}(\phi),$

where $\tilde{\Omega}\in B_{X_{2}^{p}}(1,\varepsilon)$ and
$f(\tilde{\Omega})\in B_{X_{1}^{\sigma,\delta}}(0,\varepsilon)$.

Let us now come back to $\hat{\theta}$ by inverting the adapted coordinates
using Proposition 7.3. Denote
$\tilde{\theta}(r,\vartheta)=\hat{\theta}(re^{i\vartheta})$ and then

$\displaystyle\tilde{\theta}(r,\vartheta)=$
$\displaystyle\Theta(\beta(r,\vartheta),\phi(r,\vartheta))$ $\displaystyle=$
$\displaystyle\Theta\Big{(}C_{0}(\alpha-1)r^{-1-\alpha}(1+\tilde{\beta}(r,\vartheta)),\vartheta-
C_{0}(\alpha-1)r^{-1-\alpha}(1+\tilde{\beta}(r,\vartheta))\Big{)}$
$\displaystyle=$
$\displaystyle(1+\alpha)^{-\frac{1}{2\mu}}(\alpha-1)^{\frac{1+\alpha-\gamma}{2}}r^{-(1+\alpha-\gamma)}$
$\displaystyle\times\left(\frac{\alpha-1}{1+\alpha}-\tilde{f}_{\overline{\varphi}}(\tilde{\Omega})\Big{(}C_{0}(\alpha-1)r^{-1-\alpha}(1+\tilde{\beta}(r,\vartheta)),\vartheta-
C_{0}(\alpha-1)r^{-1-\alpha}(1+\tilde{\beta}(r,\vartheta))\Big{)}\right)^{-\frac{1+\alpha-\gamma}{2}}$
$\displaystyle\times\tilde{\Omega}\Big{(}\vartheta-
C_{0}(\alpha-1)r^{-1-\alpha}(1+\tilde{\beta}(r,\vartheta))\Big{)},$

with $\tilde{\beta}\in Z^{\sigma,\alpha}$. From (18) we have that

$\theta(t,x)=\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\hat{\theta}\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right)=\frac{1}{t^{\frac{1+\alpha-\gamma}{1+\alpha}}}\tilde{\theta}\left(\frac{r}{t^{\frac{1}{1+\alpha}}},\vartheta\right),$

where $x=re^{i\vartheta}$. Hence, using that $f\in X_{1}^{\sigma,\delta}$ we
get that the initial condition agrees with

$\theta_{0}(x)=r^{-(1+\alpha-\gamma)}\tilde{\Omega}(\vartheta),$

where $\tilde{\Omega}\in B_{X_{2}^{p}}(1,\varepsilon)$.

Moreover, $\theta_{0}$ is locally integrable:

$||\theta_{0}||_{L_{\textnormal{loc}}^{1}(\operatorname{\mathbb{R}}^{2})}=||\tilde{\Omega}||_{L^{1}([0,2\pi])}\int_{0}^{R}r^{\gamma-\alpha}dr=\frac{1}{1-\alpha+\gamma}||\tilde{\Omega}||_{L^{1}([0,2\pi])}R^{1-\alpha+\gamma},$

for any $R<\infty$, where $1-\alpha+\gamma>0$. Note that $\tilde{\Omega}\in
L^{1}$ since $\tilde{\Omega}\in L^{p}$, for some $p>\frac{1}{1-\gamma}$.

###  7.3. Main result

Our main result reads as follows.

######  Theorem 7.4.

Let $\gamma\in(0,1)$ and $\alpha\in(1,1+\gamma)$. Then, there exists
$\varepsilon>0$ such that for any $\frac{2\pi}{m}$-periodic $\tilde{\Omega}\in
B_{L^{p}([0,2\pi])}(1,\varepsilon)$ and $p>\frac{1}{1-\gamma}$, the initial
condition

$\theta_{0}(re^{i\vartheta})=r^{-(1+\alpha-\gamma)}\tilde{\Omega}(\vartheta),$

describes a self-similar solution of the type (18) for
$\textnormal{(SQG)}_{\gamma}$, for any $m\geq 1$.

##  Appendix A Radial solutions

In this section, we review the expression of the stream function $\psi$
associated to a radial vorticity $\theta$.

######  Proposition A.1.

If $f=(-\Delta)^{\beta}g$, with $\beta\in(-1,0)$ and $g(x)=g(|x|)$ radial,
then

$f(x)=2\pi C_{\beta}\int_{0}^{\infty}G_{0}(r,|x|)g(r)rdr,$

where

$G_{0}(r,|x|)=\left\\{\begin{array}[]{ll}\frac{1}{\Gamma(-\beta)}\left(\frac{2}{r}\right)^{2+2\beta}\Gamma(1+\beta){}_{2}F_{1}\left(1+\beta,1+\beta,1,\frac{|x|^{2}}{r^{2}}\right),&|x|<r,\\\
\frac{1}{\Gamma(-\beta)}\left(\frac{2}{|x|}\right)^{2+2\beta}\Gamma(1+\beta){}_{2}F_{1}\left(1+\beta,1+\beta,1,\frac{r^{2}}{|x|^{2}}\right),&|x|>r.\\\
\end{array}\right.$

###### Proof.

We want to calculate the explicit kernels of

$f=(-\Delta)^{\beta}g$

whenever $-1<\beta<0$ and for general $g$ of the type

$g(r,\theta)=g_{k}(r)\cos(k\theta),\quad g(r,\theta)=g_{k}(r)\sin(k\theta).$

Note that we are indeed interested in radial functions $g$ corresponding to
$k=0$, but let us compute it in general. To do so, we will calculate the
Fourier transform:

$f(x)=\int\hat{g}(\xi)|\xi|^{2\beta}e^{ix\cdot\xi}d\xi$

We start calculating $\hat{g}(\xi)$. In the first case:

$\displaystyle\hat{g}(\xi)=\int_{\operatorname{\mathbb{R}}^{2}}e^{-ix\cdot\xi}g(x)dx=\int_{0}^{\infty}g_{k}(r)r\int_{0}^{2\pi}\cos(k\theta)e^{-ir(\cos(\theta)\xi_{1}+\sin(\theta)\xi_{2})}d\theta
dr$

Writing

$\cos(\psi)=\frac{\xi_{1}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}},\quad\sin(\psi)=\frac{\xi_{2}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}},\quad
z=\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}r,$

the inner integral becomes

$\displaystyle\int_{0}^{2\pi}e^{-iz\cos(\theta-\psi)}\cos(k\theta)d\theta=$
$\displaystyle\cos(k\psi)\int_{0}^{2\pi}e^{-iz\cos(u)}\cos(ku)du$
$\displaystyle=$ $\displaystyle 2\pi
T_{k}\left(\frac{\xi_{1}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}}\right)(-i)^{k}J_{k}(\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}r),$

where $J_{k}$ is the Bessel function of order $k$, $T_{k}$ is the $k$-th
Chebychev polynomial of the first kind. In the second case:

$\displaystyle\hat{g}(\xi)=\int_{\operatorname{\mathbb{R}}^{2}}e^{-ix\cdot\xi}g(x)dx=\int_{0}^{\infty}g_{k}(r)r\int_{0}^{2\pi}\sin(k\theta)e^{-ir(\cos(\theta)\xi_{1}+\sin(\theta)\xi_{2})}d\theta
dr$

Writing

$\cos(\psi)=\frac{\xi_{1}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}},\quad\sin(\psi)=\frac{\xi_{2}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}},\quad
z=\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}r$

the inner integral becomes

$\displaystyle\int_{0}^{2\pi}e^{-iz\cos(\theta-\psi)}\sin(k\theta)d\theta=$
$\displaystyle\sin(k\psi)\int_{0}^{2\pi}e^{-iz\cos(u)}\sin(ku)du$
$\displaystyle=$ $\displaystyle
2\pi\frac{\xi_{2}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}}U_{k-1}\left(\frac{\xi_{1}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}}\right)(-i)^{k}J_{k}(\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}r),$

where $U_{k}$ is the $k$-th Chebychev polynomial of the second kind.

We now move on to the computation of the outer integral. Switching the
integrals in $r$ and in $\xi$, and integrating first in $\xi$, in the first
case:

$\displaystyle\int_{0}^{\infty}(-i)^{k}2\pi
g_{k}(r)r\int_{\operatorname{\mathbb{R}}^{2}}T_{k}\left(\frac{\xi_{1}}{\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}}\right)J_{k}(\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}r)|\xi|^{2\beta}e^{ix\cdot\xi}d\xi
dr$ $\displaystyle=\int_{0}^{\infty}(-i)^{k}2\pi
g_{k}(r)r\int_{0}^{\infty}J_{k}(|\xi|r)|\xi|^{2\beta+1}\int_{0}^{2\pi}\cos(k\psi)e^{i|\xi|(x_{1}\cos(\psi)+x_{2}\sin(\psi))}d\psi
d|\xi|dr$
$\displaystyle=(-i)^{2k}(2\pi)^{2}\cos(k\theta)\int_{0}^{\infty}g_{k}(r)r\int_{0}^{\infty}J_{k}(|\xi|r)|\xi|^{2\beta+1}J_{k}(|x||\xi|)d|\xi|dr$

Using that $-1<\beta<0$, we get

$\displaystyle\int_{0}^{\infty}J_{k}(|\xi|r)|\xi|^{2\beta+1}J_{k}(|x||\xi|)d|\xi|\equiv
G_{k}(r,|x|)$
$\displaystyle=\left\\{\begin{array}[]{cc}\frac{1}{2\Gamma(-\beta)}\left(\frac{2}{r}\right)^{2+2\beta}\left(\frac{|x|}{r}\right)^{k}\Gamma(1+\beta+k)\Gamma(k+1)_{2}F_{1}(1+\beta,1+\beta+k,1+k,\frac{|x|^{2}}{r^{2}})&\text{
if }|x|<r\\\
\frac{1}{2\Gamma(-\beta)}\left(\frac{2}{x}\right)^{2+2\beta}\left(\frac{r}{|x|}\right)^{k}\Gamma(1+\beta+k)\Gamma(k+1)_{2}F_{1}(1+\beta,1+\beta+k,1+k,\frac{r^{2}}{|x|^{2}})&\text{
if }|x|>r\\\ \end{array}\right.$

where ${}_{2}F_{1}$ is the Gauss Hypergeometric function. This yields:

$\displaystyle f(x)=(-i)^{2k}(2\pi)^{2}\cos(k\theta)\int
g_{k}(r)G_{k}(r,|x|)rdr.$

The computation in the second case is similar, and we obtain:

$\displaystyle f(x)=(-i)^{2k}(2\pi)^{2}\sin(k\theta)\int
g_{k}(r)G_{k}(r,|x|)rdr.$

Now, taking $k=0$ we arrive to the announced result. ∎

In the following, we apply the previous proposition to our trivial solution.

######  Corollary A.2.

If $\theta(x)=|x|^{-\frac{1}{\mu}}$, then
$\psi(x)=-(-\Delta)^{-1+\frac{\gamma}{2}}\theta(x)=-C_{0}|x|^{2-\gamma-\frac{1}{\mu}}$,
where

$\displaystyle C_{0}=$ $\displaystyle\frac{\pi
2^{\gamma}\Gamma\left(\frac{\gamma}{2}\right)\mu}{\Gamma\left(1-\frac{\gamma}{2}\right)}\left\\{\frac{1}{2\mu-1}{}_{3}F_{2}\left(\left\\{\frac{\gamma}{2},\frac{\gamma}{2},1-\frac{1}{2\mu}\right\\},\left\\{1,2-\frac{1}{2\mu}\right\\},1\right)\right.$
$\displaystyle\left.+\frac{1}{1+\mu(-2+\gamma)}{}_{3}F_{2}\left(\left\\{\frac{\gamma}{2},\frac{\gamma}{2},-1+\frac{\gamma}{2}+\frac{1}{2\mu}\right\\},\left\\{1,\frac{\gamma}{2}+\frac{1}{2\mu}\right\\},1\right)\right\\}.$
(122)

###### Proof.

Use the previous proposition with $\beta=-1+\frac{\gamma}{2}$ and
$g(x)=|x|^{-\frac{1}{\mu}}$. ∎

The constant $C_{0}$ will be important in the kernel study of the linearized
operator. For that reason, let us give its expression in terms of the integral
operator $T_{k}$ defined in (105).

######  Proposition A.3.

The constant $C_{0}$ can be also written as

$\displaystyle C_{0}=$
$\displaystyle\frac{C_{\gamma}}{2\pi}\int_{0}^{\infty}\int_{0}^{2\pi}\frac{z^{\gamma-\alpha}}{\left\\{1+z^{2}-2z\cos(\eta)\right\\}^{\frac{\gamma}{2}}}$
$\displaystyle=$
$\displaystyle\frac{C_{\gamma}}{2\pi}M[T_{0}](1-\alpha+\gamma).$

###### Proof.

From the definition of the $(-\Delta)^{-1+\frac{\gamma}{2}}$ operator one has

$C_{0}=r^{-2+\gamma+1/\mu}\frac{C_{\gamma}}{2\pi}\frac{C_{\gamma}}{2\pi}\int_{0}^{\infty}\int_{0}^{2\pi}\frac{z^{\gamma-\alpha}}{\left\\{r^{2}+z^{2}-2rz\cos(\eta)\right\\}^{\frac{\gamma}{2}}}.$

By performing the change of variable $z\mapsto rz$ and using the expression of
$\mu$ in (16) and $T_{k}$ in (105) we arrive to the announced identity. ∎

##  Appendix B Special functions

This section aims to give an overview of some special functions such as the
Gauss Hypergeometric function, the Gamma function or the Digamma function. At
the end, we will also introduce the Mellin transform and some of its
properties.

Define for any real numbers
$a,b\in\mathbb{R},\,c\in\mathbb{R}\backslash(-\mathbb{N})$ the Gauss
hypergeometric function $z\mapsto F(a,b;c;z)$ on the open unit disc
$\mathbb{D}$ by the power series

$F(a,b;c;z)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{z^{n}}{n!},\quad\forall
z\in\mathbb{D},$ (123)

where the Pochhammer’s symbol $(x)_{n}$ is defined by

$(x)_{n}=\begin{cases}1,&n=0,\\\ x(x+1)\cdots(x+n-1),&n\geq 1,\end{cases}$

and verifies

$(x)_{n}=x\,(1+x)_{n-1},\quad(x)_{n+1}=(x+n)\,(x)_{n}.$

Moreover, let us recall the integral representation of the hypergeometric
function, see for instance [69, p. 47]. Assume that
$\textnormal{Re}(c)>\textnormal{Re}(b)>0,$ then

$\hskip
28.45274ptF(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\leavevmode\nobreak\
dx,\quad\forall{z\in\operatorname{\mathbb{C}}\backslash[1,+\infty)}.$ (124)

The function
$\Gamma:\operatorname{\mathbb{C}}\backslash\\{-\operatorname{\mathbb{N}}\\}\to\operatorname{\mathbb{C}}$
refers to the gamma function, which is the analytic continuation to the
negative half plane of the usual gamma function defined on the positive half-
plane $\\{\textnormal{Re}z>0\\}$, and given by

$\Gamma(z)=\int_{0}^{+\infty}\tau^{z-1}e^{-\tau}d\tau,$

and satisfies the relation $\Gamma(z+1)=z\,\Gamma(z),\ \forall
z\in\operatorname{\mathbb{C}}\backslash(-\operatorname{\mathbb{N}}).$
Moreover, the Digamma function is defined through the logarithmic of the Gamma
function as:

$\psi^{0}(z)=\frac{\Gamma^{\prime}(z)}{\Gamma(z)},$

and has the following integral representation:

$\psi^{0}(z)=\int_{0}^{1}\frac{1-t^{z-1}}{1-t}dt-\gamma_{0},$ (125)

for $\textnormal{Re}(z)>0$ and where $\gamma_{0}$ stands for the Euler
constant.

In the following lemma, we give a representation of the Gauss Hypergeometric
function in terms of some special trigonometric integrals, the proof can be
found in [44].

######  Lemma B.1.

Let $k\in\operatorname{\mathbb{N}}$, $\gamma\geq 0$ and $A>1$, then

$\bigintsss_{0}^{2\pi}\frac{\cos(k\eta)}{(A-\cos(\eta))^{\frac{\gamma}{2}}}d\eta=\frac{2\pi}{(1+A)^{\frac{\gamma}{2}+k}}\frac{\left(\frac{\gamma}{2}\right)_{k}2^{k}\left(\frac{1}{2}\right)_{k}}{(2k)!}{}_{2}F_{1}\left(k+\frac{\gamma}{2},k+\frac{1}{2};2k+1;\frac{2}{1+A}\right).$

Finally, let us introduce the Mellin transform of $f$ defined as

$M[f](s)=\int_{0}^{\infty}x^{s-1}f(x)dx.$

In the following lemma, we give some properties of the Mellin transform, we
refer to [66] for the proof and for more information about this transform.

######  Lemma B.2.

The Mellin transform of $f$ has the following properties.

  * •

If

$(f\wedge g)(x)=\int_{0}^{\infty}f(y)g\left(\frac{x}{y}\right)\frac{dy}{y},$

then

$M[f\wedge g](s)=M[f]M[g].$ (126)

  * •

$M[x^{\alpha}f](s)=M[f](s+\alpha).$ (127)

  * •

$M[f^{\prime}](s)=-(s-1)M[f](s-1).$ (128)

  * •

$M[x^{\alpha}f](s)=M[f](s+\alpha).$ (129)

  * •

$M[f(x^{a})]=|a|^{-1}M[f](s/a).$ (130)

## References

  * [1] Weiwei Ao, Juan Dávila, Manuel del Pino, Monica Musso, and Juncheng Wei.  Travelling and rotating solutions to the generalized inviscid surface quasi-geostrophic equation.  Trans. Amer. Math. Soc., 374(9):6665–6689, 2021.
  * [2] Jonas Azzam and Jacob Bedrossian.  Bounded mean oscillation and the uniqueness of active scalar equations.  Trans. Amer. Math. Soc., 367(5):3095–3118, 2015.
  * [3] Giancarlo Breschi and Marco A. Fontelos.  A note on the self-similar solutions to the spontaneous fragmentation equation.  Proc. A., 473(2201):20160740, 13, 2017.
  * [4] Alberto Bressan and Ryan Murray.  On self-similar solutions to the incompressible Euler equations.  J. Differential Equations, 269(6):5142–5203, 2020.
  * [5] Alberto Bressan and Wen Shen.  A posteriori error estimates for self-similar solutions to the Euler equations.  Discrete Contin. Dyn. Syst., 41(1):113–130, 2021.
  * [6] Tristan Buckmaster, Steve Shkoller, and Vlad Vicol.  Nonuniqueness of weak solutions to the SQG equation.  Comm. Pure Appl. Math., 72(9):1809–1874, 2019.
  * [7] Jacob Burbea.  Motions of vortex patches.  Lett. Math. Phys., 6(1):1–16, 1982.
  * [8] Marco Cannone and Liutang Xue.  Remarks on self-similar solutions for the surface quasi-geostrophic equation and its generalization.  Proc. Amer. Math. Soc., 143(6):2613–2622, 2015.
  * [9] A. Castro and D. Córdoba.  Infinite energy solutions of the surface quasi-geostrophic equation.  Adv. Math., 225(4):1820–1829, 2010.
  * [10] Angel Castro, Diego Córdoba, and Javier Gómez-Serrano.  Existence and regularity of rotating global solutions for the generalized surface quasi-geostrophic equations.  Duke Math. J., 165(5):935–984, 2016.
  * [11] Angel Castro, Diego Córdoba, and Javier Gómez-Serrano.  Uniformly rotating analytic global patch solutions for active scalars.  Annals of PDE, 2(1):1–34, 2016.
  * [12] Angel Castro, Diego Córdoba, and Javier Gómez-Serrano.  Uniformly rotating smooth solutions for the incompressible 2D Euler equations.  Arch. Ration. Mech. Anal., 231(2):719–785, 2019.
  * [13] Angel Castro, Diego Córdoba, and Javier Gómez-Serrano.  Global smooth solutions for the inviscid SQG equation.  Memoirs of the AMS, 266(1292):89 pages, 2020.
  * [14] Dongho Chae.  Nonexistence of self-similar singularities for the 3D incompressible Euler equations.  Comm. Math. Phys., 273(1):203–215, 2007.
  * [15] Dongho Chae, Peter Constantin, Diego Córdoba, Francisco Gancedo, and Jiahong Wu.  Generalized surface quasi-geostrophic equations with singular velocities.  Comm. Pure Appl. Math., 65(8):1037–1066, 2012.
  * [16] Xinyu Cheng, Hyunju Kwon, and Dong Li.  Non-uniqueness of steady-state weak solutions to the surface quasi-geostrophic equations.  Comm. Math. Phys., 388(3):1281–1295, 2021.
  * [17] Tomasz Cieślak, Piotr Kokocki, and Wojciech S. Ozański.  Well-posedness of logarithmic spiral vortex sheets.  Arxiv preprint arXiv:2110.07543, 2021.
  * [18] Tomasz Cieślak, Piotr Kokocki, and Wojciech S. Ozański.  Existence of nonsymmetric logarithmic spiral vortex sheet solutions to the 2d Euler equations.  Arxiv preprint arXiv:2207.06056, 2022.
  * [19] Peter Constantin, Ming-Chih Lai, Ramjee Sharma, Yu-Hou Tseng, and Jiahong Wu.  New numerical results for the surface quasi-geostrophic equation.  J. Sci. Comput., 50(1):1–28, 2012.
  * [20] Peter Constantin, Andrew J. Majda, and Esteban Tabak.  Formation of strong fronts in the $2$-D quasigeostrophic thermal active scalar.  Nonlinearity, 7(6):1495–1533, 1994.
  * [21] Peter Constantin, Qing Nie, and Norbert Schörghofer.  Nonsingular surface quasi-geostrophic flow.  Phys. Lett. A, 241(3):168–172, 1998.
  * [22] Diego Cordoba.  Nonexistence of simple hyperbolic blow-up for the quasi-geostrophic equation.  Ann. of Math. (2), 148(3):1135–1152, 1998.
  * [23] Diego Cordoba and Charles Fefferman.  Growth of solutions for QG and 2D Euler equations.  J. Amer. Math. Soc., 15(3):665–670, 2002.
  * [24] Diego Córdoba, Marco A. Fontelos, Ana M. Mancho, and Jose L. Rodrigo.  Evidence of singularities for a family of contour dynamics equations.  Proc. Natl. Acad. Sci. USA, 102(17):5949–5952, 2005.
  * [25] Diego Córdoba, Javier Gómez-Serrano, and Alexandru D. Ionescu.  Global Solutions for the Generalized SQG Patch Equation.  Arch. Ration. Mech. Anal., 233(3):1211–1251, 2019.
  * [26] Juan Dávila, Manuel del Pino, Monica Musso, and Juncheng Wei.  Travelling helices and the vortex filament conjecture in the incompressible Euler equations.  Calc. Var. Partial Differential Equations, 61(4):Paper No. 119, 30, 2022.
  * [27] Francisco de la Hoz, Zineb Hassainia, and Taoufik Hmidi.  Doubly connected V-states for the generalized surface quasi-geostrophic equations.  Arch. Ration. Mech. Anal., 220(3):1209–1281, 2016.
  * [28] J. Deng, T. Y. Hou, R. Li, and X. Yu.  Level set dynamics and the non-blowup of the 2D quasi-geostrophic equation.  Methods Appl. Anal., 13(2):157–180, 2006.
  * [29] David G. Dritschel.  The repeated filamentation of two-dimensional vorticity interfaces.  J. Fluid Mech., 194:511–547, 1988.
  * [30] Tarek M. Elgindi and In-Jee Jeong.  On singular vortex patches, I: Well-posedness issues.  Mem. Amer. Math. Soc., 2019.  To appear.
  * [31] Tarek M. Elgindi and In-Jee Jeong.  On singular vortex patches, II: long-time dynamics.  Trans. Amer. Math. Soc., 373(9):6757–6775, 2020.
  * [32] Tarek M. Elgindi and In-Jee Jeong.  Symmetries and critical phenomena in fluids.  Comm. Pure Appl. Math., 73(2):257–316, 2020.
  * [33] Volker Elling.  Algebraic spiral solutions of the 2d incompressible Euler equations.  Bull. Braz. Math. Soc. (N.S.), 47(1):323–334, 2016.
  * [34] Volker Elling.  Self-similar 2d Euler solutions with mixed-sign vorticity.  Comm. Math. Phys., 348(1):27–68, 2016.
  * [35] Miguel Escobedo and J. J. L. Velázquez.  On the fundamental solution of a linearized homogeneous coagulation equation.  Comm. Math. Phys., 297(3):759–816, 2010.
  * [36] M. A. Fontelos and F. de la Hoz.  Singularities in water waves and the Rayleigh-Taylor problem.  J. Fluid Mech., 651:211–239, 2010.
  * [37] Susan Friedlander and Roman Shvydkoy.  The unstable spectrum of the surface quasi-geostropic equation.  J. Math. Fluid Mech., 7(suppl. 1):S81–S93, 2005.
  * [38] F. Gancedo and R. M. Strain.  Absence of splash singularities for surface quasi-geostrophic sharp fronts and the Muskat problem.  Proceedings of the National Academy of Sciences, 111(2):635–639, 2014.
  * [39] Francisco Gancedo.  Existence for the $\alpha$-patch model and the QG sharp front in Sobolev spaces.  Adv. Math., 217(6):2569–2598, 2008.
  * [40] Francisco Gancedo and Neel Patel.  On the local existence and blow-up for generalized SQG patches.  Arxiv preprint arXiv:1811.00530, 2018.
  * [41] Claudia García.  Kármán vortex street in incompressible fluid models.  Nonlinearity, 33(4):1625–1676, 2020.
  * [42] Claudia García.  Vortex patches choreography for active scalar equations.  J. Nonlinear Sci., 31(5):Paper No. 75, 31, 2021.
  * [43] Claudia García, Taoufik Hmidi, and Joan Mateu.  Time periodic doubly connected solutions for 3D quasi-geostrophic model.  Arxiv preprint arXiv:2206:10197, 2022.
  * [44] Claudia García, Taoufik Hmidi, and Joan Mateu.  Time periodic solutions for 3D quasi-geostrophic model.  Comm. Math. Phys., 390(2):617–756, 2022.
  * [45] Claudia García, Taoufik Hmidi, and Juan Soler.  Non uniform rotating vortices and periodic orbits for the two-dimensional Euler equations.  Arch. Ration. Mech. Anal., 238(2):929–1085, 2020.
  * [46] Javier Gómez-Serrano.  On the existence of stationary patches.  Adv. Math., 343:110–140, 2019.
  * [47] Philippe Gravejat and Didier Smets.  Smooth Travelling-Wave Solutions to the Inviscid Surface Quasi-Geostrophic Equation.  International Mathematics Research Notices, 2019(6):1744–1757, 2017\.
  * [48] Zineb Hassainia and Taoufik Hmidi.  On the V-states for the generalized quasi-geostrophic equations.  Comm. Math. Phys., 337(1):321–377, 2015.
  * [49] Zineb Hassainia, Taoufik Hmidi, and Nader Masmoudi.  KAM theory for active scalar equations.  Arxiv preprint arXiv:2110.08615, 2021.
  * [50] Isaac M. Held, Raymond T. Pierrehumbert, Stephen T. Garner, and Kyle L. Swanson.  Surface quasi-geostrophic dynamics.  J. Fluid Mech., 282:1–20, 1995.
  * [51] Taoufik Hmidi and Joan Mateu.  Existence of corotating and counter-rotating vortex pairs for active scalar equations.  Comm. Math. Phys., 350(2):699–747, 2017.
  * [52] Taoufik Hmidi, Joan Mateu, and Joan Verdera.  Boundary regularity of rotating vortex patches.  Archive for Rational Mechanics and Analysis, 209(1):171–208, 2013\.
  * [53] John K. Hunter, Jingyang Shu, and Qingtian Zhang.  Global solutions for a family of GSQG front equations.  Arxiv preprint arXiv:2005.09154, 2020.
  * [54] John K. Hunter, Jingyang Shu, and Qingtian Zhang.  Global solutions of a surface quasigeostrophic front equation.  Pure Appl. Anal., 3(3):403–472, 2021.
  * [55] Philip Isett and Andrew Ma.  A direct approach to nonuniqueness and failure of compactness for the SQG equation.  Nonlinearity, 34(5):3122–3162, 2021.
  * [56] Philip Isett and Vlad Vicol.  Hölder continuous solutions of active scalar equations.  Annals of PDE, 1(1):1–77, 2015.
  * [57] Heinrich Kaden.  Aufwicklung einer unstabilen unstetigkeitsfläche.  Ingenieur-Archiv, 2(2):140–168, 1931.
  * [58] A. Kiselev, L. Ryzhik, Y. Yao, and A. Zlatoš.  Finite time singularity formation for the modified SQG patch equation.  Ann. of Math. (2), 184:909–948, 2016.
  * [59] Alexander Kiselev and Fedor Nazarov.  A simple energy pump for the surface quasi-geostrophic equation.  In Helge Holden and Kenneth H. Karlsen, editors, Nonlinear Partial Differential Equations, volume 7 of Abel Symposia, pages 175–179. Springer Berlin Heidelberg, 2012.
  * [60] Andrew J. Majda and Andrea L. Bertozzi.  Vorticity and incompressible flow, volume 27 of Cambridge Texts in Applied Mathematics.  Cambridge University Press, Cambridge, 2002.
  * [61] Ana M Mancho.  Numerical studies on the self-similar collapse of the alpha-patches problem.  Communications in Nonlinear Science and Numerical Simulation, 26(1-3):152–166, 2015.
  * [62] Fabien Marchand.  Existence and regularity of weak solutions to the quasi-geostrophic equations in the spaces $L^{p}$ or $\dot{H}^{-1/2}$.  Comm. Math. Phys., 277(1):45–67, 2008.
  * [63] Andrea R. Nahmod, Nataša Pavlović, Gigliola Staffilani, and Nathan Totz.  Global flows with invariant measures for the inviscid modified SQG equations.  Stoch. Partial Differ. Equ. Anal. Comput., 6(2):184–210, 2018.
  * [64] Koji Ohkitani and Michio Yamada.  Inviscid and inviscid-limit behavior of a surface quasigeostrophic flow.  Phys. Fluids, 9(4):876–882, 1997.
  * [65] Frank Pacard and Tristan Rivière.  Linear and nonlinear aspects of vortices, volume 39 of Progress in Nonlinear Differential Equations and their Applications.  Birkhäuser Boston, Inc., Boston, MA, 2000.  The Ginzburg-Landau model.
  * [66] R. B. Paris and D. Kaminski.  Asymptotics and Mellin-Barnes integrals, volume 85 of Encyclopedia of Mathematics and its Applications.  Cambridge University Press, Cambridge, 2001.
  * [67] Joseph Pedlosky.  Geophysical fluid dynamics.  New York and Berlin, Springer-Verlag, 1, 1982.
  * [68] D. I. Pullin.  On similarity flows containing two-branched vortex sheets.  In Mathematical aspects of vortex dynamics (Leesburg, VA, 1988), pages 97–106. SIAM, Philadelphia, PA, 1989.
  * [69] Earl D. Rainville.  Special functions.  The Macmillan Co., New York, 1960.
  * [70] Coralie Renault.  Relative equilibria with holes for the surface quasi-geostrophic equations.  J. Differential Equations, 263(1):567–614, 2017.
  * [71] Serge G Resnick.  Dynamical problems in non-linear advective partial differential equations.  PhD thesis, University of Chicago, Department of Mathematics, 1995.
  * [72] José Luis Rodrigo.  On the evolution of sharp fronts for the quasi-geostrophic equation.  Comm. Pure Appl. Math., 58(6):821–866, 2005.
  * [73] R. K. Scott and D. G. Dritschel.  Numerical simulation of a self-similar cascade of filament instabilities in the surface quasigeostrophic system.  Phys. Rev. Lett., 112:144505, 2014.
  * [74] R. K. Scott and D. G. Dritschel.  Scale-invariant singularity of the surface quasigeostrophic patch.  Journal of Fluid Mechanics, 863:R2, 2019.
  * [75] Richard K. Scott.  A scenario for finite-time singularity in the quasigeostrophic model.  Journal of Fluid Mechanics, 687:492–502, 11 2011.