We now turn to the case of linear vorticity. We set \(F(u)=au+b\) where \(a,b\in {\mathbb {R}}\). Our partial differential equation (5) becomes
$$\begin{aligned} -\Delta u+ag(x,y)u=f(x,y). \end{aligned}$$
(9)
Here
$$\begin{aligned} \begin{aligned} g(x,y)&=\frac{4(x^2+y^2)}{[e^{{\tilde{X}}_0}(x^2+y^2)+e^{-{\tilde{X}}_0}]^2}\,, \\ f(x,y)&=-\frac{4b(x^2+y^2)}{[e^{{\tilde{X}}_0}(x^2+y^2)+e^{-{\tilde{X}}_0}]^2}-8\omega \frac{[e^{{\tilde{X}}_0}(x^2+y^2)-e^{-{\tilde{X}}_0}](x^2+y^2)}{[e^{{\tilde{X}}_0}(x^2+y^2)+e^{-{\tilde{X}}_0}]^3}. \end{aligned} \end{aligned}$$
(10)
Theorem 5.1
On the domain \({\mathbb {D}}\), for \(b\in {\mathbb {R}}\) and \(a\ge 0\), the elliptic equation (9) admits a unique solution \(u\in C^{\infty }(\overline{{\mathbb {D}}})\) satisfying the boundary condition \(u\big |_{\partial {\mathbb {D}}}=0\). Moreover, for the case \(a<0\), such a unique solution u exists if \({\tilde{X}}_0\le -C_a\), where \(C_a>0\) is a constant depending on a.
Remark 1
Notice that existence of solutions for \(a<0\) depends on the location of \({\tilde{X}}_0\). In fact, we shall see in the proof that the further we move away from the Equator, the more vorticity we can allow.
Proof
We begin by rewriting (9) in its variational formulation. For any \(v\in H_0^1({\mathbb {D}})\), we have
$$\begin{aligned} \int _{{\mathbb {D}}}\nabla u\cdot \nabla v +a\int _{{\mathbb {D}}}g u v=\int _{{\mathbb {D}}}f v. \end{aligned}$$
(11)
We set
$$\begin{aligned} \beta (u,v)=\int _{{\mathbb {D}}}\nabla u\cdot \nabla v +a\int _{{\mathbb {D}}}g u v, \end{aligned}$$
(12)
as our bilinear form on \(H_0^1({\mathbb {D}})\times H_0^1({\mathbb {D}})\) and
$$\begin{aligned} j(v)=\int _{{\mathbb {D}}}f v, \end{aligned}$$
as our linear form on \(H_0^1({\mathbb {D}})\). Clearly, \(\beta (u,v)\) and j(v) are continuous (for the bilinear form, this can be shown using the Cauchy-Schwarz inequality).
Using the Poincare inequality, the bilinear form \(\beta (u,v)\) is coercive for all \(a\ge 0\). Therefore, from the Lax-Milgram theorem, we get that there exists a unique weak solution \(u\in H_0^1({\mathbb {D}})\) to (9) for all positive \(a\in {\mathbb {R}}\).
It can be shown that the operator \(L_a:=-\Delta u+ag(x,y)u=f(x,y)\) is “semi”-Fredholm (has closed range and finite-dimensional kernel). The proof follows exactly as in [14]. Moreover, by self-adjointness of the operator, the index of \(L_a\) is equal to zero for all \(a\in {\mathbb {R}}\). Consequently, we have existence of solutions for all a for which the operator is injective.
Note that since \({\mathbb {D}}\) is of class \(C^{\infty }\) and \(f\in C^{\infty }(\overline{{\mathbb {D}}})\) and \(g\in C^{\infty }(\overline{{\mathbb {D}}})\), by the theory on regularity for elliptic problems with smooth and periodic coefficients, if the solution u to (9) exists we have a unique solution \(u\in C^{\infty }(\overline{{\mathbb {D}}})\). (Uniqueness follows from Lax-Milgram, or alternatively from the maximum principle for elliptic partial differential equations on bounded domains).
We now look at the case when a is negative. Since \(g\in L^{\infty }(\overline{{\mathbb {D}}})\), by the generalized maximum principle for elliptic problems (see [18]), if we can find a function \(\sigma >0\) on \(\overline{{\mathbb {D}}}\) such that \((\Delta -ag)[\sigma ]\le 0\) in \({\mathbb {D}}\), then (9) has at most one solution.
For simplicity of notation, consistency with results in [13, 14], and to make the physical implications easier to see, we will express this \(\sigma \) in terms of the variables \(({\tilde{X}},{\tilde{Y}})\) in the strip: this can all easily be mapped into the unit disk setting using the change of variables presented at the beginning of this paper.
From [13] we know that a solution to (3) satisfying the asymptotic conditions is the Legendre polynomial (see [1, 2, 16] for more details on such special functions):
$$\begin{aligned} P_l^k(z)=\frac{(-1)^k}{2^l l!}(1-z^2)^{\frac{k}{2}}\frac{d^{k+l}}{dx^{k+l}}(z^2-1)^l, \end{aligned}$$
(13)
where \(z=\tanh ({\tilde{X}})\) and \(a=-l(l+1)\), \(l\in {\mathbb {Z}}\), with l an integer.
For \(l=1\), we set \(\sigma _1({\tilde{X}})=-\tanh ({\tilde{X}})\) which is strictly positive on \(-\infty \le {\tilde{X}}\le {\tilde{X}}_0<0\) and therefore
$$\begin{aligned} \left( \Delta -\frac{a}{\cosh ^2({\tilde{X}})}\right) [\sigma _1]=\Delta \sigma _1+\frac{2}{\cosh ^2({\tilde{X}})}\sigma _1+\frac{(-a-2)}{\cosh ^2({\tilde{X}})}\sigma _1 \end{aligned}$$
(14)
is negative if and only if \(a\ge -2\).
For \(l=2\), we set \(\sigma _2({\tilde{X}})=3\tanh ^2({\tilde{X}})-1\) which is strictly positive for \(-\infty \le {\tilde{X}}\le {\tilde{X}}_0\le -0.66\). Similarily as in (14) we get uniqueness for \(a\ge -6\).
For \(\sigma _3({\tilde{X}})=-\frac{1}{2}(5\tanh ^3({\tilde{X}})-3\tanh ({\tilde{X}}))\) we will require \({\tilde{X}}_0\le -1\) to obtain uniqueness for all \(a\ge -12\). For \(\sigma _4({\tilde{X}})=\frac{1}{8}(35\tanh ^4({\tilde{X}})-30\tanh ^2({\tilde{X}})+3)\) we will require \({\tilde{X}}_0\le -1.3\) for uniqueness for all \(a\ge -20\).
In general, assume that \(a\in [-l(l+1),-l(l-1))\) for some integer l. We now seek the associated \(\sigma _l\). Notice that for \(k=0\), the Legendre polynomial (13) can be expressed as
$$\begin{aligned} \tilde{\sigma _l}=\sum _{j=0}^{l}\begin{pmatrix} n\\ j \end{pmatrix} \begin{pmatrix} n+j\\ j \end{pmatrix} \left( \frac{\tanh ({\tilde{X}}-1)}{2}\right) ^j. \end{aligned}$$
(15)
We denote by \({\tilde{X}}_0^{(l)}\) the largest value of \({\tilde{X}}\) such that \(\tilde{\sigma _l}\) has a fixed sign on \((-\infty ,{\tilde{X}}_0^{(l)}]\). If that sign is positive, we define \(\sigma _l\) by \(\sigma _l=\tilde{\sigma _l}\) and if it is negative, we choose \(\sigma _l=-\tilde{\sigma _l}\). We denote by \(C_a\) the positive constant such that \({\tilde{X}}_0^{(l)}\le -C_a.\)
We therefore notice that the further away from 0 we are, the easier it gets to prove uniqueness. In other words, the further away we are from the Equator, the larger vorticity we can allow. The uniqueness and therefore also the existence of solutions to (9) will depend \({\tilde{X}}_0\).
We now go back to the unit disk setting.
We still need to check the asymptotic condition (7). Notice that the solution u to (9) is radial. Indeed, this can easily be shown using a moving plane argument such as in Sect. 9.5.2 in [11]. As a result, since \(u_x(0,0)\) is an odd function of x, which is smooth close to the origin, we get
$$\begin{aligned} u_x(0,0)=u_y(0,0)=0. \end{aligned}$$
This concludes the proof. \(\square \)