Bifurcation and stability for charged drops

Abstract

In this paper, we investigate the Laplace’s equation for the electrical potential of charge drops on exterior domain, and overdetermined boundary conditions are prescribed. We determine the local bifurcation structure with respect to the surface tension coefficient as bifurcation parameter. Furthermore, we establish the stability and the instability near the bifurcation point.

Keywords

Bifurcation stability charged drops

1. Introduction

The raindrop disintegration in electric field plays an important role in the formation of thunderstorms, and the related research results are widely applied in producing microscopic, uniformly sized particles and capsules, such as manufacturing pharmaceuticals and printing inks.

The research on the symmetry-breaking bifurcation and the stability of charged drops has been one of the key issues in meteorology [18], aerosol science [21], nuclear fission [2] and spray technologies [12]. Let γ be the surface tension coefficient, Q be the quantity of electric charge, $ε_{0}$ be the dielectric constant, $R_{0}$ be the radius of spherical and $\begin{array}{rcl} X = \frac{Q^{2}}{32 γ π^{2} ε_{0} R_{0}^{3}} . \end{array}$ Lord Rayleigh [17] asserted that the spherical fluid drop may become unstable if X is larger than some fixed constant, provided Q is large enough or γ is small enough.

Geoffrey Taylor [19] obtained the relationship between the major axis and the minor axis of the critical state about the droplet stability by experiment. The major axis and the minor axis were calculated by finding the pole, the point of maximum curvature and the point of minimum curvature. Miksis [11] computed the deformation of a dielectric drop in a uniform electric field numerically, his results showed that the drop remains stable, if and only if γ or $ε_{0}$ does not less than some critical value. Tsamopoulos, Akylas and Brown [20] considered the evolution of axisymmetric, charged inviscid drops near the Rayleigh limit by a combination method of domain perturbation and multiple timescale, moreover, they showed there is a bifurcation point $Q_{c}$ , such that charged drops are stable for $Q > Q_{c}$ and unstable for $Q < Q_{c}$ . Duft et al. [4] showed the evolution process of a levitating charged drop whose radius decreases with time through experiments. Other numerical results see [1,16] and references therein. In particular, by bifurcation method, Fontelos and Friedman [5] obtained a sequence of bifurcation curves and analyzed the stability of two-lobed form branch. They gave a positive answer to Rayleigh’s assertion. Muratov, Novaga and Ruffini studied a geometric variational problem modeling charged liquid drops in two space dimensions [14,15]. We also refer to the recent progress of changed drops in three dimensions space [7–9,13,15].

Since the drop is conductive, all the electric charges are distributed on the droplet surface. Outside Ω there is an electric field with a potential V vanishing at infinity, and there is a constant potential on the drop’s surface. The capillary force is proportional to the mean curvature and electrostatic repulsion of the charges. The electrostatic forces are proportional to the surface charge density and the normal component of the electric field. As in [10], the charge distributes on the surface of a conductor proportionally to the exterior normal derivative of the potential. According to all the conditions above, Fontelos and Friedman [5] obtained the following problem $\begin{matrix} (1.1) & \{\begin{matrix} Δ V = 0 & in R^{3} ∖ Ω, \\ V = C & on \partial Ω, \\ {(\frac{\partial V}{\partial ν})}^{2} = \frac{2}{ε_{0}} (γ κ - δ p) & on \partial Ω, \\ V (x) \to 0 & as | x | \to + \infty, \end{matrix} \end{matrix}$ where ν is the exterior normal unit to the $\partial Ω$ , κ is the mean curvature of the drop’s surface, $δ p$ is the pressure difference between the fluid inside Ω and the fluid outside Ω, and C is a constant.

Let Ω be $B_{R_{0}} = {x : | x | < R_{0}}$ , then $V (r) = Q / (4 π ε_{0} r)$ is a solution of problem (1.1) with $\begin{array}{rcl} C = \frac{Q}{4 π ε_{0} R_{0}} \end{array}$ and $\begin{array}{rcl} δ p = \frac{γ}{R_{0}} - \frac{ε_{0}}{2} {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2} . \end{array}$ The problem of (1.1) can be rewritten as follows $\begin{matrix} (1.2) & \{\begin{matrix} Δ V = 0 & in R^{3} ∖ Ω, \\ V = \frac{Q}{4 π ε_{0} R_{0}} & on \partial Ω, \\ {(\frac{\partial V}{\partial ν})}^{2} = \frac{2}{ε_{0}} [γ κ - \frac{γ}{R_{0}} + \frac{ε_{0}}{2} {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2}] & on \partial Ω, \\ V (x) \to 0 & as | x | \to + \infty, \end{matrix} \end{matrix}$ where we choose $C = Q / (4 π ε_{0} R_{0})$ to guarantee that $(Ω, V (r)) = (B_{R_{0}}, Q / (4 π ε_{0} r))$ is the trivial solution.

In addition, we consider the case of the perturbed boundary $\begin{array}{rcl} \partial Ω = {(r, θ, φ) : r = r (θ, φ) = R_{0} + x (θ, φ)}, \end{array}$ where $x \in X^{m + 2 + α} = {x \in C^{m + 2 + α} (Σ) : π -periodic in θ, 2 π -periodic in φ}$ with $Σ = [0, π] \times [0, 2 π]$ , $m \in N$ . Define $\begin{array}{rcl} F (x, γ) = \frac{γ}{R_{0}} - \frac{ε_{0}}{2} {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2} - γ κ + \frac{ε_{0}}{2} {(\frac{\partial V}{\partial ν} |_{\partial Ω})}^{2} . \end{array}$ Then for each $γ \in R$ , $x = 0$ is the trivial solution of $F (x, γ) = 0$ . To find nontrivial solutions of (1.2), Fontelos and Friedman [5] proved that there is a sequence of bifurcation branches with $\begin{array}{rcl} γ = γ_{l} + ε γ_{l 1} + ε^{2} γ_{l 2} + \dots, l = 2, 3, \dots \end{array}$ and the free boundary $\begin{array}{rcl} r = R_{0} + x (θ) = R_{0} + ε Y_{l, 0} (θ) + ε^{2} Λ_{l 2} (θ) + \dots, l = 0, 1, \dots \end{array}$ where $Y_{l, 0} (θ)$ is the spherical harmonics $Y_{l, m}$ with $m = 0$ and $γ_{2} > γ_{3} > \dots$ . In particular, $\begin{matrix} γ_{2} = \frac{ε_{0} Q^{2}}{2 \times {(4 π ε_{0})}^{2} R_{0}^{3}} . \end{matrix}$ Moreover, they proved that $γ_{21} : = γ^{'} (0) > 0$ , the aspheric drop is stable for $γ (ε) > γ_{2}$ and unstable for $γ (ε) < γ_{2}$ for $| ε |$ small enough.

Since $Y_{2, 0} = {(\frac{5}{16 π})}^{1 / 2} (3 {cos}^{2} (θ) - 1)$ , the fluid drop obtained in [5] are axisymmetric. If ${cos}^{2} (θ) ⩾ 1 / 3$ , $ε Y_{2, 0}$ is increasing with respect to ε, otherwise $ε Y_{2, 0}$ is decreasing with respect to ε, which indicates that the fluid drop is oblate in the y-direction if $ε > 0$ . From now on, we call the fluid drop is oblate if it is oblate in the y-direction (or $ε > 0$ ), and prolate if it is prolate in the y-direction (or $ε < 0$ ).

We find that the proof of [5, Theorem 3] is insufficient. In the calculation of ν, the $ε^{2}$ term in [5, Theorem 3] is not exactly presented, which leads to the opposite sign of $γ^{'} (0)$ . Precisely, $γ^{'} (0)$ is $- \frac{3 \sqrt{5}}{448 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{4}}$ , not $\frac{5 \sqrt{5}}{224 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{4}}$ in [5, Theorem 3]. Furthermore, the correct sign of $γ^{'} (0)$ plays an essential role in determining the stability of charged drops. Our purposes is to fix the calculation gap in [5].

Theorem 1.1.

For the bifurcation branch $(x (ε), γ (ε))$ emanating from $γ_{2}$ , $\begin{array}{rcl} γ^{'} (0) = - \frac{3 \sqrt{5}}{448 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{4}} < 0 . \end{array}$

If $γ (ε) < γ_{2}$ with $| ε | < ε_{}$ for some* $ε_{}$ small enough, then the oblate spheroids are linearly stable under small axially symmetric perturbation.*

If $γ (ε) > γ_{2}$ and $| ε | < ε_{}$ , then the prolate spheroids are linearly unstable under small axially symmetric perturbation.*

The schematic diagram of Theorem 1.1 is shown in Fig. 1, where wee represent the axis of symmetry with an arrow.

Fig. 1.
Bifurcation and stability of charge drops.

The rest of this paper is organized as follows. In Section 2, we give the exact value of $γ^{'} (0)$ . There the key step is to obtain the exact expression of $ε^{2}$ term in the exterior normal unit vector. In Section 3, we establish the stability and the instability near the bifurcation points by the Crandall–Rabinowitz principle of stability exchange.
2. Compute of $γ^{'} (0)$

Now we compute the exact value of $γ^{'} (0)$ . We rewrite the free boundary as $\begin{matrix} x = ε Y_{2, 0} + ε^{2} Λ_{2} + O (ε^{3}), \end{matrix}$ which is the bifurcation branch emanating from $γ = γ_{2}$ . It has been proved in [5] that $\begin{matrix} (2.1) & F_{x} (0, γ_{2}) Λ_{2} + γ^{'} (0) F_{x γ} (0, γ_{2}) Y_{2, 0} + \frac{1}{2} Y_{2, 0} F_{x x} (0, γ_{2}) Y_{2, 0} = 0 . \end{matrix}$ Our task is to derive $\frac{1}{2} Y_{2, 0} F_{x x} (0, γ_{2}) Y_{2, 0}$ . This contains the second Fréchet derivative of ${(\frac{\partial V}{\partial ν})}^{2}$ and the mean curvature acting on $R_{0} + ε Y_{2, 0}$ .

Set $\begin{matrix} (2.2) & V = \frac{Q}{4 π ε_{0} r} + ε V_{1} + ε^{2} V_{2} + o (ε^{2}) . \end{matrix}$ According to the boundary condition $V = \frac{Q}{4 π ε_{0} r}$ on $r = R_{0} + ε Y_{2, 0} + ε^{2} Λ_{2} + o (ε^{2})$ , from [5] we have known that $\begin{matrix} (2.3) & V_{1} = \frac{Q R_{0}}{4 π ε_{0}} r^{- 3} Y_{2, 0} \end{matrix}$ and $\begin{matrix} (2.4) & V_{2} (R_{0}) = - \frac{Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0}^{2} - V_{1}^{'} (R_{0}) Y_{2, 0} + \frac{Q}{4 π ε_{0} R_{0}^{2}} Λ_{2} . \end{matrix}$

In [5], Fontelos and Friedman used the formula of $\begin{matrix} Y_{2, 0}^{2} = \frac{1}{\sqrt{π}} [\frac{3}{7} Y_{4, 0} + \frac{1}{7 \sqrt{5}} Y_{2, 0} + \frac{1}{2} Y_{0, 0}] \end{matrix}$ to get $V_{2}$ . It contains a clerical error in this formula. The formula should be $\begin{matrix} (2.5) & Y_{2, 0}^{2} = \frac{1}{\sqrt{π}} [\frac{3}{7} Y_{4, 0} + \frac{\sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}] . \end{matrix}$ It follows from [6] that $\begin{matrix} Y_{0, 0} = \frac{1}{{(4 π)}^{1 / 2}}, Y_{2, 0} = {(\frac{5}{16 π})}^{1 / 2} (3 {cos}^{2} (θ) - 1) \end{matrix}$ and $\begin{matrix} Y_{4, 0} = \frac{1}{π^{1 / 2}} \frac{1}{16^{2}} \frac{d^{4} {(z^{2} - 1)}^{4}}{d z^{4}}, \end{matrix}$ where $z = cos (θ)$ . Direct computation implies $\begin{matrix} Y_{4, 0} = \frac{1}{π^{1 / 2}} \frac{3}{16} (35 z^{4} - 30 z^{2} + 3) . \end{matrix}$ By some elementary computations, we have that $\begin{matrix} Y_{2, 0}^{2} = \frac{5}{16 π} (9 z^{4} - 6 z^{2} + 1) \end{matrix}$ and $\begin{array}{rcl} \frac{1}{\sqrt{π}} [\frac{3}{7} Y_{4, 0} + \frac{\sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}] & = & \frac{1}{π} \frac{9}{16 \times 7} (35 z^{4} - 30 z^{2} + 3) \\ + \frac{1}{π} \frac{5}{4 \times 7} (3 z^{2} - 1) + \frac{1}{π} \frac{1}{4} \\ = & \frac{5}{16 π} (9 z^{4} - 6 z^{2} + 1) \\ = & Y_{2, 0}^{2} . \end{array}$ By (2.5), we get that $\begin{array}{rcl} V_{2} & = & \frac{2 Q}{4 π ε_{0} R_{0}^{3}} \frac{1}{\sqrt{π}} (\frac{3}{7} {(\frac{R_{0}}{r})}^{5} Y_{4, 0} + \frac{\sqrt{5}}{7} {(\frac{R_{0}}{r})}^{3} Y_{2, 0} + \frac{1}{2} \frac{R_{0}}{r} Y_{0, 0}) \\ + \frac{Q}{4 π ε_{0} R_{0}^{2}} \sum_{l = 0}^{+ \infty} \sum_{m = - l}^{l} {(\frac{R_{0}}{r})}^{l + 1} a_{l m} Y_{l, m}, \end{array}$ where $\begin{matrix} a_{l m} = \int_{0}^{2 π} \int_{0}^{π} Y_{l, m} (α, β) Λ_{2} (α, β) sin α d α d β . \end{matrix}$

We now deduce $\frac{\partial V}{\partial ν}$ which is different from [5]. The original computation in [5] contains a crucial gap. Notice that $\begin{matrix} ω = e_{r} - \frac{ε Y_{20, θ}}{R_{0} + ε Y_{2, 0}} e_{θ} \end{matrix}$ is an exterior normal vector on $r = R_{0} + ε Y_{2, 0}$ . Observe that $\begin{matrix} ω = e_{r} - \frac{ε Y_{20, θ}}{R_{0}} e_{θ} + \frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} e_{θ} + O (ε^{3}), \end{matrix}$ where $Y_{20, θ} = Y_{2, 0}^{'}$ . So the exterior normal unit vector is $\begin{matrix} ν = \frac{e_{r} - \frac{ε Y_{20, θ}}{R_{0}} e_{θ} + \frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} e_{θ} + O (ε^{3})}{\sqrt{1 + {(\frac{ε Y_{20, θ}}{R_{0} + ε Y_{2, 0}})}^{2}}} . \end{matrix}$ By the Taylor expansion again, we find that $\begin{array}{rcl} ν & = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2} + O (ε^{3})) ({\vec{e}}_{r} - \frac{ε Y_{20, θ}}{R_{0}} {\vec{e}}_{θ} + \frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} {\vec{e}}_{θ} + O (ε^{3})) \\ = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2}) {\vec{e}}_{r} + (\frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} - \frac{ε Y_{20, θ}}{R_{0}}) {\vec{e}}_{θ} + O (ε^{3}) . \end{array}$ In [5], the $ε^{2}$ term is not exactly presented, which leads to some terms missing in the computation of $\frac{\partial V}{\partial ν}$ . Moreover, the normal vector is not normalized in [5], which also leads to missing some terms. These terms eventually lead to the opposite sign of $γ^{'} (0)$ . Here we have all these terms exactly. This is the key to obtain the correct value of $γ^{'} (0)$ .

Furthermore, ν can be written as $\begin{array}{rcl} ν & = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2} + O (ε^{3})) (e_{r} - \frac{ε Y_{20, θ}}{R_{0}} e_{θ} + \frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} e_{θ} + O (ε^{3})) \\ = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2}) e_{r} + (\frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} - \frac{ε Y_{20, θ}}{R_{0}}) e_{θ} + O (ε^{3}) \end{array}$ and $\begin{array}{rcl} \frac{\partial V}{\partial ν} & = & \vec{\nabla} V \cdot ν = (\frac{\partial V}{\partial r} e_{r} + \frac{1}{r} \frac{\partial V}{\partial θ} e_{θ} + \frac{1}{r sin θ} \frac{\partial V}{\partial φ} e_{φ}) \cdot ν \\ = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2}) \frac{\partial V}{\partial r} + (\frac{ε^{2} Y_{20, θ} Y_{2, 0}}{R_{0}^{2}} - \frac{ε Y_{20, θ}}{R_{0}}) \frac{1}{r} \frac{\partial V}{\partial θ} . \end{array}$ Firstly, we show that $\begin{array}{rcl} \frac{\partial V}{\partial r} & = & - \frac{Q}{4 π ε_{0} r^{2}} + ε \frac{\partial V_{1}}{\partial r} + ε^{2} \frac{\partial V_{2}}{\partial r} + o (ε^{2}) \\ = & - \frac{Q}{4 π ε_{0} r^{2}} - ε \frac{3 Q R_{0} Y_{2, 0}}{4 π ε_{0}} r^{- 4} + ε^{2} \frac{\partial V_{2}}{\partial r} + o (ε^{2}) \\ = & - \frac{Q}{4 π ε_{0} r^{2}} - ε \frac{3 Q R_{0} Y_{2, 0}}{4 π ε_{0}} r^{- 4} + ε^{2} \frac{\partial V_{2}}{\partial r} + o (ε^{2}) . \end{array}$ By (2.2)–(2.4), one has that $\begin{array}{rcl} \frac{\partial V}{\partial r} & = & - \frac{Q}{4 π ε_{0}} (R_{0}^{2} - \frac{2 Y_{2, 0}}{R_{0}^{3}} ε + \frac{3 Y_{2, 0}^{2}}{R_{0}^{4}} ε^{2}) \\ - ε \frac{3 Q R_{0} Y_{2, 0}}{4 π ε_{0}} (R_{0}^{- 4} - 4 R_{0}^{- 5} Y_{2, 0} ε) + ε^{2} \frac{\partial V_{2}}{\partial r} + o (ε^{2}) \\ = & - \frac{Q}{4 π ε_{0} R_{0}^{2}} + ε \frac{2 Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0} - ε^{2} \frac{3 Q}{4 π ε_{0} R_{0}^{4}} Y_{2, 0}^{2} \\ - ε \frac{3 Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0} + ε^{2} \frac{12 Q}{4 π ε_{0} R_{0}^{4}} Y_{2, 0}^{2} \\ - ε^{2} \frac{2 Q}{4 π ε_{0} R_{0}^{4}} \frac{1}{\sqrt{π}} (\frac{15}{7} Y_{4, 0} + \frac{3 \sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}) + o (ε^{2}) . \end{array}$

On the other hand, it holds that $\begin{array}{rcl} \frac{1}{r} \frac{\partial V}{\partial θ} & = & \frac{1}{r} (- \frac{Q}{4 π ε_{0} r^{2}} \frac{\partial r}{\partial θ} + ε \frac{\partial V_{1}}{\partial θ} + O (ε^{2})) \\ = & - ε \frac{Q Y_{20, θ}}{4 π ε_{0}} \frac{1}{r^{3}} + ε \frac{1}{r} \frac{\partial V_{1}}{\partial θ} + O (ε^{2}) . \end{array}$ Then, (2.2), together with (2.3) and (2.4), implies that $\begin{array}{rcl} \frac{1}{r} \frac{\partial V}{\partial θ} & = & - ε \frac{Q Y_{20, θ}}{4 π ε_{0}} (R_{0}^{- 3} - 3 R_{0}^{- 4} ε Y_{20, θ} + O (ε^{2})) \\ + ε (\frac{Q R_{0}}{4 π ε_{0}} r^{- 4} Y_{20, θ} - \frac{3 Q R_{0} Y_{2, 0}}{4 π ε_{0}} r^{- 5} ε Y_{20, θ}) + O (ε^{2}) \\ = & - ε \frac{Q Y_{20, θ}}{4 π ε_{0}} R_{0}^{- 3} + ε \frac{Q R_{0} Y_{20, θ}}{4 π ε_{0}} r^{- 4} + O (ε^{2}) \\ = & - ε \frac{Q Y_{20, θ}}{4 π ε_{0}} R_{0}^{- 3} + ε \frac{Q R_{0} Y_{20, θ}}{4 π ε_{0}} (R_{0}^{- 4} - 4 R_{0}^{- 5} ε Y_{2, 0}) + O (ε^{2}) \\ = & - ε \frac{Q Y_{20, θ}}{4 π ε_{0}} R_{0}^{- 3} + ε \frac{Q Y_{20, θ}}{4 π ε_{0}} R_{0}^{- 3} + O (ε^{2}) \\ = & O (ε^{2}) . \end{array}$

Therefore, it follows that $\begin{array}{rcl} \frac{\partial V}{\partial ν} |_{r = R_{0} + ε Y_{2, 0}} & = & (1 - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2}) \frac{\partial V}{\partial r} |_{r = R_{0} + ε Y_{2, 0}} + O (ε^{3}) \\ = & \frac{\partial V}{\partial r} |_{r = R_{0} + ε Y_{2, 0}} - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2} \frac{\partial V}{\partial r} |_{r = R_{0} + ε Y_{2, 0}} + O (ε^{3}) . \end{array}$ Further, one has that $\begin{array}{rcl} \frac{\partial V}{\partial ν} |_{r = R_{0} + ε Y_{2, 0}} & = & \frac{\partial V}{\partial r} |_{r = R_{0} + ε Y_{2, 0}} - \frac{Y_{20, θ}^{2}}{2 R_{0}^{2}} ε^{2} \frac{\partial V}{\partial r} |_{r = R_{0} + ε Y_{2, 0}} + O (ε^{3}) \\ = & - \frac{Q}{4 π ε_{0} R_{0}^{2}} + ε \frac{2 Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0} - ε^{2} \frac{3 Q}{4 π ε_{0} R_{0}^{4}} Y_{2, 0}^{2} \\ - ε \frac{3 Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0} + ε^{2} \frac{12 Q}{4 π ε_{0} R_{0}^{4}} Y_{2, 0}^{2} \\ - ε^{2} \frac{2 Q}{4 π ε_{0} R_{0}^{4}} \frac{1}{\sqrt{π}} (\frac{15}{7} Y_{4, 0} + \frac{3 \sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}) \\ + ε^{2} \frac{Q}{8 π ε_{0} R_{0}^{4}} Y_{20, θ}^{2} + O (ε^{3}) \\ = & - \frac{Q}{4 π ε_{0} R_{0}^{2}} - ε \frac{Q}{4 π ε_{0} R_{0}^{3}} Y_{2, 0} + ε^{2} \frac{9 Q}{4 π ε_{0} R_{0}^{4}} Y_{2, 0}^{2} \\ - ε^{2} \frac{2 Q}{4 π ε_{0} R_{0}^{4}} \frac{1}{\sqrt{π}} (\frac{15}{7} Y_{4, 0} + \frac{3 \sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}) \\ + ε^{2} \frac{Q}{8 π ε_{0} R_{0}^{4}} Y_{20, θ}^{2} + O (ε^{3}), \end{array}$ which is different from the corresponding one of [5]. We further obtain that $\begin{array}{rcl} \frac{\partial V}{\partial ν} |_{r = R_{0} + ε Y_{2, 0}}^{2} & = & {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2} + ε \frac{2 Q^{2}}{{(4 π ε_{0})}^{2} R_{0}^{5}} Y_{2, 0} + O (ε^{3}) \\ - 2 ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} (- \frac{1}{2} Y_{2, 0}^{2} + 9 Y_{2, 0}^{2}) \\ - 2 ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} (- \frac{2}{\sqrt{π}} (\frac{15}{7} Y_{4, 0} + \frac{3 \sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0})) \\ - ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} Y_{20, θ}^{2} \end{array}$ with the fact that $\begin{matrix} Y_{20, θ}^{2} = \frac{45}{4 π} (\frac{4 \sqrt{π}}{15} Y_{0, 0} + \frac{2}{21} \sqrt{\frac{4 π}{5}} Y_{2, 0} - \frac{8}{35} \frac{\sqrt{4 π}}{3} Y_{4, 0}), \end{matrix}$ where $\begin{matrix} Y_{20, θ}^{2} = {(Y_{20, θ})}^{2} = {(\frac{d}{d θ} Y_{2, 0})}^{2} . \end{matrix}$ Furthermore, it holds that $\begin{array}{rcl} \frac{\partial V}{\partial ν} |_{r = R_{0} + ε Y_{2, 0}}^{2} & = & {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2} + ε \frac{2 Q^{2}}{{(4 π ε_{0})}^{2} R_{0}^{5}} Y_{2, 0} + O (ε^{3}) \\ - \frac{17}{\sqrt{π}} ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} (\frac{3}{7} Y_{4, 0} + \frac{\sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}) \\ + \frac{4}{\sqrt{π}} ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} (\frac{15}{7} Y_{4, 0} + \frac{3 \sqrt{5}}{7} Y_{2, 0} + \frac{1}{2} Y_{0, 0}) \\ - ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} \frac{45}{4 π} (\frac{4 \sqrt{π}}{15} Y_{0, 0} + \frac{2}{21} \sqrt{\frac{4 π}{5}} Y_{2, 0} - \frac{8}{35} \frac{\sqrt{4 π}}{3} Y_{4, 0}) \\ = & {(\frac{Q}{4 π ε_{0} R_{0}^{2}})}^{2} + ε \frac{2 Q^{2}}{{(4 π ε_{0})}^{2} R_{0}^{5}} Y_{2, 0} + O (ε^{3}) \\ - 2 ε^{2} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} \frac{1}{\sqrt{π}} (\frac{19}{4} Y_{0, 0} + \frac{4 \sqrt{5}}{7} Y_{2, 0} - \frac{3}{2} Y_{4, 0}), \end{array}$ which is still different from [5, equality (5.4)]. We further obtain that $\begin{array}{rcl} (2.6) & \frac{1}{2} Y_{2, 0} \frac{ε_{0}}{2} {(\frac{\partial V}{\partial ν} |_{\partial Ω}^{2})}_{x x} Y_{2, 0} = ε_{0} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} \frac{1}{\sqrt{π}} (- \frac{19}{4} Y_{0, 0} - \frac{4 \sqrt{5}}{7} Y_{2, 0} + \frac{3}{2} Y_{4, 0}), \end{array}$ which is different from the corresponding one of [5, equality (5.5)]. From [5, Theorem 3], it yields $\begin{array}{rcl} (2.7) & \frac{1}{2} Y_{2, 0} {(- γ_{2} κ)}_{x x} Y_{2, 0} = γ_{2} \frac{25}{16 π R_{0}^{3}} (\frac{8 \sqrt{π}}{5} Y_{0, 0} + \frac{8}{7} \sqrt{\frac{4 π}{5}} Y_{2, 0} + \frac{24 \sqrt{4 π}}{35} Y_{4, 0}), \end{array}$ which is identical to the corresponding one of [5, Theorem 3]. It follows from (2.1) that $\begin{matrix} γ^{'} (0) F_{x γ} (0, γ_{2}) Y_{2, 0} = - F_{x} (0, γ_{2}) Λ_{2} - \frac{1}{2} Y_{2, 0} F_{x x} (0, γ_{2}) Y_{2, 0}, \end{matrix}$ so $\begin{matrix} γ^{'} (0) F_{x γ} (0, γ_{2}) Y_{2, 0} = - \frac{2 γ^{'} (0)}{R_{0}^{2}} Y_{2, 0} . \end{matrix}$ Thus, $\begin{matrix} \frac{2 γ^{'} (0)}{R_{0}^{2}} Y_{2, 0} = F_{x} (0, γ_{2}) Λ_{2} + \frac{1}{2} Y_{2, 0} F_{x x} (0, γ_{2}) Y_{2, 0} . \end{matrix}$ With the fact of $F_{x γ} (0, γ_{2}) Λ_{2}$ being orthogonal to $Y_{2, 0}$ , together with (2.6) and (2.7), one has $\begin{array}{rcl} \frac{2 γ^{'} (0)}{R_{0}^{2}} & = & \frac{1}{2} Y_{2, 0} F_{x x} (0, γ_{2}) Y_{2, 0}^{2} \\ = & \frac{1}{2} Y_{2, 0} \frac{ε_{0}}{2} {(\frac{\partial V}{\partial ν} |_{\partial Ω}^{2})}_{x x} Y_{2, 0}^{2} + \frac{1}{2} Y_{2, 0} {(- γ_{2} κ)}_{x x} Y_{2, 0}^{2} \\ = & γ_{2} \frac{25}{16 π R_{0}^{3}} \frac{8}{7} \sqrt{\frac{4 π}{5}} - ε_{0} {(\frac{Q}{4 π ε_{0} R_{0}^{3}})}^{2} \frac{4 \sqrt{5}}{7} \frac{1}{\sqrt{π}} \\ = & \frac{- 3 \sqrt{5}}{224 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{6}} . \end{array}$ Then we have that $\begin{array}{rcl} γ^{'} (0) = \frac{- 3 \sqrt{5}}{448 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{4}} < 0 . \end{array}$ Since the value is different from the corresponding [5] (having opposite signs), the stability of charged drops may be affected.

3. Proof of Theorem 1.1

Let $N (T)$ and $R (T)$ denote the null space and the range of any operator T, respectively.

Definition 3.1.

Let T, K: $X \to Y$ be two bounded linear operators from a real Banach space X to another one Y. A complex number β is called a K-simple eigenvalue of T if $\begin{matrix} dim N (T - β K) = 1 = codim R (T - β K) \end{matrix}$ and $\begin{matrix} K ψ^{*} \notin R (T - β K) for 0 \neq ψ^{*} \in N (T - β K) . \end{matrix}$

We will complete the proof of Theorem 1.1 with the Crandall–Rabinowitz principle of stability exchange [3] as follows.

Theorem 3.1.

Let X and Y be two real Banach spaces and let $K : X \to Y$ be two bounded linear operators. Assume $F : R \times X \to Y$ is $C^{2}$ near $(λ_{*}, 0) \in R \times X$ with $F (λ, 0) = 0$ for $| λ_{*} - λ |$ sufficiently small. Let $T = F_{f} (λ_{*}, 0)$ be the linearization operator of F with respect to the second variable at $(λ_{*}, 0)$ . If $β = 0$ is a K-simple eigenvalue of operator T with eigenfunction $ψ^{*} \in N (T) ∖ {0}$ , then there exists locally a curve $(λ (s), f (s)) \in R \times X$ such that $\begin{matrix} (λ (0), f (0)) = (λ_{*}, 0) and F (λ (s), f (s)) = 0 . \end{matrix}$ Moreover, if $F (λ, f) = 0$ with $f \neq 0$ and $(λ, f)$ near $(λ_{*}, 0)$ , then $\begin{matrix} (λ, f) = (λ (s), f (s)) for some s \neq 0 . \end{matrix}$ Furthermore, there are eigenvalues $β (s)$ , $β_{triv} (λ) \in R$ with eigenvectors $ψ (s)$ , $ψ_{triv} (λ) \in X$ , such that $\begin{array}{l} F_{f} (λ (s), f (s)) ψ (s) = β (s) K ψ (s), \\ F_{f} (λ, 0) ψ_{triv} (λ) = β_{triv} (λ) K ψ_{triv} (λ) \end{array}$ with $\begin{array}{l} β (0) = β_{triv} (λ_{*}) = 0, ψ (0) = ψ_{triv} (λ_{*}) = ψ^{*} . \end{array}$ The bifurcation curve is $C^{1}$ if F is $C^{2}$ , with $\begin{array}{rcl} \frac{d β_{triv} (λ)}{d λ} |_{λ = λ_{*}} \neq 0, lim_{s \to 0, β (s) \neq 0} \frac{s λ^{'} (s)}{β (s)} = - \frac{1}{β_{triv}^{'} (λ_{*})} . \end{array}$

We now complete the proof of Theorem 1.1 by showing the stability of charge drops.

Proof of Theorem 1.1.

According to [5, Theorem 2], 0 is a simple eigenvalue of $F_{x} (0, γ_{l}) : = T$ and $\begin{array}{rcl} (3.1) & [F_{γ x} (0, γ_{l})] Y_{l, 0} = - \frac{1}{R_{0}^{2}} (\frac{l (l + 1)}{2} - 1) Y_{l, 0} \notin Im (T) . \end{array}$ Taking $K = F_{γ x} (0, γ_{l})$ , the above transversality conclusion implies that 0 is a K simple eigenvalue of T. With the aid of Theorem 3.1, there are eigenvalues $μ (ε)$ , $β (γ) \in R$ with eigenvectors $w (ε)$ , $ψ (γ) \in X$ , such that $\begin{array}{l} F_{x} (x (ε), γ (ε)) w (ε) = μ (ε) F_{γ x} (0, γ_{l}) w (ε), \\ (3.2) & F_{x} (0, γ) ψ (γ) = β (γ) F_{γ x} (0, γ_{l}) ψ (γ) \end{array}$ with $\begin{array}{l} μ (0) = β (γ_{l}) = 0, w (0) = ψ (γ_{l}) = Y_{l, 0} \end{array}$ and $\begin{array}{rcl} \frac{d β (γ)}{d γ} |_{γ = γ_{l}} \neq 0 and lim_{ε \to 0, β (ε) \neq 0} \frac{ε γ^{'} (ε)}{μ (ε)} = - \frac{1}{β^{'} (γ_{l})} . \end{array}$

Setting $γ (δ) = γ_{l} + δ$ , then $\begin{array}{rcl} [F_{x} (0, γ (δ))] Y_{l, 0} = λ_{l} (δ) Y_{l, 0}, \end{array}$ where $\begin{array}{rcl} λ_{l} (δ) = [\frac{γ (δ)}{R_{0}^{2}} (1 - \frac{1}{2} l (l + 1)) + \frac{ε_{0} Q^{2}}{{(4 π ε_{0})}^{2} R_{0}^{5}} (l - 1)] . \end{array}$ With (3.1) and (3.2), one has $ψ (γ) = Y_{2, 0}$ and $\begin{array}{rcl} λ_{l} (δ) = - β (γ (δ)) \frac{1}{R_{0}^{2}} (\frac{l (l + 1)}{2} - 1) . \end{array}$ Therefore, this yields $\begin{array}{rcl} β (γ (δ)) = - \frac{2 λ_{l} (δ) R_{0}^{2}}{l^{2} + l - 2} \end{array}$ and $\begin{array}{rcl} \frac{d β (γ)}{d γ} |_{γ = γ_{l}} = - \frac{2 λ_{l}^{'} (0) R_{0}^{2}}{l^{2} + l - 2} . \end{array}$ Thus $\begin{array}{rcl} lim_{ε \to 0, μ (ε) \neq 0} \frac{s γ^{'} (ε)}{μ (ε)} = - \frac{1}{β^{'} (γ_{l})} = \frac{l^{2} + l - 2}{2 λ_{l}^{'} (0) R_{0}^{2}} . \end{array}$ It follows that $\begin{array}{rcl} μ (ε) & = & ε γ^{'} (ε) \frac{2 λ_{l}^{'} (0) R_{0}^{2}}{l^{2} + l - 2} + o (ε) \\ = & - ε γ^{'} (ε) \frac{2 R_{0}^{2}}{l^{2} + l - 2} \frac{1}{R_{0}^{2}} (\frac{1}{2} l (l + 1) - 1) + o (ε) \\ = & - ε γ^{'} (ε) + o (ε) \\ = & - ε γ^{'} (0) + o (ε) . \end{array}$ In particular, for $l = 2$ , we have $\begin{array}{rcl} μ (ε) & = & \frac{3 \sqrt{5}}{448 π^{\frac{5}{2}}} \frac{Q^{2}}{ε_{0} R_{0}^{4}} ε + o (ε) . \end{array}$ Therefore, we obtain that $μ (ε) < 0$ if $ε < 0$ and $μ (ε) > 0$ if $ε > 0$ for $| ε |$ sufficiently small.

As that of [5], we define $\begin{matrix} ρ = \frac{μ (ε)}{γ_{2}} . \end{matrix}$ Setting $K = F_{γ x} (0, γ_{l})$ , then the solutions bifurcating from $γ_{2}$ are linearly stable if $ρ > 0$ , and linearly unstable if $ρ < 0$ . By the values of $μ (ε)$ and $γ_{2}$ we obtain $\begin{array}{rcl} ρ & = & \frac{3 \sqrt{5} R_{0}}{14 \sqrt{π}} ε + O (ε^{2}) . \end{array}$

Therefore, the solutions bifurcating from $γ_{2}$ are linearly stable if $ε > 0$ , and linearly unstable if $ε < 0$ . Since $γ = γ_{2} + γ^{'} (0) ε + O (ε^{2})$ and $γ^{'} (0) < 0$ , near the bifurcation point, it holds that the solutions bifurcating from $γ_{2}$ are linearly stable if $γ (ε) < γ_{2}$ , and linearly unstable if $γ (ε) > γ_{2}$ . □

Since $γ^{'} (0)$ is different from [5], the values of $μ (ε)$ and ρ are different from [5]. Then the monotonicity of graph about bifurcation branch is reversed.

Data availability statements

There is not any conflict of interest. Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

Footnotes

Acknowledgements

The research of Guowei Dai was supported in part by NSFC 11871129. The research of Ben Duan was supported in part by NSFC 12271205 and 12171498.

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