Two-uniqueness of rational ghost soliton solution and well-posedness of perturbed Einstein-Yang-Mills equations

1.   Introduction

We recall that for the spherically symmetric EYM equations, the Einstein metric is of the form

and the $SU(2)$ Yang-Mills curvature 2-form is

Here $A, C$ and $w$ are functions of $r$, and $(\tau_{1}, \tau_{2}, \tau_{3})$ form a basis for the Lie algebra $SU(2)$. Using (1.1) and (1.2), the spherically symmetric $SU(2)$ EYM equations are

and

We should point out that (1.3) and (1.4) do not involve $C$.

Neither the vacuum Einstein equations nor the pure Yang-Mills equations have nontrivial static globally regular solutions, so it is natural to conjecture that the coupled EYM equations also have no nontrivial globally regular solutions. In 1988, Bartnik and McKinnon ^[1] presented numerical evidence for the existence of a discrete family of globally regular solutions of the static EYM equations: the gravitational attraction can balance the Yang-Mills (YM) repulsive force. This prompts the mathematical theory of this conjecture came into being. However, until now, the behavior of the solution in the case of spherical symmetry is still unclear. For example, in 2018, Baxter ^[2] considered the case of spherical symmetry.

We note that Gross-Pitaevskii equation has been successfully used to deal with Bose-Einstein condensation equation by applying Hirota's method, Lie algebra structure and Backlund transformations, bilinear transform, etc (refer to ^[3,4,5] and their references), but we have not seen an example of applying these methods to Einstein-Yang-Mills equation.

In this paper, we mainly prove the following surprising properties of static spherically symmetric solutions of the $SU(2)$ Einstein-Yang-Mills equations.

Theorem 1.1. There are no any rational expression solutions for static spherically symmetric solutions to the perturbed $SU(2)$ Einstein-Yang-Mills equations

except classical Schwarzschild solutions and Reissner-Nordstrom solutions. Here we add a perturbed coefficient into (1.2), and we assume that $\lambda$ is waving near $1$.

In 1993, Smoller, Wasserman and Yau proved the existence of black hole solutions for the Einstein-Yang-Mills equations (see ^[7,8,9]). In 1995, Smoller and Wasserman ^[6] provided a rigorous proof of the existence and uniqueness of the solutions to the Einstein-Yang-Mills equations with gauge group $SU(2)$. After adding the perturbed term $\lambda$ to (1.4), we discover that the perturbed equations have the same properties as the original equations in existence and uniqueness, see ^[9].

Theorem 1.2. Let $\bar{r} > 0$ be given. Assume that $A(\bar{r}) = 0,$ and $(\bar{w}, \beta)$ satisfies

where $\Phi(\bar{r}) = \bar{r}-\frac{(1-{\bar{w}}^2)^2}{\bar{r}}\neq0.$ Then there exists a unique $C^{2, \alpha}$ solution $(A(r, \bar{w}), w(r, \bar{w}), w'(r, \bar{w}))$ of (1.6), (1.7) with the initial conditions $(A(\bar{r}, \bar{w}), w(\bar{r}, \bar{w}), w'(\bar{r}, \bar{w})) = (0, \bar{w}, \beta)$, defined on some interval $\bar{r} < r < \bar{r}+ s(\bar{w}).$ The solution is analytic on $|r - \bar{r}| < s(\bar{w}),$ and the one-parameter family $(A(r, \bar{w}), w(r, \bar{w}), w'(r, \bar{w}))$ is continuous about $r$ and $\bar w$.

Theorem 1.3. There are no any rational analytic solutions for EYM equations except classical Schwarzschild solutions or Reissner-Nordstrom solutions.

2.   Proof of Theorem 1.1

In this section, we prove Theorem 1.1. Set

where $a_{i}$, $b_{i}$, $c_{i}$, $d_{i}$ $(i\in[1, n], n\geq1)$ are constants.

To obtain a simple form, we suppose

where $p, q, h, k$ represent the corresponding polynomial of denominator and numerator. $A$ and $w$ are simplified, i.e., $q, p, h$ and $k$ are irreducible. Without loss of generality, assuming $\lambda\in (0.5, 2)$ and $\lambda\neq1$, then (1.6) and (1.7) can be simplified as

In the remainder of this paper, we use $\tilde{p}$, $\tilde{q}$, $\tilde{h}$, $\tilde{k}$ to donate the corresponding highest degree of $p$, $q$, $h$, $k$. Using $L_{1} = h^{4}r^{3}(q'p-p'q)$, $L_{2} = ph^{4}r^{2}q$, $L_{3} = 2pqr^{2}(k'h-h'k)^{2}$, $R_{1} = p^{2}h^{4}r^{2}$, $R_{2} = -p^{2}(h^{2}-k^{2})^{2}$ to facilitate the writing of (2.5), and using $\tilde{L_{1}}$, $\tilde{L_{2}}$, $\tilde{L_{3}}$, $\tilde{R_{1}}$, $\tilde{R_{2}}$ to donate the corresponding highest degree of $L_{1}$, $L_{2}$, $L_{3}$, $R_{1}$, $R_{2}$. Similarly, using $L_{21} = r^{3}h^{2}q(k''h^{3}-h''kh^{2}-2k'h^{2}h'+2hkh'^{2})$, $L_{22} = (k'h-kh')[pr^{2}h^{4}-qr^{2}h^{4}-p(h^{2}-k^{2})^{2}]$, $R_{21} = \lambda prkh^{3}(k^{2}-h^{2})$ to facilitate the writing of (2.6), and using $\tilde{L_{21}}$, $\tilde{L_{22}}$, $\tilde{R_{21}}$ to donate the corresponding highest degree of $L_{21}$, $L_{22}$, $R_{21}$.

Lemma 2.1. If the solution of the perturbed EYM equations can be written as (2.1) and (2.2). Then

or

holds. Moreover, if (2.7) holds, then $a_{n} = b_{m}$. If (2.8) holds, then $a_{n}b_{m} < 0$, $\tilde{q}-\tilde{p} = 2(\tilde{k}-\tilde{h})\geq2$. Here $A$ is the form of (2.1).

Proof. It is obvious that $p$ and $h$ can not be zero. To prove this lemma, we start with the simple case and move on to the general case.

When q = 0, i.e., $A\equiv0$, then, from (1.6) and (1.7), $w\equiv0$, $r = 1$ or $w^{2}\equiv1$, $r = 0$.

When $k = 0$, i.e., $w\equiv0$, the solution is Reissner-Nordstrom Solution.

When $q$ and $p$ are constants, i.e., $A$ is a constant, this case are included in the later discussion, where we discuss the situation that $\tilde{p} = \tilde{q}$, i.e., $p = q$, then $A\equiv1$ and $w^{2}\equiv1$.

When $h$ and $k$ are constants, i.e., $w$ is a constant, then $w^{2}\equiv1$ from (1.6) and (1.7), which is corresponding to Schwarzschild Solution.

Next, we discuss the case $k\neq0$, $q\neq0$, and $A$, $w$ may not be constants. Before observing the highest degree of (2.5) and (2.6) to each terms, we pick out some special cases as which have been discussed above: If $q'p-p'q\equiv0$, then $A'\equiv0$; If $k'h-h'k\equiv0$, then $w'\equiv0$; If $h^{2} = k^{2}$, then $w^{2}\equiv1$. Then in the following discussion, each terms in (2.5) can not be constant except $R_{2}$. Noted that

where the part on the right of colon means the highest degree.

Hint: If $\tilde{p} = \tilde{q}$, then $\tilde{L_{1}}$ $<$ $4\tilde{h}+\tilde{p}+\tilde{q}+2$. If $\tilde{p}\neq\tilde{q}$, then $\tilde{L_{1}}$ $=$ $4\tilde{h}+\tilde{p}+\tilde{q}+2$. If $\tilde{h} = \tilde{k}$, then $\tilde{L_{3}}$ $<$ $2\tilde{h}+2\tilde{k}+\tilde{p}+\tilde{q}$. If $\tilde{h}\neq\tilde{k}$, then $\tilde{L_{3}}$ $=$ $2\tilde{h}+2\tilde{k}+\tilde{p}+\tilde{q}$ and $\tilde{R_{2}}$ $=$ $\max(4\tilde{h}$, $\tilde{k}$)+2$\tilde{p}$.

When $\tilde{p} = \tilde{q}$, $\tilde{h} < \tilde{k}$, then $\tilde{R_{2}}$ $>$ $\max(\tilde{L_{1}}$, $\tilde{L_{2}}$, $\tilde{L_{3}}$, $\tilde{R_{1}}$), which leads to a contradiction; When $\tilde{p} = \tilde{q}$, $\tilde{h}\geq\tilde{k}$, then $\tilde{L_{2}}$ = $\tilde{R_{1}}$ $>$ $\max(\tilde{L_{1}}$, $\tilde{L_{3}}$, $\tilde{R_{2}}$) and $a_{n} = b_{n}$, i.e., $A = 1$, as $r\rightarrow \infty$.

When $\tilde{p}\neq\tilde{q}$, $\tilde{h} = \tilde{k}$, then $\tilde{L_{3}}\leq4\tilde{h}+\tilde{p}+\tilde{q}-2 < \tilde{L_{1}}$ = $\tilde{L_{2}}$ = $4\tilde{h}+\tilde{p}+\tilde{q}+2$, $\tilde{R_{2}}\leq4\tilde{h}+2\tilde{p} < \tilde{R_{1}}$ = $4\tilde{h}+2\tilde{p}+2$. If $\tilde{p} > \tilde{q}$, then $\tilde{R_{1}}$ $>$ $\max(\tilde{L_{1}}$, $\tilde{L_{2}}$, $\tilde{L_{3}}$, $\tilde{R_{2}}$), which leads to a contradiction. If $\tilde{p} < \tilde{q}$, $A$ and $w$ as (2.1) and (2.2), here $n-m > 0$ ($a_{n}\neq0$, $b_{m}\neq0$, $c_{s}\neq0$, $d_{t}\neq0$), then the coefficient of highest degree in ($L_{1}+L_{2}$) is

hence $\tilde{L_{1}+L_{2}} = \tilde{R_{1}}$ and $\tilde{q} = \tilde{p}$, which leads to a contradiction.

When $\tilde{p}\neq\tilde{q}$ and $\tilde{h}\neq\tilde{k}$, we have $\tilde{L_{1}} = 4\tilde{h}+\tilde{p}+\tilde{q}+2, \tilde{L_{2}} = 4\tilde{h}+\tilde{p}+\tilde{q}+2, \tilde{L_{3}} = 2\tilde{h}+2\tilde{k}+\tilde{p}+\tilde{q}, \tilde{R_{1}} = 4\tilde{h}+2\tilde{p}+2, \tilde{R_{2}} = 4\max(\tilde{h}, \tilde{k})+2\tilde{p}$.

When $\tilde{k} > \tilde{h}$, $\tilde{p} > \tilde{q}$, then $\tilde{R_{2}}$ $>$ $\max(\tilde{L_{1}}$, $\tilde{L_{2}}$, $\tilde{L_{3}}$, $\tilde{R_{1}}$), which leads to a contradiction.

When $\tilde{k} < \tilde{h}$, $\tilde{p} > \tilde{q}$, then $\tilde{R_{1}}$ $>$ $\max(\tilde{L_{1}}$, $\tilde{L_{2}}$, $\tilde{L_{3}}$, $\tilde{R_{2}}$), which leads to a contradiction.

When $\tilde{k} < \tilde{h}$, $\tilde{p} < \tilde{q}$, then $\tilde{L_{3}}$ $<$ $\tilde{L_{1}}$ = $\tilde{L_{2}}$ = $\tilde{L_{1}}+\tilde{L_{2}}$ = $4\tilde{h}+\tilde{p}+\tilde{q}+2$ and $\tilde{L_{1}}+\tilde{L_{2}}$ $>$ $\tilde{R_{1}}$ = $4\tilde{h}+2\tilde{p}+2$ $>$ $\tilde{R_{2}}$. It means that the highest degree of ($L_{1}$+$L_{2}$) can not be $4\tilde{h}+\tilde{p}+\tilde{q}+2$ from (2.1) and (2.2), here $n > m$. Calculating the coefficient of highest degree term in $L_{1}$ and $L_{2}$, one gets $n-m = -1$, which leads to a contradiction.

When $\tilde{k} > \tilde{h}$, $\tilde{p} < \tilde{q}$, we consider $\tilde{k} = \tilde{h}+1$ firstly. In this case, we have $\tilde{L_{1}} = \tilde{L_{2}} = \tilde{L_{3}} = 4\tilde{h}+\tilde{p}+\tilde{q}+2$, $\tilde{R_{2}} = 4\tilde{h}+2\tilde{p}+4 > \tilde{R_{1}}$. Letting

and calculating the coefficient of highest degree term in left, one sees that the result of coefficient can not be $0$, i.e., $\tilde{L_{1}} = \tilde{L_{2}} = \tilde{L_{3}} = \tilde{R_{2}}$, thus $\tilde{q}-\tilde{p} = 2$. Solving the both sides of coefficient of highest degree term, one gets $a_{n}b_{m} < 0$. Next, if $\tilde{k} > \tilde{h}+1$, one has $\tilde{L_{3}}$ = $2\tilde{h}+2\tilde{k}+\tilde{p}+\tilde{q}$ = $\tilde{R_{2}}$ = $4\tilde{k}+2\tilde{p}$, hence $\tilde{q}-\tilde{p} = 2(\tilde{k}-\tilde{h}) > 2$. What's more, since $R_{2} < 0$, the coefficient of highest degree term to $pq$ in $L_{2}$ and $L_{3}$ should be negative, i.e., $a_{n}b_{m} < 0$.

Thus Lemma 2.1 holds.

Lemma 2.2. If $\tilde{k} > \tilde{h}$, $\tilde{p} < \tilde{q}$, then $\tilde{k} = \tilde{h}+1$, $\tilde{q} = \tilde{p}+2$.

Proof. In order to prove this lemma, we observe (2.6). Firstly, we consider some simple cases.

When $\tilde{h} = 0$, $\tilde{k} = 1$ and $w = ar+b$, $a\neq0$, by using coefficient of variation method to (1.6), one can see the expression of $A$ as follows.

If $a^{2}\neq\frac{1}{2}$, then

where $c$ is a constant. Then we generate the results into (1.7) and discover the fact that

$(\frac{a^{5}}{2a^{2}+3}-a^{5}-\lambda a^{3})r^{3}+(\frac{4a^{4}b}{2a^{2}+2}-4a^{4}b-3\lambda a^{2}b)r^{2}+a(\frac{2a^{2}b^{2}-4ab}{2a^{2}+1}-4ab-2a^{2}b^{2}+2a^{2}+\lambda-3\lambda b^{2})r +a[\frac{(b^{2}-1)^{2}}{2a^{2}-1}-(b^{2}-1)^{2}]r^{-1}-\frac{ac}{r^{2a^{2}}}+\lambda(4a^{2}-1)(b-b^{3}) = 0$. For $(\frac{a^{5}}{2a^{2}+3}-a^{5}-\lambda a^{3}) = 0$ has solutions, one has $\lambda\leq2-\sqrt{3}$ or $\lambda\geq2+\sqrt{3}$, which leads to a contradiction.

If $a^{2} = \frac{1}{2}$, then

where $c$ is a constant. Then we generate the results into (1.7) and discover the fact that $-\frac{a(3+8\lambda)}{16}r^{3}-\frac{b(4+9\lambda)}{6}r^{2}+[(-\frac{1}{2}-3\lambda)ab^{2}-3b+(1+\lambda)a]r+a(b^{2}-1)^{2}(\ln r-1)r^{-1}-\frac{ac}{r}+\lambda(b-b^{3}) = 0$. Let the coefficient of $r^{3}$ be $0$, one gets $\lambda = -\frac{3}{8}$, which leads to a contradiction.

When $\tilde{h} = 0$, $\tilde{k}\geq2$, then $\tilde{L_{21}} = \tilde{k}+\tilde{q}+1$, $\tilde{L_{22}} = 5\tilde{k}+\tilde{p}-1$, $\tilde{R_{21}} = 3\tilde{k}+\tilde{p}+1$. Through a comparative analysis, one sees $\tilde{L_{22}}$ $>$ $\max(\tilde{L_{21}}$, $\tilde{R_{21}}$), which leads to a contradiction.

When $\tilde{h} = 1$, then $\tilde{L_{21}}$ $\leq$ $6+\tilde{k}+\tilde{q}$. If $\tilde{q}-\tilde{p}\geq4$, i.e., $\tilde{k}-\tilde{h}\geq2$, then $\tilde{L_{22}}$ = $5\tilde{k}+\tilde{p}$, $\tilde{R_{21}}$ = $3\tilde{k}+\tilde{p}+4$. Through a comparative analysis, one gets $\tilde{L_{22}}$ $>$ $\max(\tilde{L_{21}}$, $\tilde{R_{21}}$), which leads to a contradiction.

When $\tilde{h}\geq2$, if $\tilde{q}-\tilde{p}\geq4$, i.e., $\tilde{k}-\tilde{h}\geq2$, then $\tilde{L_{21}}$ $\leq$ $5\tilde{h}+1+\tilde{k}+\tilde{q}$, $\tilde{L_{22}}$ = $5\tilde{k}+\tilde{h}+\tilde{p}-1$, $\tilde{R_{21}}$ = $3\tilde{k}+3\tilde{h}+\tilde{p}+1$. By analysis, one has $\tilde{L_{22}}$ $>$ $\max(\tilde{L_{21}}$, $\tilde{R_{21}}$), which leads to a contradiction.

Thus, Lemma 2.2 holds.

Lemma 2.3. Let $w = \frac{k}{h} = \frac{a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r+a_{0}}{b_{n}r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r+b_{0}}$. If $\tilde{p} = \tilde{q}$, $\tilde{h}\geq\tilde{k}$, then $\tilde{p} = \tilde{q}$, $\tilde{h} = \tilde{k}$, and $a_{n}^{2} = b_{n}^{2}$.

Proof. When $\tilde{k} = 0$, $\tilde{h} = 1$, then $w = \frac{k}{h} = \frac{c}{r+b}$, so $\tilde{L_{22}}$ $\leq$ $5+\tilde{p}$ $<$ $\tilde{L_{21}}$ = $6+\tilde{p}$ = $\tilde{R_{21}}$. Solving both sides of the coefficient of highest degree term, one gets that $\lambda = -2$, which leads to a contradiction.

When $\tilde{k} = 0$, $\tilde{h}\geq2$, then $\tilde{L_{22}}$ $\leq$ $5\tilde{h}+\tilde{p}$, $\tilde{L_{21}}$ $\leq$ $5\tilde{h}+\tilde{p}+1$ = $\tilde{R_{21}}$. Let $k = 1$, $h = d_{n}r^{n}+d_{n-1}r^{n-1}+\cdots+d_{1}r+d_{0}$($n\geq2$), $A = \frac{q}{p} = \frac{r^{s}+a_{s-1}r^{s-1}+...+a_{1}r+a_{0}}{r^{s}+b_{s-1}r^{s-1}+\cdots+b_{1}r+b_{0}}$. Then the coefficient of highest degree term in $L_{21}$ is

and the coefficient of highest degree term in $R_{21}$ is

Solving the both sides of the coefficient of highest degree term, one has $\lambda = -n(n+1) < 0$, which leads to a contradiction.

When $\tilde{k} = 1$, $\tilde{h} = 1$, then $\tilde{L_{21}}$ $\leq$ $6+\tilde{p}$, $\tilde{L_{22}}$ $\leq$ $5+\tilde{p}$. Let $w = \frac{br+c}{r+a}$, if $b^{2}\neq1$, then $\tilde{R_{21}}$ = $7+\tilde{p}$, which leads to a contradiction. So $b^{2} = 1$, i.e., $w^{2} = 1$ $(r\rightarrow \infty)$.

When $\tilde{k} = 1$, $\tilde{h}\geq2$, then $\tilde{L_{22}}$ $\leq$ $5\tilde{h}+\tilde{p}+1$, $\tilde{L_{21}}$ $\leq$ $5\tilde{h}+\tilde{p}+2$ = $\tilde{R_{21}}$. Let $k = ar+c$, $h = d_{n}r^{n}+d_{n-1}r^{n-1}+\cdots+d_{1}r+d_{0}$ $(n\geq2)$, then the coefficient of highest degree term in $L_{21}$ is

and the coefficient of highest degree term in $R_{21}$ is

Solving both sides of the coefficient of highest degree term, one gets $\lambda = -n(n-1)$, which leads to a contradiction.

When $\tilde{k}\geq2$, and $w = \frac{k}{h} = \frac{a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r+a_{0}}{b_{m}r^{m}+b_{m-1}r^{m-1}+\cdots+b_{1}r+b_{0}}$ $(m\geq n\geq2)$, considering $\tilde{h} > \tilde{k}$ and $m > n$ firstly, then $\tilde{L_{22}}$ $\leq$ $5\tilde{h}+\tilde{p}+\tilde{k}$, $\tilde{L_{21}}$ $\leq$ $5\tilde{h}+\tilde{p}+\tilde{k}+1$ = $\tilde{R_{21}}$. The coefficient of highest degree term in $L_{21}$ is

and the coefficient of highest degree term in $R_{21}$ is

Solving both sides of the coefficient of highest degree term, one gets

In fact, by (2.13), one has

So $\lambda < 0$, which leads to a contradiction. If $\tilde{h} = \tilde{k}$, $a_{n}^{2}$ $\neq$ $b_{n}^{2}$, then $\tilde{L_{21}}$ $\leq$ $6\tilde{h}+\tilde{p}$, $\tilde{L_{22}}$ $\leq$ $6\tilde{h}+\tilde{p}-1$, $\tilde{R_{21}}$ = $6\tilde{h}+\tilde{p}+1$, which lead to a contradiction, i.e., $w^{2} = 1$ $(r\rightarrow \infty)$.

Thus, Lemma 2.3 holds.

Lemma 2.4. The case $\tilde{p} = \tilde{q}$, $\tilde{h} = \tilde{k}$ can not happen.

Proof. When $\tilde{h} = \tilde{k} = 1$, $w = \frac{r+b}{r+a}$, then $\tilde{L_{22}} < \tilde{L_{21}} = 6+\tilde{p} = \tilde{R_{21}}$, thus the coefficient of highest degree term in $L_{21}$ is

and the coefficient of highest degree term in $R_{21}$ is

Solving the both sides of the coefficient of highest degree term, we discover $\lambda = 1$, which leads to a contradiction. If let $w = \frac{-r+b}{r+a}$, also leads to a contradiction in a same way.

For $\tilde{h} = \tilde{k}\geq 2$, let $A = \frac{r^{m}+c_{m-1}r^{m-1}+\cdots+c_{1}r+c_{0}}{r^{m}+d_{m-1}r^{m-1}+\cdots+d_{1}r+d_{0}}$, $w = \frac{r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r+a_{0}}{r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r+b_{0}}$ ($w = \frac{-r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r+a_{0}}{r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r+b_{0}}$ also leads to the same result). If $a_{n-1}\neq b_{n-1}$, then $\tilde{L_{22}}$ $\leq$ $6\tilde{h}+\tilde{p}-1$ $<$ $\tilde{L_{21}}$ = $\tilde{R_{21}}$ = $6\tilde{h}+\tilde{p}$. Solving both sides of the coefficient of highest degree term, one gets $\lambda = 1$, which leads to a contradiction. So, $a_{n-1}$ $=$ $b_{n-1}$. If $a_{n-2}$ $\neq$ $b_{n-2}$, then $\tilde{L_{22}}$ $\leq$ $6\tilde{h}+\tilde{p}-2$ $<$ $\tilde{L_{21}}$ = $\tilde{R_{21}}$ = $6\tilde{h}+\tilde{p}-1$. Solving both sides of the coefficient of highest degree term, one has $\lambda = 3$, which leads to a contradiction. So, $a_{n-2} = b_{n-2}$. This reminds us that perhaps $a_{i} = b_{i}$($i\leq n-2$). In the following, we use mathematical induction to prove $a_{i} = b_{i}$. With the reduction to absurdity, assume $\exists$ $a_{i}\neq b_{i}$ and $a_{i+j} = b_{i+j}$, $j\geq1$, $0\leq i < n-2$. So $w = \frac{k}{h} = \frac{r^{n}+a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+\cdots+a_{i+1}r^{i+1}+a_{i}r^{i}+\cdots}{r^{n}+a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+\cdots+a_{i+1}r^{i+1}+b_{i}r^{i}+\cdots}$. By (2.6), one has

Solving both sides of the coefficient of highest degree term, one gets

Thus $\lambda\geq3$, which leads to a contradiction. Then $w\equiv1$, and contradicts the hypothesis. Here, if we assume $w = \frac{k}{h} = \frac{-r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r+a_{0}}{r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r+b_{0}}$ in the beginning, we will get the result that $w\equiv-1$, which is also a contradiction.

Thus, Lemma 2.4 holds.

Lemma 2.5. The case $\tilde{q} = \tilde{p}+2$, $\tilde{k} = \tilde{h}+1$ can not happen.

Proof. We consider $\tilde{h}\geq2$ firstly. Let

and $b < 0$, then $\tilde{L_{21}}\leq5\tilde{h}+\tilde{k}+\tilde{q}+1$. By calculating, the corresponding coefficient of highest degree term is $0$, thus $\tilde{L_{21}}\leq5\tilde{h}+\tilde{k}+\tilde{q}$. What's more, $\tilde{L_{22}}\leq5\tilde{h}+\tilde{k}+\tilde{q}+1$, and the coefficient of highest degree term in $5\tilde{h}+\tilde{k}+\tilde{q}+1$ is $-(a+a^{5}b)$ and $\tilde{R_{21}} = 5\tilde{h}+\tilde{k}+\tilde{q}+1$. The corresponding coefficient is $\lambda a^{3}b$. Solving both sides of coefficient of highest degree term, one gets

From (2.5), one gets $\tilde{L_{1}} = \tilde{L_{2}} = \tilde{L_{3}} = \tilde{R_{2}} = 4\tilde{h}+2\tilde{p}+4 > \tilde{R_{1}} = 4\tilde{h}+2\tilde{p}+2$. Comparing the coefficient of highest degree term in ($L_{1}+L_{2}+L_{3}$)($3b+2ba^{2}$) and $R_{2}$ ($-a^{4}b^{2}$), one gets

Calculating (2.23) and (2.24), one has

If (2.25) has solutions, then $\lambda\leq2-\sqrt{3}$ or $\lambda\geq2+\sqrt{3}$, which leads to a contradiction.

If $\tilde{h} = 0$, then $w = ar+b$. This case has been discussed in the proof of Lemma 2.2. When $\tilde{h} = 1$, $\tilde{k} = 2$, we suppose $w = \frac{k}{h} = \frac{br^{2}+b_{1}r+b_{0}}{r+c}$, $A = \frac{ar^{m+2}+a_{m+1}r^{m+1}+\cdots+a_{1}+a_{0}}{r^{m}+d_{m-1}r^{m-1}+\cdots+d_{1}r+d_{0}}$, ($a\neq0$, $b\neq0$). Solving the coefficient of highest degree term in $L_{21}$, $L_{22}$, $R_{21}$ and observing (2.6), one gets $\tilde{L_{21}}\leq10+\tilde{p}$. By calculating, the coefficient of highest degree term is $0$, thus $\tilde{L_{21}}\leq9+\tilde{p}$. What's more, $\tilde{L_{22}}\leq10+\tilde{p}$, the corresponding coefficient of highest degree term is $-b(a+b^{4})$, $\tilde{R_{21}} = 10+\tilde{p}$, the corresponding coefficient of highest degree term is $\lambda b^{3}$, then one has

Observing (2.5), one gets $\tilde{L_{1}} = \tilde{L_{2}} = \tilde{L_{3}} = \tilde{R_{2}} = 2m+8 > \tilde{R_{1}} = 2m+6$. Comparing the coefficient of highest degree term in ($L_{1}+L_{2}+L_{3}$): ($3a+2ab^{2}$) and in $R_{2}$: ($-b^{4}$), one has

Calculating (2.26) and (2.27), one can see that

Thus, when (2.28) has solutions, $\lambda\leq2-\sqrt{3}$ or $\lambda\geq2+\sqrt{3}$ holds, which leads to a contradiction.

Thus, the proof of Lemma 2.5 is completed.

By Lemma 2.1–2.5, we complete the proof of Theorem 1.1.

Remark 2.6. If we consider the case that adding the perturbed coefficient into (1.4) for the first term as follows:

we can also get the same result as the conclusion in this paper. Here $\zeta\in(0.5, 2)$ and $\zeta\neq1$. The proof of this result is similar to Theorem 1.1 if we add the perturbed coefficient $\lambda$ into (1.4), so we omit here.

3.   Proof of Theorem 1.2

In this section, we prove the existence and uniqueness of the perturbed EYM equations. The reference ^[9] provides a rigorous proof of the existence and uniqueness of the solutions to the Einstein-Yang-Mills equations with gauge group $SU(2)$. After adding the perturbed term $\lambda$ to (1.4), we discover that the perturbed equations have the same properties as the original equations in existence and uniqueness.

For the definitions of $C^{2+\alpha}(\bar{r}, \bar{r}+\varepsilon)$, $D_1,$ $D_2,$ $D_3,$ $X,$ $d(\cdot, \cdot),$ $T$, $T_1$, $T_2$ and $T_3$, we still use the definitions in ^[9]. The main work in ^[9] is to prove that $T_{1}, T_{2}$ and $T_{3}$ are contract mappings. By analysis, to prove our results, we just need to prove $T_{2}$ is a contract mapping. Recalling $u_{1}, u_{2}, w_{1}, w_{2}$ in $C^{2+\alpha}(\bar{r}, \bar{r}+\epsilon)$, $T_{2}(w, z, A) = w'(\bar{r})-\int_{\bar{r}}^{r} \frac{\lambda uw+\phi z}{s^{2}A}ds$, and denoting $\theta = (w, z, A)$, then

For convenience, denoting

then

and

Next, we estimate the above equation term by term.

For $\lambda\|\Delta(\frac{uw-\bar{u}\bar{w}}{r^{2}A})\|_{\alpha}$, denoting $\varrho_{2} = r_{2}-\bar{r}$ and by lemma 5.3 in ^[9], one can get

Due to $u_{1}, u_{2}, w_{1}, w_{2}$ in $C^{2+\alpha}(\bar{r}, \bar{r}+\epsilon)$, $(u_{1}w_{1})(\bar{r}) = (u_{2}w_{2})(\bar{r})$, $(u_{1}w_{1})'(\bar{r}) = (u_{2}w_{2})'(\bar{r})$, $(u_{1}w_{1})''(\bar{r}) = (u_{2}w_{2})''(\bar{r})$, then $\|\frac{\Delta(uw)}{\varrho_{2}}\|_{\alpha}\rightarrow0$, $\|\frac{\Delta(uw)}{\varrho_{2}}\|_{\infty}\rightarrow0$. Hence,

as $\varepsilon\rightarrow0$. By $(\frac{1}{A_{2}}-\frac{1}{A_{1}})(\bar{r}) = 0, (\frac{1}{A_{2}}-\frac{1}{A_{1}})'(\bar{r}) = 0$, $\|u_{1}w_{1}\|_{\infty}\to 0$ and $\|\bar{u}\bar{w}\|_{\infty}\to 0$, one has

as $\varepsilon\rightarrow0$. Combining (3.1) and (3.2) to get

For $\|\Delta(\frac{\phi z-\bar{\phi}\bar{z}}{r^{2}A})\|_{\alpha}$, one has

It is easy to see $\|\Delta(\frac{\phi z-\bar{\phi}z}{r^{2}A})\|_{\alpha}\rightarrow0$, as $\varepsilon\rightarrow0$. For convenience, defining

one has $n_{i}(r) = \frac{\bar{\phi} z_{i}-d\varrho\bar{\phi}-\bar{\phi}\bar{z_{i}}}{cr^{2}}$, and

Because $\frac{n_{2}}{\varrho B_{2}}-\frac{n_{2}}{\varrho B_{1}} = \frac{n_{2}}{B_{1}B_{2}}(\frac{B_{1}-B_{2}}{\varrho})$, one gets

Due to $\frac{n_{2}}{B_{1}B_{2}}\in C^{1+\alpha}$, $n_{2} = \frac{\bar{\phi}}{cr^{2}}(z_{2}-d\varrho-\bar{z})$, then

Because of $\varrho = \bar{r}-\bar{r} = 0$, $n_{2}(\bar{r}) = 0$,

Hence $(\frac{n_{2}}{B_{1}B_{2}})(\bar{r}) = 0, (\frac{n_{2}}{B_{1}B_{2}})'(\bar{r}) = 0$, furthermore

and

as $\varepsilon\rightarrow0$. Due to $(\frac{B_{2}-B_{1}}{\varrho})(\bar{r}) = 0$, one has

and then $\varepsilon\rightarrow0$,

For the last term in (3.4), denoting $h(r) = \frac{\bar{\phi}}{B_{1}r^{2}c}$ for convenience, then

Because $h\in C^{1+\alpha}, h(\bar{r}) = \frac{\bar{\phi}}{B(\bar{r})\bar{r}^{2}c}$, one has

and $(h-1)\in C^{1+\alpha}$, $(h-1)(\bar{r}) = h(\bar{r})-1 = 0$, $\|h-1\|_{\infty}\rightarrow0$. Moreover, $0 < \delta < \alpha, \|h\|_{\infty}\leq1+\delta$, then

Due to the fact that $(z_{2}-z_{1})(\bar{r}) = 0$ and $\|\frac{z_{2}-z_{1}}{\varrho}\|\leq\frac{1}{1+\alpha}\|z_{2}-z_{1}\|_{1+\alpha}$, then

and

as $\varepsilon\rightarrow0$. Let $0 < k = \frac{1+\delta}{1+\alpha} < 1$, then

Furthermore, $\lambda\bar{u}\bar{w}+\bar{\phi}\bar{z} = \bar{r}^{2}\bar{A}\bar{w}'' = 0$, then

So $T_{2}$ is a contract mapping. Thus, by Banach fixed-point theorem, in the interval $\bar{r}\leq r\leq\bar{r}+\varepsilon$, $T$ has a unique fixed point ($A(r, \bar{w}), w(r, \bar{w}), w'(r, \bar{w})$)$\in$ $X$. This completes the proof of Theorem 1.2.

4.   Conclusions

By Theorems 1.1 and 1.2, we can see that the local solution is unique. Moreover, this solution is not rational solution, but $C^{2+\alpha}$ solution. This means the results in Theorem 1.3 holds. Since the current methods of finding ghost solitons are all based on the rational fraction solution, these results show that there is no rational ghost soliton solution for the EYM equation.

Acknowledgments

The authors would like to thank the referees and the editor for their valuable comments which led to improvement of this work. The authors are supported by the National Natural Science Foundation of China (Grant Nos. 12001415, 12001276, 11961080, 11561076).

Conflict of interest

All authors declare no conflicts of interest in this paper.