Global existence of classical static solutions of four dimensional Einstein–Klein–Gordon system

Abstract

In this paper, we prove the global existence of classical static solutions of Einstein’s gravitational theory coupled to a real scalar field where spacetime admits spherically symmetry. The equations of motions can then be reduced into a single first-order integro-differential equation. First, we obtain the decay estimates of the solutions. Then, to prove the global existence, we use the contraction mapping theorem in the appropriate function spaces.

Appendix

1.1 The $\{00\}$ and $\{01\}$ components of the Einstein equations

We write the $\{00\}$ component of the Einstein equations (2.10) as follows

$$\begin{aligned} R_{00}=\frac{1}{2}g_{00}R + 8\pi T_{00}. \end{aligned}$$

(6.1)

Combining (2.2), (2.7), and (2.11), we write down Eq. (6.1) such that

$$\begin{aligned}&e^{2(F-G)}\left[ -\frac{2}{r}\partial _1 G+\frac{1}{r^2}\right] \nonumber \\&\qquad +8\pi \left[ \frac{1}{2}e^{2(F-G)}\left( \frac{\partial \phi }{\partial r}\right) ^2 - e^{2F}V(\phi )+\left( \frac{\partial \phi }{\partial u}\right) ^2-e^{F-G}\frac{\partial \phi }{\partial u}\frac{\partial \phi }{\partial r}\right] \nonumber \\&\quad =\frac{2}{r}e^{F-G}\partial _0 G. \end{aligned}$$

(6.2)

The $\{01\}$ component of the Einstein equations (2.10) yields

$$\begin{aligned} R_{01}=\frac{1}{2}g_{01}R + 8\pi T_{01}. \end{aligned}$$

(6.3)

As previously, from (2.3), (2.7), and (2.11), we write down Eq. (6.3) as follows

$$\begin{aligned}&e^{2(F-G)}\left[ -\frac{2}{r}\partial _1G+\frac{1}{r^2}\right] +8\pi \left[ \frac{1}{2}e^{2(F-G)}\left( \frac{\partial \phi }{\partial r}\right) ^2 - e^{2F}V(\phi )\right] =0. \end{aligned}$$

(6.4)

Again, since there is no term containing the second derivative with respect to u and r on F and G, this shows that F and G are no longer dynamical. Theorem 1 ensures that the metric functions F and G do globally exist.

1.2 Estimate for (3.12)

We write the estimate of (3.12) as follows:

$$\begin{aligned} |f|\le&\left| \frac{1}{2r}\left( g-\tilde{g}\right) \bar{h}\right| + \left| 4\pi g rV(\bar{h})\bar{h}\right| +\left| \frac{gr}{2}\frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \nonumber \\ =&A_1+A_2+A_3. \end{aligned}$$

(6.5)

• 1. Estimate for $A_1$ First, we estimate

  $$\begin{aligned} |\bar{h}|\le \frac{1}{r}\int _{0}^{r}|h(u,s)|{\textrm{d}}s \le \frac{x}{(k-2)}\frac{r^{k-3}}{(1+u)^{k-2}(1+r+u)^{k-2}}. \end{aligned}$$

  (6.6)

  Then, we obtain

  $$\begin{aligned} |h(u,r)-h(u,r')|\le&\int _{r'}^{r}\left| \frac{\partial h}{\partial s}(u,s)\right| ~{\textrm{d}}s\nonumber \\ \le&x\int _{r'}^{r}\frac{1}{(1+s+u)^k}~{\textrm{d}}s\nonumber \\ \le&\frac{x}{k-1}\left[ (1+r'+u)^{-k+1}-(1+r+u)^{-k+1}\right] . \end{aligned}$$

  (6.7)

  From the above estimate, we get

  $$\begin{aligned} |(h-\bar{h})(u,r)|\le \frac{1}{r}\int _{0}^{r}|h(u,r)-h(u,r')|~{\textrm{d}}r'\le \frac{xr}{(k-1)(1+u)^{k-2}(1+r+u)^{k-1}}, \end{aligned}$$

  (6.8)

  and

  $$\begin{aligned} \int _{0}^{\infty }\frac{|h-\bar{h}|^2}{r}~{\textrm{d}}r\le&\frac{x^2 }{(k-1)^2(1+u)^{2(k-2)}}\int _{0}^{\infty }\frac{s}{(1+s+u)^{2(k-1)}}~{\textrm{d}}s\nonumber \\ =&\frac{x^2 }{(k-1)^2(12-14k+4k^2)(1+u)^{4(k-2)}}. \end{aligned}$$

  (6.9)

  From (2.20), we get

  $$\begin{aligned} g(u,0)=\exp \left[ -\int _{0}^{\infty }4\pi \frac{(h-\bar{h})^2}{s}{\textrm{d}}s\right] \ge \exp \left[ -\frac{4\pi x^2 }{(k-1)^2(12-14k+4k^2)}\right] . \end{aligned}$$

  (6.10)

  Thus,

  $$\begin{aligned} \bar{g}(u,0)=&\frac{1}{r}\int _{0}^{r}g(u,0)~{\textrm{d}}r\ge \exp \left[ -\frac{4\pi x^2 }{(k-1)^2(12-14k+4k^2)}\right] . \end{aligned}$$

  (6.11)

  Now, we estimate

  $$\begin{aligned}&|g(u,r)-g(u,r')|\le \int _{r'}^{r}\left| \frac{\partial g}{\partial s}(u,s) \right| {\textrm{d}}s\le 4\pi \int _{r'}^{r}\frac{|h-\bar{h}|^2}{s}{\textrm{d}}s\nonumber \\&\quad =\frac{\pi x^2}{3k(k-1)^2(k-2)^2} \left[ \frac{(1+u)^{4-2k}}{r'^2}+\frac{4k(1+r'+u)^{1-2k}}{1-2k}\right. \nonumber \\&\qquad +\frac{4k^2(1+r'+u)^{1-2k}}{-1+2k}+\frac{4ku(1+r'+u)^{1-2k}}{1-2k}+\frac{4k^2u(1+r'+u)^{1-2k}}{(-1+2k)}\nonumber \\&\qquad +\frac{(-1+k)(1+r'+u)^{1-2k}(1+(-1+2k)r'+u)}{-1+2k}-\frac{(1+u)^{4-2k}}{r^2}\nonumber \\&\qquad -\frac{4k(1+r+u)^{1-2k}}{1-2k}-\frac{4k^2(1+r+u)^{1-2k}}{-1+2k}-\frac{4ku(1+r+u)^{1-2k}}{1-2k}\nonumber \\&\qquad \left. -\frac{4k^2u(1+r+u)^{1-2k}}{(-1+2k)}-\frac{(-1+k)(1+r+u)^{1-2k}(1+(-1+2k)r+u)}{-1+2k}\right] , \end{aligned}$$

  (6.12)

  and

  $$\begin{aligned} |(g-\bar{g})(u,r)|\le&\frac{1}{r}\int _{0}^{r}|g(u,r)-g(u,r')|{\textrm{d}}r'\nonumber \\ =&\frac{\pi x^2 r^2}{3k(k-1)^2(k-2)^2(1+u)^{-3+2k}(1+r+u)^3}. \end{aligned}$$

  (6.13)

  We also estimate

  $$\begin{aligned} \int _{0}^{r}gs^2V(\bar{h}){\textrm{d}}s\le K_0\int _{0}^{r}gs^2|\bar{h}|^{p+1}{\textrm{d}}s\le \frac{K_0x^{p+1}r^3}{3(1+u)^{4(k-2)}(1+r+u)^3}. \end{aligned}$$

  (6.14)

  Thus, we obtain

  $$\begin{aligned} \left| g-\tilde{g}\right| \le |g-\bar{g}| + \frac{8\pi }{r}\int _{0}^{r}gs^2V(\bar{h}){\textrm{d}}s\le \frac{C(x^2+x^{p+1})r^2}{(1+u)^{-3+2k}(1+r+u)^3}. \end{aligned}$$

  (6.15)

  Estimates (3.13) and (3.21) yields

  $$\begin{aligned} A_1=\left| \frac{(g-\tilde{g})}{2r}\bar{h}\right| \le \frac{C(x^3+x^{p+2})}{(1+u)^{3k-5}(1+r+u)^3}. \end{aligned}$$

  (6.16)

• 2. Estimate for $A_2$ Using (3.13), we obtain the estimate (6.17) as follows:

  $$\begin{aligned} A_2=\left| 4\pi grV(\bar{h})\bar{h}\right| \le \frac{Cx^{p+2}}{(1+u)^{5(k-2)}(1+r+u)^4}. \end{aligned}$$

  (6.17)

• 3. Estimate for $A_3$ From (3.13), we obtain

  $$\begin{aligned} A_3=\left| \frac{gr}{2}\frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \le \frac{Cx^p}{(1+u)^{3(k-2)}(1+r+u)^2}. \end{aligned}$$

  (6.18)

Combining (6.16)–(6.18) yields

$$\begin{aligned} |f|\le \frac{C(x^3+x^p+x^{p+2})}{(1+u)^{3(k-2)}(1+r+u)^2}. \end{aligned}$$

(6.19)

1.3 Estimate for (3.38)

We write the estimate of (3.38) as follows:

$$\begin{aligned} |f_1|\le&\left[ \frac{1}{2r}\left| \frac{\partial (g-\tilde{g})}{\partial r}\right| +\frac{1}{2r^2}|g-\tilde{g}|+\left| \frac{\partial g}{\partial r}\right| 4\pi r|V(\bar{h})\right. |\nonumber \\&\left. +4\pi g|V(\bar{h})|+4\pi g r\left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \left| \frac{\partial \bar{h}}{\partial r}\right| \right] \left( |{\mathcal {F}}|+|\bar{h}|\right) \nonumber \\&+\left[ \frac{1}{2r} |g-\tilde{g}|+4\pi g r|V(\bar{h})|+\frac{gr}{2}\left| \frac{\partial ^2 V(\bar{h})}{\partial \bar{h}^2}\right| \right] \left| \frac{\partial \bar{h}}{\partial r}\right| \nonumber \\&+\left[ \left| \frac{\partial g}{\partial r}\right| \frac{r}{2}+\frac{g}{2}\right] \left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \nonumber \\ =&(B_1+B_2+B_3+B_4+B_5)\left( \left| {\mathcal {F}}\right| +\left| \bar{h}\right| \right) + (B_6+B_7+B_8)\left| \frac{\partial \bar{h}}{\partial r}\right| + B_9. \end{aligned}$$

(6.20)

• 1. Estimate for $B_1$ Combining (3.13), (3.19), and (3.20) yields

  $$\begin{aligned} \left| \frac{\partial \tilde{g}}{\partial r}\right| \le&\frac{C x^2 r}{(1+u)^{-3+2k}(1+r+u)^3}\nonumber \\&\quad +\frac{8\pi K_0x^{k+1}r}{3(1+u)^{4(k-2)}(1+r+u)^3}+\frac{8\pi K_0x^{p+1}r}{(1+u)^{4(k-2)}(1+r+u)^{4}}\nonumber \\ \le&\frac{C(x^2+x^{k+1}+x^{p+1})r}{(1+u)^{-3+2k}(1+r+u)^3}. \end{aligned}$$

  (6.21)

  Then, using (2.21) and (3.15), we obtain

  $$\begin{aligned} \left| \frac{\partial g}{\partial r}\right| \le \frac{4\pi x^2r }{(k-1)^2(1+u)^{2(k-2)}(1+r+u)^{2(k-1)}}. \end{aligned}$$

  (6.22)

  Thus,

  $$\begin{aligned} B_1=\frac{1}{2r}\frac{\partial |g-\tilde{g}|}{\partial r}\le \frac{C(x^2+x^{k+1}+x^{p+1})}{(1+u)^{2(k-2)}(1+r+u)^3}. \end{aligned}$$

  (6.23)

• 2. Estimate for $B_2$ From (3.21), we obtain

  $$\begin{aligned} B_2=\frac{1}{2r^2}|g-\tilde{g}|\le \frac{C(x^2+x^{p+1})}{(1+u)^{-3+2k}(1+r+u)^3}. \end{aligned}$$

  (6.24)

• 3. Estimaste for $B_3$ We define $\frac{\partial \bar{h}}{\partial r}=\frac{h-\bar{h}}{r}$. From (3.13) and (3.15), we obtain

  $$\begin{aligned} B_3 =4\pi r\left| \frac{\partial g}{\partial r}\right| |V(\bar{h})|\le \frac{C x^{p+3}}{(1+u)^{6(k-2)}(1+r+u)^{2k}}. \end{aligned}$$

  (6.25)

• 4. Estimate for $B_4$ From (3.13), we obtain

  $$\begin{aligned} B_4=4\pi g V(\bar{h})\le \frac{C x^{p+1}}{(1+u)^{4(k-2)}(1+r+u)^4}. \end{aligned}$$

  (6.26)

• 5. Estimate for $B_5$ From (3.13) and (3.42), we obtain

  $$\begin{aligned} B_5=4\pi g r \left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \frac{|h-\bar{h}|}{r}\le \frac{C x^{p+1}}{(1+u)^{4(k-2)}(1+r+u)^{k+1}}. \end{aligned}$$

  (6.27)

• 6. Estimate for $B_6$ From (3.21), we obtain

  $$\begin{aligned} B_6=\frac{1}{2r} |g-\tilde{g}|\le \frac{C(x^2+x^{p+1})r}{(1+u)^{-3+2k}(1+r+u)^3}. \end{aligned}$$

  (6.28)

• 7. Estimate for $B_7$ From (3.13), we obtain

  $$\begin{aligned} B_7=4\pi g r V(\bar{h})\le \frac{C x^{p+1}r}{(1+u)^{4(k-2)}(1+r+u)^4}. \end{aligned}$$

  (6.29)

• 8. Estimate for $B_8$ Again, from (3.13) we have

  $$\begin{aligned} B_8 =\frac{gr}{2}\left| \frac{\partial ^2V(\bar{h})}{\partial \bar{h}^2}\right| \le \frac{C x^{p-1}r}{(1+u)^{2(k-2)}(1+r+u)^2}. \end{aligned}$$

  (6.30)

• 9. Estimate for $B_9$ From (3.13) and (3.42), we obtain

  $$\begin{aligned} B_9=\left[ \left| \frac{\partial g}{\partial r}\right| \frac{r}{2}+\frac{g}{2}\right] \left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \le \frac{C x^{p+2}r}{(1+u)^{5(k-2)}(1+r+u)^{2k}}. \end{aligned}$$

  (6.31)

Combining (6.23)–(6.31), yields

$$\begin{aligned} |f_1|\le&\frac{C(x^2+x^{k+1}+x^{p+1}+x^{p+3})}{(1+u)^{2(k-2)}(1+r+u)^3}|{\mathcal {F}}|+\frac{C(x^3+x^{k+2}+x^{p+2}+x^{p+4})}{(1+u)^{3(k-2)}(1+r+u)^4}\nonumber \\&+\frac{C(x^3+x^{p}+x^{p+2})r}{(1+u)^{3(k-2)}(1+r+u)^{k+1}}. \end{aligned}$$

(6.32)

1.4 Estimate for (4.12)

We write the estimate of (4.12) as follows:

$$\begin{aligned} |\tilde{\varphi }|\le&\frac{1}{2}\left| \tilde{g}_1-\tilde{g}_2\right| |{\mathcal {G}}_2|+\frac{1}{2r}\left| g_1-\tilde{g}_1\right| | \bar{h}_1 -\bar{h}_2|+ \frac{1}{2r}\left| g_1-\tilde{g}_1+g_2-\tilde{g}_2\right| |{\mathcal {F}}_2-\bar{h}_2|\nonumber \\&+4\pi r|g_1-g_2| |{\mathcal {F}}_1||V(\bar{h}_1)|+4\pi r|g_1-g_2||\bar{h}_1| |V(\bar{h}_1)|\nonumber \\&+4\pi r g_2\left| V(\bar{h}_1)-V(\bar{h}_2)\right| |{\mathcal {F}}_2|+4\pi r g_2|\Theta | |V(\bar{h}_1)| \nonumber \\&+ 4\pi r g_2|\bar{h}_1 - \bar{h}_2||V(\bar{h}_2)|+4\pi r g_2|\bar{h}_1|\left| V(\bar{h}_1)-V(\bar{h}_2)\right| \nonumber \\&+\frac{r}{2}|g_1-g_2||\bar{h}_1|\left| \frac{\partial ^2 V(\bar{h}_1)}{\partial \bar{h}_1^2}\right| \nonumber \\&+\frac{r}{2}g_2|\bar{h}_1-\bar{h}_2|\left| \frac{\partial ^2 V(\bar{h}_1)}{\partial \bar{h}_1^2}\right| +\frac{r}{2}g_2|\bar{h}_1|\left| \frac{\partial ^2 V(\bar{h}_1)}{\partial \bar{h}_1^2}-\frac{\partial ^2 V(\bar{h}_2)}{\partial \bar{h}_2^2}\right| \nonumber \\&{:}{=} D_1+D_2+D_3+...+D_{10}+D_{11}+D_{12}. \end{aligned}$$

(6.33)

• 1. Estimate for $D_1$ Let us denote $\Vert h_1-h_2\Vert _Y=y$. From the definition of mean value, we have

  $$\begin{aligned} |\bar{h}_1 - \bar{h}_2|\le&\frac{1}{r}\int _{0}^{r}|h_1-h_2|{\textrm{d}}s\nonumber \\ \le&\frac{1}{r}\int _{0}^{r}\frac{\Vert h_1-h_2\Vert _Y}{(1+s+u)^{k-1}}{\textrm{d}}s\nonumber \\ \le&\frac{y}{(k-2)}\frac{r^{k-3}}{(1+u)^{k-2}(1+r+u)^{k-2}}. \end{aligned}$$

  (6.34)

  Thus,

  $$\begin{aligned} |h_1-h_2-(\bar{h}_1-\bar{h}_2)|\le |h_1-h_2|+|\bar{h}_1-\bar{h}_2|\le \frac{Cy}{(1+u)^{k-2}(1+r+u)}. \end{aligned}$$

  (6.35)

  In view of (3.15) and (4.14), we have

  $$\begin{aligned} \left| |h_1-\bar{h}_1|^2-|h_2-\bar{h}_2|^2\right| \le&\left| (h_1-h_2)-(\bar{h}_1-\bar{h}_2) \right| \left( |h_1-\bar{h}_1|+|h_2-\bar{h}_2|\right) \nonumber \\ \le&\frac{Cxyr}{(1+u)^{2(k-2)}(1+r+u)^k}. \end{aligned}$$

  (6.36)

  Then, we estimate

  $$\begin{aligned} |g_1-g_2|\le&4\pi \int _{r}^{\infty }\frac{1}{s}\left| |h_1-\bar{h}_1|^2-|h_2-\bar{h}_2|^2\right| {\textrm{d}}s\nonumber \\ \le&\frac{Cxy}{(1+u)^{2(k-2)}(1+r+u)^{k-1}}. \end{aligned}$$

  (6.37)

  From the above estimate, we obtain

  $$\begin{aligned} |\bar{g}_1-\bar{g}_2|\le \frac{1}{r}\int _{0}^{r}|g_1-g_2|{\textrm{d}}s\le \frac{Cxy}{(1+u)^{3(k-2)}(1+r+u)}. \end{aligned}$$

  (6.38)

  Let us define

  $$\begin{aligned} \bar{h}_1^{p+1} - \bar{h}_2^{p+1}=(\bar{h}_1-\bar{h}_2)\int _{0}^{1}\left( t\bar{h}_1+(1-t)\bar{h}_2\right) \left| t\bar{h}_1+(1-t)\bar{h}_2\right| ^{p-1}{\textrm{d}}t, \end{aligned}$$

  (6.39)

  such that

  $$\begin{aligned} \left| |\bar{h}_1|^{p+1}-|\bar{h}_2|^{p+1}\right| \le&|\bar{h}_1-\bar{h}_2|\left( |\bar{h}_1|+|\bar{h}_2|\right) ^p\nonumber \\ =&\frac{2^px^py}{(k-2)^2(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}. \end{aligned}$$

  (6.40)

  From the above estimate, we obtain

  $$\begin{aligned}&\left| \frac{8\pi }{r}\int _{0}^{r}gs^2\left( V(\bar{h}_1)-V(\bar{h}_2)\right) {\textrm{d}}s\right| \nonumber \\&\quad \le \frac{8\pi }{r}\int _{0}^{r}s^2\frac{K_02^px^py}{(k-2)^2(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}{\textrm{d}}s\nonumber \\&\quad = \frac{K_02^{p+3}\pi x^p y r^{k-2}}{(k-2)^2(1+u)^{(k-2)(p+2)}(1+r+u)^k}. \end{aligned}$$

  (6.41)

  Combining (4.17) and (4.20), we have

  $$\begin{aligned} |\tilde{g}_1-\tilde{g}_2|\le&|\bar{g}_1-\bar{g}_2|+\left| \frac{8\pi }{r}\int _{0}^{r}gs^2\left( V(\bar{h}_1)-V(\bar{h}_2)\right) {\textrm{d}}s\right| \nonumber \\ \le&\frac{Cxy}{(1+u)^{3(k-2)}(1+r+u)}+\frac{K_02^{p+3}\pi x^p y r^{k-2}}{(k-2)^2(1+u)^{(k-2)(p+2)}(1+r+u)^k}\nonumber \\ \le&\frac{Cy(x+x^p)}{(1+u)^{3(k-2)}(1+r+u)}. \end{aligned}$$

  (6.42)

  In view of (3.47), we have

  $$\begin{aligned} D_1=&\frac{1}{2}\left| \tilde{g}_1-\tilde{g}_2\right| |{\mathcal {G}}_2|\nonumber \\ \le&\frac{Cy(x+x^p)}{(1+u)^{3(k-2)}(1+r+u)}\frac{C\left( d+x^3+x^{k+2}+x^p+x^{p+2}+x^{p+4}\right) }{\kappa ^k(1+r_1+u_1)^k}\nonumber \\&\times (1+x^2+x^{k+1}+x^{p+1}+x^{p+3})\exp \left[ C(x^2+x^4+x^{p+1})\right] . \end{aligned}$$

  (6.43)

  From (3.27), we obtain

  $$\begin{aligned} D_1\le \frac{Cy\alpha (x)}{(1+u)^{3(k-2)}(1+r+u)^{k+1}}, \end{aligned}$$

  (6.44)

  where we have denoted

  $$\begin{aligned} \alpha (x)&=C(x+x^p)\left( d+x^3+x^{k+2}+x^p+x^{p+2}+x^{p+4}\right) \\&\quad \times (1+x^2+x^{k+1}+x^{p+1}+x^{p+3})\\&\quad \times \exp \left[ C(x^2+x^4+x^{p+1})\right] . \end{aligned}$$

• 2. Estimate for $D_2$ From (3.21), we obtain

  $$\begin{aligned} D_2=&\le \frac{1}{2r}|g_1-\tilde{g}_1||\bar{h}_1-\bar{h}_2|\nonumber \\ \le&\frac{1}{2r}\frac{C(x^2+x^{p+1})r^2}{(1+u)^{-3+2k}(1+r+u)^3}\frac{y}{(k-2)}\frac{r^{k-3}}{(1+u)^{k-2}(1+r+u)^{k-2}}\nonumber \\ \le&\frac{C(x^2+x^{p+1})y}{(1+u)^{3k-5}(1+r+u)^{k}}. \end{aligned}$$

  (6.45)

• 3. Estimate for $D_3$ From (4.15), we obtain

  $$\begin{aligned} \left| g_1-g_2-(\bar{g}_1-\bar{g}_2) \right| \le&\frac{1}{r}\int _{0}^{r}\int _{r'}^{r}\left| \frac{\partial }{\partial r}(g_1-g_2)\right| {\textrm{d}}s{\textrm{d}}r'\nonumber \\ \le&\frac{4\pi }{r}\int _{0}^{r}\int _{r'}^{r}\frac{1}{s}\left| |h_1-\bar{h}_1|^2-|h_2-\bar{h}_2|^2\right| {\textrm{d}}s{\textrm{d}}r'\nonumber \\ \le&\frac{Cxyr}{(1+u)^{2k-3}(1+r+u)^{k-1}}. \end{aligned}$$

  (6.46)

  Thus,

  $$\begin{aligned}&\frac{1}{2r}\left| g_1-\tilde{g}_1-(g_2-\tilde{g}_2)\right| \nonumber \\&\quad \le \frac{1}{2r}\left[ \left| g_1-g_2-(\bar{g}_1-\bar{g}_2)\right| +\left| \frac{8\pi }{r}\int _{0}^{r}gs^2\left( V(\bar{h}_1)-V(\bar{h}_2)\right) {\textrm{d}}s\right| \right] \nonumber \\&\quad \le \frac{1}{2r}\left[ \frac{Cxyr}{(1+u)^{2k-3}(1+r+u)^{k-1}} +\frac{K_02^{p+3}\pi x^p y r^{k-2}}{(k-2)^2(1+u)^{(k-2)(p+2)}(1+r+u)^k}\right] \nonumber \\&\quad \le \frac{C(x+x^p)y}{(1+u)^{2k-3}(1+r+u)^{k-1}}. \end{aligned}$$

  (6.47)

  In view of (3.33), we have

  $$\begin{aligned} D_3&=\le \frac{1}{2r}\left| g_1-\tilde{g}_1-(g_2-\tilde{g}_2)\right| |{\mathcal {F}}_2|\nonumber \\&\le \frac{C(x+x^p)y}{(1+u)^{2k-3}(1+r+u)^{k-1}}\frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] }{\kappa ^{k-1}(1+r_1+u_1)^{k-1}}. \end{aligned}$$

  (6.48)

  From (3.27), we obtain

  $$\begin{aligned} D_3\le \frac{C y \beta (x)}{(1+u)^{2k-3}(1+r+u)^{2(k-1)}}, \end{aligned}$$

  (6.49)

  where we have denoted

  $$\begin{aligned} \beta (x)=C(x+x^p)(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] . \end{aligned}$$

• 4. Estimate for $D_4$ From (4.19) and (3.33), we obtain

  $$\begin{aligned} D_4&=4\pi r g_2\left( V(\bar{h}_1)-V(\bar{h}_2)\right) |{\mathcal {F}}_2|\nonumber \\&\le \frac{K_0 r 2^px^py}{(1+u)^{p+1}(1+r+u)^{p+1}}\frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] }{k^2(1+r_1+u_1)^2}. \end{aligned}$$

  (6.50)

  From (3.27), we obtain

  $$\begin{aligned} D_4\le \frac{Cy \gamma (x)}{(1+u)^{4(k-2)}(1+r+u)^{k+2}}, \end{aligned}$$

  (6.51)

  where we have denoted

  $$\begin{aligned} \gamma (x)=C x^p(d+x^3+x^p+x^{p+2})\exp [C(x^2+x^{p+1})]. \end{aligned}$$

• 5. Estimate for $D_5$ In view of (4.23) and (3.13), we get

  $$\begin{aligned} D_5&=\frac{1}{2r}\left| g_1-\tilde{g}_1-(g_2-\tilde{g}_2)\right| |\bar{h}_2|\nonumber \\&\le \frac{C(x+x^p)y}{(1+u)^{2k-3}(1+r+u)^{k-1}}\frac{x}{(k-2)(1+u)^{k-2}(1+r+u)}\nonumber \\&\le \frac{C(x^2+x^{p+1})y}{(1+u)^{3k-5}(1+r+u)^k}. \end{aligned}$$

  (6.52)

• 6. Estimate for $D_6$ From (3.13), (6.37), and (3.33), we have

  $$\begin{aligned} D_6&=4\pi r|g_1-g_2||V(\bar{h}_1)| |{\mathcal {F}}_1|\nonumber \\&\le Cr\frac{xy}{(1+u)^{2(k-2)}(1+r+u)^{k-1}}\frac{K_0x^{p+1}}{(k-2)^{p+1}(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}\nonumber \\&\quad \times \frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] }{\kappa ^{k-1}(1+r_1+u_1)^{k-1}}. \end{aligned}$$

  (6.53)

  Then, from (3.27) we obtain

  $$\begin{aligned} D_6\le \frac{Cy\sigma (x)}{(1+u)^{6(k-2)}(1+r+u)^{2k+1}}, \end{aligned}$$

  (6.54)

  where we have denoted

  $$\begin{aligned} \sigma (x)=Cx^{p+2}(d+x^3+x^p+x^{p+2})\exp [C(x^2+x^{p+1})]. \end{aligned}$$

• 7. Estimate for $D_7$ From (3.13) and (6.37), we obtain

  $$\begin{aligned} D_7&=4\pi r|g_1-g_2||\bar{h}_1| |V(\bar{h}_1)|\nonumber \\&\le Cr \frac{xy}{(1+u)^{2(k-2)}(1+r+u)^{k-1}}\frac{x}{(k-2)(1+u)^{k-2}(1+r+u)}\nonumber \\&\quad \times \frac{K_0x^{p+1}}{(k-2)^{p+1}(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}\nonumber \\&\le \frac{Cyx^{p+3}}{(1+u)^{7(k-2)}(1+r+u)^{3+k}}. \end{aligned}$$

  (6.55)

• 8. Estimate for $D_8$ Estimate (3.13) and (4.13) yields

  $$\begin{aligned} D_8&=4\pi r g_2|\bar{h}_1 - \bar{h}_2||V(\bar{h}_2)|\nonumber \\&\le Cr \frac{yr^{k-3}}{(k-2)(1+u)^{k-2}(1+r+u)^{k-2}}\nonumber \\&\quad \frac{K_0x^{p+1}}{(k-2)^{p+1}(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}\nonumber \\&\le \frac{Cyx^{p+1}}{(1+u)^{5(k-2)}(1+r+u)^4}. \end{aligned}$$

  (6.56)

• 9. Estimate for $D_9$ From (3.13) and (4.19), we have

  $$\begin{aligned} D_9&=4\pi r g_2\left| V(\bar{h}_1)-V(\bar{h}_2)\right| |\bar{h}_1|\nonumber \\&\le \frac{4\pi K_02^px^pyr}{(k-2)^2(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}} \frac{x}{(k-2)(1+u)^{k-2}(1+r+u)}\nonumber \\&\le \frac{Cyx^{p+1}}{(1+u)^{5(k-2)}(1+r+u)^4}. \end{aligned}$$

  (6.57)

• 10. Estimate for $D_{10}$ Again, from (3.13) and (6.37) we have

  $$\begin{aligned} D_{10}&=\frac{r}{2}|g_1-g_2||\bar{h}_1|\left| \frac{\partial ^2 V(\bar{h}_1)}{\partial \bar{h}_1^2}\right| \nonumber \\&\le Cr\frac{xy}{(1+u)^{2(k-2)}(1+r+u)^{k-1}}\nonumber \\&\quad \frac{K_0x^p}{(k-2)^p(1+u)^{p(k-2)}(1+r+u)^p}\nonumber \\&\le \frac{Cyx^{p+1}}{(1+u)^{5(k-2)}(1+r+u)^{k+1}}. \end{aligned}$$

  (6.58)

• 11. Estimate for $D_{11}$ In view of (3.13) and (4.13), we get

  $$\begin{aligned} D_{11}&=\frac{r}{2}g_2|\bar{h}_1-\bar{h}_2|\left| \frac{\partial ^2 V(\bar{h}_2)}{\partial \bar{h}_2^2}\right| \nonumber \\&\le Cr \frac{yr^{k-3}}{(k-2)(1+u)^{k-2}(1+r+u)^{k-2}}\nonumber \\&\quad \frac{K_0x^{p-1}}{(k-2)^{p-1}(1+u)^{(k-2)(p-1)}(1+r+u)^{p-1}}\nonumber \\&\le \frac{Cyx^{p-1}}{(1+u)^{3(k-2)}(1+r+u)^2}. \end{aligned}$$

  (6.59)

• 12. Estimate for $D_{12}$ From (4.19), we have

  $$\begin{aligned} \left| \bar{h}_1^{p-1}-\bar{h}_2^{p-1}\right|&\le |\bar{h}_1-\bar{h}_2|\left( |\bar{h}_1|+|\bar{h}_2|\right) ^{p-2}\nonumber \\&\le \frac{K_0yx^{p-2}}{(1+u)^{(k-2)(p-1)}(1+r+u)^{p-1}}. \end{aligned}$$

  (6.60)

  From the above estimate, combined with (3.13) we have

  $$\begin{aligned} D_{12}&=\frac{r}{2}g_2\left| \frac{\partial ^2 V(\bar{h}_1)}{\partial \bar{h}_1^2}-\frac{\partial ^2 V(\bar{h}_2)}{\partial \bar{h}_2^2}\right| |\bar{h}_1|\le \frac{r}{2}g_2\left| |\bar{h}_1|^{p-1}-|\bar{h}_2|^{p-1}\right| |\bar{h}_1|\nonumber \\&\le C r \frac{K_0yx^{p-2}}{(1+u)^{(k-2)(p-1)}(1+r+u)^{p-1}} \frac{x}{(k-2)(1+u)^{k-2}(1+r+u)}\nonumber \\&\le \frac{Cyx^{p-1}}{(1+u)^{3(k-2)}(1+r+u)^2}. \end{aligned}$$

  (6.61)

Thus, we have

$$\begin{aligned} |\tilde{\varphi }|\le \frac{Cy\left[ \alpha (x)+\beta (x)+\gamma (x)+\sigma (x)+x^2+x^{p-1}+x^{p+1}+x^{p+3}\right] }{(1+u)^{3(k-2)}(1+r+u)^2}, \end{aligned}$$

(6.62)

where $\alpha (x),\beta (x),\gamma (x),$ and $\sigma (x)$ are defined in (4.25)–(4.28) respectively.

1.5 Estimate for (5.2)

We write the estimate of (5.2) as follows:

$$\begin{aligned} \left| \frac{\partial h}{\partial u}\right|\le & {} \frac{\tilde{g}}{2}\left| \frac{\partial h}{\partial r}\right| + \frac{1}{2r}\left| g-\tilde{g}\right| \left| h\right| +\frac{1}{2r}\left| g-\tilde{g}\right| \left| \bar{h}\right| \nonumber \\{} & {} +4\pi gr\left| h\right| |V(\bar{h})|+4\pi gr|\bar{h}||V(\bar{h})|+\frac{gr}{2}\left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \nonumber \\:= & {} H_1+H_2+H_3+H_4+H_5+H_6. \end{aligned}$$

(6.63)

• 1. Estimate for $H_1$ From (3.47), we have

  $$\begin{aligned} \frac{\tilde{g}}{2}\left| \frac{\partial h}{\partial r}\right| \le \frac{K_1(x)}{\kappa ^{k}(1+r_1+u_1)^{k}}\le \frac{K_1(x)}{(1+r+u)^k}, \end{aligned}$$

  (6.64)

  where we have denoted

  $$\begin{aligned} K_1(x)= & {} C(d+x^3+x^{k+2}+x^p+x^{p+2}+x^{p+4})\nonumber \\{} & {} \times (1+x^2+x^{k+1}+x^{p+1}+x^{p+3})\nonumber \\{} & {} \times \exp \left[ C(x^2+x^4+x^{p+1})\right] . \end{aligned}$$

  (6.65)

• 2. Estimate for $H_2$ The estimate (3.33) yields

  $$\begin{aligned} \frac{1}{2r}\left| g-\tilde{g}\right| \left| h\right| \le&\frac{1}{2r} \frac{C(x^2+x^{p+1})r^2}{(1+u)^{-3+2k}(1+r+u)^3}\nonumber \\&\frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] }{\kappa ^{k-1}(1+r_1+u_1)^{k-1}}\nonumber \\ \le&\frac{K_2(x)}{(1+u)^{-3+2k}(1+r+u)^{k+1}}, \end{aligned}$$

  (6.66)

  where

  $$\begin{aligned} K_2(x)=C(x^2+x^{p+1})(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] . \end{aligned}$$

  (6.67)

• 3. Estimate for $H_3$

  $$\begin{aligned} \frac{1}{2r}\left| g-\tilde{g}\right| \left| \bar{h}\right| \le&\frac{1}{r}\frac{C(x^2+x^{p+1})r^2}{(1+u)^{-3+2k}(1+r+u)^3}\frac{xr^{k-3}}{(k-2)(1+u)^{k-2}(1+r+u)^{k-2}}\nonumber \\ \le&\frac{C(x^3+x^{p+2})}{(1+u)^{3k-5}(1+r+u)^k}. \end{aligned}$$

  (6.68)

• 4. Estimate for $H_4$ Combining (3.13) and (3.33), we obtain

  $$\begin{aligned}&4\pi g r |h||V(\bar{h})|\nonumber \\&\quad \le Cr\frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] K_0x^{p+1}}{\kappa ^{k-1}(1+r_1+u_1)^{k-1}(k-2)^{p+1}(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}\nonumber \\&\quad \le \frac{K_3(x)}{(1+u)^{4(k-2)}(1+r+u)^{k+2}}, \end{aligned}$$

  (6.69)

  where

  $$\begin{aligned} K_3(x)=Cx^{p+1}(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] . \end{aligned}$$

  (6.70)

• 5. Estimate for $H_5$

  $$\begin{aligned} 4\pi g r |\bar{h}||V(\bar{h})|&\le Cr \frac{xr^{k-3}}{(k-2)(1+u)^{k-2}(1+r+u)^{k-2}}\nonumber \\&\quad \frac{K_0x^{p+1}}{(k-2)^{p+1}(1+u)^{(k-2)(p+1)}(1+r+u)^{p+1}}\nonumber \\&\le \frac{Cx^{p+2}}{(1+u)^{(k-2)(p+2)}(1+r+u)^{k+1}}. \end{aligned}$$

  (6.71)

• 6. Estimate for $H_6$ The estimate (2.9) yields

  $$\begin{aligned} \frac{gr}{2}\left| \frac{\partial V(\bar{h})}{\partial \bar{h}}\right| \le \frac{Cx^p}{(1+u)^{p(k-2)}(1+r+u)^{p-1}}. \end{aligned}$$

  (6.72)

Combining (6.64)–(6.72), we write the estimate (5.2) as follows

$$\begin{aligned} \left| \frac{\partial h}{\partial u}\right| \le \frac{C(K_1(x)+K_2(x)+K_3+x^3+x^p+x^{p+2})}{(1+r+u)^2}. \end{aligned}$$

(6.73)