Global existence and completeness of classical solutions in higher dimensional Einstein–Klein–Gordon system

Abstract

In this paper we study the global existence and completeness of classical solutions of gravity coupled a scalar field system called Einstein–Klein–Gordon system in higher dimensions. We introduce a new ansatz function to reduce the problem into a single first-order integro-differential equation. Then, we employ the contraction mapping in the appropriate Banach space. Using Banach fixed theorem, we show that there exists a unique fixed point, which is the solution of the theory. For a given initial data, we prove the existence of both local and global classical solutions. We also study the completeness properties of the spacetime. Here, we introduce a mass-like function for $D\ge 4$ in Bondi coordinates. The completeness of spacetime along the future directed timelike lines outward to a region which resembles the event horizon of the black hole.

Appendix

1.1 Estimate for (3.84)

We write the estimate for (3.84) as follows

$$\begin{aligned} |f|\le&\frac{(D-2)}{4r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2}\tilde{g}\right| \left| \tilde{h}\right| +\frac{8\pi g r}{(D-2)}\left| V(\tilde{h})\right| \left| \tilde{h}\right| +\frac{gr}{2}\left| \frac{\partial V(\tilde{h})}{\partial \tilde{h}}\right| ,\nonumber \\ \le&A_1 + A_2 + A_3 \end{aligned}$$

(5.1)

• 1. Estimate for $A_1$

  First, we calculate

  $$\begin{aligned} |\tilde{h}|\le&\frac{1}{r^{\frac{(D-2)}{2}}}\int _{0}^{r} \frac{\Vert h\Vert _X}{(1+s+u)^{k-1}}\frac{1}{s^{\frac{(4-D)}{2}}}~ \textrm{d}s\nonumber \\ \le&\frac{2x}{(2k-D)}\frac{1}{ (1+u)^{\frac{(2k-D)}{2}}(1+r+u)^{\frac{(D-2)}{2}}}. \end{aligned}$$

  (5.2)

  Then, we obtain

  $$\begin{aligned} \left| h-\frac{(D-2)}{2}\tilde{h}\right| \le&|h| + \left| \frac{(D-2)}{2}\tilde{h}\right| \nonumber \\ \le&\frac{x}{(1+r+u)^{k-1}}+\frac{2x}{(2k-D)}\frac{1}{(1+u)^{\frac{(2k-D)}{2}}(1+r+u)^{\frac{(D-2)}{2}}}\nonumber \\ \le&\frac{Cx}{(1+u)^{\frac{(2k-D)}{2}}(1+r+u)^{\frac{(D-2)}{2}}}. \end{aligned}$$

  (5.3)

  From the above estimate, we get

  $$\begin{aligned} |g(u,r)-g(u,r')|\le&\int _{r'}^{r}\left| \frac{\partial g}{\partial s}(u,s) \right| \textrm{d}s\le \frac{8\pi }{(D-2)} \int _{r'}^{r}\frac{1}{s}\left| h-\frac{(D-2)}{2}\tilde{h}\right| ^2\textrm{d}s\nonumber \\ \le&\frac{8\pi }{(D-2)^2}\frac{x^2}{(1+u)^{2k-D}}\left[ \frac{1}{(1+r'+u)^{D-2}}-\frac{1}{(1+r+u)^{D-2}}\right] , \end{aligned}$$

  (5.4)

  and

  $$\begin{aligned} |(g-\bar{g})(u,r)|\le&\frac{1}{r}\int _{0}^{r}|g(u,r)-g(u,r')|\textrm{d}r'\nonumber \\ \le&\frac{8\pi x^2}{(D-3)(D-2)^2}\frac{1}{(1+u)^{2k-3}(1+r+u)}. \end{aligned}$$

  (5.5)

  Thus

  $$\begin{aligned} \int _{r}^{\infty }\frac{1}{s}\left| h-\frac{(D-2)}{2} \tilde{h}\right| ^2\textrm{d}s\le \frac{x^2}{(D-2) (1+u)^{2k-D}(1+r+u)^{D-2}}, \end{aligned}$$

  (5.6)

  such that we obtain

  $$\begin{aligned} |g (u,r)|=\,&\exp \left[ -\frac{8\pi }{(D-2)}\int _{r}^{\infty }\frac{1}{s} \left| h-\frac{(D-2)}{2}\tilde{h}\right| ^2\textrm{d}s\right] \nonumber \\ \ge&\exp \left[ -\frac{8\pi x^2}{(D-2)^2(1+u)^{2k-D}(1+r+u)^{D-2}}\right] . \end{aligned}$$

  (5.7)

  As a consequence, we can calculate

  $$\begin{aligned} |\bar{g}|\ge |g| + |g-\bar{g}|\ge \frac{8\pi x^2}{(D-3)(D-2)^2}\frac{1}{(1+u)^{2k-3}(1+r+u)}, \end{aligned}$$

  (5.8)

  and

  $$\begin{aligned} \frac{(D-4)}{2}\frac{\hat{R}(\hat{\sigma })}{r^{D-3}} \int _{0}^{r}|\bar{g}|s^{D-4}\textrm{d}s\le \frac{Cx^2}{(1+u)^{2k-3}(1+r+u)}. \end{aligned}$$

  (5.9)

  To proceed, we calculate

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

  (5.10)

  Now, we obtain

  $$\begin{aligned} \left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2}\tilde{g}\right| \le&|g-\bar{g}| +\frac{(D-4)}{2}\frac{\hat{R}(\hat{\sigma })}{r^{D-3}}\int _{0}^{r}|\bar{g}|s^{D-4}\textrm{d}s\nonumber \\&+\frac{8\pi }{r^{D-3}}\int _{0}^{r} gs^{D-2}|V(\tilde{h})|\textrm{d}s\nonumber \\ \le&\frac{C(x^2+x^{p+1})}{(1+u)^{2k-3}(1+r+u)}. \end{aligned}$$

  (5.11)

  Hence,

  $$\begin{aligned} A_1=\frac{(D-2)}{4r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g- \frac{(D-2)}{2}\tilde{g}\right| \left| \tilde{h}\right| \le \frac{C(x^3+x^{p+2})}{(1+u)^{\frac{6(k-1)-D}{2}}(1+r+u)^{\frac{D+2}{2}}}. \end{aligned}$$

  (5.12)

• 2. Estimate for $A_2$

  Using (1.5) and (1.3), we obtain

  $$\begin{aligned} A_2=\frac{8\pi g r}{(D-2)}\left| V(\tilde{h})\right| |\tilde{h}| \le \frac{Cx^{p+2}}{(1+u)^{\frac{(2k-D)}{2}(k+2)} (1+r+u)^{\frac{(D-2)(k+2)-2}{2}}}. \end{aligned}$$

  (5.13)

• 3. Estimate for $A_3$

  Again, using (1.3) we get

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

  (5.14)

Finally, we obtain

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

(5.15)

1.2 Estimate for (3.96)

We write the estimate of (3.96) as follows:

$$\begin{aligned} |f_1|\le&\left\{ \left| \frac{1}{2r}\left( \frac{\hat{R}(\hat{\sigma })}{(D-2)}\frac{\partial g}{\partial r}-\frac{(D-2)}{2}\frac{\partial \tilde{g}}{\partial r}\right) \right| +\left| \frac{1}{2r^2}\left( \frac{\hat{R}(\hat{\sigma })}{(D-2)}g -\frac{(D-2)}{2}\tilde{g}\right) \right| \right. \nonumber \\&\left. +\left| \frac{8\pi g r}{(D-2)} \frac{\partial V(\tilde{h})}{\partial \tilde{h}} \frac{\partial \tilde{h}}{\partial r}\right| +\left| \frac{8\pi gV(\tilde{h})}{(D-2)}\right| +\left| \frac{8\pi rV(\tilde{h})}{(D-2)}\frac{\partial g}{\partial r}\right| \right\} \left( \left| \mathcal {F}\right| +\frac{(D-2)}{2}\left| \tilde{h}\right| \right) \nonumber \\&+\left\{ \left| \frac{1}{2r} \left( \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2}\tilde{g}\right) \right| +\left| \frac{8\pi g rV(\tilde{h})}{(D-2)}\right| +\left| \frac{g r }{2}\frac{\partial ^2V(\tilde{h})}{\partial \tilde{h}^2}\right| \right\} \nonumber \\&\left| \frac{\partial \tilde{h}}{\partial r}\right| +\left| \frac{r}{2}\frac{\partial g}{\partial r}+\frac{g}{2}\right| \left| \frac{\partial V(\tilde{h})}{\partial \tilde{h}}\right| \nonumber \\ =\,&(B_1+B_2+B_3+B_4+B_5)\left( \left| \mathcal {F}\right| +\frac{(D-2)}{2}\left| \tilde{h}\right| \right) + (B_6+B_7+B_8)\left| \frac{\partial \tilde{h}}{\partial r}\right| + B_9 \end{aligned}$$

(5.16)

• 1. Estimate for $B_1$

  We use the formula $\frac{\partial \bar{g}}{\partial r}=\frac{g-\bar{g}}{r}$ to obtain

  $$\begin{aligned} \left| \frac{\partial \tilde{g}}{\partial r}\right| \le&\frac{\hat{R}(\hat{\sigma })}{(D-2)}\frac{|g-\bar{g}|}{r} +\frac{\hat{R}(\hat{\sigma })(D-4)(D-3)}{(D-2)r^{D-2}}\int _{0}^{r}| \bar{g}|s^{D-4}\textrm{d}s+\frac{\hat{R}(\hat{\sigma })(D-4)}{(D-2)}\frac{|\bar{g}|}{r}\nonumber \\&+\frac{(D-3)16\pi }{(D-2)r^{D-2}}\int _{0}^{r} \left| g\right| s^{D-2}\left| V(\tilde{h})\right| \textrm{d}s+\frac{16\pi |g|r\left| V(\tilde{h})\right| }{(D-2)}. \end{aligned}$$

  (5.17)

  The estimate (5.5) yields

  $$\begin{aligned} \frac{|g-\bar{g}|}{r}\le \frac{Cx^2r^{D-4}}{(1+u)^{2k-3}(1+r+u)^{D-2}}. \end{aligned}$$

  (5.18)

  Using (3.24), we get

  $$\begin{aligned} \frac{\hat{R}(\hat{\sigma })(D-4)(D-3)}{(D-2)r^{D-2}} \int _{0}^{r}|\bar{g}|s^{D-4}\textrm{d}s\le \frac{Cx^2}{(1+u)^{2k-3}(1+r+u)^2}, \end{aligned}$$

  (5.19)

  and

  $$\begin{aligned} \frac{(D-4)}{(D-2)}\hat{R}(\hat{\sigma })\frac{1}{r}|\bar{g}|\le \frac{8\pi x^2}{(D-3)(D-2)^2(1+u)^{2k-3}(1+r+u)^2}. \end{aligned}$$

  (5.20)

  Then, using (1.5) we have

  $$\begin{aligned} \frac{(D-3)16\pi }{(D-2)r^{D-2}}\int _{0}^{r} \left| g\right| s^{D-2}\left| V(\tilde{h})\right| \textrm{d}s\le \frac{Cx^{p+1}}{(1+u)^{k^2-D}(1+r+u)^{D-2}}, \end{aligned}$$

  (5.21)

  and

  $$\begin{aligned} \frac{16\pi gr}{(D-2)}|V(\tilde{h})|\le \frac{Cx^{p+1}r}{(1+u)^{\frac{(2k-D) (k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}. \end{aligned}$$

  (5.22)

  Thus,

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

  (5.23)

  Using (1.6) we calculate

  $$\begin{aligned} \ \frac{\hat{R}(\hat{\sigma })}{(D-2)}\left| \frac{\partial g}{\partial r} \right| \le \frac{Cx^2r}{(1+u)^{2k-D}(1+r+u)^{D}}. \end{aligned}$$

  (5.24)

  Finally, we obtain

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

  (5.25)

• 2. Estimate for $B_2$

  Using (3.27) we obtain

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

  (5.26)

• 3. Estimate for $B_3$

  Using the relation $\left| \frac{\partial \tilde{h}}{\partial r}\right| =\frac{\left| h-\frac{(D-2)}{2}\tilde{h}\right| }{r}$, we get

  $$\begin{aligned} B_3 = \left| \frac{8\pi g r}{(D-2)} \frac{\partial V(\tilde{h})}{\partial \tilde{h}} \frac{\partial \tilde{h}}{\partial r}\right| \le \frac{Cx^{p+1}}{(1+u)^{\frac{(2k-D)(k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}. \end{aligned}$$

  (5.27)

• 4. Estimate for $B_4$

  Using (1.3) we obtain

  $$\begin{aligned} B_4=\left| \frac{8\pi g V(\tilde{h})}{(D-2)}\right| \le \frac{C x^{p+1}}{(1+u)^{\frac{(2k-D)(k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}. \end{aligned}$$

  (5.28)

• 5. Estimate for $B_5$

  Combination of (1.3) and (3.39) yields

  $$\begin{aligned} B_5 = \left| \frac{8\pi rV(\tilde{h})}{(D-2)}\frac{\partial g}{\partial r}\right| \le \frac{Cx^{p+3}}{(1+u)^{3(2k-D)}(1+r+u)^{\frac{(D-2)(k+3)}{2}}}. \end{aligned}$$

  (5.29)

• 6. Estimate for $B_6$

  Using (3.27) we obtain

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

  (5.30)

• 7. Estimate for $B_7$

  The estimate (1.3) produces

  $$\begin{aligned} B_7=\left| \frac{8\pi g r V(\tilde{h})}{(D-2)}\right| \le \frac{C x^{p+1}r}{(1+u)^{\frac{(2k-D)(k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}. \end{aligned}$$

  (5.31)

• 8. Estimate for $B_8$

  Using (1.3) we have

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

  (5.32)

• 9. Estimate for $B_9$

  Again, using (1.3) we get

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

  (5.33)

Finally, we obtain

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

(5.34)

1.3 Estimate for (3.103)

We write the estimate of (3.103) as follows:

$$\begin{aligned} |\tilde{\psi }|\le&\frac{1}{2}\left| \tilde{g}_1-\tilde{g}_2\right| |\mathcal {G}_2| +\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g_1-\frac{(D-2)}{2} \tilde{g}_1\right| \frac{(D-2)}{2}\left| \tilde{h}_1 -\tilde{h}_2\right| \nonumber \\&+\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}(g_1-g_2)-\frac{(D-2)}{2} (\tilde{g}_1-\tilde{g}_2)\right| |\mathcal {F}_2|\nonumber \\&+\frac{8\pi r g_2}{(D-2)}\left| V(\tilde{h}_1)-V(\tilde{h}_1)\right| |\mathcal {F}_2|\nonumber \\&+\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}(g_1-g_2)-\frac{(D-2)}{2} (\tilde{g}_1-\tilde{g}_2)\right| \nonumber \\&\frac{(D-2)}{2}|\tilde{h}_2|+\frac{8\pi r|V(\tilde{h}_1)|}{(D-2)}|g_1-g_2||\mathcal {F}_1|\nonumber \\&+4\pi r|g_1-g_2||\tilde{h}_1| |V(\tilde{h}_1)|+4\pi r g_2|\tilde{h}_1 - \tilde{h}_2||V(\tilde{h}_2)|\nonumber \\&+4\pi r g_2 |\tilde{h}_1|\left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| \nonumber \\&+\frac{r}{2}|g_1-g_2||\tilde{h}_1|\left| \frac{\partial ^2V(\tilde{h}_1)}{\partial \tilde{h}_1^2 }\right| +\frac{r}{2}g_2|\tilde{h}_1-\tilde{h}_2|\left| \frac{\partial ^2V(\tilde{h}_1)}{\partial \tilde{h}_1^2 }\right| \nonumber \\&+\frac{r}{2}g_2|\tilde{h}_1|\left| \frac{\partial ^2V(\tilde{h}_1)}{\partial \tilde{h}_1^2 }-\frac{\partial ^2V(\tilde{h}_2)}{\partial \tilde{h}_2^2 }\right| \nonumber \\ :=\,&D_1+D_2+D_3+\cdots +D_{10}+D_{11}+D_{12} \end{aligned}$$

(5.35)

• 1. Estimate for $D_1$

  Setting $\Vert h_1-h_2\Vert _Y=y$. Using (1.5) we obtain

  $$\begin{aligned} |\tilde{h}_1 - \tilde{h}_2|\le \frac{2y}{(2k-D)}\frac{1}{(1+u)^{\frac{(2k-D)}{2}}(1+r+u)^{\frac{(D-2)}{2}}}. \end{aligned}$$

  (5.36)

  To proceed, we calculate

  $$\begin{aligned} |h_1-h_2-(\tilde{h}_1-\tilde{h}_2)|\le&|h_1-h_2|+|\tilde{h}_1-\tilde{h}_2|\nonumber \\ \le&\frac{Cy}{(1+u)^{\frac{(2k-D)}{2}}(1+r+u)^{\frac{(D-2)}{2}}}, \end{aligned}$$

  (5.37)

  and

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

  (5.38)

  As a consequence, we get

  $$\begin{aligned} |g_1-g_2|\le&\frac{8\pi }{(D-2)} \int _{r}^{\infty } \frac{1}{s}\left| |h_1-\tilde{h}_1|^2-|h_2- \tilde{h}_2|^2\right| \textrm{d}s\nonumber \\ \le&\frac{Cxy}{(1+u)^{(2k-D)}(1+r+u)^{(D-2)}}. \end{aligned}$$

  (5.39)

  Then, we calculate

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

  (5.40)

  yields

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

  (5.41)

  We estimate

  $$\begin{aligned} \frac{16\pi }{(D-2)}\frac{1}{r^{D-3}}\int _{0}^{r} g_1s^{D-2}\left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| \textrm{d}s\le \frac{Cx^py}{(1+u)^{k^2-D}(1+r+u)^{D-3}}. \end{aligned}$$

  (5.42)

  Then, we obtain

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

  (5.43)

  and

  $$\begin{aligned} \frac{(D-4)}{(D-2)}\frac{\hat{R}(\hat{\sigma })}{r^{D-3}}\int _{0}^{r}\left| \bar{g}_1-\bar{g}_2\right| s^{D-4}\textrm{d}s\le \frac{Cxy}{(1+u)^{2k-3}(1+r+u)}. \end{aligned}$$

  (5.44)

  From the definition (1.7) we get

  $$\begin{aligned} \left| \tilde{g}_1-\tilde{g}_2\right| \le&\frac{\hat{R}(\hat{\sigma })}{(D-2)}\left| \bar{g}_1-\bar{g}_2\right| + \frac{(D-4)}{(D-2)}\frac{\hat{R}(\hat{\sigma })}{r^{D-3}}\int _{0}^{r} \left| \bar{g}_1-\bar{g}_2\right| s^{D-4}\textrm{d}s\nonumber \\&+ \frac{16\pi }{(D-2)}\frac{1}{r^{D-3}}\int _{0}^{r} g_1s^{D-2}\left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| \textrm{d}s\nonumber \\ \le&\frac{Cy(x+x^p)}{(1+u)^{2k-3}(1+r+u)}. \end{aligned}$$

  (5.45)

  Finally, we obtain

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

  (5.46)

  Since $1+r(u)+u\ge \frac{1}{2}\kappa (1+r_1+u_1)$, we have

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

  (5.47)

  where $\alpha (x)=C(x+x^p)\left( d+x^3+x^{k+2}+x^p+x^{p+2}+x^{p+4}\right) (1+x^2+x^{k+1}+x^{p+1}+x^{p+3})\exp \left[ C(x^2+x^{k+1}+x^{p+1})\right] .$

• 2. Estimate for $D_2$

  Combining (3.27) and (5.36) we obtain

  $$\begin{aligned} D_2=\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{2}g_1- \frac{(D-2)^2}{4}\tilde{g}_1\right| |\tilde{h}_1-\tilde{h}_2|\le \frac{C(x^2+x^{p+1})y}{(1+u)^{\frac{6(k-1)-D}{2}}(1+r+u)^{\frac{D+2}{2}}}. \end{aligned}$$

  (5.48)

• 3. Estimate for $D_3$

  We firstly calculate

  $$\begin{aligned}&\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}(g_1-g_2)-\frac{\hat{R} (\hat{\sigma })}{2}(\bar{g}_1-\bar{g}_2)\right| \nonumber \\&\quad \le \left| g_1-g_2 -(\bar{g}_1-\bar{g}_2) \right| \le \frac{1}{r}\int _{0}^{r}\int _{r'}^{r}\left| \frac{\partial }{\partial r} (g_1-g_2)\right| \textrm{d}s~\textrm{d}r'\nonumber \\&\quad \le \frac{8\pi }{(D-2)r}\int _{0}^{r}\int _{r'}^{r}\frac{1}{s} \left| |h_1-\tilde{h}_1|^2-|h_2-\tilde{h}_2|^2\right| \textrm{d}s~\textrm{d}r'\nonumber \\&\quad \le \frac{Cxy}{r(1+u)^{2k-D}}\int _{0}^{r}\int _{r'}^{r} \frac{\textrm{d}s}{(1+s+u)^{D-1}}\textrm{d}r'\nonumber \\&\quad \le \frac{Cxyr}{(1+u)^{2k-3}(1+r+u)^{D-2}}. \end{aligned}$$

  (5.49)

  Thus,

  $$\begin{aligned}&\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}(g_1-g_2)-\frac{(D-2)}{2} (\tilde{g}_1-\tilde{g}_2)\right| \nonumber \\&\quad \le \frac{1}{2r}\left[ \left| \frac{\hat{R} (\hat{\sigma })}{(D-2)}(g_1-g_2)-\frac{\hat{R}(\hat{\sigma })}{2}(\bar{g}_1-\bar{g}_2)\right| \right. \nonumber \\&+\frac{(D-4)}{2}\frac{\hat{R}(\hat{\sigma })}{r^{D-3}}\int _{0}^{r} \left| \bar{g}_1-\bar{g}_2\right| s^{D-4}\textrm{d}s\nonumber \\&\qquad \left. +\frac{8\pi }{r^{D-3}} \int _{0}^{r} g_1s^{D-2}\left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| \textrm{d}s\right] \nonumber \\&\quad \le \frac{C(x+x^p)y}{(1+u)^{2k-3}(1+r+u)^{D-2}}. \end{aligned}$$

  (5.50)

  Finally, we obtain

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

  (5.51)

  Since $1+r(u)+u\ge \frac{1}{2}\kappa (1+r_1+u_1)$, we have

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

  (5.52)

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

• 4. Estimate for $D_4$

  Using (1.3) and (3.95) we obtain

  $$\begin{aligned} D_4=\,&\frac{8\pi r g_2}{(D-2)} \left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| |\mathcal {F}_2|\nonumber \\ \le&\frac{Crx^py}{(1+u)^{\frac{(2k-D)(k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}\frac{C(d+x^3+x^p+x^{p+2}) \exp \left[ C(x^2+x^{p+1})\right] }{\kappa ^{k-1}(1+r_1+u_1)^{k-1}}. \end{aligned}$$

  (5.53)

  Since $1+r(u)+u\ge \frac{1}{2}\kappa (1+r_1+u_1)$, we have

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

  (5.54)

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

• 5. Estimate for $D_5$

  Combination of (3.83) and (5.50) yields

  $$\begin{aligned} D_5=&\,\frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{2}(g_1-g_2)- \frac{(D-2)^2}{4}(\tilde{g}_1-\tilde{g}_2)\right| |\tilde{h}_2|\nonumber \\ \le&\,\frac{C(x^2+x^{p+1})y}{(1+u)^{\frac{6(k-1)-D}{2}}(1+r+u)^{\frac{3(D-2)}{2}}}. \end{aligned}$$

  (5.55)

• 6. Estimate for $D_6$

  We calculate

  $$\begin{aligned} D_6=\,&\frac{8\pi r}{(D-2)}|g_1-g_2||V(\tilde{h})||\mathcal {F}_1|\nonumber \\ \le&\frac{8\pi r}{(D-2)}\frac{xy}{(1+u)^{(2k-D)}(1+r+u)^{(D-2)}} \frac{K_0x^{p+1}}{(1+u)^{\frac{(2k-D)(k+1)}{2}}(1+r+u)^{\frac{(D-2)(k+1)}{2}}}\nonumber \\&\times \frac{C(d+x^3+x^p+x^{p+2})\exp \left[ C(x^2+x^{p+1})\right] }{\kappa ^{k-1}(1+r_1+u_1)^{k-1}}. \end{aligned}$$

  (5.56)

  Since $1+r(u)+u\ge \frac{1}{2}\kappa (1+r_1+u_1)$, we obtain

  $$\begin{aligned} D_6\le \frac{Cy\sigma (x)}{(1+u)^{3(2k-D)}(1+r+u)^{3(D-2)+k-2}}, \end{aligned}$$

  (5.57)

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

• 7. Estimate for $D_7$

  We use (1.3) and (3.83) to obtain

  $$\begin{aligned} D_7 = 4\pi r|g_1-g_2||\tilde{h}_1| |V(\tilde{h}_1)| \le \frac{Cyx^{p+3}}{(1+u)^{\frac{7(2k-D)}{2}}(1+r+u)^{\frac{6(2k-D)}{2}}}. \end{aligned}$$

  (5.58)

• 8. Estimate for $D_8$

  Combining (1.3) and (3.49) such that

  $$\begin{aligned} D_8 = 4\pi r g_2|\tilde{h}_1 - \tilde{h}_2||V(\tilde{h}_2)\le \frac{Cyx^{p+1}}{(1+u)^{\frac{5(2k-D)}{2}}(1+r+u)^{2(2k-D)}}. \end{aligned}$$

  (5.59)

• 9. Estimate for $D_9$

  We calculate

  $$\begin{aligned} D_9 = 4\pi r g_2 |\tilde{h}_1|\left| V(\tilde{h}_1)-V(\tilde{h}_2)\right| \le \frac{Cyx^{p+1}}{(1+u)^{\frac{5(2k-D)}{2}}(1+r+u)^{\frac{4(2k-D)}{2}}}. \end{aligned}$$

  (5.60)

• 10. Estimate for $D_{10}$

  Combination of (1.3) and (3.83) yields

  $$\begin{aligned} D_{10}=\frac{r}{2}|g_1-g_2||\tilde{h}_1|\left| \frac{\partial ^2V(\tilde{h}_1)}{\partial \tilde{h}_1^2 }\right| \le \frac{Cyx^{p+1}}{(1+u)^{\frac{5(2k-D)}{2}}(1+r+u)^{2(2k-D)}}. \end{aligned}$$

  (5.61)

• 11. Estimate for $D_{11}$

  Again, we use (1.3) and (3.49) such that

  $$\begin{aligned} D_{11} = \frac{r}{2}g_2|\tilde{h}_1-\tilde{h}_2|\left| \frac{\partial ^2V(\tilde{h}_1)}{\partial \tilde{h}_1^2 }\right| \le \frac{Cyx^{p-1}}{(1+u)^{\frac{3(2k-D)}{2}}(1+r+u)^{D-2}}. \end{aligned}$$

  (5.62)

• 12. Estimate for $D_{12}$

  We calculate

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

  (5.63)

  Hence,

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

  (5.64)

Finally, we obtain

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

(5.65)

where $\alpha (x),\beta (x),\gamma (x),$ and $\sigma (x)$ are defined in (3.104)–(3.107) respectively.

1.4 Estimate for (3.115)

We write the estimate of (3.115) as follows

$$\begin{aligned} \left| \frac{\partial h}{\partial u}\right| \le&\frac{\tilde{g}}{2}\left| \frac{\partial h}{\partial r}\right| + \frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2} \tilde{g}\right| \left| h\right| \nonumber \\&+\frac{(D-2)}{4r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2}\tilde{g}\right| \left| \tilde{h}\right| \nonumber \\&+\frac{8\pi gr}{(D-2)}\left| h\right| |V(\tilde{h})|+4\pi gr|\tilde{h}||V(\tilde{h})|+\frac{gr}{2}\left| \frac{\partial V(\tilde{h})}{\partial \tilde{h}}\right| \nonumber \\ =\,&H_1+H_2+H_3+H_4+H_5+H_6. \end{aligned}$$

(5.66)

• 1. Estimate for $H_1$

  Using (3.99) we obtain

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

  (5.67)

  where $K_1(x)=C(d+x^3+x^{k+2}+x^p+x^{p+2}+x^{p+4})(1+x^2+x^{k+1} +x^{p+1}+x^{p+3})\exp \left[ C(x^2+x^{k+1}+x^{p+1})\right] .$

• 2. Estimate for $H_2$

  Using (3.95) we obtain

  $$\begin{aligned} \frac{1}{2r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g-\frac{(D-2)}{2}\tilde{g} \right| \left| h\right| \le \frac{K_2(x)}{(1+u)^{2k-3}(1+r+u)^{k+1}}, \end{aligned}$$

  (5.68)

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

• 3. Estimate for $H_3$

  We use (3.83) and (3.27) to obtain

  $$\begin{aligned} \frac{(D-2)}{4r}\left| \frac{\hat{R}(\hat{\sigma })}{(D-2)}g- \frac{(D-2)}{2}\tilde{g}\right| \left| \tilde{h}\right| \le \frac{C(x^3+x^{p+2})}{(1+u)^{\frac{6(k-1)-D}{2}}(1+r+u)^{\frac{D+2}{2}}}. \end{aligned}$$

  (5.69)

• 4. Estimate for $H_4$

  Combination of (1.3) and (3.95) yields

  $$\begin{aligned} \frac{8\pi gr}{(D-2)}\left| h\right| |V(\tilde{h})|\le \frac{K_3(x)}{(1+u)^{2(2k-D)}(1+r+u)^{2(2k-D)+k-2}}, \end{aligned}$$

  (5.70)

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

• 5. Estimate for $H_5$

  Combination of (1.3) and (3.83) yields

  $$\begin{aligned} 4\pi gr|\tilde{h}||V(\tilde{h})|\le \frac{Cx^{p+2}}{(1+u)^{\frac{5(2k-D)}{2}}(1+r+u)^{\frac{5(4k-3D+2)-2}{2}}}. \end{aligned}$$

  (5.71)

• 6. Estimate for $H_6$

  Again, from (1.3) we get

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

  (5.72)

Finally, we obtain

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

(5.73)

where $K_1, K_2,$ and $K_3$ are defined in (3.116)–(3.118) respectively.