An Integral Formula Adapted to Different Boundary Conditions for Arbitrarily High-Dimensional Nonlinear Klein–Gordon Equations

Abstract

This chapter is concerned with the initial-boundary value problem for arbitrarily high-dimensional Klein–Gordon equations, posed on a bounded domain $\varOmega \subset \mathbb {R}^d$ for $d \ge 1$ and subject to suitable boundary conditions. We derive and analyse an integral formula which proves to be adapted to different boundary conditions for general Klein–Gordon equations in arbitrarily high-dimensional spaces.

Appendix 1. A Direct Proof of Theorem 9.6

Proof

It follows from (9.37) that

$$\begin{aligned} \begin{aligned}&\Big (\dfrac{\partial u}{\partial x}\Big )_{tt}=a^2\varDelta \Big (\dfrac{\partial u}{\partial x}\Big )+\tilde{f}(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{\prime \prime }(t_0)=&{}a^2\varDelta \varphi '_1(x_l)+\tilde{f}\big (u\big ),\\ \delta ^{\prime \prime }(t_0)=&{}a^2\varDelta \varphi '_1(x_r)+\tilde{f}\big (u\big ), \end{array}\right. \\&\Big (\dfrac{\partial u}{\partial x}\Big )^{(3)}_{t}=a^2\varDelta \Big (\dfrac{\partial u}{\partial x}\Big )_t+\tilde{f}'_t(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{(3)}(t_0)=&{}a^2\varDelta \varphi '_2(x_l)+\tilde{f}'_t\big (u\big ),\\ \delta ^{(3)}(t_0)=&{}a^2\varDelta \varphi '_2(x_r)+\tilde{f}'_t\big (u\big ), \end{array}\right. \\&\Big (\dfrac{\partial u}{\partial x}\Big )^{(4)}_{t}=a^4\varDelta ^2 \Big (\dfrac{\partial u}{\partial x}\Big )+a^2\varDelta \tilde{f}(u)+\tilde{f}^{(2)}_{t}(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{(4)}(t_0)=&{}a^4\varDelta ^2 \varphi '_1(x_l)+a^2\varDelta \tilde{f}\big (u\big )+\tilde{f}^{(2)}_{t}\big (u\big ),\\ \delta ^{(4)}(t_0)=&{}a^4\varDelta ^2 \varphi '_1(x_r)+a^2\varDelta \tilde{f}\big (u\big )+\tilde{f}^{(2)}_{t}\big (u\big ),\\ \end{array}\right. \\&\Big (\dfrac{\partial u}{\partial x}\Big )^{(5)}_{t}=a^4\varDelta ^2 \Big (\dfrac{\partial u}{\partial x}\Big )_t+a^2\varDelta \tilde{f}'_t(u)+\tilde{f}^{(3)}_{t}(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{(5)}(t_0)=&{}a^4\varDelta ^2 \varphi '_2(x_l)+a^2\varDelta \tilde{f}'_t\big (u\big )+\tilde{f}^{(3)}_{t}\big (u\big ),\\ \delta ^{(5)}(t_0)=&{}a^4\varDelta ^2 \varphi '_2(x_r)+a^2\varDelta \tilde{f}'_t\big (u\big )+\tilde{f}^{(3)}_{t}\big (u\big ),\\ \end{array}\right. \\&\Big (\dfrac{\partial u}{\partial x}\Big )^{(6)}_{t}=a^6\varDelta ^3 \Big (\dfrac{\partial u}{\partial x}\Big )+a^4\varDelta ^2 \tilde{f}(u)+a^2\varDelta \tilde{f}^{(2)}_t(u)+\tilde{f}^{(4)}_{t}(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{(6)}(t_0)=&{}a^6\varDelta ^3 \varphi '_1(x_l)+a^4\varDelta ^2 \tilde{f}\big (u\big )\\ &{}+a^2\varDelta \tilde{f}^{(2)}_{t}\big (u\big )+\tilde{f}^{(4)}_{t}\big (u\big ),\\ \delta ^{(6)}(t_0)=&{}a^6\varDelta ^3 \varphi '_1(x_r)+a^4\varDelta ^2\tilde{f}\big (u\big )\\ &{}+a^2\varDelta \tilde{f}^{(2)}_{t}\big (u\big )+\tilde{f}^{(4)}_{t}\big (u\big ),\\ \end{array}\right. \\&\Big (\dfrac{\partial u}{\partial x}\Big )^{(7)}_{t}=a^6\varDelta ^3 \Big (\dfrac{\partial u}{\partial x}\Big )_t+a^4\varDelta ^2\tilde{f}'_t(u)+a^2\varDelta \tilde{f}^{(3)}_t(u)+\tilde{f}^{(5)}_{t}(u)\\&\qquad \qquad \Rightarrow \left\{ \begin{array}{ll} \gamma ^{(7)}(t_0)=&{}a^6\varDelta ^3 \varphi '_2(x_l)+a^4\varDelta ^2\tilde{f}'_t\big (u\big ) \\ &{}+a^2\varDelta \tilde{f}^{(3)}_t\big (u\big )+\tilde{f}^{(5)}_{t}\big (u\big ),\\ \delta ^{(7)}(t_0)=&{}a^6\varDelta ^3 \varphi '_2(x_r)+a^4\varDelta ^2\tilde{f}'_t\big (u\big )\\ &{}+a^2\varDelta \tilde{f}^{(3)}_t\big (u\big )+\tilde{f}^{(5)}_{t}\big (u\big ),\\ \end{array}\right. \\&\cdots . \end{aligned} \end{aligned}$$

After an argument by induction we obtain the following results

$$\begin{aligned} \begin{aligned}&\gamma ^{(2k)}=a^{2k}\varDelta ^k\varphi '_1(x_l)+\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\gamma ^{(2k+1)}=a^{2k}\varDelta ^k\varphi '_2(x_l)+\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big ),\quad k=1,2,\dots . \end{aligned} \end{aligned}$$

(9.61)

and

$$\begin{aligned} \begin{aligned}&\delta ^{(2k)}(t_0)=a^{2k}\varDelta ^k\varphi '_1(x_r)+\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\delta ^{(2k+1)}(t_0)=a^{2k}\varDelta ^k\varphi '_2(x_r)+\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big ),\quad k=1,2,\dots . \end{aligned} \end{aligned}$$

(9.62)

Inserting the results of (9.61) and (9.62) into the Taylor expansion of $\gamma (t)$ and $\delta (t)$ at point $t_0$ yields

$$\begin{aligned} \begin{aligned} \gamma (t)&=\Big \{\phi _0\big ((t-t_0)^2a^2\varDelta \big )\varphi '_1(x)+(t-t_0)\phi _1\big ((t-t_0)^2a^2\varDelta \big )\varphi '_2(x)\\&\quad +\sum \limits _{k=1}^{\infty }\Big [\dfrac{(t-t_0)^{2k}}{(2k)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\quad +\dfrac{(t-t_0)^{2k+1}}{(2k+1)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big )\Big ]\Big \}\Big |_{x=x_l}, \end{aligned} \end{aligned}$$

(9.63)

and

$$\begin{aligned} \begin{aligned} \delta (t)&=\Big \{\phi _0\big ((t-t_0)^2a^2\varDelta \big )\varphi '_1(x)+(t-t_0)\phi _1\big ((t-t_0)^2a^2\varDelta \big )\varphi '_2(x)\\&\quad +\sum \limits _{k=1}^{\infty }\Big [\dfrac{(t-t_0)^{2k}}{(2k)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\quad +\dfrac{(t-t_0)^{2k+1}}{(2k+1)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big )\Big ]\Big \}\Big |_{x=x_r}. \end{aligned} \end{aligned}$$

(9.64)

Let

$$\tilde{F}(x,t)\triangleq \int _{t_0}^t(t-\zeta )\phi _1\big ((t-t_0)^2a^2\varDelta \big )\hat{f}\big (\zeta \big )\mathrm{d}\zeta .$$

As deduced for the Dirichlet boundary conditions in Sect. 9.4.1, it can be shown that

$$\begin{aligned} \begin{aligned}&\tilde{F}^{(2k)}_t=\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\tilde{F}^{(2k+1)}_t=\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big ),\quad k=1,2,\dots . \end{aligned} \end{aligned}$$

(9.65)

Inserting (9.65) into the Taylor expansion of $\tilde{F}(x,t)$ at the point $t=t_0$ gives

$$\begin{aligned} \begin{aligned} \tilde{F}(x,t)&=\sum \limits _{k=0}^{\infty }\dfrac{(t-t_0)^k}{k!}\tilde{F}^{(k)}_t=\sum \limits _{k=2}^{\infty }\dfrac{(t-t_0)^k}{k!}\tilde{F}^{(k)}_t\\&=\sum \limits _{k=1}^{\infty }\dfrac{(t-t_0)^{2k}}{(2k)!}\tilde{F}^{(2k)}_t+\sum \limits _{k=1}^{\infty }\dfrac{(t-t_0)^{2k+1}}{(2k+1)!}\tilde{F}^{(2k+1)}_t\\&=\sum \limits _{k=1}^{\infty }\Big [\dfrac{(t-t_0)^{2k}}{(2k)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-2)}_t\big (u\big )\\&\quad +\dfrac{(t-t_0)^{2k+1}}{(2k+1)!}\sum \limits ^{k}_{j=1}a^{2(k-j)}\varDelta ^{k-j}\tilde{f}^{(2j-1)}_t\big (u\big )\Big ]. \end{aligned} \end{aligned}$$

(9.66)

Comparing the results of (9.66) with (9.63) and (9.64) yields

$$\begin{aligned} \begin{aligned}&\gamma (t)=\Big [\phi _0\big ((t-t_0)^2a^2\varDelta \big )\varphi '_1(x)+(t-t_0)\phi _1\big ((t-t_0)^2a^2\varDelta \big )\varphi '_2(x) \\&+\int _{t_0}^t(t-\zeta )\phi _1\big ((t-\zeta )^2a^2\varDelta \big )\hat{f}\big (\zeta \big )\mathrm{d}\zeta \Big ]\Big |_{x=x_l},\\&\delta (t)=\Big [\phi _0\big ((t-t_0)^2a^2\varDelta \big )\varphi '_1(x)+(t-t_0)\phi _1\big ((t-t_0)^2a^2\varDelta \big )\varphi '_2(x) \\&+\int _{t_0}^t(t-\zeta )\phi _1\big ((t-\zeta )^2a^2\varDelta \big )\hat{f}\big (\zeta \big )\mathrm{d}\zeta \Big ]\Big |_{x=x_r}.\\ \end{aligned} \end{aligned}$$

This finishes the direct proof.    $\square$