2024年2月6日 星期二

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.2

Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.2

    1. If $F(s)$ is a $C^2$-function of the one-variable $s$, find a condition on the vector $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ so that

      $u(x_1,x_2,x_3,t)=F(\alpha_1x_1+\alpha_2x_2+\alpha_3x_3-t)$

      is a solution of (26). (Such solutions are called plane waves and are constant on the planes $\alpha\cdot x-t=\text{constants}$.)
    2. Find the relationship which must hold between the initial data $g(x)$ and $h(x)$ for a plane wave solution.
    3. Find all plane wave solutions of (26) with the initial condition $u(x_1,x_2,x_3,0)=x_1-x_2+1$.
    2. Solution
    1. Plugging the form of solution into (26), we have

      $\begin{aligned}F''(\alpha\cdot x-t)&=u_{tt}(x,t)\\&=c^2\Delta u(x,t)=c^2(\alpha_1^2+\alpha_2^2+\alpha_3^2)F''(\alpha\cdot x-t).\end{aligned}$

      Thus, the vector $\alpha$ must satisfy $|\alpha|=\sqrt{\alpha_1^2+\alpha_2^2+\alpha_3^2}=1/c$.
    2. The initial data $g$ and $h$ must satify

      $\begin{aligned} &g(x)=F(\alpha\cdot x),\\&h(x)=-F'(\alpha\cdot x).\end{aligned}$

      By differentiating $g$ with respect to $x$, we find that $g$ and $h$ satisfies the following relationship:

      $\nabla g(x)=-\nabla_xF(\alpha\cdot x)=-\alpha\cdot F'(\alpha\cdot x)=-\alpha\cdot h(x)\quad\text{for}~x\in\mathbb R^3$,

      which also means $h(x)=-c^2\alpha\cdot\nabla g(x)$ for $x\in\mathbb R^3$.
    3. Let $g(x)=g(x_1,x_2,x_3)=x_1-x_2+1$ for $x=(x_1,x_2,x_3)\in\mathbb R^3$. Then by part (b), we get

      $h(x)=h(x_1,x_2,x_3)=-c^2\alpha\cdot\nabla g(x_1,x_2,x_3)=-c^2(\alpha_1,\alpha_2,\alpha_3)\cdot(1,-1,0)=-c^2(\alpha_1-\alpha_2)\quad\text{for}~(x_1,x_2,x_3)\in\mathbb R^3$,

      which is a constant function. Now we employ the Kirchhoff's formula (37), we have

      $\begin{aligned}u(x,t)&=\frac1{4\pi}\frac{\partial}{\partial t}\left(t\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi\right)+\frac t{4\pi}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac1{4\pi}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}\!c\xi\cdot\nabla g(x+ct\xi)\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac1{4\pi}\!\int_{|\xi|=1}g(x+ct\xi)\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}\!(1-c\alpha\cdot\xi)h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac1{4\pi}\int_{|\xi|=1}\![x_1-x_2+ct(\xi_1-\xi_2)+1]\,\mathrm dS_\xi-\frac{c^2(\alpha_1-\alpha_2)t}{4\pi}\int_{|\xi|=1}\!(1-c\alpha_1\xi_1-c\alpha_2\xi_2-c\alpha_3\xi_3)\,\mathrm dS_\xi\\&=\\&=\frac1{4\pi}\int_{|\xi|=1}\!(x_1-x_2+1)\,\mathrm dS_\xi-\frac{c^2(\alpha_1-\alpha_2)t}{4\pi}\int_{|\xi|=1}\!1\,\mathrm dS_\xi\\&=x_1-x_2+1-c^2(\alpha_1-\alpha_2)t\quad\text{for}~x=(x_1,x_2,x_3)\in\mathbb R^3,\,t\geq0.\end{aligned}$

      Here we have used the symmetry of integral, i.e., $\displaystyle\int_{|\xi|=1}\!\xi_k\,\mathrm dS_\xi=0$ for $k=1,2,3$ and the surface area of unit ball in $\mathbb R^3$ is $4\pi$.

  2. Find the solution of the initial value problem

    $\left\{\begin{aligned} &u_{tt}=u_{xx}+u_{yy}+u_{zz},\\&u(x,y,z,0)=x^2+y^2,\quad u_t(x,y,z,0)=0.\end{aligned}\right.$

    1. by using (37);
    2. by using (39).
    3. Solution
    1. Let $g(x,y,z)=x^2+y^2$ and $h(x,y,z)=0$ for $(x,y,z)\in\mathbb R^3$ and $c=1$. Using the Kirchhoff's formula (37), we have

      $\begin{aligned}u(x,y,z,t)&=\frac1{4\pi}\frac\partial{\partial t}\left(t\int_{|\xi|=1}\!g(x+ct\xi_1,y+ct\xi_2,z+ct\xi_3)\,\mathrm dS_\xi\right)+\frac t{4\pi}\int_{|\xi|=1}\!h(x+ct\xi_1,y+ct\xi_2,z+ct\xi_3)\,\mathrm dS_\xi\\&=\frac1{4\pi}\int_{|\xi|=1}\!g(x+t\xi_1,y+t\xi_2,z+t\xi_3)\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}\!\xi\cdot\nabla g(x+t\xi_1,y+t\xi_2,z+t\xi_3)\,\mathrm dS_\xi\\&=\frac1{4\pi}\int_{|\xi|=1}\![(x+t\xi_1)^2+(y+t\xi_2)^2]\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}(\xi_1,\xi_2,\xi_3)\cdot(2x+2t\xi_1,2y+2t\xi_2,0)\,\mathrm dS_\xi\\&=\frac1{4\pi}\int_{|\xi|=1}\!(x^2+y^2)\,\mathrm dS_\xi+\frac{2t}{4\pi}(x\xi_1+y\xi_2)\,\mathrm dS_\xi\\&\quad+\frac{t^2}{4\pi}\int_{|\xi|=1}(\xi_1^2+\xi_2^2)\,\mathrm dS_\xi+\frac{2t}{4\pi}\int_{|\xi|=1}(x\xi_1+y\xi_2)\,\mathrm dS_\xi+\frac{2t^2}{4\pi}\int_{|\xi|=1}(\xi_1^2+\xi_2^2)\,\mathrm dS_\xi\\&=x^2+y^2+\frac{3t^2}{4\pi}\int_{|\xi|=1}\!(\xi_1^2+\xi_2^2)\,\mathrm dS_\xi=x^2+y^2+\frac{3t^2}{4\pi}\cdot\frac23\int_{|\xi|=1}\!(\xi_1^2+\xi_2^2+\xi_3^2)\,\mathrm dS_\xi\\[2mm]&=x^2+y^2+2t^2\quad\text{for}~(x,y,z)\in\mathbb R^3,\,t\geq0.\end{aligned}$

      Here we have used the symmetry of integral, i.e., $\displaystyle\int_{|\xi|=1}\!\xi_k\,\mathrm dS_\xi=0$ for $k=1,2,3$ and

      $\displaystyle\int_{|\xi|=1}\!\xi_1^2\,\mathrm dS_\xi=\int_{|\xi|=1}\!\xi_2^2\,\mathrm dS_\xi=\int_{|\xi|=1}\!\xi_1^3\,\mathrm dS_\xi=\frac{4\pi}3$.

      In addition, the integral $\displaystyle\int_{|\xi|=1}\!\xi\cdot\nabla g(x+t\xi)\,\mathrm dS_\xi$ also can be computed by the Gauss' divergence theorem as follows.

      $\begin{aligned}\int_{|\xi|=1}\!\xi\cdot\nabla g(x+t\xi_1,y+t\xi_2,z+t\xi_3)\,\mathrm dS_\xi&=\int_{|\xi|\leq1}\Delta_{\xi}[g(x+t\xi_1,y+t\xi_2,z+t\xi_3)]\,\mathrm dS_\xi\\&=t\int_{|\xi|\leq1}(\Delta g)(x+t\xi_1,y+t\xi_2,z+t\xi_3)\,\mathrm dS_\xi\\&=t\int_{|\xi|\leq1}\!4\,\mathrm dS_\xi=t\cdot4\cdot\frac{4\pi}3=\frac{16\pi t}3.\end{aligned}$

    2. Let $g(x,y)=x^2+y^2$ and $h(x,y)=0$ for $(x,y)\in\mathbb R^2$ and $c=1$. Then using (39), we have

      $\begin{aligned}u(x,y,t)&=\frac1{4\pi}\frac\partial{\partial t}\left(2t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)+\frac t{4\pi}\left(2\int_{\xi_1^2+\xi_2^2<1}\!\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&=\frac2{4\pi}\int_{\xi_1^2+\xi_2^2<1}\!\frac{(x+t\xi_1)^2+(y+t\xi_2)^2}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2+\frac{2}{4\pi}\int_{\xi_1^2+\xi_2^2<1}\!\frac{2t\xi_1(x+t\xi_1)+2t\xi_2(y+t\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\\&=\frac2{4\pi}\int_{\xi_1^2+\xi_2^2}\!\frac{(x+t\xi_1)^2+(y+t\xi_2)^2+2t[\xi_1(x+t\xi_1)+\xi_2(y+t\xi_2)]}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2.\end{aligned}$

      Using the polar coordinate, we set $\xi_1=r\cos\theta$ and $\xi_2=r\sin\theta$ for $0\leq r<1$ and $0\leq\theta<2\pi$. Then the double integral can be transformed and we have

      $\begin{aligned}u(x,y,t)&=\frac2{4\pi}\int_0^{2\pi}\!\int_0^1\!\frac{x^2+y^2+4rt(x\cos\theta+y\sin\theta)+3t^2r^2}{\sqrt{1-r^2}}\cdot r\,\mathrm dr\,\mathrm d\theta\\&=\int_0^1\!\frac{r(x^2+y^2)+3t^2r^3}{\sqrt{1-r^2}}\,\mathrm dr=(x^2+y^2)\int_0^1\!\frac r{\sqrt{1-r^2}}\,\mathrm dr+3t^2\int_0^1\frac{r^3}{\sqrt{1-r^2}}\,\mathrm dr\\&=(x^2+y^2)\cdot\left(-\sqrt{1-r^2}\right)\Big|_{r=0}^{r=1}+3t^2\cdot\left.\left(-\frac{(2+r^2)\sqrt{1-r^2}}3\right)\right|_{r=0}^{r=1}\\&=x^2+y^2+3t^2\cdot\frac23=x^2+y^2+2t^2\quad\text{for}~(x,y)\in\mathbb R^2,\,t\geq0.\end{aligned}$


  3. Use Duhamel's principle to find the solution of the nonhomogeneous wave equation for three space dimensions $u_{tt}-c^2\Delta u=f(x,t)$ with initial conditions $u(x,0)=0=u_t(x,0)$. What regularity in $f(x,t)$ is required for the solution $u$ to be $C^2$?
    4. SolutionConsider the following wave equation

    $\left\{\begin{aligned} &U_{tt}(x,t,s)-c^2\Delta U(x,t,s)=0\quad\text{for}~x\in\mathbb R^3,\,t>0,\,s>0\\&U(x,0,s)=0\quad\text{for}~x\in\mathbb R^3,\,s>0\\&U_t(x,0,s)=f(x,s)\quad\text{for}~x\in\mathbb R^3,\,s>0.\end{aligned}\right.$

    By Kirchhoff's formula, we have

    $\displaystyle U(x,t,s)=\frac t{4\pi}\int_{|\xi|=1}\!f(x+ct\xi,s)\,\mathrm dS_\xi\quad\text{for}~x\in\mathbb R^3,\,t\geq0,\,s>0$.

    Then by Duhamel's principle, the solution is

    $\displaystyle u(x,t)=\int_0^t\!U(x,t-s,s)\,\mathrm ds=\frac1{4\pi}\int_0^t\!(t-s)\int_{|\xi|=1}\!f(x+c(t-s)\xi,s)\,\mathrm dS_\xi\,\mathrm ds$

    According to this formula for solution, when $f$ is $C^2$ in $x$ and $C^0$ in $t$, $u$ will be a $C^2$-function.

  4. Let $\Omega=\{(x,y)\in\mathbb R^2\,:\,0<x<a~\text{and}~0<y<b\}$, and use separation of variables to solve the initial/boundary value problem

    $\left\{\begin{aligned} &u_{tt}=u_{xx}+u_{yy}\quad\text{for}~(x,y)\in\Omega~\text{and}~t>0,\\&u(x,y,t)=0\quad\text{for}~(x,y)\in\partial\Omega~\text{and}~t>0,\\&u(x,y,0)=\sin\frac{\pi x}a\sin\frac{2\pi y}b,\quad\text{and}~u_t(x,y,0)=0\quad\text{for}~(x,y)\in\Omega.\end{aligned}\right.$

    5. SolutionUsing the separation of variables, we suppose that the solution $u(x,t)$ is in the form

    $\displaystyle u(x,y,t)=\sum_{m,n=1}^\infty u_{m,n}(x,y,t)=\sum_{m,n=1}^\infty X_m(x)Y_n(y)T_{m,n}(t)\quad\text{for}~(x,y)\in\Omega,\,t>0$.

    Due to the superposition principle, we firstly consider $u(x,y,t)=X(x)Y(y)T(t)$ to fulfill the wave equation with the boundary condition, which gives

    $\begin{aligned} &X(x)Y(y)T''(t)=X''(x)Y(y)T(t)+X(x)Y''(y)T(t),\\&X(0)Y(y)T(t)=X(a)Y(y)T(t)=X(x)Y(0)T(t)=X(x)Y(b)T(t)=0.\end{aligned}$

    Surely, we want to find a general solution so $X$, $Y$ and $T$ are not identically zero and $X(0)=X(a)=Y(0)=Y(b)=0$. From the differential equation, we have

    $\displaystyle\frac{T''(t)}{T(t)}=\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}\quad\text{for}~(x,y)\in\Omega,\,t>0$.

    Clearly, $X''(x)/X(x)$, $Y''(y)/Y(y)$ and $T''(t)/T(t)$ are all constants. By solving the eigenvalue problems for $X''(x)=\lambda X(x)$ in $(0,a)$ with $X(0)=X(a)=0$, $Y''(y)=\mu Y(y)$ in $(0,b)$ with $Y(0)=Y(b)=0$, we obtain

    $\begin{aligned} &X_m(x)=\sin\frac{m\pi x}a\quad\text{for}~x\in(0,a),\\&Y_n(x)=\sin\frac{n\pi y}b\quad\text{for}~y\in(0,b),\end{aligned}$

    where $\lambda_m=-(m\pi/a)^2$ and $\mu_n=-(n\pi/b)^2$ for $m,n\in\mathbb N$. Then $T$ satisfies

    $\displaystyle T''(t)=(\lambda_m+\mu_n)T(t)=-\left(\frac{m^2\pi^2}{a^2}+\frac{n^2\pi^2}{b^2}\right)T(t)$,

    which gives

    $\displaystyle T_{m,n}(t)=A_{m,n}\cos(t\omega_{m,n})+B_{m,n}\sin(t\omega_{m,n})\quad\text{and}\quad\omega_{m,n}=\pi\sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}.$

    Hence the solution has the following form:

    $\displaystyle u(x,t)=\sum_{m,n=1}^\infty[A_{m,n}\cos(t\omega_{m,n})+B_{m,n}\sin(t\omega_{m,n})]\sin\frac{m\pi x}a\sin\frac{n\pi y}b\quad\text{for}~(x,y)\in\Omega,\,t>0$.

    From the intial condition, it is easy to know $A_{1,2}=1$, $A_{m,n}=0$ for $(m,n)\neq(1,2)$, and $B_{m,n}=0$ for all $(m,n)\in\mathbb N^2$. Therefore, the solution is

    $\displaystyle u(x,y,t)=\cos\left(\pi t\sqrt{\frac1{a^2}+\frac4{b^2}}\right)\sin\frac{\pi x}a\sin\frac{2\pi y}b\quad\text{for}~(x,y)\in\bar\Omega,\,t\geq0.$


  5. Find a formula for the solution $v(x,t)=v(x_1,x_2,t)$ of the Cauchy problem for the two-dimensional Klein-Gordon equation:

    $\left\{\begin{aligned} &v_{tt}=c^2\Delta v-m^2v\quad\text{for}~x\in\mathbb R^2~\text{and}~t>0,\\&v(x,0)=g(x),\quad v_t(x,0)=h(x)\quad\text{for}~x\in\mathbb R^2.\end{aligned}\right.$

    6. SolutionDefine $u:\mathbb R^3\times\bar{\mathbb R}_+$ by

    $u(x,y,z,t)=\cos(mz/c)v(x,y,t)\quad\text{for}~(x,y,z,t)\in\mathbb R^3\times\bar{\mathbb R}_+$.

    Then it can be checked that

    $\begin{aligned}u_{tt}(x,y,z,t)=\cos(mz/c)v_{tt}(x,y,t)&=\cos(mz/c)[c^2v_{xx}(x,y,t)+c^2v_{yy}(x,y,t)-m^2v(x,y,t)]\\&=c^2u_{xx}(x,y,z,t)+c^2u_{yy}(x,y,z)+c^2u_{zz}(x,y,z)=c^2\Delta u(x,y,z).\end{aligned}$

    This means $u$ satisfies the three-dimensional wave equation with the initial condition

    $u(x,y,z,0)=\cos(mz/c)g(x,y),\quad u_t(x,y,z,0)=\cos(mz/c)h(x,y)\quad\text{for}~(x,y,z)\in\mathbb R^3$.

    Then by Kirchhoff's formula (37), we have

    $\begin{aligned}u(x,y,z,t)&=\frac1{4\pi}\frac{\partial}{\partial t}\left(t\int_{|\xi|=1}\!\cos(m(z+ct\xi_3)/c)g(x+ct\xi_1,y+ct\xi_2)\,\mathrm dS_\xi\right)\\&\quad+\frac t{4\pi}\int_{|\xi|=1}\!\cos(m(z+ct\xi_3)/c)h(x+ct\xi_1,y+ct\xi_2)\,\mathrm dS_\xi.\end{aligned}$

    The soluion $v$ is

    $\begin{aligned}v(x,y,t)&=u(x,y,0,t)\\&=\frac1{4\pi}\frac\partial{\partial t}\left(t\int_{|\xi|=1}\!\cos(mt\xi_3)g(x+ct\xi_1,y+ct\xi_2)\,\mathrm dS_\xi\right)+\frac t{4\pi}\int_{|\xi|=1}\!\cos(mt\xi_3)h(x+ct\xi_1,y+ct\xi_2)\,\mathrm dS_\xi\\&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_{\xi_1^2+\xi_2^2<1}\frac{\cos\left(mt\sqrt{1-\xi_1^2-\xi_2^2}\right)g(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&\quad+\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{\cos\left(mt\sqrt{1-\xi_1^2-\xi_2^2}\right)h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\\&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_0^{2\pi}\!\int_0^1\!\frac{r\cos\left(mt\sqrt{1-r^2}\right)g(x+ctr\cos\theta,y+ctr\sin\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta\right)\\&\quad+\frac t{2\pi}\int_0^{2\pi}\!\int_0^1\!\frac{r\cos\left(mt\sqrt{1-r^2}\right)h\left(x+ctr\cos\theta,y+ctr\sin\theta\right)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta.\end{aligned}$


    1. For $n=3$, suppose that $g,h\in C_0^\infty(\mathbb R^3)$ and consider the solution of (26), (27) given by (37). Show that there is a constant $C$ so that $|u(x,t)|\leq C/t$ for all $x\in\mathbb R^3$ and $t>0$.
    2. Is a similar result true for $n=2$?
    7. Solution
    1. Since $g$ and $h$ are in $C_0^\infty(\mathbb R^3)$, there exist two positive constants $R_g$ and $R_h$ such that $\text{supp}(g)\subseteq\bar B_{R_g}(0)$ and $\text{supp}(h)\subseteq\bar B_{R_h}(0)$. By the extreme value theorem, the continuous function must attain its maximum value in any given compact set, which means there exist two positive constants $M_g$ and $M_h$ such that $|g(x)|\leq M_g$ and $|h(x)|\leq M_h$ for $x\in\mathbb R^3$. Moreover, the support of $|\nabla g|$ is a subset of one of $g$, and $|\nabla g|$ is also a continuous function, which means there exists a positive constant $M_{\nabla g}$ such that $|\nabla g(x)|\leq M_{\nabla g}$ for $x\in\mathbb R^3$. Then we use (37) and the change of variable $y=x+ct\xi$ (with $\displaystyle\mathrm dS_\xi=(ct)^2\,\mathrm dS_y$) to get

      $\begin{aligned}u(x,t)&=\frac1{4\pi}\int_{|\xi|=1}\!(g(x+ct\xi)+th(x+ct\xi))\,\mathrm dS_\xi+\frac t{4\pi}\int_{|\xi|=1}\!c\xi\cdot\nabla g(x+ct\xi)\,\mathrm dS_\xi\\&=\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!(g(y)+th(y))\,\mathrm dS_y+\frac t{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!\frac{y-x}t\cdot\nabla g(y)\,\mathrm dS_y\\&=\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!(g(y)+th(y))\,\mathrm dS_y+\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!(y-x)\cdot\nabla g(y)\,\mathrm dS_y.\end{aligned}$

      Now we can estimate $|u(x,t)|$ as follows.

      $\begin{aligned}|u(x,t)|&\leq\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!(|g(y)|+t|h(y)|)\,\mathrm dS_y+\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)}\!|y-x||\nabla g(y)|\,\mathrm dS_y\\&=\frac1{4\pi c^2t^2}\int_{\partial B_{ct}(x)\cap B_{R_g}(0)}\!|g(y)|\,\mathrm dS_y+\frac1{4\pi c^2t}\int_{\partial B_{ct}(x)\cap B_{R_h}(0)}\!|h(y)|\,\mathrm dS_y+\frac1{4\pi ct}\int_{\partial B_{ct}(x)\cap B_{R_g}(0)}\!|\nabla g(y)|\,\mathrm dS_y\\&\leq\frac{M_g}{4\pi c^2t^2}\mathcal H^2(\partial B_{ct}(x)\cap B_{R_g}(0))+\frac{M_h}{4\pi c^2t}\mathcal H^2(\partial B_{ct}(x)\cap B_{R_h}(0))+\frac{M_{\nabla g}}{4\pi ct}\mathcal H^2(\partial B_{ct}(x)\cap B_{R_g}(0)),\end{aligned}$

      where $\mathcal H^2$ denotes the two-dimensional Hausdorff measure in $\mathbb R^3$. Here we have used the fact that $|y-x|=ct$ for $y\in\partial B_{ct}(x)$.

      Now we claim that $\mathcal H^2(\partial B_{ct}(x)\cap B_R(0))\leq4\pi R^2$ for $R>0$. When $ct\leq R$, it is easy to know $\partial B_{ct}(x)\cap B_R(0)\subseteq\partial B_{ct}(x)$ and

      $\mathcal H^2(\partial B_{ct}(x)\cap B_R(0))\leq\mathcal H^2(\partial B_{ct}(x))=4\pi(ct)^2\leq4\pi R^2$.

      When $ct>0$, the value of $\mathcal H^2(\partial B_{ct}(x)\cap B_R(0))$ can be attained as $\partial B_{ct}(x)$ passes the equatorial of $B_R(0)$. As the radius of $\partial B_{ct}(x)$ increases, the intersection $\partial B_{ct}(x)\cap B_R(0)$ would become more flat, which means $\mathcal H^2(\partial B_{ct}(x)\cap B_R(0))\geq\pi R^2$, the area of the circle formed by the equatorial of $B_R(0)$. On the other hand, as the radius of $\partial B_{ct}(x)$ decreases to $R$ and $\partial B_{ct}(x)$ passes the equatorial of $B_R(0)$, the intersection $\partial B_{ct}(x)\cap B_R(0)$ may become $\partial B_R(0)$ and $\mathcal H^2(\partial B_{ct}(x)\cap B_R(0))\leq\mathcal H^2(\partial B_R(0))=4\pi R^2$. This completes the proof of Claim.

      If $t\geq1$, we can yse Claim to get

      $\begin{aligned}|u(x,t)|&\leq\frac{M_gR_g^2}{c^2t^2}+\frac{M_hR_h^2}{c^2t}+\frac{M_{\nabla g}R_g^2}{ct}=\frac1{t}\left(\frac{M_{\nabla g}R_g^2}c+\frac{M_hR_h^2}{c^2}+\frac{M_gR_g^2}{c^2t}\right)\\&\leq\frac1t\left(\frac{M_{\nabla g}R_g^2}c+\frac{M_hR_h^2}{c^2}+\frac{M_gR_g^2}{c^2}\right):=\frac{C_1}t,\end{aligned}$

      where $\displaystyle C_1=\frac{M_{\nabla g}R_g^2}c+\frac{M_hR_h^2}{c^2}+\frac{M_gR_g^2}{c^2}$. On the other hand, if $0<t<1$, we have $t^2<t<1$ and

      $\begin{aligned}|u(x,t)|&\leq\frac{M_g}{4\pi c^2t^2}\mathcal H^2(\partial B_{ct}(x))+\frac{M_h}{4\pi c^2t}\mathcal H^2(\partial B_{ct}(x))+\frac{M_{\nabla g}}{4\pi ct}\mathcal H^2(\partial B_{ct}(x))\\&=\frac{M_g}{4\pi c^2t^2}(4\pi c^2t^2)+\frac{M_h}{4\pi c^2t}(4\pi c^2t^2)+\frac{M_{\nabla g}}{4\pi ct}(4\pi c^2t^2)\\&<\frac{M_g}{4\pi c^2t^2}(4\pi c^2t)+\frac{M_h}{4\pi c^2t}(4\pi c^2)+\frac{M_{\nabla g}}{4\pi ct}(4\pi c^2)\\&=\frac{M_g+M_h+cM_{\nabla g}}t:=\frac{C_2}t,\end{aligned}$

      where $C_2=M_g+M_h+cM_{\nabla g}$. Hence we choose $C=\max\{C_1,C_2\}$, which is desired.
    2. No. To find a counterexample, we consider the initial condition $u(x,y,0)\equiv0$ and $u_t(x,y,0)=h(x,y)$, where $h$ satisfies $h\equiv1$ in $B_1(0)$, $0\leq h\leq1$ in $B_2(0)-B_1(0)$ and $h\equiv0$ in $\mathbb R^2-B_2(0)$. The existence of function $h$ follows from Urysohn's lemma. Suppose by contradiction that there exists a positive constant $C$ such that $|u(x,y,t)|\leq C/t$ for $(x,y)\in\mathbb R^2$. By (39), we have

      $\begin{aligned}u(x,y,t)&=\frac1{4\pi}\frac\partial{\partial t}\left(2t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)+\frac t{4\pi}\left(2\int_{\xi_1^2+\xi_2^2<1}\!\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&=\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{h(x+ct\xi_1,y+ct\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2=\frac1{2\pi c^2t}\int_{B_{ct}(x,y)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct)^{-2}((\eta-x)^2+(\mu-y)^2)}}\,\mathrm d\eta\,\mathrm d\mu.\end{aligned}$

      It is clear that $B_1(0)\subseteq B_{ct+1}(ct,0)$ for $t\geq0$. Then we have

      $\begin{aligned}u(ct,0,t+c^{-1})&=\frac1{2\pi c(ct+1)}\int_{B_{ct+1}(ct,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&\geq\frac1{2\pi c(ct+1)}\int_{B_1(0,0)}\!\frac{h(\eta,\mu)}{\sqrt{1-(ct+1)^{-2}((\eta-ct)^2+\mu^2)}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_{B_1(0,0)}\!\frac1{\sqrt{(ct+1)^2-(\eta-ct)^2-\mu^2}}\,\mathrm d\eta\,\mathrm d\mu\\&=\frac1{2\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1-r^2+2ct(1+r\cos\theta)}}\,\mathrm dr\,\mathrm d\theta\\&\geq\frac1{4\pi c}\int_0^{2\pi}\!\int_0^1\!\frac r{\sqrt{1+4ct}}\,\mathrm dr\,\mathrm d\theta=\frac1{4\sqrt{1+4ct}}.\end{aligned}$

      Thus, by $|u(x,y,t)|\leq C/t$ for all $(x,y)\in\mathbb R^2$ and $t>0$, we get

      $\displaystyle\frac1{4\sqrt{1+4ct}}\leq\frac Ct\quad\text{for all}~t>0$,

      which is equivalent to $t^2-64c^3t-16c^2\leq0$ for all $t>0$. This leads a contradiction. Thereofre, it is impossible to find a positive constant $C$ such that $|u(x,y,t)|\leq C/t$ for all $(x,y)\in\mathbb R^2$ and $t>0$.

  7. Suppose $n=2k+1$ for a positive integer $k$ and $u$ satisfies (26).
    1. Show that $V^x(r,t)$ given by (40) satisfies (32).
    2. Derive (42) from (35) and (41); note $c_n=1\cdot3\cdots(n-2)$.
    8. Solution
    1. Firstly, we prove the following identity by mathematical induction:

      $\displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^k(r^{2k}\phi'(r)),$

      where $k\in\mathbb N$ and $\phi\in C^{k+1}(\mathbb R)$. Then we claim that
      Proof of ClaimWhen $k=1$, it is easy to verify that

      $\begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}(r\phi(r))=\frac{\mathrm d}{\mathrm dr}(r\phi'(r)+\phi(r))=r\phi''(r)+2\phi'(r)\\&=\frac{r^2\phi''(r)+2r\phi'(r)}r=\frac1r\frac{\mathrm d}{\mathrm dr}(r^2\phi'(r)).\end{aligned}$

      This shows the identity holds true for $k=1$. Now we suppose the identity holds true for $k=n$, i.e.,

      $\displaystyle\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r)).$

      Then we can observe that

      $\begin{aligned}\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}\phi(r))&=\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}\left[\frac{(2n+1)r^{2n}\phi(r)+r^{2n+1}\phi'(r)}r\right]\\&=(2n+1)\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}\phi(r))+\frac{\mathrm d^2}{\mathrm dr^2}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n-1}(r^{2n-1}(r\phi'(r)))\\&=(2n+1)\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}\phi'(r))+\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n}(r\phi'(r))')\\&=\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n(r^{2n+1}r^{2n}\phi'(r)+r^{2n}(r\phi''(r)+\phi'(r)))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^n\frac1r\frac{\mathrm d}{\mathrm dr}(r^{2n+2}\phi'(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{n+1}(r^{2n+2}\phi'(r)).\end{aligned}$

      Hence by the mathematical induction, the identity holds true for all $n\in\mathbb N$.

      Now we can use (28), (40) and Claim to obtain

      $\begin{aligned}\frac{\partial^2V^x}{\partial t^2}(r,t)&=\frac{\partial^2}{\partial t^2}\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\left(\frac1r\frac{\partial}{\partial r}\right)^{k-1}(r^{2k-1}\frac{\partial^2M_u}{\partial t^2}(x,r,t))\quad\text{(by (28))}\\&=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\left[r^{2k-1}\cdot c^2\left(\frac{\partial^2}{\partial r^2}+\frac{n-1}r\frac\partial{\partial r}\right)M_u(x,r,t)\right]\quad\text{(by (40))}\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\left(r^{n-1}\frac{\partial^2}{\partial r^2}+(n-1)r^{n-2}\frac\partial{\partial r}\right)M_u(x,r,t)\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^{k-1}\cdot\frac1r\frac\partial{\partial r}\left(r^{n-1}\frac\partial{\partial r}\right)M_u\\&=c^2\left(\frac1r\frac\partial{\partial r}\right)^k\left(r^{2k}\frac{\partial M_u}{\partial r}(x,r,t)\right)\\&=c^2\frac{\partial^2}{\partial r^2}\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))\quad\text{(by Claim)}\\&=c^2\frac{\partial^2V^x}{\partial r^2}(x,r,t)\quad\text{(by (28))}.\end{aligned}$

      The proof is complete.
    2. For any $k\in\mathbb N$ and $\phi\in C^{k-1}(\mathbb R)$, we prove the following identity by the mathematical induction:

      $\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r)$,

      where the constants $\beta_j^k$ ($j=0,\dots,k-1$) are independent of $\phi$.
      Proof of Claim 1When $k=1$, it is easy that

      $\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))=r\phi(r)=\beta_0^1\cdot r^{0+1}\phi^{(0)}(r)$,

      where $\beta_0^1=1=c_3$. This shows the claim holds true for $k=1$. Now we suppose the claim holds true for $k=m$, i.e., there exists $\beta_j^m$ ($j=0,1,\dots,m-1$) such that

      $\displaystyle\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}\phi(r))=\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\phi^{(j)}(r)$,

      where $\beta_0^m=c_{2m+1}$. Then we can observe that

      $\begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\frac1r\frac{\mathrm d}{\mathrm dr}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{m-1}(r^{2m-1}(r^2\phi(r)))\\&=\frac1r\frac{\mathrm d}{\mathrm dr}\sum_{j=0}^{m-1}\beta_j^mr^{j+1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\frac1r\sum_{j=0}^{m-1}\beta_j^m\left[(j+1)r^j\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+r^{j+1}\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\right]\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=0}^{m-1}\beta_j^mr^j\frac{\mathrm d^{j+1}}{\mathrm dr^{j+1}}(r^2\phi(r))\\&=\sum_{j=0}^{m-1}(j+1)\beta_j^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\sum_{j=1}^m\beta_{j-1}^mr^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))\\&=\beta_0^mr\phi(r)+\sum_{j=1}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))+\beta_{m-1}^mr^{m-1}\frac{\mathrm d^m}{\mathrm dr^m}(r^2\phi(r)).\end{aligned}$

      • For $j=1$, $\displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=2r\phi(r)+r^2\phi'(r)$;
      • for $j\geq2$, $\displaystyle\frac{\mathrm d^j}{\mathrm dr^j}(r^2\phi(r))=r^2\phi^{(j)}(r)+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r)$.
      Thus, we have

      $\begin{aligned}\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^m(r^{2m+1}\phi(r))&=\beta_0^mr\phi(r)+(2\beta_1^m+\beta_0^m)(2r\phi(r)+r^2\phi'(r))\\&\quad+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j-1}[r^2\phi^{(j)}+2jr\phi^{(j-1)}(r)+j(j-1)\phi^{(j-2)}(r)]\\&\quad+\beta_{m-1}^mr^{m-1}(r^2\phi^{(m)}+2mr\phi^{(m-1)}(r)+m(m-1)\phi^{(m-2)}(r))\\&=(3\beta_0^m+4\beta_1^m)r\phi(r)+(2\beta_1^m+\beta_0^m)r^2\phi'(r)+\sum_{j=2}^{m-1}[(j+1)\beta_j^m+\beta_{j-1}^m]r^{j+1}\phi^{(j)}\\&\quad+\sum_{j=1}^{m-2}2[(j+2)(j+1)\beta_{j+1}^m+(j+1)\beta_j^m]r^{j+1}\phi^{(j)}(r)\\&\quad+\sum_{j=0}^{m-3}[(j+3)(j+2)(j+1)\beta_{j+2}^m+(j+2)(j+1)\beta_{j+1}^m]r^{j+1}\phi^{(j)}(r)\\&\quad+m(m-1)\beta_{m-1}^mr^{m-1}\phi^{(m-2)}(r)+2m\beta_{m-1}^mr^m\phi^{(m-1)}(r)+\beta_{m-1}^mr^{m+1}\phi^{(m)}\\&=\sum_{j=0}^m\beta_j^{m+1}r^{j+1}\phi^{(j)}(r),\end{aligned}$

      where $\beta_j^{m+1}$ are given by

      $\begin{aligned} &\beta_0^{m+1}=3\beta_0^m+6\beta_1^m+6\beta_2^m,\\&\beta_1^{m+1}=\beta_0^m+4\beta_1^m+18\beta_2^m+24\beta_3^m,\\&\beta_j^{m+1}=\beta_{j-1}^m+3(j+1)\beta_j^m+3(j+2)(j+1)\beta_{j+1}^m+(j+3)(j+2)(j+1)\beta_{j+2}^m\quad\text{for}~2\leq j\leq m-3,\\&\beta_{m-2}^{m+1}=\beta_{m-3}^m+2(m-1)\beta_{m-2}^m+3m(m-1)\beta_{m-1}^m,\\&\beta_{m-1}^{m+1}=3m\beta_{m-1}^m+m\beta_{m-1}^m,\\&\beta_m^{m+1}=\beta_{m-1}^m.\end{aligned}$

      Therefore, we complete the proof of Claim 1.

      Then we claim that $\beta_0^k=c_{2k+1}=c_n$ for $k\in\mathbb N$.
      Proof of Claim 2Clearly, the claim holds true for $k=1$. From Claim 1, a direct computation gives

      $\begin{aligned}\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\phi^{(j)}(r)=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-1}(r^{2k-1}\phi(r))\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-2}[(2k-1)r^{2k-3}\phi(r)+r^{2k-2}\phi'(r)]\\&=\left(\frac1r\frac{\mathrm d}{\mathrm dr}\right)^{k-3}[(2k-1)(2k-3)r^{2k-5}\phi(r)+\cdots]\\&=\cdots=(2k-1)(2k-3)(2k-5)\cdots3\cdot r^1\phi(r)+\cdots.\end{aligned}$

      It is clear that $\beta_0^k=(2k-1)(2k-3)(2k-5)\cdots3\cdot1=c_{2k+1}=c_n$. The proof is complete.

      By (40) and Claims 1 and 2, we have

      $\displaystyle V^x(r,t)=\left(\frac1r\frac\partial{\partial r}\right)^{k-1}(r^{2k-1}M_u(x,r,t))=\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t).$

      Then we use (30) and (35) to get

      $\begin{aligned}u(x,t)&=\lim_{r\to0^+}M_u(x,r,t)\quad\text{(by (30))}\\&=\frac1{c_n}\lim_{r\to0^+}(c_nM_u(x,r,t))\\&=\frac1{c_n}\lim_{r\to0^+}\frac1r\sum_{j=0}^{k-1}\beta_j^kr^{j+1}\frac{\partial^jM_u}{\partial r^j}(x,r,t)\\&=\frac1{c_n}\lim_{r\to0^+}\frac{V^x(r,t)}r\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{G^x(r+ct)+G^x(r-ct)}{2r}+\frac1{2cr}\int_{r-ct}^{r+ct}\!H^x(\rho)\,\mathrm d\rho\right]\quad\text{(by (41))}\\&=\frac1{c_n}\lim_{r\to0^+}\left[\frac{(G^x)'(r+ct)+(G^x)'(r-ct)}2+\frac{H^x(r+ct)-H^x(r-ct)}{2c}\right]\quad\text{(by L'Hospital's Rule)}\\&=\frac{(G^x)'(ct)+(G^x)'(-ct)}{2c_n}+\frac{H^x(ct)+H^x(-ct)}{2c_nc}.\end{aligned}$

      From (21) and (41) with replacing the variable $r$ by $t$, we note that

      $\begin{aligned}G^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_g(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}=\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+t\xi)\,\mathrm dS_\xi,\\H^x(t)&=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}(t^{2k-1}M_h(x,t))=\left(\frac1t\frac\partial{\partial t}\right)^{k-1}\left(\frac{t^{2k-1}}{\omega_n}\int_{|\xi|=1}h(x+t\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+t\xi)\,\mathrm dS_\xi.\end{aligned}$

      This implies

      $\begin{aligned}(G^x)'(ct)&=\frac1{\omega_n}\left(\frac\partial{\partial(ct)}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi\right)\\&=\frac1{\omega_n}\frac\partial{\partial t}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!g(x+ct\xi)\,\mathrm dS_\xi,\\H^x(ct)&=\frac1{\omega_n}\left(\frac1{ct}\frac\partial{\partial(ct)}\right)^{\frac{n-3}2}(ct)^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi\\&=\frac c{\omega_n}\left(\frac1t\frac\partial{\partial t}\right)^{\frac{n-3}2}t^{n-2}\int_{|\xi|=1}\!h(x+ct\xi)\,\mathrm dS_\xi.\end{aligned}$

      Also note that $(G^x)'(ct)=(G^x)'(-ct)$ and $H^x(ct)=-H^x(-ct)$. Therefore, we obtain (42) by combining above equations.
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