Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.3

1. Let $\Omega$ be a smooth, bounded domain in $\mathbb R^n$. For a $C^2$ function $u(x,t)$ of the wave equation $u_{tt}=c^2\Delta u$ for $x\in\Omega$ and $t>0$, define the energy to be $\displaystyle\mathcal E_\Omega(t)=\frac12\int_\Omega\!(u_t^2+c^2|\nabla u|^2)\,\mathrm dx$. If $u$ satisfies either the boundary condition $u(x,t)=0$ or $\partial u/\partial\nu(x,t)=0$ for $x\in\partial\Omega$, where $\nu$ is the exterior unit normal, then show that $\mathcal E_\Omega(t)$ is constant.
Solution
Note that if $u=0$ on $\partial\Omega$, we have $u_t=0$ on $\partial\Omega$. Thus, $\displaystyle u_t\frac{\partial u}{\partial\nu}=0$ on $\partial\Omega$ when either $u=0$ or $\displaystyle\frac{\partial u}{\partial\nu}=0$ on $\partial\Omega$. A direct computation gives
$\begin{aligned}\frac{\mathrm d\mathcal E_\Omega}{\mathrm dt}&=\frac{\mathrm d}{\mathrm dt}\left[\frac12\int_\Omega\!(u_t^2+c^2|\nabla u|^2)\,\mathrm dx\right]=\int_\Omega\!\left(u_tu_{tt}+c^2\sum_{i=1}^nu_{x_i}u_{x_it}\right)\,\mathrm dx\\&=\int_\Omega\!\left(u_tu_{tt}-c^2\sum_{i=1}^nu_{x_ix_i}u_t\right)\,\mathrm dx+c^2\int_{\partial\Omega}\!u_t\frac{\partial u}{\partial\nu}\,\mathrm dS_x\\&=\int_\Omega\![u_t(u_{tt}-c^2\Delta u)]\,\mathrm dx+c^2\int_{\partial\Omega}u_t\frac{\partial u}{\partial\nu}\,\mathrm dS_x=0.\end{aligned}$
Therefore, $\mathcal E_\Omega(t)=\mathcal E_\Omega(0)$ for all $t\in[0,\infty)$.


2. Use the previous exercise to show uniqueness of the solution for the (nonhomogeneous) wave equation $u_{tt}=\Delta u+f(x,t)$ in a smooth, bounded domain $\Omega\subset\mathbb R^n$ with either (a) Dirichlet condition $u=g$, or (b) Neumann condition $\partial u/\partial\nu=h$ on $\partial\Omega$.
Solution
1. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\tilde u_{tt}=\Delta\tilde u+f(x,t)\quad\text{for}~x\in\Omega,~t>0,\\&\tilde u(x,t)=g(t)\quad\text{for}~x\in\partial\Omega,~t>0,\\&\tilde u(x,0)=h_1(x),\quad\tilde u_t(x,0)=h_2(x)\quad\text{for}~x\in\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &u_{tt}=\Delta u+f(x,t)\quad\text{for}~x\in\Omega,~t>0,\\&u(x,t)=g(t)\quad\text{for}~x\in\partial\Omega,~t>0,\\&u(x,0)=h_1(x),\quad u_t(x,0)=h_2(x)\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Define $w=\tilde u-u$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &w_{tt}=\Delta w\quad\text{for}~x\in\Omega,~t>0,\\&w(x,t)=0\quad\text{for}~x\in\partial\Omega,~t>0,\\&w(x,0)=0,\quad w_t(x,0)=0\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Consider the energy $\mathcal E_\Omega$ of $w$ defined by

   $\displaystyle\mathcal E_\Omega(t)=\frac12\int_\Omega\!(w_t^2+|\nabla w|^2)\,\mathrm dx\quad\text{for}~t\geq0$.

   Then by Exercise 1, $\mathcal E_\Omega(t)=\mathcal E_\Omega(0)$ for $t\geq0$. By the initial condition, $\mathcal E_\Omega(0)=0$, which means $w_t\equiv0$ and $\nabla w\equiv0$ in $\Omega\times(0,\infty)$, i.e., $w\equiv0$ in $\Omega\times[0,\infty)$. Therefore, we obtain $\tilde u\equiv u$ in $\Omega\times[0,\infty)$.
2. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\tilde u_{tt}=\Delta\tilde u+f(x,t)\quad\text{for}~x\in\Omega,~t>0,\\&\frac{\partial\tilde u}{\partial\nu}(x,t)=h(t)\quad\text{for}~x\in\partial\Omega,~t>0,\\&\tilde u(x,0)=h_1(x),\quad\tilde u_t(x,0)=h_2(x)\quad\text{for}~x\in\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &u_{tt}=\Delta u+f(x,t)\quad\text{for}~x\in\Omega,~t>0,\\&\frac{\partial u}{\partial\nu}(x,t)=h(t)\quad\text{for}~x\in\partial\Omega,~t>0,\\&u(x,0)=h_1(x),\quad u_t(x,0)=h_2(x)\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Define $w=\tilde u-u$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &w_{tt}=\Delta w\quad\text{for}~x\in\Omega,~t>0,\\&\frac{\partial w}{\partial\nu}(x,t)=0\quad\text{for}~x\in\partial\Omega,~t>0,\\&w(x,0)=0,\quad\tilde w_t(x,0)=0\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Consider the energy $\mathcal E_\Omega$ of $w$ defined by

   $\displaystyle\mathcal E_\Omega(t)=\frac12\int_\Omega\!(w_t^2+|\nabla w|^2)\,\mathrm dx\quad\text{for}~t\geq0.$

   Then by Exercise 1, $\mathcal E_\Omega(t)=\mathcal E_\Omega(0)$ for $t\geq0$. By the initial condition, $\mathcal E_\Omega(0)=0$, which means $w_t\equiv0$ and $\nabla w\equiv0$ in $\Omega\times(0,\infty)$, i.e., $w\equiv0$ in $\Omega\times[0,\infty)$. Therefore, we obtain $\tilde u\equiv u$ in $\Omega\times[0,\infty)$.


3. Use (48) to prove (50).
Solution

For any $a,b\in\mathbb R$ and $\varepsilon>0$, it is clear that $\displaystyle\left(\frac1{\sqrt\epsilon}a-\sqrt\varepsilon b\right)^2\geq0$, which implies

$\displaystyle2ab\leq\frac1\varepsilon a^2+\varepsilon b^2$.　(Young's inequality with $\varepsilon$ or Peter-Paul inequality)

Then for $a=u_t$, $b=\nabla\cdot\nu=u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n$ and $\varepsilon=c\sqrt{1+c^2}$, we have

$\begin{aligned}2u_t(u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n)&\leq\frac 1{c\sqrt{1+c^2}}u_t^2+c\sqrt{1+c^2}(\nabla u\cdot\nu)^2\\&\leq\frac1{c\sqrt{1+c^2}}u_t^2+c\sqrt{1+c^2}|\nabla u|^2|\nu|^2\\&=\frac1{c\sqrt{1+c^2}}u_t^2+\frac c{\sqrt{1+c^2}}|\nabla u|^2\quad\text{(by (48))}.\end{aligned}$



4. The partial differential equaiton $u_{tt}=c^2\Delta u-q(x)u$ arises in the study of wave propagation in a nonhomogeneous elastic medium: $q(x)$ is nonnegative and proportional to the coefficient of elasticity at $x$.
   1. Define an appropriate notion of energy for solutions.
   2. Verify the corresponding energy inequality.
   3. Use the energy method to prove that solutions are uniquely determined by their Cauchy data.
Solution
1. Mutiplying the equation by $u_t$ and integrating it $\Omega$, it yields the energy of solutions:

   $\displaystyle\mathcal E_\Omega(t)=\frac12\int_\Omega\!(u_t^2+c^2|\nabla u|^2+q(x)u^2)\,\mathrm dx\quad\text{for}~t\geq0$.

   This is an appropriate notion of energy of solutions because $\displaystyle\frac{\mathrm d\mathcal E_\Omega}{\mathrm dt}=0$, i.e., $\mathcal E_\Omega$ is a constant on $[0,\infty)$.
2. Let $(x_0,t_0)\in\mathbb R^n\times(0,\infty)$. For any time $\tau\in[0,t_0]$, let $\bar B_\tau=\{x\in\mathbb R^n\,:\,|x-x_0|\leq c(t_0-\tau)\}$ (see Figure 1 at page 92 in the textbook). Consider the local energy function

   $\displaystyle\mathcal E_{x_0,t_0}(\tau)=\frac12\int_{B_\tau}(u_t^2+c^2|\nabla u|^2+q(x)u^2)\Big|_{t=\tau}\,\mathrm dx\quad\text{for}~0\leq \tau\leq t_0$.

   From Figure 1, we employ the following sets

   $\begin{aligned} &\Omega_\tau=\{(x,t):\,|x-x_0|<c(t_0-t),\,0<t<\tau\},\\&C_\tau=\{(x,t):\,|x-x_0|=c(t_0-t),\,0<t<\tau\}.\end{aligned}$

   It is clear that $\partial\Omega_\tau=C_\tau\cup(\bar B_0\times\{0\})\cup(\bar B_\tau\times\{\tau\})$, where the unions are disjoint. Moreover, the exterior unit normal $\nu$ on $\partial\Omega_\tau$ is given on $\bar B_\tau\times\{\tau\}$ by $\nu=\langle0,\dots,0,1\rangle$, and on $\bar B_0\times\{0\}$ by $\nu=\langle0,\dots,0,-1\rangle$. On $C_\tau$, the normal $\nu=\langle\nu_1,\dots,\nu_n,\nu_{n+1}\rangle$ satisfies (48), i.e.,

   $\displaystyle\nu_1^2+\cdots+\nu_n^2=\frac{\nu_{n+1}^2}{c^2}=\frac1{1+c^2}$.

   Now we define the vector field

   $\vec V=\langle2c^2u_tu_{x_1},\dots,2c^2u_tu_{x_n},-(u_t^2+c^2|\nabla u|^2+q(x)u^2)\rangle$.

   Then we compute the divergence of $\vec V$ in variable $(x,t)$ to get

   $\begin{aligned}\text{div}(\vec V)&=2c^2\sum_{i=1}^n(u_{tx_i}u_{x_i}+u_tu_{x_ix_i})-\frac\partial{\partial t}(u_t^2+c^2|\nabla u|^2+q(x)u^2)\\&=2c^2\sum_{i=1}^nu_{tx_i}u_{x_i}+2c^2u_t\Delta u-2u_tu_{tt}-2c^2\sum_{i=1}^nu_{x_i}u_{x_it}+q(x)\cdot2uu_t\\&=2u_t[c^2\Delta u-u_{tt}+q(x)u]=0.\end{aligned}$

   Hence by the Gauss' divergence theorem, we have

   $\displaystyle\int_{\partial\Omega_\tau}\!\vec V\cdot\nu\,\mathrm dS=\int_{\Omega_\tau}\!\text{div}(\vec V(x,t))\,\mathrm d(x,t)=0$.

   On $C_\tau$, we can use (50) to derive

   $\begin{aligned}2u_t(u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n)&\leq\frac c{\sqrt{1+c^2}}|\nabla u|^2+\frac1{c\sqrt{1+c^2}}u_t^2\\&\leq\frac c{\sqrt{1+c^2}}|\nabla u|^2+\frac1{c\sqrt{1+c^2}}(u_t^2+q(x)u^2).\end{aligned}$

   Hence the flux of $\vec V$ through $C_\tau$ is $\displaystyle\int_{C_\tau}\!\vec V\cdot\nu\,\mathrm dS\leq0$ because

   $\vec V\cdot\nu=2c^2u_t(u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n)-(u_t^2+c^2|\nabla u|^2+q(x)u^2)\nu_{n+1}\leq0\quad\text{on}~C_\tau$.

   Therefore, we have

   $\begin{aligned}\mathcal E_{x_0,t_0}(0)-\mathcal E_{x_0,t_0}(\tau)&=\int_{B_0}\!(u_t^2+c^2|\nabla u|^2+q(x)u^2)\Big|_{t=0}\,\mathrm dx-\int_{B_0}\!(u_t^2+c^2|\nabla u|^2+q(x)u^2)\Big|_{t=\tau}\,\mathrm dx\\&=\int_{B_0\times\{0\}}\!\vec V\cdot\nu\,\mathrm dS+\int_{B_\tau\times\{\tau\}}\!\vec V\cdot\nu\,\mathrm dS=-\int_{C_\tau}\!\vec V\cdot\nu\,\mathrm dS\geq0,\end{aligned}$

   which gives the energy inequality.
3. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\tilde u_{tt}=c^2\Delta\tilde u-q(x)\tilde u\quad\text{for}~x\in\Omega,\,t>0,\\&\tilde u(x,t)=f(t)\quad\text{for}~x\in\partial\Omega,~t>0\\&\tilde u(x,0)=g(x),\quad\tilde u_t(x,0)=h(x)\quad\text{for}~x\in\Omega;\end{aligned}\right.\quad\left\{\begin{aligned} &u_{tt}=c^2\Delta u-q(x)u\quad\text{for}~x\in\Omega,~t>0,\\&u(x,t)=f(t)\quad\text{for}~x\in\partial\Omega,~t>0\\&u(x,0)=g(x),\quad u_t(x,0)=h(x)\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Define $w=\tilde u-u$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &w_{tt}=c^2\Delta w-q(x)w\quad\text{for}~x\in\Omega,~t>0,\\&w(x,t)=0\quad\text{for}~x\in\partial\Omega,~t>0,\\&w(x,0)=0,\quad w_t(x,0)=0\quad\text{for}~x\in\Omega.\end{aligned}\right.$

   Then the energy $\mathcal E_\Omega$ of $w$ is identically zero because $\mathcal E_\Omega(t)=\mathcal E_\Omega(0)=0$ for all $t\geq0$. Since $q(x)\geq0$, we get $w\equiv0$, i.e., $\tilde u=u$. The proof is complete.
   Warning: The question does not mention the boundary condition so for the sake of simplicity, we impose the Dirichlet boundary condition. In addition, we also may consider the problem in whole domain $\mathbb R^n$ with initial data which has compact supports.


5. Consider the $n$-dimensional wave equation with dissipation

   $\left\{\begin{aligned} &u_{tt}-c^2\Delta u+\alpha u_t=0\quad\text{for}~(x,t)\in\mathbb R^n\times\mathbb R_+,\\&u(x,0)=g(x),\quad u_t(x,0)=h(x)\quad\text{for}~x\in\mathbb R^n,\end{aligned}\right.$

   where $g$ and $h$ have compact support and $\alpha\geq0$ is a constant. Define the energy $\mathcal E(t)$ by (43).
   1. Prove a domain of dependence result to conclude that solutions have a finite propagation speed.
   2. Show that $\mathcal E(t)$ is nonincreasing in $t>0$.
   3. Use the energy method to prove that solutions are uniquely determined by their Cauchy data.
Solution
1. Let $(x_0,t_0)\in\mathbb R^n\times(0,\infty)$. For any time $\tau\in[0,t_0]$, let $\bar B_\tau=\{x\in\mathbb R^n:|x-x_0|\leq c(t_0-\tau)\}$ (see Figure 1 at page 92 in the textbook). Consider the local energy function

   $\displaystyle\mathcal E_{x_0,t_0}(\tau)=\frac12\int_{B_\tau}\!(u_t^2+c^2|\nabla u|^2)\Big|_{t=\tau}\,\mathrm dx\quad\text{for}~0\leq \tau\leq t_0$

   (see (46)). From Figure 1, we employ the following sets

   $\begin{aligned} &\Omega_\tau=\{(x,t):\,|x-x_0|<c(t_0-t),\,0<t<\tau\},\\&C_\tau=\{(x,t):\,|x-x_0|=c(t_0-t),\,0<t<\tau\}.\end{aligned}$

   It is clear that $\partial\Omega_\tau=C_\tau\cup(\bar B_0\times\{0\})\cup(\bar B_\tau\times\{\tau\})$, where the unions are disjoint. Moreover, the exterior unit normal $\nu$ on $\partial\Omega_\tau$ is given on $\bar B_\tau\times\{\tau\}$ by $\nu=\langle0,\dots,0,1\rangle$, and on $\bar B_0\times\{0\}$ by $\nu=\langle0,\dots,0,-1\rangle$. On $C_\tau$, the normal $\nu=\langle\nu_1,\dots,\nu_n,\nu_{n+1}\rangle$ satisfies (48), i.e.,

   $\displaystyle\nu_1^2+\cdots+\nu_n^2=\frac{\nu_{n+1}^2}{c^2}=\frac1{1+c^2}$.

   Now we define the vector field

   $\vec V=\langle2c^2u_tu_{x_1},\dots,2c^2u_tu_{x_n},-(u_t^2+c^2|\nabla u|^2)\rangle.$

   Then we compute the divergence of $\vec V$ in variable $(x,t)$ to get

   $\begin{aligned}\text{div}(\vec V)&=2c^2\sum_{i=1}^n(u_{tx_i}u_{x_i}+u_tu_{x_ix_i})-\frac\partial{\partial t}(u_t^2+c^2|\nabla u|^2)\\&=2c^2\sum_{i=1}^nu_{tx_i}u_{x_i}+2c^2u_t\Delta u-2u_tu_{tt}-2c^2\sum_{i=1}^nu_{x_i}u_{x_it}\\&=-2u_t[u_{tt}-c^2\Delta u]=2\alpha u_t^2\geq0.\end{aligned}$

   Here we have used the fact that $\alpha\geq0$. Hnece by the Gauss' divergence theorem, we have

   $\displaystyle\int_{\partial\Omega_\tau}\!\vec V\cdot\nu\,\mathrm dS=\int_{\Omega_\tau}\!\text{div}(\vec V(x,t))\,\mathrm d(x,t)\geq0$.

   On $C_\tau$, we recall (50), i.e.,

   $\displaystyle 2u_t(u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n)\leq\frac c{\sqrt{1+c^2}}|\nabla u|^2+\frac1{c\sqrt{1+c^2}}u_t^2$.

   Hence the flux of $\vec V$ through $C_\tau$ is $\displaystyle\int_{C_\tau}\!\vec V\cdot\nu\,\mathrm dS\leq0$ because

   $\vec V\cdot\nu=2c^2u_t(u_{x_1}\nu_1+\cdots+u_{x_n}\nu_n)-(u_t^2+c^2|\nabla u|^2)\nu_{n+1}\leq0$.

   Therefore, we have

   $\begin{aligned}\mathcal E_{x_0,t_0}(0)-\mathcal E_{x_0,t_0}(\tau)&=\int_{B_0}\!(u_t^2+c^2|\nabla u|^2+q(x)u^2)\Big|_{t=0}\,\mathrm dx-\int_{B_0}\!(u_t^2+c^2|\nabla u|^2+q(x)u^2)\Big|_{t=\tau}\,\mathrm dx\\&=\int_{B_0\times\{0\}}\!\vec V\cdot\nu\,\mathrm dS+\int_{B_\tau\times\{\tau\}}\!\vec V\cdot\nu\,\mathrm dS=-\int_{C_\tau}\!\vec V\cdot\nu\,\mathrm dS\geq0,\end{aligned}$

   which gives the energy inequality. When $g$ and $h$ are identically zero in $\bar B_0$, we have $\mathcal E_{x_0,t_0}(\tau)=0$ for $0\leq\tau\leq t_0$, which implies $u_t\equiv0$ and $\nabla u$ in $\Omega_{t_0}$, i.e., $u$ is a constant in $\Omega_{t_0}$. Since $u(x,t)=0$ in $\bar B_0$, we get $u\equiv0$ in $\Omega_{t_0}$, and hence $u(x_0,t_0)=0$. This shows the domain of dependene. Moreover, this implies the finite propagation speed because the domain of dependence indicates that the impact of the initial condition cannot exceed the speed $c$.
2. Due to (a), the energy $\mathcal E$ is well-defined. Then a direct computation gives

   $\begin{aligned}\frac{\mathrm d\mathcal E}{\mathrm dt}&=\frac{\mathrm d}{\mathrm dt}\left[\frac12\int_{\mathbb R^n}\!(u_t^2+c^2|\nabla u|^2)\,\mathrm dx\right]=\int_{\mathbb R^n}\!\left(u_tu_{tt}+c^2\sum_{i=1}^nu_{x_i}u_{x_it}\right)\,\mathrm dx\\&=\int_{\mathbb R^n}\!\left(u_tu_{tt}-c^2\sum_{i=1}^nu_{x_ix_i}u_t\right)\,\mathrm dx\quad\text{(integration by parts and using finite propagation)}\\&=\int_{\mathbb R^n}\!u_t[u_{tt}-c^2\Delta x]\,\mathrm dx=-\alpha\int_{\mathbb R^n}\!u_t^2\,\mathrm dx\leq0.\end{aligned}$

   Thus, energy $\mathcal E$ is nonincreasing in $t>0$.
3. Let $\tilde u$ and $u$ be solutions of

   $\left\{\begin{aligned} &\tilde u_{tt}-c^2\Delta\tilde u+\alpha\tilde u_t=0\quad\text{for}~(x,t)\in\mathbb R^n\times\mathbb R_+,\\&\tilde u(x,0)=g(x),\quad\tilde u_t(x,0)=h(x)\quad\text{for}~x\in\mathbb R^n;\end{aligned}\right.\quad\left\{\begin{aligned} &u_{tt}-c^2\Delta u+\alpha u_t=0\quad\text{for}~(x,t)\in\mathbb R^n\times\mathbb R_+,\\&u(x,0)=g(x),\quad u_t(x,0)=h(x)\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$

   Define $w=\tilde u-u$. Clearly, $w$ satisfies

   $\left\{\begin{aligned} &w_{tt}-c^2\Delta w+\alpha w_t=0\quad\text{for}~(x,t)\in\mathbb R^n\times\mathbb R_+,\\&w(x,0)=0,\quad w_t(x,0)=0\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$

   Consider the energy $\mathcal E$ of $w$ defined by

   $\displaystyle\mathcal E=\frac12\int_{\Omega}\!(w_t^2+c^2|\nabla w|^2)\,\mathrm dx\quad\text{for}~t\geq0$.

   By (b), we have $\mathcal E(t)\leq\mathcal E(0)=0$, which means $w_t\equiv0$ and $\nabla w\equiv0$ in $\mathbb R^n\times\mathbb R_+$, i.e., $w\equiv0$ in $\mathbb R^n\times\mathbb R_+$, and hence $\tilde u\equiv u$ in $\mathbb R^n\times\mathbb R_+$. The proof is complete.


6. Consider a flexible beam with clamped ends at $x=0$ and $x=1$. Small wave motion in the beam satisfies

   $\left\{\begin{aligned} &u_{tt}+\gamma^2u_{xxxx}=0\quad\text{for}~0<x<1,\,t>0,\\&u(0,t)=0=u(1,t)\quad\text{for}~t>0,\\&u_x(0,t)=0=u_x(1,t)\quad\text{for}~t>0,\end{aligned}\right.$

   where $\gamma^2$ is a constant depending on the shape and the material of the beam. Show that the energy $\displaystyle\mathcal E=\frac12\int_0^1\!(u_t^2+\gamma^2u_{xx}^2)\,\mathrm dx$ is conserved.
Solution
From the boundary conditions, we differentiate them with respect to $t$ to note that
$u_t(0,t)=0=u_t(1,t),\quad u_{xt}(0,t)=0=u_{xt}(1,t)\quad\text{for}~t>0.$
Using the integration by parts, we find
$\begin{aligned}\frac{\mathrm d\mathcal E}{\mathrm dt}&=\frac{\mathrm d}{\mathrm dt}\left[\frac12\int_0^1\!(u_t^2+\gamma^2u_{xx}^2)\,\mathrm dx\right]=\int_0^1\!(u_tu_{tt}+\gamma^2u_{xx}u_{xxt})\,\mathrm dx\\&=\int_0^1(\!u_tu_{tt}-\gamma^2u_{xxx}u_{xt})\,\mathrm dx+\gamma^2(u_{xx}(1,t)u_{xt}(1,t)-u_{xx}(0,t)u_{xt}(0,t))\\&=\int_0^1\!(u_tu_{tt}+\gamma^2u_{xxxx}u_t)\,\mathrm dx-\gamma^2(u_{xxx}(1,t)u_t(1,t)-u_{xxx}(0,t)u_t(0,t))\\&=\int_0^1\!u_t[u_{tt}+\gamma^2u_{xxxx}]\,\mathrm dx=0.\end{aligned}$
This means energy $\mathcal E$ is conserved.