Let $x\in[0,L]$ and consider the following equation, $$\varepsilon \left( t \right)=\frac{1}{2}\int_{0}^{L}{({{\rho }_{1}}{{\left| {{\varphi }_{t}} \right|}^{2}}+{{\rho }_{2}}{{\left| {{\varphi }_{t}} \right|}^{2}}+{{\rho }_{1}}{{\left| {{\omega }_{t}} \right|}^{2}}}+b{{\left| {{\psi }_{x}} \right|}^{2}}+k{{\left| {{\varphi }_{x}}+\psi +lw \right|}^{2}}+{{k}_{0}}{{\left| {{\omega }_{x}}-l\varphi \right|}^{2}}+\theta _{1}^{2}+\theta _{2}^{2})dx$$
where $\varphi$, $\psi$ and $\omega$ are functions while θ1 and θ2 are constants. Moreover, $k_0$,$k$,$b$ and ${{\rho }_{1}},{{\rho }_{2}}$ with $l=1/R$ are all positive constants where $R$ is the radius of curvature
In this task we are interesting to find the derivative of $\varepsilon(t)$ w.r.t to time $t>0$ . I think first to use Leibtiz's rule to do this derivative presented in this Leibniz integral this leads me to some thing wrong and not similar to the result I have in this paper Paper.Page 2
$$\frac{d\varepsilon }{dt}=-{{\int_{0}^{L}{({{\gamma }_{1}}{{\left| {{\left( {{\varphi }_{x}}+\psi +l\omega \right)}_{t}} \right|}^{2}}+{{\gamma }_{2}}{{\left| {{\psi }_{xt}} \right|}^{2}}+{{\gamma }_{0}}{{\left| {{\omega }_{xt}}-l{{\varphi }_{t}} \right|}^{2}}+\left| {{\theta }_{1}}_{x} \right|}}^{2}}+{{\left| {{\theta }_{2x}} \right|}^{2}})dx$$ where all $\gamma$'s are viscosity coeficients. How they derived it to get the above result ?
