u_t + - Delta _p^su^m = fu其中0 s 1在等式的左右两边同乘以u^m_t对变量x和变量t积分可以的到什么样的等式能量泛函是多少

左右两边同乘以[{({u^m})_t}]得到： [{({u^m})_t}{u_t} + {({u^m})t}{( - {\Delta p})^s}{u^m} = {({u^m})t}f(u)] 对变量[x]和变量[t]积分，得到： [\int{\Omega} {({u^m})t}{u_t}dxdt + \int{\Omega} {({u^m})t}{( - {\Delta p})^s}{u^m}dxdt = \int{\Omega} {({u^m})t}f(u)dxdt] 由于[{({u^m})t} = mu^{m-1}{u_t}]，所以第一项可以化简为： [\int{\Omega} {({u^m})t}{u_t}dxdt = \frac{m}{m+1}\int{\Omega}{u^{m+1}t}dxdt] 根据分部积分公式，第二项可以化简为： [\begin{aligned} \int{\Omega} {({u^m})t}{( - {\Delta p})^s}{u^m}dxdt &= \int{\Omega} {u^{2m-1}}{( - {\Delta p})^s}{u_t}dxdt \ &= \int{\Omega} {u^{2m-1}}{(-1)^s}{\Delta^s}{u_t}dxdt \ &= \int{\Omega} {u^{2m-1}}{(-1)^s}{\frac{\partial}{\partial t}}({\Delta^s}{u})dxdt \ &= \int{\Omega} {u^{2m-1}}{(-1)^s}{\frac{\partial^2}{\partial x_i^2}}({\Delta^{s-1}}{u}{\frac{\partial}{\partial x_i}}{\Delta^{s-1}}{u})dxdt \end{aligned}] 因此，原方程可以化为： [\frac{m}{m+1}\int{\Omega}{u^{m+1}t}dxdt + \int{\Omega} {u^{2m-1}}{(-1)^s}{\frac{\partial^2}{\partial x_i^2}}({\Delta^{s-1}}{u}{\frac{\partial}{\partial x_i}}{\Delta^{s-1}}{u})dxdt = \int{\Omega} {({u^m})t}f(u)dxdt] 能量泛函为： [E(u) = \frac{1}{m+1}\int{\Omega}{u^{m+1}}dx + \frac{(-1)^s}{2}\int{\Omega} {({\Delta^{s-1}}{u})^2}dx\