Consider the following equation $$ y_{tt} - y_{xx} - y_{tx} - y_{txx} = 0 , \ \ (x,t) \in (0,1) \times (0,\infty) $$ with $y(0,t) = y(1,t) = 0$, $t \in (0,\infty)$. I want to find the energy of system.
For this, I am trying to go on usual: Multiplying the equation for $ \overline{y}_{t} $ and integrating about $ (0.1) $, applying parts integration and and taking the real part
$$ \dfrac{d}{dt}\bigg(\dfrac{1}{2}\int_{0}^{1}|y_{t}|^{2}dx + \dfrac{1}{2}\int_{0}^{1}|y_{x}|^{2}dx\bigg) + \text{Re}\int_{0}^{1}y_{t}\overline{y}_{tx}dx + \dfrac{1}{2}\int_{0}^{1}|y_{tx}|^{2}dx= 0 $$ Then $$ E(t) = \dfrac{1}{2}\int_{0}^{1}|y_{t}|^{2}dx + \dfrac{1}{2}\int_{0}^{1}|y_{x}|^{2}dx $$ And $$ E^{\prime}(t) = -\text{Re}\int_{0}^{1}y_{t}\overline{y}_{tx}dx - \dfrac{1}{2}\int_{0}^{1}|y_{tx}|^{2}dx $$
Am I right? How can I rewrite the term $$ \int_{0}^{1}y_{t}\overline{y}_{tx}dx ?? $$
I want to show $E^{\prime}(t) \leq 0$.
If the title of the question was not cool, I am sorry
The differential operator in this problem is real and linear, so $\operatorname{Re}[y]$ and $\operatorname{Im}[y]$ will also satisfy the equation if $y$ does. That means it's sufficient to solve the problem for real functions, then combine the results at the end if $y$ actually is complex.
Multiplying through by $\partial_t y$ is the correct way to proceed. You get $$ (\partial_t y)(\partial_{tt}y) - (\partial_t y)(\partial_{xx}y) - (\partial_t y)(\partial_{xt}y)-(\partial_t y)(\partial_{xxt}y) = 0 $$ Now we try to write it in terms of squares of $y$ and its derivatives, and exact $x$ derivatives. Some finagling gives $$ \frac{\partial}{\partial t}\left[\frac{(\partial_t y)^2+(\partial_x y)^2}{2}\right]-\frac{\partial}{\partial x}\left[(\partial_t y)(\partial_x y)+\frac{(\partial_t y)^2}{2} + (\partial_t y)(\partial_{xt}y)\right]+\frac{(\partial_{xt} y)^2}{2} = 0. $$ Because $\partial_t y = 0$ on the boundary, the exact $x$ derivative term will vanish when integrating over $[0,1]$, and we have $$ \frac{d}{dt}\left[\int_{0}^1\frac{(\partial_t y)^2+(\partial_x y)^2}{2}dx\right]+\int_{0}^1\frac{(\partial_{xt} y)^2}{2}dx=0. $$ Notice that in the second integral, the integrand is always nonnegative. Therefore the value of the integral itself is nonnegative. Moving it over to the other side gives $$ E'(t) = \frac{d}{dt}\left[\int_{0}^1\frac{(\partial_t y)^2+(\partial_x y)^2}{2}dx\right]=-\int_{0}^1\frac{(\partial_{xt} y)^2}{2}dx\le 0 $$ If $y = y_r + i y_i$ is a complex function, we can add the energies of the real and imaginary parts together get \begin{eqnarray} E'(t) &=& \frac{d}{dt}\left[\int_{0}^1\frac{(\partial_t y_r)^2 + (\partial_t y_i)^2+(\partial_x y_r)^2+(\partial_x y_i)^2}{2}dx\right] = \frac{d}{dt}\left[\int_{0}^1\frac{|\partial_t y|^2 +|\partial_x y|^2}{2}dx\right] \\&=&-\int_{0}^1\frac{(\partial_{xt} y_r)^2+(\partial_{xt} y_i)^2}{2}dx=-\int_{0}^1\frac{|\partial_{xt} y_r|^2}{2}dx\le 0. \end{eqnarray} So the result still holds.