Find the energy of a partial differential equation 
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
 A: 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.
