Suppose we have a function $u(t,x): A \times B \to R$, where $A$ and $B$ are bounded intervals. We know, the derivative up to order $1$ exists and is continuous, furthermore, the derivative with respect to $x$ exists up to order 2 and is continuous. In addition it holds $u_{tx} \in L^2(A \times B)$ then the following is valid:
$\frac{1}{2} \frac{d}{dt} \|u_x(t)\|_{L^2(B)}^2 = \int_{B} u_x(t) u_{xt}(t) dx.$
I have no idea, why this equation should hold.