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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.

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1 Answer 1

We have $$\frac 12\frac d{dt}\lVert u_x(t)\rVert^2_{L^2(B)}=\frac 12\lim_{n\to +\infty}h_n^{—1}\int_B(u_x(t+h_n,x)^2-u_x(t,x)^2)dx,$$ where $\{h_n\}$ is a sequence of non-zero reals which converges to $0$. This gives \begin{align}\frac 12\frac d{dt}\lVert u_x(t)\rVert^2_{L^2(B)}&=\lim_{n\to +\infty}h_n^{-1}\int_B\int_t^{t+h_n}u_{xt}(s,x)u_x(s,x)dsdx\\ &=\lim_{n\to +\infty}h_n^{-1}\int_B\int_0^{h_n}u_{xt}(t+s,x)u_x(t+s,x)dsdx\\ &=\lim_{n\to +\infty}\int_B\int_0^1u_{xt}(t+sh_n,x)u_x(t+sh_n,x)dsdx. \end{align}

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