Integration by part in Sobolev spaces Let $x \to f(x,t) \in H_0^1(0,L)$ my equation is about the integration by part formula in this cas. I want to evaluate the quantity $$\int\limits_0^1 {{\partial _{xx}}f(x,t){\partial _t}f(x,t)} dx$$
so
$$\int\limits_0^1 {{\partial _{xx}}f(x,t){\partial _t}f(x,t)} dx = {\partial _x}f(L,t){\partial _t}f(L,t) - {\partial _x}f(0,t){\partial _t}f(0,t) - \int\limits_0^1 {{\partial _x}f(x,t){\partial _{tx}}f(x,t)} dx$$
Do I have $${\partial _x}f(L,t){\partial _t}f(L,t) - {\partial _x}f(0,t){\partial _t}f(0,t)=0$$?
Thank you.

Probably you need to be more precise about where your funtions live, for example to be able to define the left hand side of your equation you might want $f \in H^2$ and $ \partial_t f \in L^2.$
In any case the answer is yes, because for any fixed time $f(L,t) = f(0,t) =0.$ Where this statement can be made precise in terms of the trace operator. Hence - although a little bit informally: $\partial_t f(L,t) = \partial_t f(0,t) = 0.$
On the other hand for functions in $H^1_0$ you know that they wanish on the boundary, but you can´t say the same about the derivatives, so you can´t say that $\partial_xf(L,t) = 0$. Think about the fuction: 
$$f(t)=
\begin{cases} 
       t+1 & -1 \le t \le 0 \\  1-t & 0 \leq t \le 1 
   \end{cases}
$$
I argue that $f \in H^1_0 (-1,1) $. In fact this function can be approximated in $H^1(-1,1)$ by the sequence of functions: 
$$f_{\epsilon}(t)=
\begin{cases} 
       \frac{t}{1 - \epsilon}+1 & -1+ \epsilon \le t \le 0 \\  1-\frac{t}{1 - \epsilon} & 0 \leq t \le 1 - \epsilon \\ 0 &  \text{ elsewhere } 
   \end{cases}
$$
Now it should be clear that $f_{\epsilon} \in H^1_0.$ It follows that $f \in H^1_0.$
What is the value of $\partial f$ on the boundary?
