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.
 A: 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?
