How do I integrate $\int _0 ^1 uu_xu_{xx} dx$ where $u = u(x,t)$

Question

I know I must integrate $\int _0 ^1 uu_xu_{xx} dx$ by parts. I have $u = u$, $du = du$, $dv = u_xu_{xx}dx$, $v = \frac{1}{2} u^2_x$. With this, I end up with $\frac{1}{2}uu_x^2 |^1_0 - \frac{1}{2}\int u_x^2du$, but I have a note saying that the second term should be $u_x^3$, instead of $u_x^2$.

Also, was my process correct? I haven't integrated by parts in a very long time and it seems unfamiliar considering that $u=u(x,t)$.

Answer 1

You have done: $$ \int u u_xu_{xx}dx=\int u\cdot\,d\left(\frac{1}{2}u_x^2\right)= \frac{1}{2}uu_x^2-\int\frac{1}{2}u_x^2 du=\frac{1}{2}uu_x^2-\int\frac{1}{2}u_x^2 u_xdx $$ and the variable $t$ has no rule in this integration.