I am trying to understand the proof of the following theorem:
Theorem (Uniqueness): If the functions $\alpha$, $\beta$, and $f$ are sufficiently smooth to ensure that $u$, $u_{t}$, $u_{x}$, and $u_{xx}$ are continuous in $G$ and up to the boundary of $G$, including the two "corner" points, then the. following IBVP has at most one solution:
$u_{t}(x,t) = ku_{xx}(x,t)+q(x,t)\,\,$ $0<x<L$, $t>0$ (PDE)
$u(0,t) = \alpha(t)$, $u(L,t)=\beta(t)$, $t>0\,\,$ (BCs)
$u(x,0)=f(x)$, $0<x<L\,\,$ (IC).
$G$ is a one-dimensional uniform rod with sources, insulated lateral surface, and temperature prescribed at both endpoints.
Now, the proof starts out with having us consider two solutions $u_{1}$ and $u_{2}$ to the IBVP. Then, because of linearity, $u = u_{1}-u_{2}$ must also be a solution to the fully homogeneous form of the IBVP:
$u_{t}(x,t) = k u_{xx}(x,t)$, $0<x<L$, $0<t<T$
$u(0,t)=0$, $u(L,t)=0$, $0<t<T$,
$u(x,0)=0$, $0<x<L$,
where $T>0$ is any fixed number.
So, then, the proof goes on to have us multiply the homogeneous PDE by $u$ and integrate over $[0,L]$. This is the part that I need help with!
I have that $0 = \int_{0}^{L}\left(u u_{t} - k u u_{xx}\right)dx$.
Now, let's disregard the limits of integration for a second, and just consider the indefinite integral $0 = \int \left(u u_{t} - k u u_{xx}\right)dx$ for simplicity's sake.
Splitting this up into two integrals, we obtain $0 = \int u u_{t} dt - k \int u u_{xx} dx$.
Now, the second integral is easy: integration by parts yields us that $ k \int u u_{xx} dx = kuu_{x} - k\int u_{x}^{2}dx $.
The other integral I have absolutely no idea how to do.
According to the proof, I'm supposed to get $\displaystyle 0 = \int_{0}^{L}\left(u u_{t} - k u u_{xx}\right)dx = \int_{0}^{L} u u_{t} dt - k \int_{0}^{L} u u_{xx} dx \\ \displaystyle = \int_{0}^{L} \left( \frac{1}{2}\left(u^{2} \right)_{t} + ku_{x}^{2}\right)dx-k \left[ u u_{x}\right]_{x=0}^{x=L}$.
But, for the life of me, I am unable to figure out how they got $\displaystyle \int_{0}^{L} \frac{1}{2}\left(u^{2} \right)_{t}dx$.
I realize that asking for this sort of thing is generally frowned upon on MSE, but could somebody actually work out that part of the integral in full detail to show me where the $\mathbf{\displaystyle \int_{0}^{L} \frac{1}{2}\left(u^{2} \right)_{t}dx}$ came from?
Thank you!
No integration by parts is needed for that. It's mere product rule: $$(u^2)_t=2u_tu$$