Integrate $u_t - \Delta u = 0$ to get $\frac12 \frac{d}{dt} \int_{\Omega} u^2 + \int_{\Omega}|\nabla u|^2 = 0$? In my PDE class, my instructor wrote the following notes:
Consider equations $u_t - \Delta u = 0$ in $\Omega$, where $\Omega \subset \mathbb{R}^n$ is bounded. Suppose boundary conditions $u = u_0(x)$ at $t=0$, and $u=0$ at $\partial \Omega$.
Multiplying the equation by $u$ and integrating by parts gives
$$\frac12 \frac{d}{dt} \int_{\Omega} u^2 + \int_{\Omega}|\nabla u|^2 = 0$$ 
(with no boundary terms, using the bc). This already shows
$$\frac{d}{dt} \int_{\Omega}u^2 \leq 0 \tag{1}$$
which gives uniqueness, since the problem is linear (so the difference of two solutions has initial data 0).
My Questions:


*
*I'm not seeing the integration by parts here, can someone talk me through it?
*How are we deducing (1)?
*How does (1) show uniqueness?

Thanks for your help, the details are eluding me here.
 A: Multiply by $u$:
$$uu_t -u\Delta u = 0$$
Integrate
$$\int uu_t -\int u\Delta u = 0$$
Rewrite as
$$\frac{1}{2}\frac{d}{dt}\int u^2+\int |\nabla u|^2= 0$$
where I used Green's theorem on the second term. I pulled out the $\frac{d}{dt}$ since the integral is over space only. Move the second term onto the RHS.
$$\frac{1}{2}\frac{d}{dt}\int u^2=-\int |\nabla u|^2 \leq 0$$
with the inequality because the integrand is always positive, and the minus sign outside makes it negative.
For uniqueness: suppose there are 2 solutions $u_1$ and $u_2$ solving the problem. So the difference satisfies
$$(u_1-u_2)_t - \Delta (u_1-u_2) = 0$$
$$u_1-u_2 = u_0 - u_0 = 0 \qquad\text{at $t=0$}$$
and $u_1-u_2=0$ on $\partial \Omega.$ Then it is clear, that the estimate (1) also holds with this difference:
$$\frac{d}{dt}\int_{\Omega} (u_1(t)-u_2(t))^2 \leq 0$$
Now the thing you're differentiating is a positive function, which is zero at $t=0$. Its derivative is always negative, so it must zero almost everywhere (draw a graph). Hence $u_1 = u_2$ for almost every $t$.
