# 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:

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

Thanks for your help, the details are eluding me here.

• $1.$: apply Green's first identity. $2.$: $(1)$ follows from the equation above it: $|\nabla u|^2 \ge 0$ so clearly the term on the left has to be less than or equal to zero. Sep 7, 2013 at 21:17
• ${ \nabla\cdot\left(u\nabla u\right) = \nabla u\cdot\nabla u + u\nabla\cdot\left(\nabla u\right) = \left\vert\nabla u\right\vert^{2} + u\nabla^{2}u }$ Sep 14, 2013 at 1:33

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 $t$.
• Thanks! I'm looking up Green's identity, and I'm not finding anything about $|\nabla f|^2$ as such. Is it the first, second, or third identity? en.wikipedia.org/wiki/Green's_identities Sep 7, 2013 at 21:24
• I would say: notice that $u\Delta u = \nabla \cdot (u\nabla u) - |\nabla u|^2$ (to see this, use the product rule on the first term on the RHS). Then integrate: the first term on the RHS becomes $\int_{\Omega} \nabla \cdot (u\nabla u) = \int_{\partial \Omega}u\nabla u \cdot \nu$ where $\nu$ is normal vector. But this is zero cause $u$ is zero on the boundary. Actually this is Divergence Theorem! Sep 7, 2013 at 21:28
• BTW you get uniqueness because the estimate (1) applies to the difference of two solutions $u_1$ and $u_2$ by linearity. Notice that the thing you're differentiating is always $\geq 0$, and that at $t=0$, it is zero. So (1) says the derivative of a positive function which starts at zero is negative, so it must be zero everywhere. Draw a graph to see it. Sep 7, 2013 at 21:30
• I don't understand this: "I pulled out the $\frac{d}{dt}$ since the integral is over space only." We should justify how we can interchange the limits. I really want to know how it is plausible. I think we need some regularity conditions, however my main problem is with weak solution. Thanks Apr 18, 2018 at 9:30