# Uniqueness of solution to $u_{t} - \Delta u + |\nabla u|^{2} = 0$

The problem I am working on is as follows:

Let $\Omega$ be a connected bounded domain in $\mathbb{R}^{n}$ with smooth boundary and let $f, g: \mathbb{R}^{n} \rightarrow R$ be smooth. Show that there is at most one smooth solution $u(x, t)$ such that \begin{align*} \begin{cases} u_{t} - \Delta u + |\nabla u|^{2} = 0 & \text{ in } \Omega \times (0, \infty)\\ u(x, t) = g(x) & \text{ on } \partial \Omega \times (0, \infty)\\ u(x, 0) = f(x) & \text{ in } \Omega. \end{cases} \end{align*}

This seems like an energy method type problem. Suppose $u_{1}, u_{2}$ are 2 smooth solutions of the above equation. Let $w := u_{1} - u_{2}$. Then \begin{align*} \begin{cases} w_{t} - \Delta w + \nabla w \cdot \nabla(u_{1} + u_{2})= 0 & \text{ in } \Omega \times (0, \infty)\\ w(x, t) = 0 & \text{ on } \partial \Omega \times (0, \infty)\\ w(x, 0) = 0 & \text{ in } \Omega. \end{cases} \end{align*}

My question is: how can I define an $E(t)$ such that $E(t) \leq 0$ for all time $t \geq 0$? As a first attempt, I tried defining $E(t) := \frac{1}{2}\int_{\Omega} w^{2}\, dx$, but I get $$E'(t) = \int_{\Omega}w(\Delta w - \nabla w \cdot \nabla (u_{1} + u_{2}))\, dx.$$ However, I am not sure how to work with the $\nabla (u_{1} + u_{2})$ term. Any hints on how to get around this?

• Call $g=\nabla (u_1+u_2)$, then $w$ satisfies the heat equation $$\partial_t w-\Delta w + g\cdot \nabla w=0.$$ By the maximum principle we get $w\leq 0$ everywhere, and by the minimum principle $w=0$. Commented Jan 6, 2015 at 0:06
• The comment above works. Also, you can approach the problem directly with the maximum principle. Try showing that $w-\varepsilon t$ cannot have an interior maximum for any $\varepsilon>0$ (just use necessary conditions for a maximum). This shows that $u_1 \leq u_2$. Then swap the roles of $u_1$ and $u_2$.
– Jeff
Commented Aug 31, 2016 at 21:30