The following problem comes from an old exam:
Consider \begin{cases} u_t – \Delta u + |u_{x_1}| = 0 \text{ in } \mathbb{R}^{n} \times (0,\infty) \\ u(x,0)=g(x) \text{ in } \mathbb{R}^n \end{cases} where $g$ is continuous and has compact support, and assume as $|x| \rightarrow \infty$, $u \rightarrow 0$. Show there is at most one solution that vanishes as $|x| \rightarrow \infty$.
My attempt:
I was able to prove that from the growth condition on $u$ and that as $u$ is a sub-solution to the heat equation that \begin{equation*} u(x,t)\leq \max(\sup_{(x,t) \in \mathbb{R}^{n} \times (0,\infty)}g(x),0) \leq M \end{equation*} where the final inequality is obtained since $g$ is continuous and has compact support.
I feel like this lemma would be helpful for an energy argument for showing uniqueness, but the problem is that the equation is non-linear, so if I let $u,v$ be two solutions that vanish and let $w:=u-v$, then $w$ satisfies \begin{cases} w_t – \Delta w + |u_{x_1}| – |v_{x_1}| = 0 \Rightarrow w_t – \Delta w \leq |w_{x_1}| \\ w(x,0) = 0 \end{cases}
Then I attempted to use an energy argument with the bounds above with \begin{equation} E(t) := \frac{1}{2} \int_{\mathbb{R}^n} w^2 \end{equation} Then when I differentiated with respect to $t$ \begin{equation} E'(t) := \int_{\mathbb{R}^n} w(\Delta w + |v_{x_1}| – |u_{x_1}|) \leq \int_{\mathbb{R}^n} -|Dw|^2 + |w||w_{x_1}| \end{equation} (the inequality is from integration by parts and triangle inequality) I feel that it'll be hard to show that this integral is negative, so I thought of using Gronwall's Inequality, which would require me to bound $|w_{x_1}|$ in terms of $|w|$, which I am unable to do. Any hints or tips on how to approach this problem would be useful!
Best Answer
You should use a maximum principle argument. Suppose $u$ and $v$ are two solutions that vanish as $|x|\to \infty$. For $\varepsilon>0$, let $(x_0,t_0)$ be a point where $u-v-\varepsilon t$ attains its maximum value over $\mathbb{R}^n\times [0,T]$ for fixed $T>0$ (the existence of such as point follows from assuming $u$ and $v$ vanish at $\infty$). Then at $(x_0,t_0)$ we have $u_t \geq v_t + \varepsilon$, $u_{x_1}=v_{x_1}$ and $\Delta u \leq \Delta v$ provided $0 < t_0 \leq T$. Thus
$$0=u_t - \Delta u +|u_{x_1}| \geq v_t + \varepsilon - \Delta v + |v_{x_1}| = \varepsilon$$
at $(x_0,t_0)$, provided $t_0>0$. This is a contradiction, so the max is attained at $t_0=0$ where $u=v$, so $u-v-\epsilon t \leq 0$. Therefore
$$u \leq v + \epsilon t$$
for all $t>0$. Sending $\varepsilon\to 0$ we get $u \leq v$. Then swap the roles of $u$ and $v$.